module2 gmat quantitative

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MODULE II QUANTITATIVE ASSESSMENT

CONCEPTS REVIEW AND EXERCISES

MAT Quantitative assessment measures your quantitative

reasoning skills and tests your ability to make logical interpretation of information that is presented to you, or conceptually known to you. You will then use the ‘logical interpretation’ to make ‘logical decisions’. A ‘logical interpretation’ is one that considers all valid explanations of agiven information. For example, if the information is that X is an integer, then a logical interpretation of this statement will be X could be any positive whole number, any negative whole number, or Zero. Similarly, if the information presented to you is |X| = 2, then the logical interpretation of this information is that X is either 2 or –2. Likewise, if the information given to you reads that the ‘average of X, Y, and Z is 10’, your ‘logical interpretation’ of this statement will read: “X, Y, and Z can be any value they could be as long as they ADD UP to 30’. MAKING LOGICAL DECISIONS USING LOGICAL INTERPRETATION OF INFORMATION GIVEN TO YOU The process of making ‘logical decisions’ involves ‘testing for consistency across all logical interpretations of the given information’. For example, if the question was: If X2 = 4, is X positive? We will have to read this question as follows: Given that X2 = 4, can you determine with logical certainty that X is greater than Zero? Our logical interpretation of the information X2 = 4

tells us that X is either 2 or –2. If X is 2, then it is positive; if it is –2, another value for X consistent with our logical interpretation of the ‘evidentiary’ information, then X is NOT positive. Can you see that we have two likely values for X consistent with our logical interpretation of the ‘evidentiary statement’ (X2 = 4) but there is no ‘consistency’ with respect to the decision we can arrive at across the two different ‘scenarios’? Obviously, we cannot make a ‘logical decision’ about whether X is positive or not on the basis of the ‘evidence’ X2 = 4. On the GMAT, you will treat as ‘evidence’ any ‘IF’ information that is given to you or any information that is known to you as ‘concept’. You will then be required to interpret this ‘evidence’ in a logical fashion, and then use the logical interpretation to decide whether or not any decision having logical consistency is feasible. RULE: If the logical interpretation of an ‘evidence’ presents more than one valid scenario, then neither scenario alone is a logical interpretation. For example, if the ‘evidence’ is ‘X times Y is positive’, our ‘logical interpretation’ of this statement presents two ‘valid’ scenarios: X and Y are both positive or X and Y are both negative. Neither X nor Y can be zero. Notice that our ‘logical interpretation’ gave rise to three valid scenarios about the values for X and Y, and neither scenario alone is a logical interpretation. For example, if you made a ‘decision’ by using the scenario that ‘X and Y are positive so that X times Y is positive’, then your ‘decision’ is ‘not logical’.

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Your logical decision-making process must consider all ‘three’ valid scenarios and test for ‘consistency’ across all valid scenarios. If the question was: “If X times Y is positive, is X+Y positive?” you will read this question as “Can you determine with logical certainty whether X+Y is positive, given that X times Y is positive?’ (Remember that the ‘decision questions’ on the GMAT data sufficiency do not use English language statements. They use the variant of ‘cling-on’, and you must ‘translate’ the question into English language in order to grasp the real meaning of decision questions. ‘IS X + Y positive’ is cling-on for ‘can you determine with logical certainty whether X + Y is positive, given X times Y is positive?’) Our logical interpretation of the information that ‘X times Y is positive’ gave rise to three valid scenarios: X and Y are both positive, or X and Y are both negative, and Neither X nor Y can be Zero.

In the first ‘valid’ scenario, X + Y is positive. In the second ‘valid’ scenario, X + Y is negative. In the third valid interpretation of the evidence, X + Y cannot be Zero. Therefore, our ‘logical decision’ is that X + Y can be positive or negative, and neither decision alone is a logical decision. We must conclude that we cannot determine with logical certainty whether X + Y is positive, given than X times Y is positive. If we were tempted to do so, we would be ‘guilty’ of making ‘illogical decisions’ because we

‘assumed’ that X and Y cannot both be negative so that X times Y is positive. Notice that this process of making ‘logical decisions’ is analogous to our ‘logical analysis’ of an ‘illogical argument’ presented to us in the Analytical Writing Assessment or in the Critical Reasoning Section of the GMAT. IF the evidence was: “John made a GMAT score of 500 out of 800”, one interpretation of this information is that John did not take our prep course. Another interpretation could be that John was not well on the day of the test. Or, there could be any number of other interpretations, and neither interpretation alone is a valid interpretation. We applied the same rule when we analyzed an Issue. If the issue was: Why do you figure that courtesy is a vanishing virtue in an urban environment?, one explanation could be that the urban dwellers are prone to higher levels stress brought on by the aggressive and competitive urban environment. Another explanation could be that there is a distrust of strangers in an urban area. A third explanation could be that the principle of reciprocity or ‘pay back’ does not work well in an urban environment because the ‘good deed’ done to a stranger is not ‘returned’ to the urban dwellers, with the result that the urban denizens have less motivation to actively practice courtesy. As you can see, our ‘logical interpretation’ of the evidence that ‘courtesy is a vanishing virtue in an urban area’ is open to at least three different ‘valid’ explanations, and neither explanation alone is a logical explanation. (That is why we asked you to come up with at least two different valid and logical explanations for the ‘issue’ in




Module I so that the reader will know that you are not making ‘illogical’ explanations.) QUANTITATIVE ASSESSMENT is the first of the two ‘adaptive’ test sections on the GMAT. The ‘adaptivity’ is a process by which the GMAT test continuously adjusts the difficulty level of the questions presented to you. This ‘adjustment’ is made using ‘artificial intelligence’ that judges your comfort level by the way you are picking answer choices to the problems presented to you. This manual is constantly updated in order to reflect changes taking place on the GMAT and to feature GMAT problems that are truly typical of the recent GMAT questions. Therefore, work through this manual, not just ‘read through’ it. Be sure to practice with all the ‘examples’, not merely read through them. Because you will be taking the test in a timed-environment, you need to make sure that your recognition of the type of problem presented to you and your ability to set up the problem expeditiously are up to snuff. This goal can be accomplished through practice, and through several repetitions. As the saying goes, excellence is not an act but a habit, and habit is a matter of repetition of the same act over time. Also, use this MODULE along with the Quantitative Preprogram assignment file and with Quantitative Concepts in a nutshell file. Assessment of your quantitative reasoning skills will be made by asking you to deal with about 18 to 19 problems in Problem Solving and another 18 to 19 questions in Data Sufficiency format. Out of the 37 questions given to you, only 28 will be scored, and the other 9 questions will be experimental questions. This means that exactly 14 problems will

be scored in each of the two sections: Problem Solving and Data Sufficiency. PROBLEM SOLVING Even though the section is titled ‘Solving of Problems’, not all ‘problems’ will involve crunching of numbers in the course of ‘solving a problem’. Some problem solving questions will simply ask you to do quantitative reasoning and make a logical decision. Consider the following problem: What is the units’ digit of 1499? The exponent of 99 on 14 looks ominous and daunting but the simple rule of GMAT is that the more daunting and ominous a problem appears, the more likely it is a simple reasoning problem. You will engage in your reasoning process by taking some ‘baby steps’: 141 = 14 The units’ digit is 4 142 = XX 6 The units’ digit is 6. Notice that we did not care about the tens or the hundreds’ digits here because the problem is not about digits other than the units’ digit. 143 = xxx4 Again, we are concerned about the units’ digit and not the other digits. We already begin to see a pattern developing. The units’ digit cycles between 4 and 6, with 4 as the units’ digit for exponents 1 and 3, and with 6 as the units’ digit for exponents 2 and 4. We will see the same pattern throughout and because the exponent 99 is an odd integer, we will ‘logically conclude’ that the units’ digit of 1499 will be 4. Notice that we did not do a significant amount of ‘number crunching’ in this ‘problem’




but relied on a ton of reasoning to arrive at a logical conclusion. Consider another problem in which you will be asked to do similar ‘reasoning’ and to recognize a pattern in order to reach a logical conclusion. Consider the following PROBLEM: In the sequence 1, 2, 4, 8, 16, 32,……… what is the sum of 25th term and 27th term? We notice that the pattern is such that every value after the first one is twice the preceding but because the problem asks us to ‘add’ values 25 and 27, we figure that there must be a ‘better way’ to deal with the problem than to extend the series all the way down to 27th term. One of the skills that will come in handy when you work on the GMAT is your ability to size up the problem and to ‘transform’ it to an equivalent form. If we transformed the sequence as a sequence of exponents, we will be able to deal with the problem more effectively. The transformed sequence is 20, 21 , 22 , 23 , 24 , 25, …………

We now notice that the value of the exponent in the First term is 0, the value of the exponent in the second term 1, the value of the exponent in the third term 2, and so on. We can see that the value of the exponent is 1 less than the ordered positional value of the term. We can now ‘logically conclude’ that the 25th term will be 224 and the 27th term will be 226.

Therefore, the sum of the 25th and 27th term will be 224 + 226 = 224 (20 + 22 ) = 5•224 Did we do any number crunching here? We did some when we did some ‘factoring’ of the terms in the final steps but most of the problem was about reasoning and about recognizing a pattern in order to make a logical extrapolation. Of course, there will be some problems that will require some reasoning and some crunching of numbers, but you will never get a problem (at a high difficulty level) that does not involve any reasoning at all. DATA SUFFICIENCY GMAT Data Sufficiency is about making a logical interpretation of the evidence presented to you in two different statements and then to use the logical interpretation to decide whether a logical decision involving ‘logical certainty’ can be made about the ‘decision question’ posed in the question stem. Each Data Sufficiency problem will consist of a Question Stem, which may also contain some ‘evidence’, and two different statements constituting the evidence to be examined in the course of making a decision.




FORMAT 1: {STEM EVIDENCE}, [DECISION QUESTION] STATEMENT 1 STATEMENT 2

If you come across a Data Sufficiency problem having the above format, you should read it as follows: [DECISION QUESTION] {STEM EVIDENCE}, STATEMENT1 {STEM EVIDENCE}, STATEMENT2 REMEMBER: THE ‘STEM EVIDENCE’ separately belongs to each statement, and must be used as ‘additional evidence’ presented. EXAMPLE: “IF X IS AN INTEGER, IS X AN EVEN INTEGER?” 1. X > 0 2. X < 3 You should read the above problem as: “IS X AN EVEN INTEGER?” 1. X IS AN INTEGER AND X > 0 2. X IS AN INTEGER AND X < 3.

FORMAT 2 [DECISION QUESTION] STATEMENT 1 STATEMENT 2 EXAMPLE: “How old is John?” 1. John is 3 years older than Samantha. 2. 3 years from now, John will be twice

as old as Samantha. Each Data Sufficiency problem will have FIVE answer choices, which are arranged in the SAME ORDER for every problem. ST 1 ALONE IS SUFFICIENT BUT

ST 2 ALONE IS NOT. ST 2 ALONE IS SUFFICIENT BUT

ST 1 ALONE IS NOT. ST 1 AND ST 2 COMBINED IS

SUFFICIENT BUT NEITHER ALONE IS.

EACH STATEMENT ALONE IS SUFFICIENT.

NEITHER STATEMENT ALONE OR COMBINED IS SUFFICIENT TO MAKE A LOGICAL DECISION.

Notice that Statement 1 alone is considered to be GOOD for making a LOGICAL decision with in options 1 and 4. Statement 2 alone is considered to be GOOD for making a LOGICAL decision with in options 2 and 4.




DATA SUFFICIENCY TYPES OF QUESTIONS 1. UNIQUE VALUE DECISION 2. TRUE OR FALSE DECISION UNIQUE VALUE decision problems in Data Sufficiency typically start in WHAT or How, and the TRUE OR FALSE DECISION problems typically start in IS EXAMPLE: UNIQUE VALUE DECISION: “What is X?” 1. | X | = 2 2. X < 0

You should keep in mind that the DECISION QUESTION in the stem is not worded in English language. It is a mutant form of the ‘klingon’ language we are used to hearing on Star Trek. You must, therefore, TRANSLATE the ‘klingon’ statement to English statement before you process the question. Mutant KLINGON: “What is X?” ENGLISH: “Can you determine with logical certainty a UNIQUE VALUE FOR X by using a LOGICAL INTERPRETATION of the EVIDENCE presented in statement 1 alone and on the basis of evidence presented in statement 2 alone, or , if need be, on the basis of the COMBINED information presented in statements 1 and 2?” Now that we have ‘translated’ the ‘klingon’ statement to ‘English’ statement, let us proceed to ‘attack’ the problem.

The question is: Can you determine a unique value for X on the basis of a logical interpretation of the evidence presented in statement 1 alone first? What is the logical interpretation of the evidence presented in statement 1? If the ‘absolute value’ of X is 2, then X could be 2 or – 2. Is that a UNIQUE value? By no means. Therefore, we must conclude that the evidence presented in statement 1 alone does not allow us to make a unique value decision about the value for X. IF Statement 1 alone is not sufficient for a unique decision, then neither of the two options in the answer choices, option 1 and option 4, can be good answer choices. We must, therefore, eliminate options 1 and 4. We have three options remaining at this stage of analysis. We must now proceed to examine statement 2, and determine whether, on the basis of a logical interpretation of the evidence presented in statement 2, we can make a unique value decision with respect to the value for X. What is the LOGICAL INTERPRETATION of the evidence presented in statement 2? Statement 2 tells us that X is a negative value. How many LIKELY negative values for X can we think of? Unlimited. Is it possible to arrive at a UNIQUE value decision for the value of X on the basis of a logical interpretation of the evidence presented in statement 2? Hardly. WE must, therefore, conclude that statement 2 alone is also not sufficient to make a UNIQUE VALUE decision about X. Statement 2 is considered to be good for making a




logical decision in options 2 and 4. We have already eliminated option 4 when we concluded that statement 1 alone was not sufficient to make a logical decision about the value for X. We must now eliminate option 2. We have two answer choices remaining: option 3 and option 5. If the ‘combined’ evidence of statements 1 and 2 is sufficient to allow us to make a unique value decision for the value of X, then we will pick option 3. If, after examining the combined evidence of the statements 1 and 2, we cannot determine a UNIQUE value for X, then we must choose option 5. Let us COMBINE the two statements and see whether the combined evidence allows us to make a unique value decision about X. The combined evidence reads as follows: 3. X is either 2 or – 2, AND X is

negative.

Looks like we have a winner here. If X is negative and X is either 2 or negative 2, then it must be – 2. A positive value for X will not satisfy the statement 2 and must be discarded as not a value for X. We can see that the COMBINED evidence of statements 1 and 2 allows us to make a unique value decision about X, and we must pick option 3.

NOTE: You must ‘COMBINE’ the two statements ONLY WHEN neither statement ALONE is sufficient to make a unique value or a logical certainty decision. If any one statement alone was sufficient, you will NOT proceed to combine the two statements. Example of a problem in which we will not combine the information in statements 1 and 2: If the area of a square S is equal to the area of the rectangle R, what is the perimeter of the square region? 1. The length of ONE side of the

rectangle R is twice the length of the side of the square S.

2. The area of the rectangle R is 49. (Translation: WE have STEM EVIDENCE that must belong to each statement independently. The question is: Can you determine a UNIQUE VALUE FOR THE PERIMETER of the square region S on the basis of a logical interpretation of the evidence presented in statement 1 alone and on the on the basis of statement 2 alone, OR on the basis of the ‘combined’ evidence of statements 1 and 2?)




Because the Stem Evidence must be treated as part of each statement alone, we have the two statements read as follows: 1. The area of the rectangle R is equal

to the area of the square S, AND the length of one side of the rectangle R is twice that of the square S.

2. The area of the rectangle R is equal to the area of the square S, AND the area of the rectangle is 49 square units.

Be sure to process the stem evidence in the manner indicated above so that you will not FORGET to use stem evidence. A fatal mistake committed by the test-takers is that they forget about the stem information and make some bad decisions. Let us now begin to examine statement 1 alone and see whether a ‘logical interpretation’ of the evidence as ‘redefined’ above to include the stem evidence is sufficient to make a unique value decision about the perimeter of the square region S. ( A unique numerical value for ‘4a’, the perimeter of the square region of side S.) STATEMENT 1 If ‘a’ is the value of each side of the square S, and if L is the length and W, the width of the rectangular region R, then we have a2 = L•W and L = 2a. Our ‘logical conclusion’ of the above evidence is that W must be ½ of a. Notice that there are 3 variables: a, L, and W, and we cannot uniquely solve for any of them on the basis of TWO simultaneous equations. We must,

therefore, decide that statement 1 alone is NOT sufficient to make a logical decision about the unique value for the perimeter of the square region because we cannot determine a unique value for the length of the side ‘a’ of the square. We must now eliminate options 1 and 4, and proceed to examine statement 2 alone and decide whether, on the basis of statement 2 alone, we can come up with a unique value for ‘a’ and, therefore, for ‘4a’. STATEMENT 2 On the basis of the redefined statement 2 that combines the stem evidence with the additional information presented in statement 2, we have a2 = L•W = 49 Does this allow us to determine a unique value for the side of the square region? Yes, it does. WE can uniquely determine that the square region must be 7 units on each side, and the perimeter must be a unique 28 units. Notice that although the values for a turn out to be 7 and – 7, we have to discard the negative value because the value for the side of a square cannot be a negative value. WE must now conclude that we can make a UNIQUE value decision for the perimeter of the square region S on the basis of the redefined evidence in statement 2 alone. We must, therefore, pick option 2. Because any one statement alone was sufficient to make a logical and unique-value decision about the perimeter of the square region S, we will NOT combine the evidence of statements 1 and 2.




TRUE OR FALSE DECISIONS in DATA SUFFICIENCY Data Sufficiency questions that start off in ‘is’ are typically ‘true or false’ decision questions. EXAMPLE: IS X a positive integer? 1. X2 > 4 2. X3 < - 8 Once again, we need to TRANSLATE the ‘mutant klingon’ decision question in the stem to English statement. MUTANT KLINGON: “Is X a positive integer”? ENGLISH TRANSLATION: “Can you determine with logical certainty whether or not X is positive whole number by using the logical interpretation of the evidence presented in statement 1 alone and by using the logical interpretation of the evidence presented in statement 2 alone, OR, if need be, by using the COMBINED evidence of statements 1 and 2?” Of course, any ‘stem evidence’, if given, must be independently considered part of each statement. In this problem, there is no stem evidence and we will have to work with the evidence presented in the statements alone.

Let us proceed to examine statement 1 alone first and decide whether the logical interpretation of the evidence presented in statement 1 alone is sufficient to make a decision having ‘logical certainty’ about whether or not X is a positive integer. LOGICAL CERTAINTY DECISION is one that is CONSISTENT across all valid interpretations of the evidence. STATEMENT 1 A logical interpretation of evidence in statement 1 tells us that X could be greater than 2 or less than negative 2. What are the values for X consistent with the above interpretation? X could be 3 or 3¼ or – 3 or – 3¼ . As you can see, we can make the EVIDENCE in statement 1 work for both POSITIVE INTEGER VALUES OF X, NEGATIVE INTEGER VALUES OF X, POSITIVE NON-INTEGER VALUES OF X AND NEGATIVE NON-INTEGER VALUES OF X. Note that the values we ‘chose’ to ‘test’ for ‘likely values of X’ were all consistent with a logical interpretation of the statement 1. Yet, we have the following decision outcomes across those valid ‘scenarios’. If X is 3, then it is a positive integer. If X is 3 ¼, then it is NOT a positive integer. (it is positive but not an integer). If X is – 3, then it is not a positive integer. (It is an integer, but not positive). If X is – 3¼ , then it is neither positive nor an integer.




Is there any ‘CONSISTENCY’ across these ‘valid’ scenarios, consistency with respect to the ‘true or false’ decision about whether or not X is a positive integer? No, there isn’t. We must, therefore, conclude that strictly on the basis of a logical interpretation of the evidence presented in statement 1 alone, we cannot make a ‘true or false’ decision having ‘logical certainty’ about whether or not X is positive whole number. We must now eliminate options 1 and 4, and proceed to examine statement 2. STATEMENT 2 X3 < - 8. Our ‘logical interpretation’ of this evidence is that X is negative, and must be less than negative 2. ‘Valid’ values for X, consistent with the above logical interpretation, are – 2 ½, - 3, and so on. What does this interpretation tell us about whether or not X is a positive whole number? It tells us that X MUST NOT BE A POSITIVE WHOLE NUMBER, and we are able to make a LOGICAL CERTAINTY DECISION on the basis of the evidence presented in statement 2 alone. Statement 2 allows us to definitively conclude that X CANNOT BE A POSITIVE WHOLE NUMBER. We must pick option 2 corresponding to the decision that a logical certainty decision can be made solely on the basis of the information presented in statement 2 alone.

REMEMBER: You do not have to answer in the affirmative in order to be able to make a logical certainty decision. The question is: Is it possible for us to make a unique decision, either confirming that X is a positive integer or confirming that X must not be a positive integer. This decision must be firm across all valid scenarios that are consistent with a logical interpretation of the evidence used in the statements given to us. In order to understand this process better, consider the decision along the following lines: Statement 2 alone allows us to make a logical decision about whether or not X is a positive integer. And the decision is that X cannot be a positive integer.




Let us consider another problem and understand how to make logical decisions about ‘true or false’ questions in data sufficiency. “If a and b are integers, is b an even integer?” 1. 3a + 4b is even integer. 2. 3a + 5b is even integer. Once again, our first step is to ‘translate’ the Decision question in mutant klingon into English statement. Also, notice that the problem gives us ‘stem evidence’ that must be treated as part of each statement taken alone. ENGLISH language Decision QUESTION: “Can you determine with logical certainty whether or not ‘b’ is an even integer, given that a and b are both integers, by using a logical interpretation of the redefined evidence in statement 1 (along with stem evidence) alone and by using a logical interpretation of the redefined evidence in statement 2 (including the stem evidence) alone, OR, if need be, by using a logical interpretation of the combined evidence of statements 1 and 2, taken with the stem evidence?” Our REDEFINED evidence items in statements 1 and 2 become: 1. a and b are integers, and 3a + 4b is

an even integer. 2. a and b are integers, and 3a + 5b is

an even integer.

Let us start with the evidence in STATEMENT 1 first. Our logical interpretation of the evidence in statement 1 tells us that ‘a’ MUST BE an even integer and ‘b’ could be odd integer or even integer in order to make the statement 1 work. Notice that we are adding two values to get an even integer. What are the different ways in which we can add two integer values to get an even sum? Each integer could be odd integer or each must be even integer. In the evidence, the term ‘4b’ must be even regardless of whether ‘b’ is even or odd. (IF you multiply any integer by an EVEN integer such as 4, the result is always an EVEN integer). For instance, ‘b’ could be 3 and 4b is 12, an even integer. ‘b’ could also be 2 and 4b is 8, an even integer. WE must logically conclude that the term 4b cannot be an odd integer because the factor 4 will make the term ‘4b’ uniquely even integer. This means that the other term ‘3a’ must be an even integer in order that 3a + 5b will be an even integer. How do we make the term ‘3a’ an even integer? By making ‘a’ an even integer. Notice that ‘a’ CANNOT be an odd integer because, if it were, then ‘3a’ will be odd and 3a + 4b will be an odd integer. That will not be consistent with a logical and literal interpretation of the evidence presented in statement 1. Therefore, we logically conclude, on the basis of a logical interpretation of the evidence presented in statement 1 alone, that ‘a’ must be an even integer but the value ‘be’ could be an odd integer or an even integer in order to make the evidence in statement 1 work. Do we have consistency across the two valid




interpretations of the likely values for ‘b’? No. In one interpretation, ‘b’ could be an odd integer and make the evidence in statement 1 work; in another, ‘b’ could be an even integer and also make the evidence in statement 1 work. Our logical interpretation of statement 1, in sum, is that ‘a’ must be an even integer but ‘all bets are off’ about ‘b’: ‘b’ could be an odd integer or an even integer, and be consistent with a logical interpretation of the statement 1. We must now conclude that the evidence presented in the statement 1 (taken with the stem evidence) alone is not sufficient to make a logical determination of whether or not ‘b’ is an even integer. WE must now also eliminate options 1 and 4, and move on to examine statement 2. STATEMENT 2 (taken with the stem evidence) reads: ‘a’ and ‘b’ are integers, AND 3a + 5b is an even integer. Notice that we have two valid scenarios: 3a and 5b are both odd integers so that their sum is an even integer. OR 3a and 5b are both even integers so that their sum is even integer. In order to make the first scenario work, ‘a’ and ‘b’ must be both odd integers so that 3a will be odd integer, and 5b also an odd integer so that 3a + 5b will be an even integer.

In the second scenario, ‘a’ and ‘b’ must be both even integers so that 3a will be even integer and 5b will also be an even integer so that 3a + 5b will be an even integer. In one of the two valid scenarios, b is odd integer; in another valid scenario, b is an even integer. Is there consistency with respect to the decision about whether ‘b’ is even or not across the two valid scenarios? Hardly. If there is no consistency with respect to the decision across at least two valid scenarios consistent with a logical interpretation of the evidence, then we must conclude that no logical decision can be made on the basis of a logical interpretation of the evidence. Therefore, we must now conclude that statement 2 alone is also not sufficient to make a logical decision about whether or not ‘b’ is an even integer. We must now eliminate option 2, and proceed to examine the combined evidence of statements 1 and 2, taken with the stem evidence. THE COMBINED EVIDENCE READS: STEM EVIDENCE ‘a’ and ‘b’ are integers. STATEMENT 1 INTERPRETATION ‘a’ is even integer and ’b’ could be even or odd integer. STATEMENT 2 INTERPRETATION ‘a’ is odd integer AND ‘b’ is odd integer, OR ‘a’ is even integer and ‘b’ is even integer. The only value for ‘a’ that will satisfy the evidence in statement 1 and statement 2




is an ‘even’ integer value for ‘a’ . According to the logical interpretation of statement 2, if ‘a’ is even integer, then ‘b’ must be an even integer. WE can see that ‘a’ an even integer and ‘b’ an even integer will satisfy the stem evidence, evidence in statement 1, and the evidence in statement 2. Even though statement 1 allowed us to conclude that ‘b’ could be an odd integer or an even integer, but ‘a’ must be even integer, the only valid scenario that will satisfy both scenarios is an even integer value for ‘a’ and even integer value for ‘b’. We can conclude on the basis of the combined evidence in statements 1 and 2, taken with the stem evidence, that both ’a’ and ‘b’ must be even integers in order to satisfy both statements. We must pick option 3. We will revisit the data sufficiency drill later on in this file. But for now, understand that the GMAT will ask you to deal with problem solving and with data sufficiency. In data sufficiency, there are two kinds of decision problems you will come across: unique value decision, and ‘true or false’ decision. The procedure for making logical decisions in data sufficiency is as follows: 1. Make a logical interpretation of the

evidence presented in statements 1 and 2 (taken with stem evidence, if applicable).

2. Use the logical interpretation of the evidence to make a decision involving logical certainty about the decision question.

3. Do not start with the ‘decision question’ and ‘back’ into the evidence.

NEED TO KNOW set of information When you deal with data sufficiency, you must predetermine what pieces of information will help you solve the puzzle or answer the question posed in a unique manner. Let us call it the N.T.K list or the Need-to-Know list. Let us say that the question posed is: How much commission did Jane make on her sales in the first half of 1999? What pieces of information will help us answer this question? We need to know the amount of sales she made in the first half of 1999, and the commission rate at which she was remunerated. We have to establish this N.T.K list and then look for these two specific pieces of information in the statements 1 and 2. If we do not have these two pieces of information, we cannot answer the question posed in a unique fashion. Similarly, if the data sufficiency question asks you: “Is X equal to Y?”, our N.T.K list will read: • I need to know the value of X and

the value of Y so that I can compare the values and determine whether they are equal or not.

If the question posed reads: What is the average speed for travel between Chicago and Indianapolis?, our N.T.K list will seek out the distance between the two cities and the time of travel between the two cities. In real-life, we go through the same process, don’t we? If we want to buy a new car, we need to know whether we can afford the monthly payments for the car and the insurance, and the parameters determining affordability - may be our existing




financial commitments, our income, and our future obligations for which we must save now. If we want to marry someone, we need to know whether the other person has temperamental compatibility, compatible life style habits, and ability to make adjustments. In the corporate world, if we want to launch a new product, we would like to know whether the target market has a need or want for the product, whether the price is something that the market will bear, and whether we have the required distribution channels and resources to do a successful job of marketing the product. We might want to call this aspect of our decision-making “doing the home-work”. It is no different when it comes to data sufficiency. What is the danger if we do not predetermine requirements? We will be like a philosopher, who is defined as a blind person looking for a black cat in a dark room when it is not there. The probability of finding the darn cat is the same as the probability of survival for a snowflake in hell,. You will have the same chance of finding the correct answer if you do not predetermine what you want to know and establish your N.T.K list in advance before you go to examine statements 1 and 2 independently. In Data Sufficiency, you will be presented a question along with or without any additional information and two statements. Your task is to examine the statements 1 and 2 independently and determine whether the question posed can be answered in a unique fashion on the basis of your NTK list.

You will follow the procedure below each time you deal with a data sufficiency problem. • Establish your N.T.K list and be clear

about what information will help you answer the question posed.

• Associate choices A and D with statement 1, and associate Choices B, C, and E with statement 2. Although the real test will not identify the choices as A, B, C, D, and E, you will call the first choice ‘Choice A’, the second choice ‘Choice B’, and so on.

• Examine statement 1 and look for your NTK information. If statement 1 provides your NTK information in toto, then you will consider statement 1 alone as sufficient for now. In this event, you will eliminate the choices B, C, and E (associated with statement 2) out of hand. On the other hand, if you do not find the NTK information in statement 1, you will eliminate choices A and D associated with that statement.

• If statement 1 was good, then you will carry forward choices A and D to the second statement. If statement 1 was not good for a unique answer, you will deal with the remaining three choices associated with statement 2.

• You can see that at this stage, your odds of picking the right answer have improved to 33% or to 50% from the initial 20%. Not a bad deal, huh?

• SCENARIO 1: Statement 1 was good. We carry forward A and D, and kill the other three choices, B, C, and E.




• If you find that statement 2 also has

the information in your NTK list, then you must select D, corresponding to “Each statement alone is sufficient”. On the other hand, if statement 2 does not have the NTK information, then you must select A.

• SCENARIO 2: Statement 1 was NOT good to begin with. We will kill choices A and D, and keep choices B, C, and E to deal with.

• IF Statement 2 has our NTK information, then we must select choice B corresponding to “Statement 2 alone is sufficient but statement 1 alone is not”. If Statement 2 also does not have the NTK information, then we must eliminate choice B and proceed to combine statements 1 and 2. We have choices C and E remaining at this stage.

• SCENARIO 3: COMBINE STATEMENTS 1 AND 2

• If the combined information provides our NTK information, then we must select choice C, which corresponds to “Statements 1 and 2 combined are sufficient but neither one alone is sufficient.”

• If, after combining the two statements, we still do not have the NTK information we are seeking, then we must select the choice E, which corresponds to “I tried every trick in the book but it does not work. Leave me alone” or words to that effect (Neither statement alone or combined is sufficient).

The key to doing well in the data sufficiency section is your ability to proceed methodically in the manner

described above. If you do not have the method suggested behind your madness, you are going to be one disappointed individual at the end of the test when the unofficial score report pops up on the screen. The rewards for proceeding methodically are great, and begin with the popping sound of champagne cork. We will deal with data sufficiency questions as we move along and also later on as a specific area of attention in this file. For now, familiarize yourself with the rules and the methodical procedures starting with establishing the NTK list. Problem Solving questions are designed to test your basic mathematical skills and your ability to reason quantitatively and solve quantitative problems. Your ability to conceptualize verbal descriptions of situations and solve problems presented using basic arithmetic, elementary algebra and commonly known and used concepts of geometry will be put to the test in these sections. “Data sufficiency” section asks you to use the same conceptual skills as do problem solving questions. However, decision-making does not involve full implementation and verifying results. Data sufficiency is like a jig-saw puzzle, and you are only required to determine whether you can complete the picture by using the puzzle pieces in statements one and two. Let us see how the same question can be presented to us either format:




Problem Solving:

“What is X, if X + 2 = 4?” We know that we can solve for X from the given information: X = 2. We will pick an answer choice that states a value of 2 for X. How can the same problem be presented in data sufficiency context? In Data sufficiency, you will be asked a question, and your task is to determine whether you can answer the question from the additional information in two different statements independently. We stated earlier on in this discussion that data sufficiency is about making a decision, and not about actual implementation. Decision-making involves establishing your Need To Know information. In addition to establishing your N.T.K., you will also be required to use any information you may know, information such as the properties of a triangle, or the number of integers qualifying as “digits”, or the number of hours in a day. Also, be sure to use any information that may be provided along with the question itself. For example, the question might read: “If X is a prime integer less than 7, what is X?” We should know that there are only 3 prime integers less than 7: 2, 3, and 5, and that X should be one of these three values. Last, but not least, you should be able to redefine the information in statements 1 and 2 in the manner indicated below.

Let us take an example and see how we can establish our N.T.K list and how we can rephrase the question posed. “If X is a prime integer less than 7, what is X?”

1. X2 = 9. 2. 2 < X < 5

The questions we will ask are: ♦ What do I know about prime

integers? ♦ What else do I need to know in order

to make a decision? The answer to the first question is that prime integers are whole numbers that have exactly two different positive divisors, 1 and itself. We also know that there are only 3 prime integers less than 7: 2, 3, or 5. What else do we need to know in order to make a decision about what X is? We will need some additional information such as an equation involving X or some such thing. Having established our N.T.K., let us rephrase the question posed in the statements 1 and 2 independently, one at a time, starting with statement 1 first. We must read the statement 1 as: “IF X2 = 9, and if X is either 2, 3, or 5 (a prime integer less than 7), can we determine one way or the other what X is?” We notice that if we take the square root on both sides, X could be +3 or –3. But given that X is a prime integer and less than 7, X cannot be negative integer. Therefore, -3 is not a likely value for X.

The answer choices in Data Sufficiency will always be arranged in the same order for all questions. The first choice corresponds to a decision in which you conclude that Statement 1 alone is sufficient but Statement 2 alone is not. Choice 2 corresponds to a decision in which statement 2 alone is sufficient but statement 1 alone is not. Choice 3 corresponds to a decision that Statements 1 and 2 TOGETHER are sufficient but neither alone is. Choice 4 means that either statement alone is sufficient. Choice 5 means that the statements alone or combined are not sufficient to make a unique decision.




The only likely value for X is +3. Is this a unique value? Yes. Can we make a unique determination about what X is on the basis of Statement 1 alone? Yes. We will keep the choices A and D, and kill options B, C, and E. We will need to similarly rephrase that statement 2 as follows: “If X is a prime integer having one of the following three values: 2, 3, 5, and if 2 < X < 5, can we determine what X is?” You bet. The only integer value that lies in the range between 2 and 5 is 3. We can make a unique determination of what X is on the basis of statement 2 alone also. We notice that each statement alone is sufficient to answer the question posed. The choice corresponding to this decision is Choice D.

et us see how we can establish our N.T.K and redefine the question

posed by looking at another example. “What is the average speed for the trip between City A and City B?”

1. The distance between City A and City B is 300 Miles.

2. It took 6 hours to go from City A to City B.

We need to ask ourselves:

♦ What do we know about Distance, Time, and Speed?

♦ What do we need to know in order to answer the question posed?

We know that Distance = Speed X Time.

In order to compute the average speed, we need to know the total distance between City A and City B, and the time of travel between the two cities. Let us rephrase the question posed in the light of statement 1 first: 1. If the distance between City A and

City B is 300 Miles, can we determine the average speed of travel between the two cities? We cannot because we do not know how long it took to get from A to B. Therefore, we cannot answer the question on the basis of statement 1 alone. We will kill choices A and D, and keep B, C, and E only.

2. Let us examine statement 2 by redefining it: If it took 6 hours to go from City A to City B, can we determine what the average speed was? We cannot, because we do not know the distance between the two cities. Remember that you cannot use the information you learned in statement 1 when you examine the statement 2 on its own merits first. Let us kill choice B and keep C and E only as viable options at this juncture.

3. Because we cannot make a unique decision on the basis of Statement 1 only and also on the basis of Statement 2 only, we MUST now combine the two statements and see whether we can make a decision on the basis of the combined information. You must bear in mind that if any one of the two statements was sufficient to answer the question posed in a unique way, you should not go to step 3 and combine the two statements.

L




4. Let us rephrase the combined

information: “If the distance between City A and City B is 300 miles, and If it took 6 hours to go from City A to City B, can we uniquely determine the average speed?” Of course, we can. We have both pieces of information we predetermined in our N.T.K. Therefore, the combined information is good for making a unique decision, and we must pick choice 3 or C.

Your ability to define the Need-To-Know information and to “rephrase” the question posed in the light of statements 1 and 2 is a critical factor in your ability to answer Data Sufficiency questions correctly. Let us take a look at another Data Sufficiency question and hone this skill. DATA SUFFICIENCY: “How many more men than women are in the club X?”

1. The ratio of men to women in the club is 4 to 3.

2. If 4 more women join the club, the ratio of men to women will be 1.

Our Need to Know information will be some information that either tells us the number of men and women separately or one that tells us the difference between the two values in some fashion. Let us take a look at Statement 1. We will read the statement as: If the ratio of men to women in the club X is 4 to 3, can we determine the difference between the men and the women numbers? We will see later on in this file, “ratio” is a “proportionality” concept, and is true for values such as 40 men and 30 women or

400 men and 300 women. As you can see, the difference between the men and women numbers is 10 in one scenario but 100 in the second scenario, and each scenario is consistent with the same ratio of 4 to 3. Are we able to determine the difference between the men and women numbers in a unique way on the basis of the redefined statement 1? No, we are not. We should consider Statement 1 deficient on its own, and kill the two choices A and D associated with Statement 1. We have just 3 options: B, C, and E left. Let us move on to redefine the question in the light of statement 2. If we know that 4 more women will make the ratio of men to women 1 (or make the number of men and women equal), can we determine the difference between men and women numbers in the club? Of course, we can. The statement 2 information tells us that there are 4 more men than women in the club so that if 4 more women came on board, the ratio will be 1 or the number of men will be equal to the number of women making the ratio equal to 1. We can determine the answer unique fashion by using statement 2 alone while Statement 1 alone was deficient. Of the 3 remaining choices, B, C, and E, we must pick choice B to correspond to this decision scenario. Let us take a look at a couple of more data sufficiency problems and get comfortable with the process of predetermining the “NTK” and of redefining the question posed in the light of statements 1 and 2.

A,D

B,C,E




TRUE OR FALSE DECISIONS IN DATA SUFFICIENCY Data Sufficiency will ask you to determine whether you can uniquely determine on the basis of information provided in the statements whether something is definitely true or definitely false. The best way to deal with such problems is to try to ‘create a CONFLICT’ across two scenarios using the information provided in the statements so that you can ‘bail out’ of the statement. Consider the following Statement: “If 3X = 4Y, Is X > Y?” The evidence we have is that 3X = 4Y. Notice that this is a ratio statement involving X and Y. We can write the above ‘evidence’ as X/Y = 4/3. We will learn later in this book that a ratio statement is a ‘multiples statement’. 4/3 is the same ratio as 8/6 or as –4/-3. If X/Y=4/3, then X = 4 and Y = 3. In this scenario, X > Y. But X = -4 and Y=-3 will also satisfy the above ‘evidentiary information’. In the latter scenario, X < Y. Notice that we have two different scenarios, each satisfying the provided statement but one in which X > Y and the other in which X < Y. Do we have a conflict across scenarios here? You bet. Can we make a ‘logical decision’ as to whether X > Y or not? We cannot. We must conclude that the information 3X = 4Y is not sufficient to make a unique determination of which is true: X > Y or Y >X. Consider another example:

“If X is a prime integer that lies in the range between 30 and 40 inclusive, is X+2 a prime integer?” Once again, our objective is to try to create a CONFLICT across scenarios, if we can. If we are unable to create a conflict, we may be able to come up with a unique determination of whether X + 2 is a prime integer or not. We know that X must be either 31 or 37. These are the only two prime integers in the range between 30 and 40, inclusive. If X = 31, then X+2 = 33, which is not a prime integer. If X = 37, then X+2 = 39, which is also not a prime integer. Notice that we cannot create any other scenarios because the values for X are limited to two: 31 and 37. For either values of X, we can see that X+2 is NOT a prime integer. Therefore, the provided information is sufficient to uniquely determine that X+2 is NOT a prime integer. Consider another example of this type of scenarios testing. “IF R and T are integers such that R•T = 12, is T positive?” You must read the ‘is T positive’ as ‘is it possible to uniquely determine whether T is positive or not positive using the statement information?’ If R•T = 12, it means that both R and T are positive or both R and T are negative. It is not possible for one of the




two values to be positive and the other negative because in such a scenario, the product of R and T cannot be positive. If R = 3 and T = 4, then R•T = 12. In this scenario, T is positive. If R = -6 and T = -2, then R•T = 12. In this scenario, T is negative. Can you see that T is positive in one scenario and negative in the other, but both scenarios will satisfy the evidence provided: R•T = 12. We have ‘major conflict’ across scenarios, and must conclude that we cannot uniquely determine whether T is positive or negative on the basis of the evidence provided. Remember: Your objective is to create a conflict across two scenarios using the same statement information so that you can ‘bail out’ of the statement and conclude that the statement is not sufficient to make a unique determination of whether it is definitely true or definitely false that T is positive. Now that you understand the procedure, let us hone our understanding by looking at a complete data sufficiency problem providing a question and two statements. Data Sufficiency:

“Is 5k < 600?”

A, D (1) k is a positive integer. B, C, E (2) 5k+1 < 3000 In order to make a unique decision about whether 5k is less than 600 or not, we need to get a handle on the likely values for k, the only unknown in the statement.

Let us now examine statement 1. Statement 1 tells us that k is positive. Is this enough information to make a unique decision about whether 5k is less than 600 or greater than 600? It is not. How do we arrive at this decision? By creating conflict across two scenarios, each scenario satisfying the statement that k is positive. If k = 3, a positive integer, 5k = 53 = 125, and 5k < 600. But if k = 4, another positive integer, 5k = 54 = 625, which is greater than 600. Notice that we have managed to create conflict across two scenarios and in each scenario, we picked values for k consistent with the statement 1 information. Are we in a position to uniquely determine whether 5k is less than or greater than 600? We are not. We must conclude that statement 1 alone is not sufficient to make a unique and logical decision about 5k in terms of whether it is less than or greater than 600. We must eliminate the options A and D because neither one of these two choices can be picked if statement 1 is not sufficient to make a unique decision. Statement 2 tells us that 5k+1 < 3000. We also notice that 5k+1 = 5k•51 < 3000 Let us divide both sides by 5 to get rid of the factor 51 on the left side of the inequality. Notice that we must do it so that we will have an inequality in terms of 5k about which we are attempting to make a unique decision. We get: (5k•51) / 5 < 3000/5 Or 5k < 600.




Statement 2 tells us exactly what we are attempting to determine. We must conclude that statement 2 alone is sufficient whereas statement 1 alone was not. We must pick option B, which means that the statement 2 alone is sufficient but statement 1 alone is not. WE do not proceed to combine the two statements because ‘combining the two statements’ is a ‘desperate act’ required only when each statement alone is not sufficient. If either statement alone is sufficient or each statement is sufficient, then we do not have to combine the two statements. Consider another data sufficiency problem asking us to make a ‘true or false’ decision in a definite manner.

Is Lisa older than John? A, D (1) Lisa is 15 years of age. B,C,E (2) If Lisa were 5 years older, she

would be twice as old as John. Our NTK list says that we need to know the age of Lisa and of John so that we can compare the ages and determine who is older. If, on the basis of Statement 1, we know that Lisa is 15 years old, can we uniquely determine who is older: John or Lisa? We cannot because we do not know how old John is. This statement tells us nothing about John, and we cannot compare Lisa’s age with an unknown quantity. We must eliminate choices A and D, and proceed to examine statement 2. We have just three choices remaining: B, C, or E. Statement 2 needs to be translated into a mathematical equation, which will be:

L + 5 = 2J

Can we find out the ages of L and J from this equation? We cannot. We have two unknowns and one equation. We require two simultaneous equations in order to be able to uniquely determine the values for L and J. We conclude that statement 2 alone is also not sufficient. We must eliminate option B, keep C and E, and proceed to combine the statements because neither statement alone is sufficient to make a unique decision. When we combine the statements, we will read: If L = 15 and if L + 5 = 2J, can we determine L and J and make a comparison? We notice that we have two unknowns and two equations so that we can find out the values of L and J in a unique fashion. We can now answer the question in a unique fashion by combining the two statements, and we must select choice C. Notice how we did not bother to solve for L and J. When you are dealing with Data Sufficiency, do not bother to solve equations as long as you recognize that you have all the required information with which to get the answers. If you were asked the same question in problem solving, you must, of course, solve for L and J by substituting the value of 15 for L in the second equation. If L = 15, then 15 + 5 = 2J, or J = 10. Obviously, Lisa is older than John but we need not bother to solve for L and J as long as we know that we have the required pieces of information that helps us answer the question in a unique sort of way.




Consider another problem that requires ‘testing scenarios’ and ‘creating conflict across scenarios’ in the process of making a unique ‘true or false’ decision. “Is X a multiple of 12?” 1. X is a multiple of 4. 2. X is a multiple of 6. The question is: Can we make a ‘definitely true’ or ‘definitely false’ decision about whether X is a multiple of 12 or divisible by 12 using the information in the statements provided? (We will see more about ‘divisibility’ in the chapter on numbers and factors coming up later in this book). Statement 1 tells us that X is a multiple of 4. A ‘multiple of 4’ is to be read ‘an INTEGER multiple of 4’. X could be 4, 8, 12, 16, 20, 24, 28, 32, and so on. Notice that all the ‘likely’ values for X are ‘evenly’ divisible by 4. Can we use this statement alone to make a unique decision about whether X is divisible by 12? We cannot, because if X = 8, a multiple of 4 value, then 8/12 is not an integer. But X could also be 24, another multiple of 4, which is divisible by 12 because 24/12 is an integer. Do you see a ‘conflict’ across scenarios here? You bet. Can we make a unique decision using this statement about whether X is a multiple of 12 or not? We cannot. We must conclude that because statement 1 alone is not sufficient to make a unique decision

about ‘divisibility of X by 12’, we cannot pick either option A or option D. Let us now take a look at statement 2 and see whether we can make a unique decision using this statement alone about whether X is divisible by 12 or not. If X is a multiple of 6, then X is an ‘integer’ multiple of 6. X could be 6, 12, 18, 24, 30, 36, and so on, all multiples of 6. If X = 6, a multiple of 6 value, then X is not divisible by 12 because 6/12 is not an integer. But if X = 12, which is also a multiple of 6, then X is divisible by 12 because 12/12 is an integer. Do we have a conflict across scenarios here? Yes, we do. Can we make a unique decision in the face of conflict? We cannot. We must conclude that statement 2 alone is also not sufficient to make a unique decision about whether X is indeed a multiple of 12 or not. We must eliminate option B and proceed to combine the two statements, as we must when either statement alone is not sufficient. The combined information of statements 1 and 2 tells us that X is a multiple of 12. Why did we make this ‘logical conclusion’? Because if X is a multiple of 4 and also a multiple of 6, then X must be a multiple of 12, the least common multiple of 4 and 6. Notice that in the set of likely values that are multiples of 4 and 6, 12 is the smallest number that is common to both sets.




If our ‘logical conclusion’ on the basis of the combined information in statements 1 and 2 is that X is an INTEGER multiple of 12, then X could be 12, 24, 36, 48, and so on. Are we in a better position to make a unique decision about whether X is a multiple of 12 or not? We are because all the values that are multiples of 12 must be divisible by 12. The combined information is sufficient to uniquely determine that X must be divisible by 12, if X is divisible by 4 and by 6. We must choose option C. Get used to the ‘testing scenarios’ process, and be sure to ‘create conflict’ across scenarios, if you can. In fact, your goal is to create ‘conflict’ across scenarios in the process of making ‘must be true’ decisions, because if you can create conflict across scenarios, then the condition that you are testing ‘could be true’ but is not necessarily ‘must be true’. “MUST BE TRUE” type of problems can also be tested in Problem solving, and a later chapter is devoted to this type of problems. We have seen in the discussion on Data Sufficiency thus far that the focus is on reasoning and on predetermining what information will help us answer the question posed in a unique fashion. You will also be expected to make ‘logical decisions’ by examining all possible explanations of the stated evidence. This process is similar to the ‘critique of the argument’ process tested in the

Analytical Writing Assessment section, where you tried to come up with a bunch of ‘may be’ scenarios in the process of looking at ‘alternative explanations’ for the stated evidence. Objectivity is about making decisions after examining all possible explanations of a given information. If the evidence is: X2 = 4, then a conclusion that states X = 2 is illogical because it ignores the possibility that X could be –2 in order to satisfy the same evidence. Let us consider another data sufficiency problem and see how we can make a decision by examining the different interpretations of a stated information.

“Is X = 4?” 1. X2 = 4X 2. X is an even integer. We are required to make a definitely true or definitely false decision about whether X is 4 or not. Statement 1 tells us that X2 = 4X. People schooled in algebra will be tempted to cancel X on both sides and conclude that X = 4. But we must pause and examine whether another interpretation of the same evidence exists. The statement; X2 = 4X can be satisfied by X= 4 and by X=0. If we ‘rush to a decision’ stating that statement 1 alone is sufficient to make a unique determination of whether X is 4 or not, we will be wrong. Because X could be 4 or 0 in order to satisfy the evidence presented in statement 1, we will consider this statement to produce two ‘different interpretations of the likely values of X’, and must consider that statement 1 alone is not sufficient to make a unique decision. We must now eliminate options A and D.




Let us proceed to examine statement 1 on its own merit. Statement 2 tells us that X is an even integer. X could be 0, 2, 4, 6, 8, …, -2, -4, -6 and so on. Are we looking at a unique decision on the basis of statement 2 here? We are not. We must conclude that statement 2 alone is also not sufficient to make a unique determination of whether X is 4 or not. We must eliminate option B and proceed to combine the two statements. On the basis of the combined information, we see that X could be 0 or 4, both even integers. The values 0 and 4 will satisfy BOTH statements, and are a non-unique set of values. How are we to be sure that X is 4 when it is only a 50% chance that it is. There is an equal chance that X is not 4 and equal to 0. Do you see a conflict across scenarios here? We have tried all the procedural steps: Examine the statements alone and combined, and are unable to make a unique decision about whether X is 4 or not. We must choose option E, corresponding to a decision that no unique determination of whether X is 4 or not can be made on the basis of the information provided in the statements 1 and 2. The question was: Must X be 4? The short answer is: X could be 4 but does not have to be. The best we could have done on the basis of the information provided in the two statements to us is to conclude that X COULD be 4, but does not have to be.

Get used to this type of analysis. GMAT data sufficiency section will test your skill in this area. Let us do another one before we move on to the next concept area. DATA SUFFICIENCY “If X and Y are non-zero integers, is X equal to 2”? 1. X+Y = X•Y 2. X = Y Notice that the question stem provides one piece of information about X and Y, information that must be used in conjunction with the information in the statements. We know that neither X nor Y can be 0. Statement 1 tells us that X+Y=X•Y. There are only two sets of values that will satisfy this relationship: Either X and Y are each equal to 0 or each equal to 2. Because we have been told that X and Y cannot be 0, our ‘logical conclusion’ here must be that X and Y must be each equal to 2. We can make a unique decision on the basis of information in the statement 1 alone and confirm in the affirmative that X is indeed 2. We must keep options A and D for now, and move on to examine statement 2. Statement 2 tells us that X and Y are equal. X and Y could be each equal to 1, or 2, or 3, or 4, or 100 or any value at all. Statement 2 is not good for a unique determination of whether X is equal to 2 or not, but statement 1 alone is sufficient to make this unique determination. We must choose option A. The ‘buzz phrases’ are ‘unique determination’ and ‘alternative explanations or scenarios’.




SEQUENCES AND PATTERNS

GMAT will test your ability to recognize patterns and iterations, and to determine the value for an element in a sequence. ‘Recognition’ of a pattern is a critical skill that you will require in real life too. If you consistently fail to win the lottery week after week, there is a pattern of picking a bad set of numbers. If a sales person consistently fails to meet his quota, there is a pattern of non-performance. If one’s lover consistently fails to send flowers on Valentine’s day, there is a pattern of ‘couldn’t-care-less’ aspect to his or her behavior. Normally, we will ‘recognize’ such patterns and make a ‘federal case’ out of them. GMAT problems will ask you to do the same in the context of numbers and variables. Let us say that we have a sequence of numbers as follows: 2, 3, 5, 7, 11, 13, 17, 23, X, ………….. We can see that the above is a sequence of positive prime integers. The next prime integer after 23 must be 29. Therefore, we can logically conclude that X must be 29. Let us take a look at another sequence of values: 1, 2, 4, 8, 16, 32, X,….. We can see that each successive value is double the preceding one. We must logically conclude that X must be twice 32 or 64.

Consider the following problem: “In a recurring sequence, -4, -2, 0, 2, 4 are values that repeat indefinitely in the same order. What is the sum of the 27th value and the 29th value in the sequence?” Because the 5 values specified repeat ad infinitum in the same order, we can see that the 26th value must start in –4. The 27th value will be –2. The 28th value will be 0. The 29th value is 2. The sum of the 27th value and the 29th value is –2+2 = 0. We must choose the option that corresponds to a value of 0 for the sum of the 27th and the 29th terms. Consider another ‘higher difficulty level problem’: “If in a sequence T1 = 2, T2 = 5, T3 = 11, T4 = 23, T5 = 47, What is T7? What is the expression for TN in terms of the preceding values in the sequence?” Do you get the feeling that you just walked out of a hot tub in a sauna room? You will not be alone and will have plenty of company in this regard. But you must bear in mind that no GMAT problem is meant to be difficult. You must also bear in mind that GMAT problems test your ‘reasoning’ ability and your ‘ability to recognize patterns’. What do we see is going on here?




We see that T2 is one more than TWICE the preceding term T1. T3 is one more than twice the preceding term T2. T4 is one more than TWICE the preceding term T3. T5 is one more than TWICE the preceding term T4 and so on. Can you see a pattern emerging here? We can set up the Formula as follows: TN = 1 + 2 (TN-1) The value for T7 will be 1 more than TWICE T6. The value for T6 will be 1 more than TWICE T5. Knowing that T5 is 47, we can conclude that T6 = 1 + 2(T5)

= 1 + 2(47) = 95 Therefore, T7 = 1 + 2(T6) T7 = 1 + 2 (95) = 1 + 190 = 191 NOTE: We could have also set up the FORMULA representing the pattern as follows: T1 = 2 T2 = T1 + 3•20 T3 = T2 + 3•21 T4 = T3 + 3•22 T5 = T4 + 3•23 The pattern is as described in the rows above. We can ‘extrapolate’ the above sequence and make the following ‘logical determinations’. Each successive term after the first term is the sum of the preceding term and a multiple of 3 value. Notice that T5 is the sum of the preceding term T4 and the exponent on the base of 2 is two less than the suffix of the term T5. You can see that this pattern holds true for all the values after the first value. We can, therefore, conclude that TN must be equal to:

TN = TN-1 + 3•2N-2 We can test this equation by replacing N with 2 and by checking to see whether we get a value of 5 for the Term 2. Similarly, replace N with 3 and check to see whether you get a value of 11 for the term 3. You will. Now we can determine T7 by simply replacing N with 7 in the above relationship. T7 = T6 + 3•25 = T6 + 3•32 = T6 + 96 T6 = T5 + 3•24 = 47 + 3•16 = 95 Therefore, T7 = 95 + 96 = 191 Notice that the second approach is infinitely more abstruse and GMAT or GRE is likely to have you come up with the FORMULA as discussed in the first scenario. But, you will not get both the formulas because they are both correct and you cannot pick one over the other. Can you see how this ‘pattern’ problems are dealt with? Let us do a couple of more so that you can get ‘real comfortable’ with this ‘pattern stuff’. (Never mind the fact that we used an adjective ‘real’ to modify another adjective, ‘comfortable’. We are not dealing with Sentence Correction here, are we? If it were sentence correction, we would have used ‘really comfortable’)




PROBLEM 1 “If F(X) = 1/X – 1/(X+1), What is the sum of F(1)+F(2)+F(3)+F(4)+…….+F(100)?” The problem looks ‘intimidating’ but we must bear in mind that no problem on the test is meant to be difficult. Challenging, perhaps, but never ‘difficult’. We notice that the original statement is in terms of X, but the summation involves real integers. Let us try to replace the variable X with the integers and see whether there is a recognizable pattern emerging. F(1) = 1/1 – ½ = ½ F(2) = ½ - 1/3 = 1/6 F(3) = 1/3 – ¼ = 1/12 F(4) = ¼ - 1/5 = 1/20 We notice that the problem will be ‘very complex’ if it involves adding fractions such as ½, 1/6, 1/12, 1/20, and so on. GMAT is not a slave-driver but a test that asks you to work through the problems efficiently and smartly. We must say to ourselves that ‘adding fractions’ is not the way to go. Let us ‘size up’ the problem and see whether there is a pattern emerging. We do see a pattern. We notice that the –1/2 is the mirror image of ½; -1/3 is the mirror image of 1/3 and so on. When we add F(1), F(2), etc., each successive term will cancel out leaving just the first term of the first function and the last term of the last function. Notice that the successive terms cancel out diagonally.

F(1) = 1 – ½ + F(2) = ½ -1/3 + F(3) = 1/3 – ¼ + F(4) = ¼ - 1/5 + F(5) = 1/5 – 1/6 + F(6) = 1/6 – 1/7 + * * * * + F(99) = 1/99 – 1/100 + F(100) = 1/100 – 1/101 The required sum = 1/1 – 1/101 = 101/101 – 1/101 = 100/101 Our ability to recognize an emerging ‘pattern’ saved the day for us, and got us to the answer. Train yourself to recognize patterns. The test will test your ability to do so. Also, the above problem tested your ability to ‘associate’ values with X. Given that F(X) is in terms of X, we ‘associated’ X with 1 when we computed F(1); X with 2 when we computed F(2), and so on. If we are given that F(X,Y) = X3 + Y2, we can compute F(-3, -2) by ‘associating’ X with –3 and Y with –2. F(-3, -2) =(-3)3 + (-2)2 = -27 + (-4) = -31




Let us take a look at another problem asking us to recognize patterns. “In a sequence, the first three values are the first three positive prime integers in ascending order, and each of the successive values is the sum of all of the preceding values. What is the ratio of the 30th term to the 25th term?” We have the sequence going as follows: 2, 3, 5, (2+3+5), (2+3+5+2+3+5), …… Notice that a pattern begins to emerge. The fifth term is twice the fourth, and we will expect the same pattern to continue. Let us verify that it is indeed so. The sequence is: 2, 3, 5, 10, 20, (20+10+5+3+2) 40, (40+20+10+5+3+2) 80, ……….. Each successive term after the third term is twice the preceding term, and the same iteration will continue. Therefore, the term 26th will be twice term 25th and so on. Term26 = 2• T25 T27 = 2•T26 = 2•2•T25 T28 = 2•T27 = 2•2•2•T25 T29 = 2•T28 = 2•2•2•2•T25

T30 = 2•T29 = 2•2•2•2•2•T25 = 25•T25 Therefore, the ratio of T30 to T25 is 25 = 32/1. Take a look at another problem: “If S1 = 9, S2 = 99, S3 = 999, S4 = 9999, ……….What is SN in terms of N?”

What is the pattern we recognize here? We notice that all the values are in terms of digit ‘9’, which repeats as many times as the value of the suffix. S1 is a single digit number 9. S2 is a two digit number 99 and so on. We can determine that the digit 9 must recur N times in the value for SN. We also see another pattern: 9 is 10-1 99 is 102 – 1 999 is 103 – 1 9999 is 104 – 1 We can clearly see that SN must be 10N – 1. If you are unable to leap forward to this conclusion, you can ‘back-solve’ by looking at the options and by replacing N with 1 and checking to see whether you get a value of 9; Verify by replacing N with 2 and see that you get 99, and so on. Recognition of ‘patterns’ can also be tested in data sufficiency problems. Take a look at the above problem in the following data sufficiency setting.




DATA SUFFICIENCY “If the set S contains subsets S1 through SN such that S1 = 9, S2 = 99, S3 = 999, …. SN = 10N – 1, are all the subset values divisible by the prime integer P?” 1. P < 5 2. Subset S5 is divisible by P. The information provided is mind-numbing for sure, but we must repeat the mantra that GMAT problems are not supposed to be ‘difficult’ but ‘challenging’. Also, keep in mind the simple rule that the harder the problem sounds, the less complex it will be. How do we know that this problem is not as complex as it seems on first blush? Our reasoning says so. Remember that all problems will involve some degree of reasoning effort on your part. We notice a pattern in the subsets: They are all in terms of the recurring digit 9. Given that P is a prime integer, the only prime integer that will divide ALL of the subsets is 3. No other prime integer will evenly divide the values in the subsets. In fact, if the single-digit sum of the digits in an integer is 9, then the integer must be divisible by at least 1, 3, and 9. Knowing that 1 and 9 are not prime integers, we need to seek information telling us whether P is equal to 3 or not. Notice how we ‘transformed’ a complex problem to a ‘manageable’ one through a simple reasoning exercise. Let us examine the statements one by one. Statement 1 tells us that P < 5. Given that P is a prime integer, P could be 2 or 3. If P is 2, then none of the values of the subsets is divisible by P. On the other

hand, if P is 3, then all the subset values are divisible by P. Because we do not know whether P is 2 or 3, we are unable to make a unique determination of whether all the subset values are divisible by P or not. We have conflicting outcomes across two different scenarios when we test P = 2 and P = 3. We must conclude that Statement 1 alone is not sufficient, and eliminate options A and D. These two choices cannot be picked if the statement 1 turns out to be not sufficient. Statement 2 is somewhat intimidating but we must once again tell ourselves that the more intimidating the statement looks, the less complex it is likely to be. We know that S5 = 99999. How do we know this? The pattern tells us so. If S5 is divisible by P, we can logically conclude that P must be 3. No other prime integer will divide an integer with 5 recurring 9’s as digits. Knowing that P is 3, it stands to reason that all the subsets will be divisible by P because all the subsets are made up of the same digit 9 in a recurring pattern. Statement 2 alone tells us that the subset values are all divisible by the prime integer, which must be equal to 3. Statement 2 alone is sufficient but statement 1 alone is not. We must pick option A. The upshot of the above analysis is the following: “Do not get intimidated by how the problem looks and sounds. Just think through it, and remember: the more difficult and intimidating the problem seems, the less complex it is likely to be.”




Let us plow through another Patterns problem and increase our comfort level with such problems. The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72 B. 74 C. 75 D. 78 E. 80

In order to be able to see a pattern, we need to take some ‘baby-steps’ and set up a few more values of the ‘sequence’. a5 = a5-4= a1 = 2 a6 = a6-4 = a2 = -3 a7 = a7-4 = a3 = 5 a8 = a8-4 = a4 = -1 a9 = a9-4 = a5 = a1= 2 As you can see, the sequence reverts to the first value ‘2’ after a set of 4 values, and the sequence carries on in this fashion. We have seen a clear pattern of ‘ad infinitum’ repetition of the first 4 values. We can logically conclude that the 97th value will revert to 2. In the first 96 values, there are 24 sets of the first four values. The first 4 values add up to 3. The sum of the first 96 values is 24 times 3 = 72. If we add the 97th value to this total of 72, we get 74. Therefore, the sum of the first 97 values is 74. Choice B. Let us take a look at another ‘pattern’ problem.

A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32

B. 43 C. 64

D. 100

E. 128

This is a relatively simpler problem than the ones we have been previously grappling with. We can see that if k = 3, then a3 = a2+2a1 = 1 + 2(1) = 3 a4 = a3 + 2a2 = 3 + 2(1) = 5 a5 = a4 + 2a3 = 5 + 2(3) = 11 a6 = a5 + 2a4 = 11 + 2(5) = 21 a7 = a6 + 2a5 = 21 + 2(11) = 43. Option B is the answer. Take another pattern problem for a test-drive.




If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =

A. 219 B. 220 C. 221 D. 220 - 1 E. 221 - 1

Let us take some ‘baby-steps’ and set up a few values of the sequence so that we can begin to see a clear ‘pattern’. X1 = 3 X2 = 2x1 – 1 = 2(3) – 1 = 5 X3 = 2x2 – 1 = 2(5) – 1 = 9 X4 = 2x3 – 1 = 2(9) -1 = 17 X5 = 2x4 – 1 = 2(17) – 1 = 33 Notice that the question is about the ‘difference’ of a particular set of consecutive values. On the basis of the above representation of sequence, let us see whether any ‘clear’ pattern emerges about the ‘difference’ of consecutive values. x2 – x1 = 5 – 3 = 21 x3 – x2 = 9 – 5 = 22 x4 – x3 = 17 – 9 = 23

x5 – x4 = 33 – 17 = 24 We are now able to see a clear pattern: the difference of the first two values is 2, that of next two is 22, that of the next two is 23, and so on. We also see that the exponent corresponds to the subscript of the value being subtracted. We can logically conclude that x20 – x19 must be equal to 219, with the exponent on 2 coinciding with the subscript of the value x19 being subtracted. We must choose option A.

Pattern problems test your ability to observe and to see a clear and recurring connection among the values . If, in the above problem, the question was: what is x20?, then we will use the sequence 3, 5, 9, 17, 33, …. to see the following pattern: (21+1), (22 + 1), (23+)1, (24+)1, (25+)1, and so on. We can see that the exponent 1 corresponds to value 1, exponent 2 to value 2, and so on. We can logically conclude that x20 must be (220+ 1).




As some wise guy said, if there are no numbers, then there is no math. You should expect more than a few questions about numbers and integers in the GMAT. Let us say that I have a number in my hat, and want you to bet on it. Would you? I would not, if I were you. Because, I need to know 4 things about the number so that I can get closer to the reality. Specifically, I need to know whether the number is a positive number or a negative number, and whether the number is a whole number or “not a whole number”. Let us say that I have a number N. Unless I define the number to be a positive number, you will not assume that it is positive. Unless the number is defined as an integer or a whole number, you will not assume that it is a whole number. If I simply state that N is a number, then you will understand this statement to mean an “all-bets-are-off” information. “All bets are off” because the number could be a positive or a negative number, an integer (whole number) or not an integer. Convention dictates that we omit the plus sign in front of the number, if that number is deemed positive. For example, we do not say that the car costs plus 20,000 dollars. We simply say that the car costs $20,000. Similarly, we do not say that we make plus 5,000 dollars in wages per month. We simply state that our wages are $5,000 a month. This conventional thinking should not force you to consider N to be positive just because a plus sign is omitted. In the GMAT, N could be positive or negative unless it is specifically defined one way or the other. If you are told that there are two numbers X and Y such that X + Y is an integer (whole number), then you will make sense of this information by asking a question: Do we know anything else about X and Y in terms of whether these numbers are positive or negative, or integers or non-integers? If the answer is that we do not, then we will take this to be an “all-bets-are-off”

scenario in which X and Y could be anything they want to be as long as the sum of the two numbers is a whole number. You will also notice that the fact the sum of X and Y is a whole number does not mean that the sum is a positive whole number. The sum of X and Y could be a negative whole number, because the sum is not specified as a positive whole number. What are the possible sets of values for X and Y such that X + Y will be an integer? X : +1 - ½ - ¾ + 23

4

Y : -2 -212 - 31

4 + 114

X + Y -1 -3 -4 +4 As you can see, X and Y need not be integers, positive or negative so that X + Y can be an integer. All numbers can be represented on a number line such that 0 is smack in the middle, and all positive values are to the right of Zero, and all negative values are to the left of Zero. A rational number is one that can be positioned precisely on a number line. Example of a rational number are: +2, - 2 ½ , + 7 ¾ , and so on. You will notice that ½ and ¾ can be converted to terminating decimals, and we can position these rational numbers precisely on a number line. An irrational number is a number that has a non-terminating decimal. Examples of irrational numbers are: 6, 1/3, 5/7, 4/9, and so on. You will notice that these numbers are non-terminating decimals, and cannot be precisely located on a number line. We can only find an approximate location for these irrational numbers. All numbers to the right of Zero will be steadily increasing values; all numbers to the left of Zero are steadily decreasing values.




The absolute value simply measures the distance from Zero of a number on a number line. Since “distance” is a positive quantity, the absolute value is always positive. Consider X and Y as shown on the number line below: X 0 Y - 21

2 0 + 212

As you can see from the above representation, X and Y have the same distance from Zero on the number line, and will have the same absolute values. We will denote this information by the following notation:

| X | = | Y | We will also notice that X and Y are not equal to each other because they lie on the opposite sides of Zero, albeit the same distance from Zero. If we did not have the benefit of the visual representation above, and if we were simply told that | X | = | Y |, we cannot determine one way or the other whether X and Y are equal to each other. What are the possible scenarios? Scenario 1: X 0 Y - 21

2 0 + 212

OR, (Scenario 2) Y X X 0 Y - 21

2 0 + 212

X and Y could lie on the opposite sides of Zero the same distance from Zero. In this scenario, X and Y have the same absolute values but are not equal to each other. Or, X and Y could lie the same distance from Zero on the same side of Zero. In this scenario, X and Y not only have the same absolute values but also are equal to each other. But we do not know what scenario we must set store by, do we? Therefore, if the question in data sufficiency read: Is X = Y ?

(1) | X | = | Y | (2) X2 = Y2

Can we answer the question on the basis of information in statement 1 alone? We cannot because, as we have seen earlier in this section, there are two possible scenarios: one in which X and Y are not equal and the other in which X and Y are equal. Since we are dealing with more than one unique scenario, we must conclude that statement 1 is not sufficient to answer the question. You should also know that X2 = Y2 is the same information as | X | = | Y | , and is another way of saying the same thing. For example, if X is +2 and Y is –2, X2 = 4 and Y2 = 4 but X and Y are not equal. Alternatively, X and Y could be both +2 or both –2, in which case X2 = Y2 and X = Y. Once again, two different scenarios, and not a unique solution or answer to the question.




Also, the absolute value statement does not give us a clue as to where the number lies on the number line in terms of the Zero point. Let us say that the question is: IS A > 0?

(1) A = - B We will read the statement 1 to mean that A and B have the same absolute values, and that A and B lie on the opposite sides of Zero. But do we know which of these two values lies to the right of Zero and which lies to the left of Zero? We do not. We simply know that if A is to the right of Zero, then B is to the left of Zero, and vice versa. In other words, if B is +2, then A is –2. If B is –3, then A is +3. Two different scenarios, one in which A is positive, and the other in which A is negative. Not a unique solution. We must conclude that we cannot answer the question on the basis of statement 1 alone.

The important thing to keep in mind is that when you deal with absolute values, you know the distance of that value from 0 on the number line but you do NOT know whether the value lies to the right of zero or to the left. You must be sure to check out scenarios pertaining to: • What if the value is to the right of zero?,

and • What if the value is to the left of zero on the

number line?

Let us see how we can use these two scenarios when we deal with some problems from the absolute value concept area.

EXAMPLE 1: “IF | X – 3 | = 9, WHICH OF THE FOLLOWING COULD BE A VALUE FOR X – 5?” • 6 • -1 • -11 • -15 • -21 We know that the absolute value for X – 3 is 9. This means that X – 3 as a value lies either 9 units to the right of zero or 9 units to the left of zero. WE have, therefore, TWO potential values for X – 3 to contend with. X – 3 = 9 or X – 3 = -9 OR, X = 12 OR X = - 6 IF X = 12, then IF X = -6, then X – 5 = 7 X – 5 = -11 Let us see which one of these two likely values for X – 5 can we find in the choices. We have –11, and we must select that choice. EXAMPLE 2: “IF | X3 | = 125, WHICH OF THE FOLLOWING COULD BE A VALUE FOR X + 1?” • - 6 • - 4 • 4 • 8 • 10 IF the absolute value of X3 is 125, then there are two possible values for X3. X3 = 125 OR X3 = -125 Or, X = 5 X = -5 X + 1 = 6 X + 1 = - 4 +6 as a likely value for X is not one of the answer options. We must, therefore, choose the second answer that corresponds to the other likely value that X can have.




Remember: IF X IS RAISED TO AN ODD POWER SUCH AS 3. 5, 7, ETC., then the value for X will have the same sign as the original sign. For example, if X5 = -32, then X = -2, because we need to go with –2 five times as factors in order to get a result of –32 as a product. However, if X is raised to an EVEN power such as 2, 4, 6, 8, etc., then X can have both positive and negative values. For example, if X2 = 25, then X could be +5 or –5. Similarly, if X4 = 81, then X = 3. Also, if X is raised to an even power, the value for that expression cannot be negative. For example, X2 = -9 does not have real roots or real values for X. Therefore, the equation X2 + 9 = 0 does not have real roots. You should also remember the values in the following table in order to improve your speed while taking the test:

X X2 X3 X4

2 4 8 16

3 9 27 81

4 16 64 256

5 25 125 625

6 36 216 1,296

7 49 343 2,401

EXAMPLE 3: “IF | X | = 3 AND | X | = 4, WHICH OF THE FOLLOWING CANNOT BE A VALUE FOR X + Y?” • -7 • -1 • 0 • 1 • 7

We must conclude that X could be +3 or –3. Likewise, Y could be + 4 or – 4. Can we get a –7 by using a sum of the likely values of X and those of Y? Yes. We can combine –3 as a value for X and –4 as a value for Y to get a likely –7. The first choice IS a likely value. The question asks us to pick a choice that is NOT a likely value. Can we get a –1 for the sum of X and Y? Yes. WE can combine +3 for X and –4 for Y to get –1. Can we get a 0 for the sum of X and Y? We cannot because there is no way we can combine ± 3 and ± 4 to get a zero sum. We must select this choice as representing an unlikely or impossible value for X + Y. We can see that X + Y could be +1 or +7 because we can put together –3 for X and +4 for Y to get +1; and +3 for X and +4 for Y to get +7 for X + Y. EXAMPLE 4: DATA SUFFICIENCY “WHAT IS THE VALUE FOR | X + Y |?” 1. | X3 | = 125 2. Y2 = 36 Let us associate choices A and D with statement 1, and B,C, and E with the statement 2. Our N.T.K involves values for X and Y so that we can determine the distance of X + Y from Zero on the number line. Statement 1 tells us that X is +5 or –5. We know that X is 5 units away from Zero on the number line, but do we know anything about Y using this statement? NO. We must conclude that statement 1 is not sufficient and kill the choices A and D. We have just three choices to contend with now. Statement 2 tells us that Y is + 6 or –6. Statement 2 tells us nothing about X. We must eliminate choice B. When we combine the two statements, we see that X + Y could be either 1 unit away from Zero or 11 units away




from zero because X + Y could be +6+5 = 11 or +6 – 5 = +1 or –6 +5 = - 1 or –6 – 5 = -11. We are getting two likely distances for X + Y from zero on the number line, and two different absolute values as likely distances for X + Y. We said that the statements examined independently or in a combined fashion must be good for a unique determination. Can we make a unique determination using the combined information? We cannot. We must, therefore, go with choice E, or the last answer-option in the GMAT-CAT.

INTEGERS An integer is a whole number. There are four sub-classifications of integers: • ODD INTEGERS are by default not

divisible by 2. Odd integers could be positive or negative. Examples of odd integers are: -11, -5, +3, +111, and so on.

• EVEN INTEGERS are by default divisible by 2. EVEN integers could be positive, negative, or ZERO, which is neither positive nor negative. Examples of even integers are: -20, -16, -2, 0, +4, +116, and so on. An EVEN INTEGER must end in one of the following digits: 0, 2, 4, 6, or 8. You must remember that ZERO is an integer and is an even integer.

• PRIME INTEGERS are by definition integers that have just 2 different positive factors: 1 and itself. PRIME INTEGERS are always positive, and the least value for a prime integer is 2. The integer 2 is the ONLY even integer that is a prime integer. All other prime integers are ODD integers. You should know the prime integers in the range between 2 and 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, and so on.

• DIGITS are integers having a value between 0 and 9. A digit could be positive or zero. A digit cannot be negative. –5 is a number using 5 as a digit.

You should also know that the consecutive ODD and EVEN integers are 2 values apart on the number line. Examples of consecutive EVEN integers are: -8, -6, -4, -2, 0, 2, 4, 6, 8, 10, 12, 14, etc. Examples of consecutive ODD integers are: -7, -5, -3, -1, 1, 3, 5, 7, 9, etc.

IF N is an integer, then for all values of N, 2N ± 1 is an ODD integer, and 2N will be an EVEN integer.

WE can represent consecutive positive ODD integers as follows: 2N + 1, 2N+3, 2N+5, 2N+7, etc.. Consecutive positive EVEN integers are: 2N, 2N+2, 2N+4, 2N+6, 2N+8, ETC. Consecutive integers are spaced 1 apart. Examples of consecutive integers are: -7. –6, -5, -4, -3, -2, -1, 0, +1, +2, +3, and so on. For any integer N, we can write the consecutive integers as: N, N+1, N+2, N+3, N+4, N+5, and so on. Just remember the following “combination” rules as they apply to integers>:




For any integer N, we saw that we can represent an odd integer as (2N ± 1). Notice how we turned the integer N into an even integer by multiplying it by 2. For all integer values of N, 2N will be an even integer, and when we add an odd integer to (or subtract an odd integer from) an even integer, the outcome is an odd integer. For all integer values of N, we can represent a series of consecutive odd integers as: 2N ± 1, 2N ± 3, 2N ± 5, 2N ± 7, 2N ± 9, and so on. Similarly, for any integer value N, consecutive even integers can be represented as:

2N, 2N ± 2, 2N ± 4, 2N ± 6, and so on. You might have noticed that any two consecutive odd integers are spaced 2 apart, and any two consecutive even integers are also spaced 2 apart. Example 1: The sum of a set of six consecutive positive odd integers is 216. How much greater is the largest integer in the set than the smallest integer? Solution: We will start by defining the consecutive positive odd integers as 2N + 1, 2N + 3, 2N + 5, 2N + 7, 2N + 9, and 2N + 11. The sum of these values is 12N + 36. Our equation is: 12N + 36 = 216 Or 12N = 180 or N = 15. The largest number in the set is 2N + 11 or 41. The smallest integer in the set is 2N + 1 or 31 The difference of the two integers is 10.

Example 2: IF X is an integer, is Y an integer?

(1) X = Y + 1 (2) X + Y = Integer

Statement 1 tells us that X and Y are consecutive integers because X is 1 greater than Y. Remember: Consecutive integers are spaced 1 apart. We can answer the question on the basis of the information in statement 1 alone. Let us move on to examine statement 2 and see whether we can answer the question on the basis of statement 2 alone also. Statement 2 tells us that X + Y is an integer. Because X is defined as an integer in the stem, we conclude that Y must be an integer as well. (If Y is not an integer, then X + Y cannot be an integer, given that X is an integer). So. We conclude that either statement is sufficient to answer the question, and we will pick the 4th choice in the answer section. Example 3: IF X and Y are consecutive integers, is X > Y?

(1) X – 1 and Y + 1 are consecutive integers.

(2) X and Y + 2 are consecutive integers.

You should understand the question to mean: Is X = Y + 1 ? IF X and Y are consecutive integers, then the only way that X can be greater than Y is if X = Y + 1. Let us examine statement 1. If X is less than Y, then X = Y –1 X – 1 will be Y – 2. Statement 1 tells us that X –1 and Y + 1 are consecutive integers. Because X – 1 is Y –2 under this scenario, we notice that Y – 2 and Y




+ 1 cannot be consecutive integers. We conclude that X cannot be less than Y. If X is greater than Y, then X = Y + 1. X – 1 will be Y. X –1 and Y + 1 will be consecutive integers because X – 1 is Y and Y and Y+1 are consecutive integers. We conclude that statement 1 lets us determine precisely that X is greater than Y. Statement 1 alone is sufficient. Let us move on to examine statement 2. IF X is less than Y, then X = Y – 1 Statement 2 tells us that X and Y+2 are consecutive integers. We notice that X is Y –1 under this scenario, and Y –1 and Y + 2 cannot be consecutive integers. Therefore, we conclude that X cannot be less than Y. IF X is greater than Y, then X = Y + 1 Statement 2 tells us that X and Y + 2 are consecutive integers. Since X is Y + 1 under this scenario, we conclude that Y + 1 and Y + 2 are indeed consecutive integers, and that X must be 1 greater than Y. We can answer the question independently using either statement. We will pick the 4th choice.

Example 4: If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?

I. c – a = 2 II. b = a + 1 = c – 1 III. abc is an even integer. θ I only θ II only θ III only θ I and III only θ I, II, and III

Solution: If a, b, and c are consecutive integers, and a < b < c , then we should know that a = b – 1 or b = a + 1 and b = c – 1 or a + 1 = c – 1 or c = a + 2

We can see that conditions I and II must be valid. What do we think of condition III ? If we are dealing with three consecutive integers, at least one of them must be an even integer, and any integer multiplied by an even integer will give an even integer as the outcome. We conclude that condition III must be true as well. We must pick the last choice.

NUMBER OF INTEGERS IN A GIVEN RANGE OF

VALUES

If N1 and N2 are integers such that N2 > N1 , then the number of integers between N1 and N2, inclusive, is N2 – N1 + 1 . If N1 and N2 are not integers, then the number integers in the range is the absolute integer value difference of N1 and N2, provided that the difference is an integer. For example, the number of integers between 1 ½ and 5 ½ is the difference of the two values. How many odd integers exist between 123 and 2451, inclusive? Step 1: Find out the number of integers in the range, inclusive. Number of integers in the range is (2451 – 123 + 1) = 2329 We notice that the number of integers in the range is an odd value. This tells us that the end values in the range are both the same kind – odd and odd or even and even. In our problem, we notice that the end values are both ODD integers – 123 and 2451 are both odd integers. This means that the number of ODD integers in this range is exactly ONE more than the number of EVEN integers in the same range. In our problem, we will find the number of odd integers by adding a 1 to the number of integers and dividing the sum by 2. Number of odd integers =

½ •(2329 + 1) = 1165.




Number of even integers in the same range is ONE less than the number of ODD integers or 1164. Notice that the sum of the two must add up to the number of integers in the range. If the end values in the range are such that one is odd and the other is even, then the number of odd integers and the number of even integers in the range will be each equal to ½ of the number of integers in the range. For example, the number of odd integers and the number of even integers will be each equal to ½. Of the number of integers in the range between 24 and 473, inclusive. We notice that the number of integers between 24 and 473, inclusive, is (473 – 24 + 1) = 450. The number of odd integers = ½ • 450 = 225. = Number of even integers in the range.

Number of multiples in a range:

You may come across questions in the GMAT asking you to compute the number of multiple values of a given integer in a range of specified values. The number of integers in the range will usually be the ratio of the difference of the end values to the multiple value itself. But then, we may have to make some adjustments to this ratio in the following manner. Let us say that the question asked you: How many integers that are divisible by 4 exist in the range between N1 and N2, inclusive? The answer is: Integer value of (N2 – N1) + 1, 4 if either N1 or N2 or both is evenly divisible by 4. We will also add a 1 if the least of the end values is one less than an integer multiple of 4 and the other is one more than an integer multiple of 4. For example, if the problem asked us to compute the number of integers in the range between 100 and 200, inclusive, we will take the difference of 100 and 200 and divide the difference by 4. We get an integer quotient value of 25. We then take a look at the extreme values in the range and notice that both 100 and 200 are divisible by 4. We need to add a 1 to the quotient value of 25 we obtained, and state the number of

integers as 26, including the integers 100 and 200 – both multiples of 4. REDEFINE THE RANGE WITHIN THE SPECIFIED RANGE IN ORDER TO DETERMINE THE NUMBER OF MULTIPLES OF N within the specified range. Let us say that the problem asked us to compute the number of multiples of 4 in the range between 99 and 201. As you can see, the FIRST value in the specified range that is a multiple of 4 is 100 and the last value in the range that is a multiple of 4 is 200. We can ‘redefine’ the question to ask: HOW MANY MULTIPLES OF 4 are there between 100 and 200, inclusive? The answer is ¼ (200 – 100) + 1. We add a 1 because both 100 and 200 in the defined range are multiples of 4. Let us say that the question was: How many integers that are multiples of 3 exist between 100 and 300, inclusive? The difference of 300 and 100 is 200, and 200 divided by 3 will give us 66. But then, we notice that 300 is a multiple of 3 and we must add 1 to the number 66 for a total of 67 multiples of 3 in the range between 100 and 300, inclusive. You could have also arrived at the same result by ‘redefining’ the range in terms of the first value that is a multiple of 3 and the last value in the range that is also a multiple of 3. The ‘redefined’ range is between 102 and 300, inclusive. The number of multiples of 3 in this ‘redefined’ range within the specified range is 1/3 (300 – 102) + 1 or 67 multiples of 3. We added a 1 to the quotient of ½(300-102) because both values, 300 and 102, are multiples of 3. IF the problem asked you: How many multiples of 3 exist between 25 and 80, inclusive? Redefine the range to one having values between 27 and 78, inclusive. 27 is the first value in the range that is a multiple of 3 and 78 is the last value in the SPECIFIED range and a multiple of 3. The number of multiples of 3 between 27 and 78, inclusive, is 1/3 (78 – 27) + 1= 28 multiples of 3. Again, we added a 1 to the quotient because both 27 and 78 in the ‘redefined’ range are divisible by 3.




Example 1: How many integers exist between 234 and 37, 456, inclusive? The answer is: (37456 – 234) + 1 or 37222 + 1 = 37223 There are 37,223 integers between 234 and 37,456, inclusive. Example 2: How many integers that are divisible by 5 exist between 20 and 200 inclusive? We notice that the difference of 20 and 200 is 180, and 180 divided by 5 is 36. But then, both 20 and 200 are multiples of 5, and we need to add a 1 to 36 to get a total of 37 multiples of 5 in the range between 20 and 200, inclusive. Example 3: How many integers that are divisible by 3 exist between 100 and 200, inclusive? We notice that neither 100 nor 200 is a multiple of 3. Also, 100 is not 1 less than a multiple value of 3 and 200 is not 1 more than a multiple value of 3. Therefore, the number of multiples of 3 in the range is the quotient of (200 – 100)/3 or 33.

Example 4: If m and n are numbers such that n > m and n – m = 6, how many integers exist between m and n, inclusive? θ 6 θ 5 θ 4 θ 3 θ Cannot be determined You will notice that m and n are not defined as integers in this problem, and we are not going to assume that they are. We must be sure to check the scenarios: What if they are

whole numbers and what if they are not? Will the answer be the same? We know that the difference of n and m is 6. If m and n are integers such that m is 1 and n is 7 so that the difference is 6, then there are 7 integers in the range between 7 and 1, inclusive. If we do not consider the end values, then the number of integers in the range is only 5. Let us check the scenario in which m and n are not integers. If m is 1.2 and n is 7.2, then there are only 6 integers in the range between m and n, inclusive. We notice that if m and n are integers, then the number of integers in the range is 7 (if we consider inclusive situation) or 5 (if m and n are not included) but if m and n are NOT integers, then the number of integers is the same as the integer value of difference between the two end values. Two different scenarios, and two different values. We must pick the last choice: Cannot be determined. NOTE: If the problem does not specify “inclusive”, there is a good chance that the end values in the range are NOT integers, and you must be sure to consider the scenarios pertaining to non integer end values. IF the end values are integers, and we need to compute the number of integers excluding the end values, then we need to subtract 1 from the difference of the end values. For example, the number of integers between 25 and 100 excluding 25 and 100 is 100 – 25 – 1 = 74.




You will understand a Factor to be a number that will divide a given a number a whole number of times. If Y is a factor of X, then Y must divide X a whole number of times. Or X / Y must be an integer. We can write this information in mathematical form as: X = k, or X = k. Y Y Where k is an integer. The following statements are interchangeably used, and mean the same thing. Y is a factor of X. X is a multiple of Y. X is proportional to Y Y is a divisor of X Y is a factor of X.

Any of the above statements will translate to a mathematical form: X = k. Y If X is proportional to Y, you will read this statement to mean that Y is a factor of X. In mathematical terms, you will write: X = k.Y If X is inversely proportional to Z, you will read this statement to mean that the reciprocal of Z, viz. 1/Z, is a

factor of X. In mathematical terms, the equation is: X = p. 1/Z where p is an integer. ( 1/Z will divide X an integer number of times to quality as a factor). If X is proportional to Y, and inversely proportional to Z, you will read this information to mean that Y/Z is a factor of X. The mathematical representation of this information is: X = n. Y/Z where n is an integer. In other words, Y/Z will divide X an integer number of times. Example 1: If X is proportional to Y and inversely proportional to Z, and if X is 4 when Y is 2 and Z is 8, then X in terms of Y and Z is:

θ Y/Z θ 4Y/Z θ 8Y/Z θ 16Y/Z θ 32Y/Z

Step 1 is to translate English into math. If X is proportional to Y and inversely proportional to Z, we understand this statement to mean that Y/Z is a factor of X or X = k. Y/Z where k is an integer. Step 2 is to find a value for k. In order to help us do this, the problem tells us that X is 4 when Y is 2 and Z is 8. Let us plug in these values for X, Y and Z to get the value for k. We have: 4 = k. 2/8 or k = 16




Step 3 is to re-write the original equation by replacing k with a numerical value of 16. We have X = 16. Y/Z or X = 16Y/Z We can see that the best answer is the 4th choice. X = 16Y/Z Example 2: If X is proportional to Y, and Y is a multiple of Z, then

1. X is proportional to Z 2. X – Y is proportional to Z. 3. X + Y is proportional to Z.

If X is proportional to Y, we can write: X = k. Y (k is an integer) If Y is a multiple of Z (read this to mean that Z is a factor of Y), then, Y = p.Z (p is an integer) We have X in terms of Y and Y in terms of Z. Can we express X in terms of Z? We can. Let us write pZ for Y in X = k.Y to get X = k.p.z. We notice that this statement means that X is proportional to Z. So we have X = k.p.Z And Y = p. Z What is X – Y ? X – Y = k.p.Z – p.Z = (k.p – p)Z We notice that Z is a factor of X – Y because (k.p – p) is an integer. We can see that this information tells us that X – Y is proportional to Z. What is X + Y ? X + Y = k.p.Z + p.Z = (k.p + p) Z We can clearly see that Z is a factor of X + Y because k.p + p is an integer. In other words, (X + Y) is proportional to Z. We can see that all three conditions are valid.

We could have dealt with this problem by using real-life numbers, instead of algebra. If X is proportional to Y, then we know that X is a multiple of Y. If Y is a multiple of Z, then Z happens to be the least value of the three values. If Z is 1, then Y could be a multiple of 1, say 2. If Y is 2, then X is a multiple of 2, say 4. So, we have X = 4, Y = 2, and Z = 1, and these values satisfy the conditions specified in the problem. Let us test the condition 1. IF X is 4, and Z is 1, is 4 a multiple of 1? You bet. Therefore, condition 1 is valid. Let us test condition 2. IF X is 4 and Y is 2, then X – Y is 2. Is 2 a multiple of 1? You bet. Condition 2 is also valid. Let us test condition 3. If X is 4 and Y is 2, then X + Y is 6. If 6 proportional to or a multiple of 1? Yes indeed. Therefore, all three conditions are valid. We could have chosen an alternate set of values of X, Y, and Z consistent with the way things are defined and still come out with the same results. If you find working with algebra as an abstraction intimidating, then you can choose to work with real-life numbers. But then, you must know that 2 is a multiple of 1. If you have this basic understanding, then you can work with the most intimidating of GMAT problems with ease.




Factors with a Wrinkle. If, when X is divided by Y, we get a remainder of R, then (X-R) must be divisible by Y. In other words, Y must be a factor of (X – R). In mathematical terms, we can write this information as: (X – R) = k (k is an integer) Y Or (X – R) = k. Y Or X = k.Y + R In real life terms, if 32 is divided by 5, we get a remainder of 2. We must conclude that (32 – 2) must be divisible by 5. In mathematical terms, this information will read: 32 = 6.5 + 2 Example 1. If a is divided by 7, we get a remainder of 2 and if b is divided by 7, we get a remainder of 4, then which of the following cannot be a value for (a+b) ?

θ 20 θ 48 θ 76 θ 97 θ 110

The mathematical representation of the information that a, when divided by 7, gives a remainder of 2 is: a = k.7 + 2 Similarly, we can write b = p.7 + 4 What is a + b ? a + b = k.7 + 2 + p.7 + 4 Or a + b = (k+p). 7 + 6 This information tells us that if (a + b) is divided by 7, we should get a remainder of 6. Or, (a + b – 6) must be divisible by 7. Now take a look at the answer choices. You can see that the values in choices 1 through 4 all give a remainder of 6 when divided by 7. 110 cannot be a value for (a+b) because when 110 is divided by 7, we get a remainder of 5, and not 6. We must pick the last choice. AS A RULE, IF X AND Y ARE TWO VALUES, AND EACH OF THESE VALUES IS DIVIDED BY A COMMON FACTOR PRODUCING SEPARATE REMAINDERS R1 AND R2 RESPECTIVELY, THEN (X + Y) WHEN DIVIDED BY THE SAME FACTOR MUST YIELD A REMAINDER EQUAL TO THE SUM OF THE TWO SEPARATE REMAINDERS. In the above problem, a+b, when divided by 7, MUST yield a remainder of 2 + 4 =6. Our task is to look for a choice value that will not provide this remainder value when divided by 7. The value 110 does not produce a remainder of 6 when divided by 7.




Example 2: If we get a remainder of 3 when k is divided by 7, then which of the following will give a remainder of 6 when divided by 7? I. 2k II. 3k + 4

III. k + 3 When k is divided by 7, we get a remainder of 3. The mathematical form for this information is: k = p.7 + 3 Let us test the conditions one by one. I. 2k = (2p).7 + 6 We can see that when 2k is divided by 7, we get a remainder of 6 II. 3k + 4 = (3p).7 + 9 + 4 3K + 4 = (3p).7 + 7 + 6 3k + 4 = (3p+1).7 + 6 We can see that 3k + 4 will yield a remainder of 6 when divided by 7. III k + 3 = p.7 + 3 + 3 = p.7 + 6 We can see that k+3 will also yield a remainder of 6 when divided by 7. We must pick a choice that says “I, II, and III”. We could have dealt with this problem by using real life numbers. What is the least possible value for k consistent with the way the number is defined in the problem? It is 7 + 3 = 10. What is 2k? It is 20. Does 20 give a remainder of 6 when divided by 7? Yes, it does. Similarly, if k is 10, what is 3k + 4? It is 34. Does 34 produce a remainder of 6 when divided by 7? Yes, it does. Lastly, if k is 10, what is k + 3? It is 13. Does 13 produce a remainder of 6 when divided by 7? You bet it does. We can see that all three conditions yield a remainder of 6 when divided by 7. We must pick a choice that says that all three conditions are good.

If you find working with real-life numbers is less messy, feel free to use this approach. Be sure to select values in keeping with the way the problem is described Example 3: N! represents the product of consecutive positive integers from 1 through N. For example, 3! Is the product of 1.2.3. If X = 21! + 17, which of the following cannot be a factor of X?

I. 21 II. 17 III. 13

π I only π II only π III only π I and III only π I, II and III Solution: 21! Represents the product of consecutive positive integers from 1 through 21. One of the factors of 21! Will be 17. Because X is 21! Plus 17, 17 is the only value that will divide X a whole number of times. 21 and 13 will not yield integer values when they divide X because 17, when divided by 21 or 13, does not yield an integer. The answer is I and III cannot be factors of X. We pick the 4th choice.




Example 4: If n is an integer such that 91.k.n is the product of 2.3.5.7.11.13, which of the following cannot be a value for k? π 22 π 30 π 55 π 60 π 66 Solution: If n is an integer, then 91.k must be a factor of 2.3.5.7.11.13. n = 2•3•5•7•11•13 91.k (91.k divides 2.3.5.7.11.13 a whole number of times; therefore, 91.k must be a factor of 2.3.5.7.11.13) Let us try to simplify the expression to a more manageable form. We notice that 7 and 13 on top will cancel out the 91 at the bottom leaving a simpler equation: n = 2•3•5•11 k The value for k must be such that the value will divide the expression on the numerator a whole number of times. We notice that k cannot be 60 because if k is 60, n will not be an integer. We must pick the 4th choice. All other values in the choices 1 through 3 and 5 will yield a whole number value for n.

How would a simple problem asking you to find the factors of a given integer appear in the GMAT? In the following fashion: “If N is an integer such that K•N = 87, how many positive integer values of K exist?” We notice that if N is an integer, we can rearrange the given equation in terms of N as follows: N = 87 / K = Integer Our definition of factors tells us that K must be a factor of 87 because it divides 87 an integer number of times. In fact, K could be any of the values that are positive factors of 87. How many positive factors of 87 can we count? The approximate square root value of 87 is 9. Let us check out all integers between 1 and 9 and see whether any of them or all of them could be factors of 87.

Factors of 87: 1, 87, 3, 29, That is it. There are just four positive factors of 87, and K could be any of these values. Therefore, there are four positive integer values of K exist.




Let us say that we want to find how many factors exist for 145 and what they are. We will do the following drill: 1. We will find out the

approximate square root value of 145. The square root of 145 is roughly 12.

2. We will check out all integer values from 1 through 12 to see if the values in this range will divide 145 a whole number of times.

3. Remember: 1 and the number itself are always factors of any number.

4. 1 will divide the given number 145 times. We will take 1 and 145 as factors.

5. 2 cannot divide 145 a whole number of times. 2 cannot be a factor of 145.

6. Can 3 divide 145 a whole number of times? No. 3 cannot be a factor of 145.

7. Can 4 divide 145 a whole number of times? No. Therefore, 4 cannot be a factor of 145.

8. 5 will divide 145 twenty-nine times. Therefore 5 and 29 are factors of 145.

9. 6 through 12 will not divide 145 a whole number of times, and

these values cannot be factors of 145.

10. Therefore, the factors of 145 are: 1, 145, 5, 29

What are the factors of 256? We will check out all integer values in the range from 1 through 16 (square root of 256). Remember: We will take pairs of values as factors such that the two values will multiply to give the original number. If a factor repeats itself, we will take just one value as the factor. In this example, since the square root of 256 is exactly 16, 256 will be a perfect square, and any number that is a perfect square will have odd number of factors. If you apply the drill outlined earlier in this section, you will find that the factors of 256 are: 1, 256, 2, 128, 4, 64, 8, 32, 16 What are the factors of 171? We must check out all integer values in the range from 1 through 13 (the closest approximation to the square root of 171). The factors are: 1, 171, 3, 57, 9, 19 Remember: If the digits of a number add up to a single digit value of 9, then the number must be divisible by 3 and 9, if the integer is an odd integer; if the integer is an even integer, and the digits add up to a single digit sum of 9, then the number must be divisible by 3, 6, and 9. If the digits of an odd integer add up to 6, then the integer is divisible by




3; if the digits of an even integer add up to 6, then the integer is divisible by both 3 and 6. A number to be divisible by 5 must end in a digit of value 5 or 0. An even integer divisible by 2 must end in 0. 2. 4. 6. Or 8. If an integer has just two positive, and different factors : 1 and the number itself, then the integer is a prime number. Remember: A prime number must have exactly two positive, and different factors. For this reason, 1 is not a prime number. The smallest and the only even integer that qualifies as a prime number is 2. All other prime numbers are odd. Examples of other prime numbers are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, and so on. Remember: All but one prime numbers are odd integers. But all odd integers are not prime numbers. Examples of odd integers that are not prime numbers are: 15, 21, 25, 27, 33, and so on. Example 1: If X and Y are prime integers such that X + Y is also a prime integer, then X or Y must be equal to

2 3 4 5 cannot be determined

This problems tests your understanding of prime integers and of the properties of integers. We know that any prime integer greater than 2 must be an odd integer. We are adding two prime integers X and Y to get another prime integer. This must mean that X + Y is an odd integer. How do we get an odd integer by adding two integers? Only when one of them is an odd integer, and the other an even integer. (If both were odd integers, the sum of X and Y will be even, and cannot qualify as a prime integer). What is the only even integer that qualifies as a prime integer? 2 . Therefore, we pick the first choice as the answer. Did you know that any non-prime integer can be expressed as the product of prime integers, not necessarily distinct? For example, 6 can be expressed as the product of 2 and 3, prime numbers. 27 can be expressed as the product of 3, 3 and 3.




In order to express an integer as a product of prime integers, we will follow the drill suggested here: Let us say that we want to express 910 in terms of its prime factors. We will start dividing 910 by the smallest prime number value that will divide it a whole number of times. In this case, 2 will divide 910 an integer number of times: 455. But 455 can no longer be divided by 2. Let us try the next prime number: 3. 3 cannot divide 455 a whole number of times. The next prime integer that will divide 455 a whole number of times is 5. 5 Will divide 455 Ninety-one times. 91 can no longer be divided by 5. The next prime number that will divide 91 a whole number of times is 7. 7 will divide 91 and yield 13. We notice that 13 is a prime integer, and is no longer divisible. We can, therefore, express 910 as a product of prime integers as: 910 = 2.5.7.13 Similarly, 330 can be expressed as the product of 2.3.5.11 170 can be expressed as the product of 2.5.17 NOTE: If the problem specifies that a given integer is the product of more than 2 factors, each greater than 1, then we must deal with the problem as “prime factors” problem. Example 1.

If 910 is the product of m.n, r, and p such that 1 < m < n < r < p, What is the value of mnp ? π 70 π 130 π 182 π 455 π cannot be determined. We can express 910 as a product of 4 prime integers: 2, 5, 7, and 13. Because of the relationship specified between m, n, r, and p, we must conclude that m=2, n=5, r=7 and p = 13. The product of m, n and p is 130. We pick the second choice as the answer. EXAMPLE 2: If 390 is a product of four integers a, b, c, and d such that 1 < a < b < c < d, then what is ab/cd? If the problem tells us that a given integer can be expressed as a product of more than two integers, we must treat it as a prime factors problem. We can set up 390 in terms of its prime factors as 390 = 2•3•5•13. Because 1 < a < b < c < d , we know that a = 2, b = 3, c = 5, and d = 13. Therefore, ab/cd = 6/65.




We learned that if Y is a factor of X, then X is a multiple of Y. In real-life terms, if 5 is a factor of 30, then 30 is a multiple of 5. (Read ‘multiple of’ as ‘any integer times’). Let us say that X is a multiple of 4. X could be any of the following values, starting with 4 as the least positive value ( 1 times 4; then it goes as 2 times 4, 3 times 4 and so on): 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, and so on. Let us also say that X is a multiple of 5. X could be any of the following values consistent with this definition. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, and so on. We notice that if X is a multiple of 4 and a multiple of 5, then X must be a multiple of 20, the least common multiple value of 4 and 5. If X is a multiple of 20, then it can have values starting with 20 and moving in steps of 20: Likely values are 20, 40, 60, 80, 100, 120, and so on. Do we know precisely what value X has? We do not. All that we know is that X can be any of the values that are multiples of 20.

IF the problem asked you whether X is a multiple of 30 given that it is a multiple of 4 and 5, the answer will be a resounding “may be”. If X is 20, it is not a multiple of 30. If X is 40, it is not a multiple of 30. But if X is 60, then X is a multiple of 30. In fact, if you take a look at the list of likely values of X starting with 20 and moving on in steps of 20, we notice that X is a multiple of 30 for 1 out of every 3 values. Therefore, the probability that X will be a multiple of 30 is 1/3. Any probability of less than 1 or not equal to zero is good for a “may be” answer and not for a definite yes or no answer. IN data sufficiency, if they asked us the following: IS X A MULTIPLE OF 20? 1. X IS A MULTIPLE OF 5 2. X IS A MULTIPLE OF 6 Our N.T.K is that we need to know whether X is divisible by 20 for all likely values of X or we need to know the actual value of X in order to determine whether it is a multiple of 20 or not. Statement 1 tells us that X could be any of the following values: 5, 10, 15, 20, 25, 30, 35, 40, 45, and so on. If X is 5, then it s not divisible by 20. If X is 20, then it is. We notice that for every four likely values of X, only one of them is a multiple of 20. The probability that X is a multiple of 20 given the first statement is ¼, which is good for a “may be” answer. Not good enough. We must kill choices A and D. Statement 2 tells us that X can be any of the following values: 6, 12, 18, 24, 30, 36, 42, 48,




54, 60, 66, and so on. If X is 6, it is not divisible by 20. If X is 60, then it is. In fact, for every 10 likely values of X, only one of them is divisible by 20. The probability that X will be divisible by 20, given that it is a multiple of 6, is a minuscule 1/10. Not good for a definite yes or no answer. We must kill choice B and move on to combine the two statements. We have left two choices: C and E. When we combine the two statements, the least common multiple value that is in both sets of likely values is 30. This means that X is a multiple of 30, and goes in steps of 30 starting with 30. Likely values of X are: 30, 60, 90, 120, 150, 180, and so on. Can we conclude that X is a multiple of 20 for all likely values of X, given that X is a multiple of 30? We cannot. Because, if X is 30 then it is not divisible by 20. If X were 60, then it is. As you can see from the list of likely values of X, X is a multiple of 20 for every other likely value of X. This translates into a probability of ½ that X is divisible by 20. A probability of less than 1 is good for a “may be” answer, and we have learned that in data sufficiency we will not accept “may be” for an answer. We will say that we cannot tell definitely one way or the other whether X is a multiple of 20 unless we know the actual value of X. We must choose E as the answer. When you come across problems asking you to deal with multiples, be sure to proceed methodically and systematically by employing the above reasoning. You cannot go wrong.

Let us try another example. If X is a multiple of 4 and 6, then which of the following cannot be a value for X? π 24 π 36 π 48 π 120 π 150 IF X is a multiple of 4, then it can be any of the following values: 4, 8, 12, 16, 20, 24, 28, and so on. If X is a multiple of 6, then it can be any of the following values: 6, 12, 18, 24, 30, 36, 42, and so on. We notice that the smallest value that is consistent with both specifications is 12. We must conclude that X is a multiple of 12. X could be any of the following values: 12, 24, 36, 48, 60, 72, .. 120, 144, 156 and so on. We notice that X cannot have a value of 150 because 150 is NOT a multiple of 12. We must pick the last choice as the answer. Remember: If X is a multiple of 12, then X is divisible by 12.




Example 2: Is N divisible by 24? (1) N is divisible by 4 (2) N is divisible by 6 π Statement 1 alone but Statement 2 alone is not. π Statement 2 alone but statement 1 alone is not. π Statement 1 and 2 together π Either Statement 1 or statement 2 alone is sufficient π Neither statement, alone or combined, is sufficient. This is a data sufficiency question. The question is: Is N a multiple of 24? Can we answer the question on the basis of statement 1 alone? We cannot because statement 1 tells us that N could be any of the following values: 4, 8, 12, 16, 20, 24, 28, 32, 36,….. If N is 4 or 16 or 28, then it is not divisible by 24. If N is 24 or 48, then it is. More than one possibility here. Not a unique solution. Let us kill choices A and D, and move on to examine statement 2 and see whether we can answer the question in a unique fashion on the basis of statement 2 alone.

Statement 2 tells us that N could be any of the following values: 6, 12, 18, 24, 30, 36, 42, 48…….. Once again, if N is 12 or 18 or 30, it is not divisible by 24. If N is 24 or 48, then it is. The probability that X will be a multiple of 24 given that it is a multiple of 6 is ¼ or not good enough. Let us kill choice B and move on to combine the two statements and see whether we can come up with a winner. When we combine the two statements, we notice that N must be a multiple of 12, the least common multiple of 4 and 6. IF N is a multiple of 12, then N can be any of the following values: 12, 24, 36, 48, 60, 72,…….. Can we conclude that N is a multiple of 24? We cannot. The fact that N is a multiple of 12 does not mean that N is automatically a multiple of 24. The smallest values that N can have is 12, and 12 is not divisible by 24. Or N could be 24, which is a multiple of 24 as well. If N is 24 (or 48 or 72), then N is a multiple of 24. The probability that N is a multiple of 24, given that it is a multiple of 12 is ½, which is good for a “may be” answer, but not for a definite yes or no answer. We must give up and pick the last choice: Neither statement alone or combined is sufficient to answer the question definitively ( yes or no).




Example 3: Exactly ¼ of the boys in a school walk to the school. Exactly 1/5 of the boys in the same school play soccer. Which of the following values cannot be the value for the number of boys in the school? π 20 π 60 π 150 π 240 π 320 Solution: If exactly ¼ of the boys walk to the school, then the number of boys in the school must be a multiple of 4. The number of boys could be any of the following: 4, 8, 12, 16, 20, 24, and so on. Of exactly 1/5 of the boys play soccer, then the number of boys must be a multiple of 5 as well. The number of boys consistent with this definition could be any of the following values: 5,10, 15, 20, 25, 30, 35, …… When we take the two lists of likely values into account, we conclude that the number of boys must be a multiple of 20, the least common multiple in both sets. The only value that is not a multiple of 20 in the answer choices is 150. We must pick the third choice as the answer.

DIGITS OF A

NUMBER A number in a decimal notation consists of digits in specific locations. A digit is an integer having a value between 0 and 9. Consider any decimal number. Let us say that we have 35.467. We can write this number as: 35.467 = (3 X 10) + (5 X 1) + (4 X 110) + ( 6 X 1100)

+ (7 X 11000) As we can see, 3 takes 10 as the coefficient, and is called the “tens’” digit. Likewise, 5 is called the units’ digit; 4 the tenth digit; 6 the hundredth digit; and 7 the thousandth digit. If we multiply the number by 10, each of the coefficients attached to the digits gets multiplied by 10, and the digits become the digit of one higher order. If 35.467 is multiplied by 10, we get 354.67. The 3 is now the hundreds’ digit, the 5 the tens’ digit; 4 the units’ digit; 6 the tenth digit; and 7 the hundredth digit. IF we divide the number by 10, each of the coefficients attached to the digits gets divided by 10, and the digits of this new number become digits of one lower order.




Example 1: If ∇, @, #, $, ! and & represent non-zero digits, and if ∇ @ + # $ $ & ! then $ is π 1

π 2 π 3 π 4 π 5

Solution: If two two-digit numbers are added, and the result is a 3-digit number, then the hundreds’ digit of the 3-digit number cannot be more than 1. (In the extreme case, let us add 99 to 99. The sum is 198, and the hundreds’ digit is 1). Since the digits are specified as non-zero digits, we conclude that $ must be equal to 1.

EXAMPLE 2: DATA SUFFICIENCY If @, #, and $ represent three different positive digits such that @ + # = $, what is the value of #? 1. $ = 4 2. @ > 1

Let us associate choices A and D with statement 1, and B, C, and E with statement 2. We know that the three digits we are looking at – @, #, and $ - are non-zero values (they are defined as positive), and different. Statement 1 tells us that @ could be a 1 or a 3 and # must be a 3 or a 1 so that the sum will be a 4. But then, we have two possible values for # and @, and we cannot answer the question definitely using statement 1 alone. Let us kill choices A and D. Statement 2 tells us that @ is more than 1. That does not explain what # is about. Let us kill choice B, and proceed to combine the two statements. When we combine the two statements, we notice that @ cannot be 1 because a value of 1 will not satisfy statement 2. The only other value possible for @ is 3. If @ is a 3, then # MUST be a 1 so that the sum will be a 4, the value for $. The combined information helps us sort through the mess and determine a unique value for #. We must choose C as the answer. (In computer format, C corresponds to the third choice in the list of choices).




AS a manager-wanna-be, you must demonstrate that you have a good sense of proportion. In fact, GMAT is a Factors, Ratios and proportions test, and they must re-christen the test FRAP test. Example 1: If X is inversely proportional to r2, and if r is halved, then X is: π halved π increased two fold π decreased two fold π increased four fold π decreased four fold Solution: If X is inversely proportional to r2, then we can write: X = k. 1r

2

IF r becomes r/2, then X = k. 1/(r/2)2 = k.4/r2 As we can see, if r is

halved, then X is increased 4 times. We pick the 4th choice.

You will deal with these “sense of proportion” questions when you go to “solid objects” in geometry.

Example 2: The cost, C, of manufacturing an item is given by the formula: C = 0.49• r•s•t2 Where r, s and t are three different component values. If, as a result of plant modernization and labor agreements, r increases by 50%, s increases by 20% and t decreases by 30%, then C π is unchanged π decreases by 12% π increases by 12% π decreases by 20% π increases by 20% Solution: If r increases by 50%, then r becomes 1.5r If s increases by 20%, then s becomes 1.2s If t decreases by 30%, then t becomes (0.7t) And t2 becomes (0.49t2) Let us plug in the changes to the formula: C = 0.49. (1.5r)(1.2s)(0.49t2) C = (0.49) r•s•t2 • ( 0.88) (1.5 times 1.2 times 0.49 gives 0.88) We can see that C is 88% of the original value or a decrease of 12%. We pick the 2nd choice. You can choose to deal with this problem by thinking up real life numbers. For example, R could be 2 initially and change to 3 after the increase. S could be 5 initially and change to 6 after the increase. T could be 10 initially and decrease to 7 after the decrease. Do this and see whether you get the same answer as above.

If the problem requires that you express the answer as a ratio or as a proportion or as a percent value, you can choose to work with your own ‘base-line’ values. If the problem tells us that the radius of a circle doubles, and asks how many times the area is increased, we can choose to work with values of 1 and 2 for R. If R=1, then the area is PI. If R becomes 2, the area is PI•22 = 4•PI The ratio of the new value to the old value is 4/1. Remember: “Y is how many times X” means the same to a mathematician as the statement ‘what fraction of X is Y?”. Both questions will ask you to compute the ratio Y/X.




PROBLEM 2: “The volume of a cylinder is obtained by using the formula Volume = PI•R2•H, where PI is a constant, R is the radius of the cylinder and H its height. If the radius doubles and the height becomes 1/3rd its initial value, the new volume is how many times the old volume?” Notice that ‘how many times’ statement is an invitation to making a ratio statement. We are required to compute the ratio of the new Volume to the old Volume. Because the answer will be in the form of a ratio, we can set our own baseline values and work through the problem. Remember: You can set your own baseline values as long as the final answer is in the form of a ratio. If the problem asked you to find out the value of the ‘new volume’, you cannot set your own baseline values because the answer is not required in a ratio format. We will say that R is initially 1 and becomes 2 after the change; H is initially 3 and becomes 1 after the change. Why did we pick these values? Because we wanted to work with reasonably small values and because we wanted to continue to work with integer values. If we had chosen a value of 1 for the Height initially, then we will have to use 1/3 as the new value after the change. Instead, we chose to work with 3 and 1, integer values. Remember: You can process integers more efficiently and easily than you can process fractions. Also remember to work with as small a value as possible. 3 and 1 are more easy to process and deal with than are the values 6 and 2, or 18 and 6.

Also notice that we will ‘ignore’ the constant PI because the final answer will be in the form of a ratio, and the constant is unaffected by the change. When we do the ratio, the constant, which is unaffected by the change, will cancel out. Old Volume (R=1 and H=3) = 12•3 = 3 New Volume (R=2 and H=1) = 22•1 = 4 New Volume/Old Volume = 4/3 The new volume is 33% more than was the old volume. PROBLEM 3: “Container X is twice the capacity of container Y. If X is half-filled and Y is 3/4th filled, and if all the liquid in cylinder Y is transferred to cylinder X, to what fraction of its capacity is X filled?” Notice that the problem asks us to express the volume of liquid in X as a fraction of its capacity. The answer is required in a ‘ratio’ format, and we will set our own ‘baseline’ values. When we pick our baseline values, we must be sure to factor in the fractions specified in the problem, as otherwise we will be dealing with some crazy fractions along the way. The two fractions specified are ½ and ¾. Let us simply multiply the denominators and say that X has capacity 8 gallons and Y 4 gallons. X Y CAPACITY 8 4 VOLUME 4 3 AFTER TRANSFER 4+3=7 0 X is filled to 7/8th of its capacity. You will get the same answer no matter what baseline values you chose to work with.




Exercise: If x and y are each two digit numbers with the same two digits but in the reverse order, is x - y divisible by 9? One approach to solving this problem is to think up two 2-digit numbers: Say, 27 and 72. When you take the difference, we get 72 - 27 = 45. We know that 45 is divisible by 9. How do you work with algebraic expressions to get the same result and reach the same conclusion? Let us set up x as x = 10 t + u .... (1) Let us set up y as y = 10 u + t .....(2) What is x - y ? x - y = 10 t + u -(10 u + t ) = 10t + u - 10u - t x - y = 9t - 9u = 9 ( t - u) Is (x - y) divisible by 9 ? You bet. FRACTIONS, RATIOS, PROPORTIONS, AND PERCENTAGES ALL REFER TO PRETTY MUCH THE SAME INFORMATION. A fraction of ½ represents a ratio of 1 to 2 or a percentage

value of 50%. Proportionality will be tested big time in the GMAT, and questions appearing in the GMAT will have the following genre: EXAMPLE 1: “ABC company employed 537 more people in 1999 than they did in 1998. If the additional number of employees in 1999 represented an increase of 30% on the number for 1998, how many employees were there in 1998?” We can see that a number value of 537 corresponds to a percent value of 30%. The question is: What is 100%? Our proportionality statement in an equation form will be: 30/537 = 100/ X1998 Where X1998 represents the number of employees in 1998. Let us cross multiply and rearrange the equation to get: X1998 = 100•537/30 = 1,790 The company employed 1,790 employees in 1998. The value for 1999 will be 537 more than 1,790, or 2,327.

TRIVIA: If X and Y are two integers such that X and Y are made up of the same digits, not in the same order, then the difference of X and Y MUST be divisible by 9. For example, if X is 65 and Y is 56, then the difference is 9, which is divisible by 9. If X is 123 and Y is 312, then the difference is 189, which is divisible by 9. IF X is 2576 and Y is 7625, then the difference is 5049, which is divisible by 9. Now, go ahead and try to impress someone you know with this trivia.




EXAMPLE 2: “50 fish were caught, tagged, and returned to the pond. During a second catch of 50 fish, 2 were found to be tagged. If the percentage of tagged fish in the second catch approximates the percentage of all tagged fish in the entire pond, how many fish are in the pond?” We know that the pond has 50 tagged fish. In the second catch, the percent of tagged fish is 2/50 or 4% of the catch. The problem tells us that 4% represents all tagged fish in the entire pond. This means that 4% corresponds to a value of 50. The question is: What is 100%? Our set-up will read: 4/50 = 100/X or, X = 50•100/4 = 1,250 The number of fish in the pond is 1,250 and the number of tagged fish – 50 – represents 4% of this total. Our verification shows that the value we obtained makes sense. Understand the problem information and then proceed to set it up as a simple proportionality equation. EXAMPLE 3:

“A train engine’s maximum speed is 120 m.p.h when no cars are attached, and the maximum speed decreases in proportion to the square root of the number of cars attached. IF 4 cars are attached, the maximum speed drops by 30 m.p.h to 90 m.p.h. How many cars of the above type need to be attached so that the engine will stop moving?” The question is: How many cars should be attached so that the speed reduction will be in the order of 120 m.p.h. We know that a speed reduction of 30 corresponded to square root of 4. The question is: A speed reduction of 120 corresponds to how many cars? Because the speed reduction is in proportion to the square root of the number of cars attached, our set-up for this problem will look like this: 30 /sqrt(4) = 120/sqrt(N) where N is the number of cars that will bring the engine to a grinding halt. Sqrt (N) = 120• sqrt(4) / 30

= 4• 2 = 8 Let us square both sides to get: N = 82 = 64. We need to attach 64 cars in order to bring the engine to a halt.




Assignment: Integers 1. Integer 9 is not a divisor of which of the following values of y? (A) 63 (B) 108 (C) 36 (D) 200 (E) 126 Your Answer:_______________ 2. Even integer when added to an Odd integer results in __________ integer. 3. Even integer when multiplied by an even integer results in __________integer. 4. An odd integer when subtracted from an even integer results in _____________ integer. 5. A prime number n is an integer that has exactly ________ different _________ divisors and the divisors are ________________________________. 6. Which of the following numbers cannot be prime number(s) and why? (A)3 (B)169 (C) 32 (D) 83 (E) 59 Your Answer:_____________________ _________________________________ _________________________________ 7. The only even prime number is_______. 8. The expression n/n=1 is true for all values of n except _________________

9. How many factors or divisors do the following integers have? What are they? (A) 81 ____________ _________ (B) 140____________ __________ (C) 256____________ ___________ (D) 171____________ ____________ 10. The difference of two consecutive even integers is _________________ 11. The difference of two consecutive odd integers is __________________ 12. What is the next integer in the following sequence of even integers? 2n, 2n+2, 2n+4, 2n+6,_________ 13. Which of the following are odd, even or indeterminate integers? (n is not 0) Odd Even Can’t Say (A) 2n+3 O O O (B) n+5 O O O (C) 2n+24 O O O (D) 2n-3 O O O (E) 2(n-7) O O O (F) 3(2n-1) O O O (G) (2n+15) - (2n-4) O O O (H) (2n+64) + (3n+5) O O O (I) n(2n-11)÷n O O O (J) (2n+1)(2n+6) O O O Hint: n takes on all values (except 0), positive and negative. 14. If x and y are prime numbers and (x+y) is a also a prime number, then x or y must be equal to ______________




Fractions

fraction is obtained by taking the ratio of two integers, and has the form: n/d, where n

is the numerator and d the denominator. The smallest value for the fraction is obtained when the numerator is the smallest and the denominator the largest values. The largest value for the fraction is obtained by making the numerator the largest it can be and the denominator the smallest it can be. Let us say that we have 5 integers in a set: 3, 13, 17, 23, 31. If we are required to make the smallest fraction using the values in the set, it will be 3/31. The largest fraction we can set up using the same values in the set will be 31/3. You will be dealing with a lot of fractions in the GMAT, and you are advised to deal with fractions as given instead of turning them into decimal numbers. GMAT is, in the final analysis, a factors test, and you will be able to simplify expressions by canceling out common factors contained in the fractional expressions. If you are required to add two fractions, you will be required to bring both to a common denominator value and add the numerators. Example: Add 2/7 and 3/11. The trick is to multiply the numerator and the denominator of the first fraction by the denominator of the second fraction, and the numerator and the denominator of the second fraction by the denominator of the first fraction. Let us multiply the numerator and the denominator of the first fraction by 11. We get: 2/7 = 22/77 Let us multiply the numerator and the denominator of the second fraction by 7. We get: 3/11 = 21/77 As you can see, we have expressed the fractions 2/7 and 3/11 in terms of a common denominator 77. What remains to be done is to add the numerators 21 and 22 while keeping the same denominator.

We get: 2/7 + 3/11 = 22/77 + 21/77 = 43/77 Example 2:

Add a/b and c/d. Let us multiply the numerator and the denominator of the first fraction by d, and those of the second fraction by b so that both fractions will be expressed in terms of a common denominator. We get:

a/b + c/d = ad/bd + bc/bd = (ad + bc)/bd We will follow the same procedure when we have to subtract two fractions. Example:

2/9 – 3/7 = 14/63 – 27/63 = -13/63

Multiplication and Division: Multiplication of two fractions is a simple process: We multiply the numerators and the denominators separately and form the new fraction. Before we begin to multiply, we must try to cancel out common factors so that we can have a fraction in its simplest form. Example:

3/8 • 4/7 = 3/2 • 1/7 = 3/14 Division of a fraction by another fraction is done by taking the inverse of the denominator fraction, and multiplying the numerator fraction. Example:

(3/5) ◊ (5/9) = 3/5 • 9/5 = 27/25

Example:

(a/b)◊(c/d) = a/b • d/c = ad/bc

Sometimes, we may have to deal with a mixed number that contains an integer and a fraction.

A




Example: 1 ¼ = 1 + ¼ = 4/4 + ¼ = 5/4

GMAT will test your comfort level with

fractions. Take a look at the following examples.

Example 1:

What number is 3/5 times as far away from 2/3

as 6 4/7 is from 5/4?

The problem is simply one of finding a value for the following expression: 2/3 + 3/5 • (6 4/7 – 5/4) Let us work on the expression within the parenthesis first. (6 4/7 – 5/4) = 46/7 – 5/4 = 184/28 – 35/28 = 149/28

Let us now deal with the given expression:

2/3 + 3/5 • (6 4/7 – 5/4) = 2/3 + 3/5 • 149/28

= 2/3 + 447/140

= 280/420 + 1341/420

= 1621/420

= 3 361/420 (ANSWER)

Example 2:

“If a, b, and c are three different digits and can

have any of the following values: 3, 5 or 7, what

is the smallest possible value for the expression

(a/b) / c?

The expression (a/b) / c is equivalent to

(a/b) / (c/1), which translates to:

a/b • 1/c = a/b•c

Because we are the looking for the least possible

value for the expression, we must assign the

smallest of the three possible values to a and

assign the other two values to b and c.

The least possible value for the given expression

is, therefore, 3/35.

IF the problem specified that we get the

maximum possible value for the given

expression, then we must assign 7 to a and the

other two values to b and c. The greatest

possible value for the given expression is,

therefore, 7/15.

Example 3:

Which of the following expressions has the highest value?

5/22•52

6/22•53

62/23•53

32/22•53

15/2•53

Solution:

We notice that the fraction with the greatest

numerator in the choice expressions has the

denominator reading 23•53

Let us write all choices in terms of the base

denominator value of 23•53. (This means that we

have to multiply the numerator and the

denominator of the first choice by 2•5, the

numerator and the denominator of the second and

the fourth choices by 2 and those of the final

choice by 2•2).

The choices begin to look as shown below after we

have manipulated the numerators and the

denominators so that the denominators of all

choices will be the same value:

50/23•53

12/23•53

62/23•53

64/23•53

60/23•53

The answer is option D, which has the greatest value for the numerator. Remember to express all fractions in terms of a common denominator when you are asked to compare fractions.




Example 4: The price of a stock at opening bells was 67

8, and the price of the same stock at closing bells was 73

7. By what percent did the price of the stock change during the trading day? % Change =

(change in value)/(opening value) •100% % Change = [(52/7 – 55/8) + 55/8 ]• 100% = [(416/56 – 385/56) +55/8 ]• 100% = [31/56 • 8/55] • 100% = 31/7 • 1/55 • 100% (8 cancels 56

seven times)

= 31/7 • 1/11 • 20% (100 and 55 have a

common factor of 5)

= 620/77% = 8% (approximately) The price increased by 8% over the opening value.

Note: A fraction is turned into a percent value simply by the process of multiplication of the fraction by 100. Example 5:

5/100 is what percent of 5/1000? 5/100 is the same as 50/1000. 50/1000 is ten times 5/1000, or 10•100% = 1000% of 5/1000.

Note: 5/1000 is 10% or 1/10th of 5/100 (50/1000).

Example 6: 10/0.0005 = ?

We notice that 0.0005 is the equivalent of 5/10,000,

and 10/(5/10,000) = 10 • (10,000/5) = 20,000.

We could have obtained the same results by

multiplying the top and the bottom by 10000 so

that we can turn 0.00005 into 5. The numerator

becomes 100,000 and when 100,000 is divided

by 5, we get 20,000.

Example 7:

1 ¾ + 6 7/ 8 = ?

6 + 1

¾ + 5/9

The numerator expression simplifies to:

7/4 + 55/8 = 56/32 + 220/32 = 276/32 = 69/8

The denominator expression simplifies to:

6 + 1 = 6 + 36/47

27/36 + 20/36

= 282/47 + 36/47 = 318/47

The given expression is the ratio of:

69/8 + 318/47

= 69/8 • 47/318

= 3243/2544 = 1081/848 (We killed the common factor of 3 from the top and the bottom).




RATIOS A ratio specifies the relative proportion of the component values in a set. If the ratio of men to women to children in a population is 5:4: 4, we can conclude that for every 5 men in the population, there are 4 women and 4 children respectively. Do we know anything about the size of the population or the numbers of men, women, and children in the population? We do not. All that we know is that the numbers of men, women, and children have a common multiple that will cancel out leaving behind the ratio as specified above. For example, the population could consist of 5000 men, 4000 women and 4000 children; or, the population could consist of 5 million men, 4 million women, and 4 million children. All that we can say from the above ratio is that there are 5N men, 4N women, and 4N children in the group. (N is a common multiple, which will cancel out when a ratio is taken). Total value for the population is = 5N + 4N + 4N = 13N Men are 5N/13N • 100% = 38% of the population. Women are 4N/13N • 100% = 31% of the population Children are 4N/13N • 100% = 31% of the population Example: “Mary and Cathy have marbles in the ratio 11: 7. If Mary gifts away 3 of her marbles to Cathy, then the new ratio of marbles will be 16:11. How many more marbles does Mary have than Cathy does after the gift?” Solution: Step 1: Because we do not know how many marbles that each lady has, we will start by saying that May has 11N marbles and Cathy has 7N marbles so that we have a ratio of 11:7. After the gift, Mary will have 11N-3 marbles.

After receiving 3 marbles, Cathy will have 7N+3 marbles. The new ratio, in terms of N, after the gift is: (11N – 3) (7N+3) The numerical value of this ratio is 16/11. Our equation is: (11N – 3) = 16 (7N + 3) 11 Let us cross multiply and set up the equation: 11•( 11N-3) = 16•(7N + 3) 121N – 33 = 112N + 48 9N = 81, or N = 9 Let us plug in the value for N to get the number of marbles Mary and Cathy have after the gift. Mary has 11N – 3 = 96 marbles. Cathy has 7N + 3 = 66 marbles. Answer: Mary has 96-66 = 30 more marbles than Cathy does after the gift. Learn to set things up in the form of a matrix as shown below so that you can see things in clear perspective.

Before After

Mary 11N 11N-3

Cathy 7N 7N+3

Ratio 11/7 (11N-3)/(7N+3) = 16/11

NOTE: As you can see, we cannot determine the new ratio if the problem specifies that Cathy loses 3 marbles and Mary gains 3 marbles because we do not know to or from what initial values we are adding or subtracting the values. But, if we are told the new ratio after the exchange of marbles, we can determine the values initially and the values after the exchange.




Ratio of more than two quantities and how to get a

combined ratio? If X to Y is 2 to 3 and Y to Z is 3 to 5,

what is X:Y:Z? WE can see that for every 2 X’s, we have 3 Y’s and for every 3 Y’s, we have 5 Z’s. The ratio of X:Y:Z is 2:3:5. If X to Y is 2 to 3 and Y to Z is 5 to 7,

what is X:Y:Z? We notice that the proportional values for Y in the two statements are not the same. We need to bring them to a common value so that we can express the ratio of X to Y to Z. The drill is as follows: X Y Z 2 : 3 5 : 7 We notice that we have two values for Y’s in the two ratios, but these values do not have anything in common. We NEED to bring the two Y values to the same level so that we can tell for a given number of Y’s how many of X’s we have and how many of Z’s we have. Let us do the trick we will apply when we deal with two fractions, with different denominators: we bring them to a common denominator value. Let us do the same in this case. After all, a ratio is nothing but a fraction. (The ratio of X to Y is the same as the value of X/Y). Multiply the first ratio by the proportional value of Y from the second ratio: In this case, 5 The first ratio becomes X: Y =10:15 Multiply the second ratio by the proportional value of Y from the first ratio: In this case 3. The second ratio becomes X:Y = 15:21 We notice that for every 10 X’s, we have 15 Y’s and for every 15 Y’s, we have 21 Z’s. The ratio of X to Y to Z is 10:15:21

Let us do one more to understand this drill: If the ratio of men to women in a club is 3 to 4 and the ratio of women to children in the same club is 5 to 2, what is the ratio of men to women to children in the club? We notice that the “women” values in the two ratios are not the same. The drill is that we multiply the first ratio by the “women” value in the second ratio. The first ratio becomes 15:20 We also multiply the second ratio by the “women” value of the first ratio. The second ratio becomes 20:8 We can see that for every 15 men, there are 20 women and for every 20 women, there are 8 men. The ratio of men to women to children is 15:20:8 Exercise:

If x = 4y and 3y = 7z, what is x:y:z? We notice that the first statement tells us that the ratio of x to y is 4:1. The ratio of X to Y is the same as the value of X/Y. If X = 4Y, then X/Y = 4/1 or the ratio is 4 to 1. The second statement tells us that the ratio of y to z is 7:3 because if 3Y = 7Z, then Y/Z = 7/3. We notice that Y has two values 1 and 7 but we need to bring them to the same common value so that we can tell for a given number of Y’s, how many X’s and how many Z’s exist. Let us multiply the first ratio by 7 to get the ratio of X to Y as 28:7 Let us multiply the second ratio by 1 to get the ratio of Y to Z as 7:3

X Y Z 28 : 7 7 : 3

We notice that for every 7 y’s we have 28 x’s and 3 z’s. The ratio of x:y:z is 28:7:3




Exercise: The ratio of men to women in a club is 2 to 3, and the ratio of women to children in the same club is 4 to 3. If a total of 580 members are in the club, how many more women are there than children? The ratio of Men to women is 2 to 3 and that of women to children is 4 to 3. We notice that the “women” numbers are the same in both statements, and we need to bring them to the same value. Let us multiply the first ratio by 4 and the second by 3. The ratio of Men to women becomes 8 to 12. The ratio of Women to children becomes 12 to 9. We can see that the ratio of men to women to children is 8:12:9 We can see that there are 8N men, 12N women and 9N children for a total of 29N. We are given that 29N = 580 or N = 20. Let us plug the value of N back into the numbers to get: Number of men in the club is 160. Number of women in the club is 240. Number of children in the club is 180. There are 60 more women than there are children in the club. Assignment: (Please show your set up.) 1. If the ratio of biographies to novels in a

library is 5 to 7 and the ratio of novels to science books is 3 to 8, how many more science books than the other two types combined are in the library if the library has a total collection involving these three categories of 1,656 books?

Answer: 360 2. A town has passenger cars of the following

three descriptions: sedans, minivans, and sport utility vehicles, and the total population of these vehicles is 1,590,000. The ratio of sedans to minivans is 2 to 5 and the ratio of minivans to sport utility vehicles is 4 to 5. How many more sedans and minivans combined are in the town than there are sport utility vehicles?

Sedans: Minivans: SUV’s

2 8 5 20

4 20 5 25

Ratio of Sedans:Vans:SUV is 8:20:25 Number of sedans = 8N Number of Vans = 20N Number of SUV’s = 25N Total number of all types of vehicles = 53N = 1,590,000 N = 30,000 (Sedans + Minivans) – SUV’s = 28N – 25N = 3N = 90,000 vehicles. (Answer) 3. The ratio of the number of men and women

in a club is 3 to 4. If 6 men resign their membership and 2 more women join the club, the new ratio becomes 3 to 5. How many more women than men are there in the club after the change?

Before After

Men 3N 3N-6

Women 4N 4N+2

Ratio ¾ (3N-6) /(4N+2) = 3/5

Complete the problem using the above set-up. Answer: 20 4. Adam, Bob, and Chris agree to divvy up the

lottery winning in the ratio of 3: 4: 5. If the combined share of Bob and Chris is $180,000, What is the share, in dollars, of Adam?

Answer: $60,000 5. A car dealership has 150 new cars in its lot.

The cars are in two sizes: Mid-size and Large. If there are three large sized cars for every two mid-sized vehicles in the lot, and if there are no other sizes of cars in the lot, how many large sized and mid sized cars, respectively, are in the lot?

Answer: 90 and 60




6. Two-thirds of the 23.4 million air travelers from Chicagoland used Chicago’s O’Hare airport. If the number using Chicago’s O’Hare was twice the number using the Mid-way airport, and if the number using Gary, Indiana’s municipal airport was one-third the number using Chicago’s Mid-way airport, approximately how many million passengers used the Gary’s municipal airport?

Answer: 2.6 million We saw earlier in this discussion that if 3X = 4Y, then the ratio of X to Y is 4 to 3. As a rule, if MX = NY, then the ratio of X to Y is the ratio of coefficient of Y to that of X, or N to M. If 2X = 5Y, then the ratio of X to Y is 5 to2. If 6X = 4Y, then the ratio of X to Y is 4 to 6. Try this one for size: 7. “If the ratio of Men to Women is A to B,

and that of men to children is C to D, what is the ratio of men to women to children?"

Hint: Set up the problem as follows: MEN WOMEN CHILDREN

A B

C D

Let us bring the two values A and C to a common “denominator” value of AC by multiplying the first ratio by C and the second ratio by A. We get: MEN WOMEN CHILDREN

AC BC

AC AD

What do we see here? For every BC women, there are AC men and AD children. Therefore, the ratio of men to women to children is: Answer: AC:BC:AD 8. “If the ratio of passenger cars to mini-vans

in a town is X to Y, and that of passenger cars to S.U.V’s is Y to Z, the ratio of passenger cars to mini-vans to S.U.V’s is what?” PASSENGER CARS

MINI VANS

S.U.V’s

X Y

Y Z

Answer: XY : Y2 : XZ 9. There are twice as many apples in a basket

as oranges, and three times as many oranges as pears. If there are 100 fruits of the above three description, How many more apples than oranges are there in the basket?”

Hint: The ratio of apples to oranges is 2 to 1,and that of oranges to pears is 3 to 1. Answer: 30 NOTE: If we have to write as an equation the statement “there are twice as many apples as oranges”, we would write A = 2•O, and read

the ratio of apples to oranges as 2 to 1. “Three times as many oranges as pears” will be written: O = 3•P or the ratio of oranges to

pears will be 3 to 1. You should have the flexibility to deal with the same information in either manner. GMAT is about flexibility

and not about rigid thinking.




PROBABILITY IS A PROPORTION STATEMENT

Probability is another way of expressing a proportion information. If the problem specifies a ‘ratio’, it is possible for us to use the ratio information to compute what fraction of the total population is each of the values in the ratio. Let us say that the ratio of males to females in a population is 2 to 3. A ratio is a ‘conditional statement’ and tells us that IF there are 2 males, there must be 3 females for a total group size of 5 people. Therefore, males are 2/5th of the total group, and females are 3/5th of the total group. If we were to randomly pick a person from the population, the probability that it will be male is the same value as what fraction of the total is made up males. In this problem, the probability of randomly picking a male is 2/5. Similarly, the probability of randomly picking a female is 3/5. Let us sum up the procedure for computing probability using a ratio statement: If the ratio of males to females in a group is X to Y, then the probability of randomly picking a male is X/(X+Y). The probability of randomly picking a female is Y/(X+Y). Let us say that a State Congress is made up of Republicans and Democrats. If the ratio of Republicans to Democrats is 3 to 4, then we know that IF there are 3 Republicans, there must be 4 Democrats for a total Congress size of 7 people. Therefore, the probability of randomly picking a Republican is the same as what fraction of the total Congress is made up of Republicans. It is 3/7. Similarly, the probability of randomly picking a Democrat from the population is 4/7. Take a look at the following problem: “If a box contains 10 Red balls, 13 green balls, and 17 yellow balls, and if a ball is to be drawn randomly from the box, what is the probability that it is NOT Red?” “Probability of NOT RED” is the same as the probability of ‘green or yellow’. We are required to express the number of Green and Yellow balls as a fraction of all balls in the box. Green and Yellow add up to 30, and

there are 40 balls in the box. Therefore, the probability of not picking a Red is the same as the proportion of Green and Yellow balls in the box. It is 30/40 or 3/4th. Some problems will test your ability to apply a fraction on another fraction. Such problems take on the connotation of ‘conditional probability’, which is nothing but a ‘fraction of a fraction’ problem. Consider the following problem: “In a population, the ratio of males to females is 3 to 5. Among the males, the ratio of smokers to non-smokers is 2 to 5. If a person is to be randomly chosen from the population, what is the probability that a male-smoker will be picked?” IF the ratio of Males to Females is 3 to 5, then we know that the probability of picking a male is 3/(3+5) = 3/8, and that of picking a female is 5/8. IF the ratio of smokers to non-smokers among men is 2 to 5, we know that 2/7th all males are smokers and 5/7th of all males are non-smokers. The probability of picking a male smoker from the total group = 2/7•3/8 = 3/28. This means that 2/7th of 3/8th of all people in the group are male-smokers. Take a look at another problem. “Students pick one of the three majors: English, French, or Spanish. If the ratio of English majors to French majors in a school is 5 to 3, and if the ratio of French majors to Spanish majors is 4 to 7, and if a person is to be randomly chosen from this group, what is the probability that a Spanish major was picked?” We need to express the combined ratio of English to French to Spanish. Notice that French is a common value to both ratios, and before we can use this common value to express a combined ratio of three values, we must make the values for French equal in both ratios. The least common multiple for 3 and 4 is 12, and we must write the French value in terms of 12, and all other values using the same multiples.




We get: E F S 5 20 3 12

4 12 7 21 We can see that for a group of 12 French majors, there are 20 English majors and 21 Spanish majors. The combined ratio is 20:12:21 for a total of 53 units. English majors are 20/53 of the total. French majors are 12/53 of the total. Spanish majors are 21/53 of the total. Therefore, the probability of randomly picking a Spanish major is 21/53. PROBLEM 3: “A State Legislature is made up of Hottentots and Rottentots. If the ratio of Hottentots to Rottentots is 5 to 6, and if the ratio of Catholics to non-Catholics in the group is 4 to 5, and if a person is to be randomly chosen from the Legislature, what is the probability that a Hottentot subscribing to the Catholic faith will be picked?” If the ratio of H to R is 5 to 6, then the proportion of H in the group is 5/11 and that of R is 6/11. If the ratio of Catholics to non-Catholics is 4 to 5, then Catholics are 4/9th of the total group, and Non-Catholics 5/9th of the total group. We should expect 4/9th of 5/11th of the total group to be made up of Catholic Hottentots. Therefore, the probability of randomly picking a Catholic Hottentot is 20/99 or approximately 1/5.

PROBLEM 4: “A box contains 7 Red balls, 5 Blue balls, and 6 Green balls. Three balls are drawn from the box sequentially, without replacement. What is the probability that AT LEAST one Red ball was drawn?” We are required to compute the probability that at least 1 Red was drawn. “At least one” means “one or more”. The easiest approach to dealing with a problem that specifies ‘at least one’ of a type of outcome is to find out the probability of an undesired outcome and subtract it from the total probability of 1. Remember that a probability of 1 is a certainty. In this problem, we do not want an outcome in which NONE of the three balls is Red. This outcome is possible when each of the three balls drawn is either Blue or Green. The probability of drawing a Blue or Green in each of the three successive draws without replacement = 11/18•10/17•9/16 = 55/272 Notice that we had 11 balls of Blue or Green color out of a total of 18 when we were about to draw the first ball. Because the ball was not replaced, the total number of balls is 17 for the second draw, and 16 for the third draw. Also, we need to subtract the total number of blue and Green balls by 1 for each successive draw because we are assuming that the ball drawn is not red. Probability of picking a blue or a green in each of the three successive draws = 55/272 Notice that this is the probability of an undesired outcome. Probability of picking at least 1 Red ball = 1 – probability of picking no red ball = 1 – 55/272 = 272/272 – 55/272 = 217/272 Remember: PROBABILITY IS NOTHING BUT A PROPORTION STATEMENT. Probability of picking X in a random selection is the same as what fraction of the total group is represented by X. Conversely, if the probability of picking a male in a random selection is 1/3, then 1/3rd of the total group is made up of males.




Averages The average of a group of values is the ratio of the sum of all the values to the number of values in the group. The sum of a set of values is the product of the average of the set and the number of values in the set. Example: “If the set X contains the values n, n+1, 2n+1, 3n+1, and 4n+1, what is the average of this set of values?” Average = (Sum of all values) / (No. of values) = (n+n+1+2n+1+3n+1+4n+1) / 5 = (11n + 4) / 5

Example 1: “If x + y = 11z, what is the average, in terms of z, of 3x, 3y, and 6z?” The average of 3x, 3y, and 6z is (3x+ 3y + 6z) = x + y + 2z 3 We are required to get the answer in terms of z only. This requirement mandates that we get rid of x and y in the answer. How can we do it? We are given that x + y = 11z. Let us rewrite the answer by replacing (x + y) with 11z. We get: The average of 3x, 3y, and 6z, in terms of z, is 11z + 2z = 13z (Answer) Example 2: “The average of six consecutive positive odd integers is 20 less than twice the largest integer in the set. What is the average of the set of integers?”

Let us define the consecutive positive odd integers as 2n+1, 2n+3, 2n+5, 2n+7, 2n+9, and 2n+11. The sum of the six integers is 12n + 36. The average of the six integers is 2n + 6. The equation for the information that “the average of six consecutive integers is 20 less than twice the largest integer in the set” is: 2n + 6 = 2(2n+11) - 20 2n + 6 = 4n + 22 - 20 2n = 4 or n = 2 The average of the integers is: 2n + 6 = 10 Example 3: “The sales in the first half of the year were X dollars. The sales in the second half of the year were on an average $10,000 more per month than in the first half. What is the expression, in terms of X, for the average sales per month during the year?” Sales in the First half = X dollars. Sales in the Second half = X + 6•$10,000 = X +$60,000 Total sales for the year = X + X + $60,000 = 2X + $60,000 Average sales per month during the year = (2X + $60,000) / 12 = X/6 + $5,000 We should look for a choice that contains the expression: X/6 + $5,000.




Example 4: ABC couriers charge 75 cents for the first pound and 50 cents for each additional pound of package sent. What is the average cost per pound, in dollars, of a package weighing p pounds? (1 dollar = 100 cents) Cost of sending the first pound = 75 cents Cost of sending the additional (p-1) pounds = (p – 1)• 50 cents Total cost of sending the package = 75+(p-1)•50 = (25 + 50•p) cents Average cost per pound =(25 + 50•p) / p cents Average cost per pound in dollars = (25 + 50•p)/100•p dollars = 25(1+2p)/100•p dollars = (1 + 2p)/4p dollars Example 5: “N students agree to share equally the cost of a gift to their teacher. The gift cost X dollars. If M students later failed to contribute their share, and the total cost of the gift was shared equally by the remaining students, how many more dollars does each contributing student pay as their share of the cost of the gift?” Scenario 1: Cost of the gift = X dollars Number of students contributing = N Share of each student = Cost / Number sharing the cost = X/N dollars Scenario 2: Cost of the gift = X dollars Number of contributing students = (N-M) Share of each student = X / (N-M) dollars Additional share of each student under the scenario 2 = [X/(N-M)] - X / N = X(N - N + M) / N•(N-M) = MX N•(N-M)

Example 6 If x+y = 3z, what is the average of x, 3x, 4y, and 12z, in terms of z? The average of any set of values is the sum of the values divided by the number of values in the set. We have four values: x, 3x, 4y, and 12z. The average of the four values is: (x+3x+4y+12z) / 4 = (4x+4y+12z)/4 = x + y + 3z We notice that we are required to state the answer in terms of z only. Can we find a way to get rid of x + y from the above answer? Yes, we can. We are given that x + y = 3z. Let us replace x + y with 3z to get: The average of 4 values = 3z+3z = 6z. Example 7: “The average of Sam’s three consecutive weekly savings was $80. If Sam’s first weekly savings was one half that in the second week and one third that in the third week, how much did he save the second week?” That the average of three week’s savings was $80 tells us that the total savings in the three weeks are $240. Let x be the savings in week1, y that in week2, and z that in week3. We have : x + y + z = $240 ………. (1) We also have: x = ½ • y or y = 2x And x = 1/3 • z or z = 3x Let us rewrite (1) by replacing y and z in terms of x only. We get: x + 2x + 3x = $240 or x = $40 Because y = 2x, we conclude that y = $80. If the problem asked you to find the value for z, you will go back to z = 3x and conclude that z = $120




Example 8: Data Sufficiency: “If the average of a set of 4 integers is 45, how many integers in the set are greater than 45?” 1. Two of the integers in the set are

50 and 35. 2. One of the integers in the set is 20. We notice that the sum of the four integers in the set is 4 times 45 or 180. 3. Statement 1 tells us that two

integers add up to 85. This information tells us that the remaining two must add up to 180-85 = 95. The remaining two numbers could be both bigger than 45 (say, Two of the integers in the set are 50 and 35.

4. One of the integers in the set is 20. We notice that the sum of the four integers in the set is 4 times 45 or 180. Statement 1 tells us that two integers add up to 85. This information tells us that the remaining two must add up to 180-85 = 95. The remaining two numbers could be both bigger than 45 (say, 47 and 48) or just one integer could be more than 45 (say, 80) and the other less than 45 (say, 15). We are looking at two conflicting scenarios here and must conclude that the statement 1 is not good for a unique answer. We have eliminated choices 1 and 4 as possible contenders. We will have to choose one of the three remaining choices. The statement 2 tells us that one of the values is 20. This means that the remaining three add up to 160. All three could be more than 45 each, or just two are more than 45 or just one. (say, 100, 35 and 25). Once again, multiple conflicting scenarios here. We

have eliminated choice2 as a viable choice at this point. We have to choose one of the remaining two: choice 3 or choice 5. When we combine the two statements, we have information pertaining to three of the four values, and we can determine that the fourth value must be 75. We can answer the question definitively now, using the combined information. We must pick choice3. Example 8: “A group consists of 20 men and 15 women. If the average I.Q. of men in the group is 125 and that of women in the group is 130, what is the average I.Q. of all members of the group?” This is a weighted average problem. We need to determine the total I.Q values for men and women in the group, add them, and divide the total by the total number of men and women in the group. Average I.Q. = (20•125 + 15•130) / (20+15) = (2500+1950) / 35 = 4450/35 = 127 (approx.) The average I.Q of all members in the group is about 127. 10. “If the average of (a+4), (b+5), (c+6), and (d-11) is 11, what is the average of (a + b) and (c + d)?” The question is: What is ½ • (a + b + c + d)? The only unknown quantity is the value for (a + b + c + d). Can we determine that by using the first piece of information? Yes, we can. The set up for the first piece of information is: ¼ • (a + 4 + b + 5 + c + 6 + d – 11) = 11 Or, a + b + c + d + 4 = 44 Or, (a + b + c + d) = 40. The average of (a + b) and (c + d) is ½ • 40 = 20.




Example 9: “The average price per dozen of donuts during the month of July was $5.95, and that during the month of August was $6.40. If twice as many dozens were sold in August as in July, what was the average price per dozen for the two month period?” This is a weighted average problem. If twice as many dozens were sold in August as in July, then we know that the ratio of sales in August to that in July is 2 to 1. We can set the quantities as X for July and 2X for August or simply take quantity 1 for July and quantity 2 for August. Remember, we are dealing with proportions here and whether we take X and 2X or 1 and 2, or 5 and 10 is immaterial to the answer, which will be same regardless of how we set it up. Because we do not want to deal with numbers that look like Bill Gate’s networth or Amazon.com’s cumulative losses since their I.P.O, let us deal with simple 1 and 2. Our set up will look like this:

Month Avg. Price

Quantity Sold

Revenue

July 5.95 1 5.95

August 6.40 2 12.80

Total 3 18.75

The average price per dozen is the ratio of the total revenue to the total

quantity sold in the two-month period.

The average price/dozen = 18.75/3

= $6.25 per dozen.

Example 10:

“If John bought X books at $5 each and Y books at $7 each, what is the average price paid by John per book in terms of X and Y?” This is another weighted average problem. We need to find out how much money John spent buying books and then divide the money spent by the number of books bought by John. John paid 5X dollars to buy X books at $5 each. John paid 7Y dollars to buy Y books at $7 each. John paid a total of 5X + 7Y dollars to buy (X + Y) books. Average price paid by John is

(5X + 7Y) / (X + Y)

Example 11 DATA SUFFICIENCY “Is the average of X and Y greater than 10?”

1. The average of X + 5, Y + 7, and 10 is 10.

2. X = 3Y Our N.T.K is that we need to know the value for ½ •(X + Y) so that we can compare this value to 10 and decide whether the average is bigger than 10 or not. The statements must help us get a number value for this average. Statement 1 helps us determine the value for X + Y, and therefore the average. The set up for statement 1 is: 1/3 • (X + 5 + Y + 7 + 10) = 10 Or, X + Y + 22 = 30 or X + Y = 8 We can see that it is possible to compute the average by knowing the value for X + Y and compare it to 10, and answer the question with a definite yes or no. Statement 1 is good. Let us kill B, C, and E. We retain choices A and D. Is statement 2 good for a number value for the average? No. We can only get the average in terms of X or Y, and there is no way we can compare it to 10. We must choose Choice A.




Standard Deviation Standard Deviation is a concept that is used to indicate the distribution of the values in the population around the arithmetic mean, or the average. Let us say that a thousand applicants vie for admission to a graduate school of business, and the average GMAT score of all applicants is 550. Let us also say that the scores are normally distributed and that two-thirds of the applicants submit scores that are one standard deviation away from the average score and that 95% of applicants submit scores that are within two standard deviations away from the average score. If we know that the Standard Deviation for the population is 35, we can conclude that the GMAT scores of two-thirds of the applicants lie in the range between (550-35)=515 and (550 + 35) = 585. This is how we interpret the information that the scores of two-thirds of applicants lie one standard deviation away from the average score. How do we read the information that the GMAT scores of 95% of applicants are 2 standard deviations away from the mean? We will read this information to mean that the GMAT scores of 95% of applicants lie in the range between (550- 2•35) = 480 and (550+2•35) = 620. If the problem specified further that 99% of all applicants had their scores within three standard deviations of the mean, we will read this information to mean that the scores of nearly all applicants were between (550 – 3•35) = 445 and (550 + 3•35) = 655. Standard Deviation is a concept that should invoke the familiar “bell-shaped” curve when the name is dropped. You should understand this concept thoroughly because you will expect to get some questions relating to this concept. Example: Students at a local town took the SAT1, and the results were normally distributed, with 67% scoring within one standard deviation of the mean and with 95% scoring within two standard deviations of the mean. If the mean score was

500 and the standard deviation for this population is 40, approximately what percentage of the students scored between 540 and 580 on the test? Solution: The problem tells us that 67% scored between (500-40)=460 and (500+40)=540. The problem also tells us that 95% scored between (500-80)=420 and (500+80)=580. The inference we can make is that 95%-67%=28% scored in the following two ranges: 420-460 and 540-580. We are told that the scores are normally distributed, which means that the percent of those scoring in the 420-460 range is the same as that of those scoring in the range 540-580. And the value is ½ of 28% or 14%. The answer is that 14% of test-takers scored in the range between 540 and 580.

How do we find the standard deviation of a group of

values? Step 1: Find the average of the group of values. Step 2: Find the deviation from the average for each of the values in the group. Step 3: Square the individual deviations from the mean, and add the squared deviations. Step 4: Divide the sum of the squared deviations by the number of values in the group. Step 5: Take the square root of the value in step 4. You will have the standard deviation for the group looking at you.




Example: A group of GMAT test-takers has the following scores submitted by the test-takers: 550, 590, 620, 650, and 660. What is the standard deviation for this group of test-takers?

Value Deviation from mean (X – 614)

Squared Deviation from mean.

550 -64 (-64)2 = 4096

590 -24 (-24)2 = 576

620 6 62 = 36

650 36 362 = 1296

660 46 462 = 2116

Mean = 3070/5 = 614

Sum of the squared deviations (Sum of all the values in the third column) = 8120 Let us carry out the step 4 described in the preceding page, and divide 8120 by the number of values, namely, five. We get: 8120 + 5 = 1624 The standard Deviation for this group of values is the square root of 1624, or 40 (approximately). (We know that 40-squared is 1600 and 1624 is pretty close to 1600). In a larger population with normally distributed values, 67% of the values may be distributed one standard deviation away from the mean and 95% of the values may be distributed within 2 standard deviations of the mean. The “scaled scores” in the GMAT bring the values within the specified percentage ranges on the bell-curve, and different editions of the GMAT may involve different “bell-curving”, if you know what we mean.

FORMULAS FOR COMPUTING Standard

deviation

The standard Deviation could be computed by using any one of the following formulas:

THERE ARE DIFFERENT WAYS OF COMPUTING THE STANDARD DEVIATION OF A GROUP OF ‘N’ VALUES. ALL THREE FORMULAS PROVIDED BELOW ARE DIFFERENT WAYS OF EXPRESSING THE STANDARD FORMULA (see formula 3 below). If we know the SUM OF THE VALUES OR THE AVERAGE OF THE VALUES, and THE SUM OF THE SQUARES OF THE VALUES, WE CAN COMPUTE THE STANDARD DEVIATION. REMEMBER THIS. GMAT WILL TEST YOUR UNDERSTANDING OF THESE FORMULAS




MEDIAN The median of a group of values is the middle value of an ordered set of values. We must arrange the values in an ascending or a descending order and pick the middle value such that there is an equal number of values on either side of the median value. This is possible only if we have an odd number of values. When we have an even number of values, we must pick the middle two values such that there is an equal number of values on either side of the middle two values, and take the average of the middle values. Example: What is the median value of the following GMAT scores submitted by a group of applicants to an M.B.A program? 670,590,470,700,690,680,670,600,610 Let us arrange the values in an order, starting from the smallest to the largest. (You could do it the other way around too). 470, 590, 600, 610, 670, 670,680,690,700 We can clearly see that 670 is the median value because it occurs smack in the middle of an ordered set of values. If we added one more value to the list, 450, then the list is going to look as shown below: 450, 470, 590, 600, 610, 670, 670,680,690,700 The median of this set of values is the average of 610 and 670, or 640. Example 2: The median of a set of six consecutive positive odd integers is how much greater than the mean of the same group of values? Let us define the values as 2n+1, 2n+3, 2n+5, 2n+7, 2n+9, and 2n+11. The Median is the average of 2n+5 and 2n+7 = 2n+6. The average of the group of values = (12n+36)/6 = 2n+6 As you can see, the mean and the median of any consecutive set of integers will be equal and the difference will be ZERO.

Example 3: The point value of a dive is determined by taking the median score of all scores allotted by the judges and multiplying it by the degree of difficulty. If a dive with a degree of difficulty of 5 received the following scores from the 5 judges: 4.0, 4.2, 3,8, 3.7, and 4.3, what was the point value of the dive? The median of the five scores is obtained by arranging the values in an order: 3.7, 3.8, 4.0, 4.2, 4.3 The median value is 4.0 The point value is obtained by multiplying the median value by the degree of difficulty, which is 5.

The point value is 4•5 = 20 (Answer)

MEDIAN TRIVIA

For a set of consecutive integers (including consecutive odd, consecutive even, or any set of values having the same distance between values), the MEAN and the MEDIAN are the same. For example, if we consider the set 24, 26, 28, 30, 32, 34, the median of this set is the average of 28 and 30, and equal to 29. This is also the value for the average of the set.

The sum of any consecutive set of values is the average or the median times the number of values in the set. In the above example, the sum of the set of values listed will be 29 times 6 or 174.




Mode The mode is the most frequently occurring value or values in a group. It is possible for a group of values to have multiple modes. For example, a school can receive 100 applications each with GMAT scores of 670 and 690. In this scenario, the set of values has two modes: 670 and 690. Example: The average GMAT score of 7,000 applicants to the Stanford M.B.A program was 610. If 600 applicants submitted the GMAT score of 650, and if 340 applicants submitted the GMAT score of 680, what is the average GMAT score of the remaining 6,060 applicants? (We notice that the modal value could be 650 unless there is another group exceeding 600 and submitting the same GMAT score). The total score of 600 applicants each submitting a score of 650 = 390,000 The total score of 340 applicants each submitting a score of 680 = 231,200 Total score submitted by 940 applicants = 390,000 + 231,200 = 621,200 Total score submitted by all 7,000 applicants = 7,000 • 610 = 4,270,000 Total score submitted by 7000-940 = 6060 applicants = 4,270,000 – 621,200 = 3,648,800 The average score of 6,060 applicants = (3,648,800) + 6060 = 600 (approx.)

Radicals

The square root sign is a radical, and you should remember the following rules: Rule 1: √x + √y is not equal to √ x + y Rule 2: √x • √y = √xy Rule 3: √x divided by √y = √x/y

Rule 4: √x2 + y2 is not equal to x + y IF we can find a way to express the values within the radical sign in terms of factors that are perfect squares, we will be able to take the numbers out of the radical sign. Example: √32 = ? We notice that 32 is not a perfect square and we can tell that the answer will be more than 5 and less than 6. In the GMAT, you will be asked to recognize an equivalent value. Can we find a way to express 32 in terms of two factors, at least one of which is a perfect square? We notice that we can write 32 as 2•16, and that 16 is a perfect square. (42) √32 = √2.16 = √2.42 = 4√2 4√2 is a called a mixed radical, and the answer choice will be usually in this form. You should be looking for a choice that reads 4√2.




Example 2:

√243 - √108 = ? We notice that neither 243 nor 108 is a perfect square. Let us do the next best thing: Express 243 and 108 in terms of two factors, at least one each of which is a perfect square. √243 = √3•81 = √3•92 = 9√3 √108 = √3.36 = √3•62 = 6√3

√243 - √108 = 9√3 - 6√3 = 3√3 (Answer) Example 3:

√632 + 362 = ? We notice that 632 and 362 are not factors within the radical sign, and we cannot take them out as 63 and 36. When you come across problems of this type, look for a standard form, as we will see with algebra later on, or look for the greatest common factor for the values provided. The greatest common factor of 63 and 36 is 9. 632 = (7•9)2 = 72•92

362 = (4•9)2 = 42•92 √632 + 362 = √92•72 + 92•42 = √92•(72+ 42) = 9√49 + 16 = 9√65 ϕ 9•8 ϕ 72 (approx.) Example 4:

√772 + 1102 = ?

We notice that 772 and 1102 are not factors within the radical sign. We cannot take 77 and 110 out of the radical sign. Let us do the next best thing: let us express 77 and 110 in terms of their greatest common factor of 11. 772 = (7•11)2 = 72•112

1102 = (10•11)2 = 102•112

√772 + 1102 = √ 112•(72 + 102) = 11•√149 ϕ 11• 12 ϕ 132 (approx.) We will revisit the radicals after we have become familiar with some algebraic formulas.

Perfect Square A perfect square is an integer that can be expressed as the square of an integer. For example, 16 is a perfect square because it can be expressed as 42. A perfect cube is an integer that can be expressed as the cube of an integer. For example, 27 is a perfect cube because it can be expressed as 33. Example: Data Sufficiency: If n is an integer, is n a perfect square? 1. If p is an integral factor of n, so is p2. 2. √n is an integer. Statement 1 tells us that p2 is a factor of n. But we are clue-less about the other factors of n. For example, n could be 12, in which case we can write n as n = 3•22. (Here p could be 2). We cannot write n as a perfect square because 3 is not a perfect square. But then, n could also be 36, in which case we can write n = 32•22, where p could be 2 or 3 and n could be expressed as a perfect square. Statement 1 is not good for a unique solution. Our choices are limited to choice 2, choice 3 or choice 5. Statement 2 tells us that square root of n is an integer. This can mean only one thing: that n is the square of an integer, or a perfect square. (If √n = integer, then n = (integer)2). Statement 2 tells us in unambiguous terms that n is a perfect square. We conclude that statement 1 alone is not sufficient but statement 2 alone is sufficient to answer the question. We must pick the second choice.




ALGEBRA Algebra is what you deal with when you set up equations for verbose information in word problems. We have seen that an integer can be represented in terms of its factors. For example, 16 can be represented in terms of 1 and 16 or 2 and 8 and so on. We can also express an algebraic expression in terms of its algebraic factors. Let us say that we have two algebraic factors: (2x+1)•(3x-7) We are allowed to break up a binomial in terms of two monomial factors and write: (2x+1)•(3x-7) = 2x•(3x-7) + 1•(3x-7) or, (2x+1)•(3x-7) = 6x2 – 14x + 3x – 7

(2x+1)•(3x-7) = 6x2- 11x – 7 In the GMAT, you will be required to proceed in reverse gear from a given quadratic expression or equation and rewrite it in terms of its factors. As a rule, try to look for a simpler and more manageable expression every time you are dealing with an algebraic expression. All GMAT algebraic expressions will be amenable to simplification and will lead to a simpler, and more easy-to-deal-with form. Example 1: 3x•(x-7) – 2x+14 = ? (3x – 2) We notice that the first set of items is already factored in terms of 3x and (x-7), and, as a rule, We will not mess with factored expressions and will leave them be. Let us take a look at –2x + 14. What is common to both –2x and 14? We see a common factor of 2. We can write –2x + 14 as –2•(x-7) Let us rewrite the numerator expression as: 3x•(x-7) – 2x+14 = 3x•(x-7) –2•(x-7) We can see that (x-7) is factor that can be taken out so that the remaining two items can be grouped together.

3x•(x-7) – 2x+14 = (x-7)•(3x-2) As you can see, we have managed to write the numerator expression in terms of two factors. You can also see that one of the factors on the numerator will cancel out the denominator factor, leaving a simpler expression. 3x•(x-7) – 2x+14 = (x-7)•(3x-2) = (x-7) (3x – 2) (3x-2) You should be looking for a choice that says (x-7).

GMAT UNPLUGGED We have seen during the programs that some participants, who have been through the Kaplan process, randomly choose numbers for x and try to see whether any of the choices will give the same result as the original expression when the value is plugged in. You cannot do that. The only time when you can choose to assign numbers to variables specified in the problem is when a relationship is specified between the variables, relationship such as x = 2y = 3z. You can set z = 2 and that value will determine the values for y and x. If z = 2, then y is 3 and x is 6. The proportional values for x, y and z are consistent with the original definition, and will be an acceptable methodology. IF the problem tells that if X is divided by 5, we get a remainder of 4, we can choose to work with a number value of 9 for X. If the problem told us that a + b = c, then we can choose to assign 1 for a, and 2 for b so that the value for c will be 3. If the problem defines a ratio, such as A/B= 2/3, and asks to verify another ratio involving the same two variables such as (A+B)/(A-B), we can assign 2 for A and 3 for B and verify the new ratio. But if the problem specifies that A/B = 2/3, and asks what is A+B?, then it is all-bets-are-off situation we have. If the problem merely stated that x, y and z are three values and does not specify a relationship, you should not be tempted to use plug-in process because you will be wasting a lot of time in trying to reach an unreachable conclusion.




Example 2: Let us take an example of another “clumsy”

algebraic expression:

4x•(x-2) - 2x + 4 = ?

(4x-2)

We notice that -2x+4 in the numerator can be

written -2•(x-2), and we recognize that (x-2) is

sitting as a factor in the first part of the numerator

expression.

We can write the numerator as:

4x•(x-2)-2•(x-2) = (x-2)•(4x-2)

What did we do here? We noticed that (x-2) was a

common factor, and took it out. We were left with

4x from the first part and -2 from the second part,

and we grouped them together to get (x-2)(4x-2)

Let us rewrite the expression in terms of these

factors as:

(x-2)•(4x-2) = (x-2)

(4x-2)

This is the summary of the drill we performed:

1. We recognized common factors in parts of an

expression, and re-wrote a part of the expression in

terms of its factors.

2. We then took out the common factor to both parts

of the expression, and grouped the remaining values

to write an expression in terms of two factors.

3. We found to our pleasant surprise that one of the

factors in the numerator expression was the same as

the denominator expression. These values canceled

each other out, leaving behind a simpler, more

manageable expression.

Now try the following on your own, and see if you can simplify the each of the expressions to a more manageable form: (Explained at the back of this file) 1. 6x3+3x2+6x+3 = (x2+1) 2. 5x3+7x2-25x-35 = (x2-5) 3. 7x2(6x+5)-18x-15 = (7x2-3) Example 3: Data Sufficiency:

What is (a + b)?

1. ad + bd + ac + bc = 15 2. c + d = 5

4. 3x(x2+2)-7x2-14 = (3x-7)




Statement 1 is a “clumsy” algebraic expression, and our first “instinctual reaction” must be to try and simplify it. ad + bd + ac + bc = d•(a+b) + c•(a + b) = 15 = (a + b)•(c + d) = 15 We notice that we have been able to express the given expression in terms of two factors but we cannot find the value for a + b unless we know what c + d is. Our conclusion is that statement 1 alone is not sufficient to answer the question. But you must go the distance we have indicated before you can come to this determination. Statement 2 alone is also not sufficient to answer the question because statement 2 tells us nothing about a + b. We must now combine the two statements and see whether the combined information will produce a winner. The combined information does. If we know the value of c + d, then we can determine the value of a + b. We must pick choice3 in the test. Remember: you must “play around” with the algebraic expressions before you can begin to make sense of the information. Usually, such “manipulation” will give you better insights into the nature of the problem. Example 4: Data Sufficiency: What is X?

1. 2x + 1 = 0 2. x•(2x+1) – 14x – 7 = 0

(x-7)

where x is not equal to 7. Statement 1 is a no-brainer. We can solve for x using the equation specified. Our choices narrow to choice1 or choice 4 in the test. Statement 2 needs “manipulation”. We notice that (–14x – 7) has 7 as a common factor and can be written as –7•(2x+1) The numerator expression becomes X•(2x+1) –7•(2x+1) = (x-7)•(2x+1) Because there is a “factor” of (x – 7) in the denominator, we can see that the expression in statement 2 will simplify to: x•(2x+1) – 14x – 7 = (x – 7)• (2x + 1) = 0 (x-7) (x – 7)

WE get: 2x + 1 = 0 We notice that this equation is the same as the one provided in statement 1, and we can determine the value for x using this statement alone also. We conclude that we can answer the question using either statement alone independently, and we must pick choice 4.




Roots of an equation The root of an algebraic equation is simply the value for the variable in the equation satisfying the equation. In the equation x + 2 = 0, the value –2 for x will satisfy the equation and will be called the root of the equation. In the equation 2x + 3 =0, the value –3/2 for x will satisfy the equation and will be referred to as the root of the equation. In the GMAT, you will encounter many situations in which you will be required to find the roots of a quadratic equation of the form 1•x2 + bx + c = 0 Notice that the coefficient of the x2 terms is 1 and b and c could be any values, usually integers. The standard form for a quadratic equation is ax2 + bx + c = 0 The quadratic equation will have two roots, and the formula for computing the roots is x12 = -b ± sqrt(b2 – 4ac) 2a

We said earlier in this discussion that in the GMAT, the quadratic equation you will encounter would be amenable to a form in which the coefficient of x2 terms is a 1. If this is the case, then you need not apply the formula to compute the two roots of the equation. We will take recourse to “factoring” so that we can find the roots of the given equation. We have to re-write the quadratic equation x2 + bx + c = 0 as (x + r1)•(x + r2) = 0 where the values for r1 and r2 are such that c = r1•r2 and b = r1+ r2 After we have written the given equation in terms of its factors, we can determine the roots of the equation as x1 = -r1 or x2 = -r2 Let us take a look at an example and understand this process. Example:




What are the roots of the equation 2x2 + 6x – 56 = 0? We notice that the coefficient of x2 terms is not 1, and we must look for a way to bring it down to a 1. We notice that 2 is a common factor of all the coefficients in the equation, and we can divide both sides of the equation by 2 to eliminate the 2 in front of the x2 term. The equation becomes:

X2 + 3x – 28 = 0 Now, let us factor the “c” value, the stand-alone number in the equation, in terms of two integers that will multiply to give –28 and, at the same time, will yield +3 when added together. We notice that we can pick the factors +7 and –4 so that the product is –28 and the sum of these two factors is +3. We will now rewrite the given equation as:

(x + 7) • (x – 4) = 0 or x = -7 or +4. -7 and +4 are the two roots of the given equation. Example 2: Find the roots of the quadratic equation 3x2 – 12x – 135 = 0. Step 1: Let us find a way to make the coefficient of x2 term equal to 1. We notice that all coefficients have 3 as a factor. Let us divide the equation by 3 to get: x2 – 4x – 45 = 0 step 2: Let us pick the two factors of –45 such that those two values will add to give –4. We

hone in on –9 and +5. The product is –45 and the sum is –4. Step 3: Let us rewrite the equation as (x – 9) • (x + 5) = 0 Step 4: The two roots are +9 and –5. (because one of the two factors is equal to 0. This means that either x – 9 = 0 and x = 9 or x + 5 = 0 and x = -5).

EXAMPLE 3:

“If 3 is one root of the equation: X2 + 7X – P = 25, what is the other root?”

We know that if 3 is a root, then X – 3 is a binomial factor of the given expression X2+7X-(P+25). We notice that the middle value is +7 and that is the sum of the two factors of the stand-alone number. We know that –3 is one factor of the stand alone number and the other must be +10 so that +10 – 3 = +7. The other binomial factor will be (X + 10) or the other root will be –10. You can also determine the other root by first finding the value of P by writing 3 for X. You will get P = 5. If you move the number 25 to the left and wrote the equation in standard form for quadratic equation, you will get: X2 + 7X – 30 = 0 We need to pick two factors of –30 that will add to give us +7. –3 and +10 will. We must write: (X-3)(X+10) = 0 or X = 3 or –10.




Finding the positive root of the variable (when we know that the variable in the equation cannot have a negative value) You will have lots of opportunities for creating quadratic “situations” when you deal with some word problems such as the distance problems and when you deal with geometry. Let us say that the problem specifies that in a right triangle, one side is 3 more than the other, and the area is 27 square units. We will set the sides as x and (x + 3) and use the area formula: Area of a triangle = ½ • x • (x + 3) = 27 When we multiply both sides by 2 so that we can get rid of the ½ on the left, we get: x • ( x + 3) = 54 We know that the value for the side of a triangle cannot be negative. If we are aware of this “subtlety” here, we will not bother to turn x•(x + 3) = 54 into a quadratic mess. We will simply look for the two positive factors of 54 such that they are 3 apart and will multiply to give us 54. We must conclude that x is equal to 6 and x + 3, by default, must be equal to 9. Time is of the essence in the GMAT, and if you can eliminate some needless steps, you will ensure that you are moving along briskly through the test. Example: The area of a rectangle is the product of its length and width. If the length of a rectangle is 4 more than its width, and if the area is 77 square units, what is the value for the length? IF the width is X units, then the length is X + 4 units. Area = X • (X + 4) = 77 We also know that the value for X cannot be negative. Therefore, we will not bother to turn this into a "quadratic mess”. Instead, we will simply look for the two positive factors of 77 such that they are 4 apart and will multiply to

give 77. We must conclude that X must be 7 and the length must be X + 4 or 11 units.

Algebraic Formulas to remember:

You will be required to memorize and remember the following algebraic formulas, and recognize expressions that are in the same standard forms as the expressions in the formulas. You may want to write down these formulas on your scratch-paper during the tutorial phase of the test. (The good news is, the tutorial phase is not timed, and you should use this opportunity to “boot up” or load all the information you have learned into your memory or brain cells so that you will be in a state of readiness for the test.) (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2

(a + b)•(a – b) = a2 – b2

or a2 – b2 = (a + b)•(a – b) Example:

What is the value of Sqrt(632 – 372)? We notice that the value inside the radical sign is in the standard form a2 – b2. Sqrt(632 – 372) = Sqrt[(63+37)•(63-37)] = Sqrt (100•26) = Sqrt(102•26) We notice that we have managed to express the value inside the radical in terms of two factors, a step crucial to our ability to take numbers out of the radical sign. We also notice that 102 can be taken out of the radical as 10, leaving the value of 26 inside the radical. Sqrt(632 – 372) = 10•Sqrt(26) We notice that the square root of 26 is approximately 5, and the approximate value for the expression is: Sqrt(632 – 372) = 10•Sqrt(26) ϕ 10•5 ϕ 50 Recognition is a critical requirement for success in the GMAT, and by recognizing the standard form, we have been able to get to the answer a lot sooner than if we had tried to expand each value and do multiple arithmetic operations. Remember: Each additional operation entails the potential for careless mistakes.




Practice Exercises:

Exercise1: “If (a + b)2 = a2 + b2, which of the following must be true?” I a = 0 II b = 0 III ab = 0 IV (a – b)2 = a2 + b2 Let us expand (a + b)2: a2 + 2ab + b2 = a2 + b2

Or, 2ab = 0 or, ab = 0 We can see that the equation simplifies to ab = 0. But, do we know which value is 0? We do not. We do know that one of the values is a 0, because the product is a 0. We can eliminate conditions I and II as not necessarily true. Condition III is true because ab must be equal to 0. Condition IV needs to be checked after playing around with the expression (a – b)2. We know that (a – b)2 = a2 – 2ab + b2 If ab = 0, then the middle term –2ab must be equal to 0, and (a – b)2 = a2 + b2. We can clearly see that condition IV must be valid as well. So, we must look for a choice that says “III and IV only”. Exercise 2: If x + 1/x = 3, what is x2 + 1/x2 ? Let us square both sides of (x + 1/x) = 3 and see what we get. (x + 1/x)2 = 32 = 9 x2 + 1/x2 + 2•x•1/x = 9 x2 + 1/x2 + 2 = 9 Let us move the number 2 to the right. We get:

x2 + 1/x2 = 9 – 2 = 7. (Answer)

Exercise 3:

What is the value of (√2 + √8)2 ? We recognize the form (a + b)2, and we know how to deal with this form. (√2 + √8)2 = (√2)2 + (√8)2 + 2.√2.√8 = 2 + 8 + 2.√16 = 10 + 2•4 = 18

(√2 + √8)2 = 18 Exercise 4:

Is (a + b)2 > a2 + b2 ?

1. a > 0 2. b < 0

Let us expand the information in the stem. We get: Is a2 + 2ab + b2 > a2 + b2 ? Or, is 2ab > 0 ? Or, is ab > 0? The question is simply one of whether the product of a and b is positive. When will the product of two values be positive? When both are positive or when both are negative. Therefore, the information we are going to be looking for will have to deal with signs of a and b. Statement 1 tells us that “a” is positive. That does not help us at all, because we are in the dark about the value for “b”. Statement 2 tells us that “b” is negative. Once again, statement 2 fails to provide the complete answer alone. We must proceed to combine the two statements. When we combine the two statements, we notice that a and b do not have the same sign and the product is not going to be positive. We can answer the question definitively, using the combined information. Remember: We should be able to answer the question one way or the other, and not necessarily in the affirmative. We must pick choice 3.




Exercise 5: “IF x/y = 3/2, what is

1. x + y ? y

2. x + y ? x – y Let us deal with the first part of the question. We have x/y = 3/2 Let us add a 1 to both sides. We get: x/y + 1 = 3/2 + 1 Let us write y/y for 1 on the left and 2/2 for 1 on the right so that the fractions will have the same denominators. x/y + y/y = 3/2 + 2/2 We get: (x + y)/y = 5/2. ……………. (1) And that is the answer. Notice how we recognized that we can go from x/y to (x +y)/y by simply adding a 1 to the expression. Recognition is key to unsolving the mysteries of the GMAT. Now, let us deal with the second part of the question. We notice that the expression (x +y) / (x –y) is obtained by taking the ratio of (x +y)/y to (x –y)/y. We have found the value for (x +y) / y. Now let us find the value for (x –y) / y. We recognize the form and conclude that we can go from x / y to (x –y) / y by simply subtracting a 1 from x/y. We have: x/y = 3/2 Let us subtract a 1 from both sides. We get: x/y - 1 = 3/2 – 1 We get: (x – y)/y = ½ ……………. (2) Now, let us take the ratio of the values in equations (1) and (2) to get: (x +y) / (x –y) = (5/2) / (1/2) = 5 We could have dealt with the same problem by assigning a value of 3 for X and 2 for Y. You will find that your life would have been a lot easier. Try assigning 3 for X and 2 for Y and see

whether you get the same results. You will. In the GMAT, know that we can go from X/Y to (X+Y)/Y by adding a 1 to the fraction. But for sake of time, work with numbers as suggested. Remember: We can work with real-life numbers as long as a relationship between variables is specified, as we find is true in this problem. Exercise 6: (x2 + xy) / xy is equivalent to what? Let us divide each term on the numerator by the denominator. We get: (x2 + xy) / xy = (x2 / xy) + (xy / xy)

(x2 + xy) / xy = (x/y) + 1 (Answer)

Algebra in word problems Algebra is going to come back to haunt you every time you are looking at information specified in terms of letters and alphabets. GMAT would like you to be as comfortable with algebraic notations as with real life numbers. Example 1: “If Pamela was x years of age y years ago, how old will Pamela be z years from now?” Pamela’s age today is (x + y) years of age. z years from now, Pamela will be (x + y + z) years of age.




Example 2: “Manchester United won exactly 2 more games than it lost in the league games. IF the team played a total of N games, and if there were no ties, find an expression for the games won in terms of N.”

Let W be the number of games won, and Let L be the number of games lost. Number of games won + lost = N i.e. W + L = N (1) And W = L + 2 or L = W – 2. (2) Let us replace L with W –2 in (1)so that we can write the equation in terms of W only. (1) becomes:

W + W – 2 = N Or, 2W = N + 2 Or, W = (N + 2) / 2 And that is the expression, in terms of N, that represents the number of games won.

EXAMPLE 3: DATA SUFFICIENCY: “What is (A + B) (C + D)?”

1. A (C + D) = 10 2. B (C + D) = 15.

We notice that Statement 1 uses one of the monomial terms of the first binomial (A + B) as a factor with the second binomial taken as a whole. Obviously, we cannot use statement 1 alone to answer the question. We must kill choices A and D. We have just 3 choices remaining: B, C, or E. Statement 2 is also another attempt at using one of the monomial terms of the first binomial (A + B) as a factor with the second binomial. Again, we cannot answer the question using statement 2 alone. Let us kill choice B. If we combine the two statements, we notice that (A + B) (C + D) = A (C + D) + B (C + D) We can answer the question by using the combined information. We must go with choice C.

EXAMPLE 4: “Which of the following has at least one root in common with the equation X2 – 5X + 4 = 0?” • X2 + 1 = 0 • X3 + 2X2 – 7X – 14 = 0 • X3 + 1 = 0 • X5 – 1 = 0 • X = - 4 We can determine that the given equation has the following roots: 1 and 4. We can do one of two things: We can try to replace X with 1 and 4 in the choice equations and see which one will be satisfied by at least one of these two values. The other option is to quickly determine the roots of the choice equations because most of them are pretty easy to deal with. Choice 1 is not worth messing with, because X2 cannot be equal to –1. Choice 2 can be expressed in terms of two binomial fators as: X2(X + 2) – 7 ( X + 2) = 0 Or, (X2 – 7) ( X + 2) = 0 Or, X = 7 or X = ± sqrt(7) We notice that none of these three roots has anything in common with 1 or 4 Choice 3 gives us a root of –1 for X. We are looking for +1 or +4. Choice 4 is right on the money. If X5 = 1, then X must be 1. We must choose this option. Choice 5 is not good because –4 is not the same as +4. Remember: 1) Learn to recognize standard forms. 2) Always “play around with” or

“manipulate” algebraic expressions, and look for a simplified form. When you deal with data sufficiency involving algebraic information, do some minimal implementation and get to a simplified form, which is amenable to making a decision.

3) Learn to redefine the question when asked in data sufficiency with algebraic information.




ALGEBRA TESTED AS RADICALS

You may be expected to apply some algebraic formulas when you deal with some problems involving radicals. Example: (√7 – 5) (√ 7 + 5) = ? Can we associate the above binomial factors with any standard algebraic rule we know of? Sure. The expression is in the form: (a + b) (a – b), and we know that (a + b) (a – b) = a2 – b2 In our problem, a is √ 7 and b is 5 What is a2 – b2 ?

(√ 7 – 5) (√ 7 + 5)

= (√ 7)2 – 52

= 7 – 25 = -18

EXAMPLE 2:

SQRT(632 – 372) = ?

The expression within the radical sign is of the form: (a + b) (a – b). Here a is 63 and b is 37. SQRT(632 – 372) = SQRT[(63 + 37)(63-37)]

=SQRT[(100)(26)]

=10•SQRT(26)

= 50 (APPROXIMATELY)

When you deal with radicals, check to see whether you can apply any standard algebraic formula and deal with the problem. If you cannot, then you may have to deal with factoring of values in terms of integers, at least one of which is a perfect square and can be isolated as a common factor. For example, take a look at the following problem: SQRT[812 + 542] = ? Our first instinct will be to look for a standard algebraic formula with which we can associate this expression. Unfortunately, we cannot. If we were to use the form (a + b)2, then we need another term 2ab, which is missing. Let us do the next best thing, and try to find the greatest common factor for 81 and 54. 81 can be written as 3 times 27 and 54 can be written as 2 times 27. We have found the greatest common factor between 81 and 54 – the value of 27. Let us rewrite the given expression in terms of this common factor and see where we go from there. SQRT[812 + 542 ] = SQRT[32•272 + 22•272] = SQRT[272•(32 + 22) We can take 272 out of the square root sign as 27. We get: SQRT[812 + 542 ] =27• SQRT(32 + 22) = 27• SQRT(13) And that is the answer you should be looking for. Remember: Try to look for any standard algebraic formula you can use in the context of a problem using radicals. If you do not associate the expression with any standard algebraic form, do the next best thing: Look for the greatest common factor between the values specified under the radical.




MISCELLANEOUS PROBLEMS USING ALGEBRA

EXAMPLE:

“If a + b = c, which of the following is equal to 1?”

I (c - b)/a II (c – a)/b III (a + b)/c This is a problem that tests your ability to manipulate algebraic equations. If a + b = c, then a = c – b. If we divide both sides by a, we get 1 = (c – b) /a We can see that condition I stacks up. Let us play with the equation some more. IF a + b = c, then b = c – a Let us divide both sides by b, and get: 1 = (c –a)/b We can see that condition II also stacks up. Let us do one more thing with the equation. If a + b = c, then let us divide both sides by c. We get: (a + b) / c = c/c = 1 We know that condition III is valid too. We must pick a choice that says “I, II, and III”. IF you were not up to speed with algebraic manipulation, you could have answered this problem by thinking up real-life numbers because there is a relationship specified between the variables. IF we set a = 1 and b = 2, then c must be 3. Now test the conditions I, II, and III by using these values and see whether the expressions specified give a value of 1. They will.

EXAMPLE: “IF X is a multiple of Y and Y is a multiple of Z, which of the following is not necessarily an integer?” (A) (X + Y) / Z (B) (X – Y) / Z (C) X•Y/Z (D) (X + Z) / Y (E) X•Y•Z If X is a multiple of Y, we can write: X = K•Y where K is an integer. Similarly, if Y is a multiple of Z, we can write: Y = P•Z where P is another integer. Notice that we did not choose to use the same integer multiple value of K for both equations because X and Y could be different multiples of Y and Z respectively. You should also be able to combine the two statements and express X in terms of Z. Can you? If X = K•Y and Y = P•Z, then let us replace Y with P•Z in the first equation. We get: X = K•P•Z and Y = P•Z We can see that both X and Y are multiples of a common variable Z. WE can now continue to deal with the problem algebraically or choose to work with numbers because we have a relationship between the variables specified. IF we choose to work algebraically, we will notice that both X and Y are multiples of Z. We can see that choices A, B, and C use X and Y are values on the numerator and Z as a denominator. Because both X and Y are multiples of Z, the Z at the bottom will cancel out the Z on the numerator. For example, let us examine choice A. (X + Y) / Z = (K•P•Z + P•Z) / Z = K•P + P Because K and P are both integers, K•P + P MUST Be an integer. We can similarly confirm that choices B, C and E will yield integer values always. What about choice D? We notice that we have a Y at the bottom of the expression, and that spells trouble.




(X + Z) / Y = (K•P•Z + Z) / P•Z = (K•P + 1) / P = K + 1/P We notice that while K is an integer, 1/P does not have to be an integer unless P happened to be 1. Therefore, choice D does NOT HAVE TO BE AN INTEGER, because P could be any integer and not necessarily a 1. We must pick choice D. Notice how we were required to use the concepts of factors, and of algebraic factoring in this problem. When we dealt with the expression (X + Z) / Y = (K•P•Z + Z) / P•Z We tried to separate the common factor of Z and see where we went with the exercise. (X + Z) / Y = (K•P•Z + Z) / P•Z = Z•(K•P + 1) / P•Z Z will cancel each other out on top and bottom leaving (K•P + 1) / P Could you have done this problem by thinking up real-life numbers consistent with the relationships specified. You could have. But before we could begin to use real-life numbers, we should have reached the following stage: X = K•P•Z and Y = P•Z We can see that both X and Y are multiples of Z. Therefore, Z will be have to be the least value. IF we let Z be equal to 2, then Y could be 4 and X could be 12. Notice that values of 2 for z, 4 for Y and 12 for X are consistent with the relationship specified between the variables. Now use these real-life numbers to test the choice expressions and see which one will NOT necessarily give an integer value. You will also see that we could have chosen 1 for all X, Y, and Z because we will still be working within the parameters defined. IF X, Y, and Z were each equal to 1, then all choices will give rise to an integer value. But the question is: “not necessarily an integer”. X , Y, and Z do not have to be necessarily each equal to 1. Remember, you can work with real-life numbers if a relationship is specified but then you should

know how to assign values in keeping with the way things are specified. That does call for some understanding of the concept of multiples. EXAMPLE 3: Concentration in parts per million of a chemical at any depth X feet of a solution is given by the formula: Concentration = 4 / SQRT(3X – 2) At what depth in feet will the concentration of the chemical be equal to 3 parts per million?

The question is, What should be the value of X, if the concentration is 3?

Our set up will read: 3 = 4 / SQRT (3X – 2) Let us cross multiply and turn this into an equation. We get: 3• SQRT(3X – 2) = 4 Let us square both sides to get rid of the square root sign on the left. We get: 9•(3X – 2) = 16 27X – 18 = 16 27X = 34 or X = 34/27 feet = 1 7/24 feet That is the answer we should be looking for. EXAMPLE 4: The equation R = - 0.028t + 20.8 can be used to predict the world record in the 200 meters dash, in which R stands for the record in seconds and t stands for the number of years since 1920. In what year will the record be 18.0 seconds? If R = 18, then what is 1920+ t ? Let us replace R with 18 and write the equation as: 18 = - 0.028t + 20.8 Let us isolate the numbers on one side and the variable on the other, and solve for t. -0.28t = -2.8 t = 2.8/0.028 = 100 Since t represents the number of years since 1920, we conclude that the record of 18 seconds will be achieved in the year 1920+100=2020.




Exponents Exponent is a value that denotes how many times the base repeats itself as a factor. For example, 22 tells us that 2 occurs as a factor twice. If we have an exponential value of xn, we know that x occurs n times as a factor. When we take the square root of an exponent, we can take out of the radical one factor for every two we see inside. For example, the square root of 216 will be 28 because the original exponential value tells us that 2 occurs as a factor 16 times inside the radical, and we can take one 2 out of the radical for every two we see inside. What happens when we multiply two exponential values, each with the same base? Let us say that we would like to multiply x2 by x3.

x2•x3 = x•x•x•x•x = x5 We notice that when we multiply two exponential values with the same base, we keep the same base and simply add the exponent. This “discovery” brings us to the first rule of exponents, and to many more as shown below:

Rules of Exponents:

1) xa•xb = x(a+b)

2) xa/xb = x(a-b) = 1/(Xb-a)

3) (xa)b = xab = (xb)a COROLLARY: XAYA = (XY)A

4) 1/xa = x-a

5) IF xa = xb, then a = b IF XP = YQ , AND IF X IS ODD AND Y EVEN OR VICE VERSA, THEN P = Q = 0. Notice that any exponent on an odd integer will yield an odd integer; any exponent on an even integer will yield an even integer. The only way the equation will work is if the exponents are equal to 0.

6) X0 = 1

Example 1:

If 21•22•23•24•••2k = 255, what is k? We notice that we are dealing with multiplication of exponents with the same base. We apply the rule No. 1, and get:

21+2+3+4+5+•••••+k = 255

We now apply the rule number 5, and get:

1 + 2 + 3 + •••••• + k = 55 The question is: how far do we have to go when we add consecutive integers so that the sum will be 55? We need to go up to 10 so that the sum of consecutive integers will be 55. We can get the same result by applying the formula for the sum of consecutive integers. The formula for the sum of consecutive integers from 1 through n is n•(n+1)/2. In this problem, sum is equal to 55 We get: n•(n+1)/2 = 55 or n•(n+1) = 110 Because we know that n is positive, and we do not bother to turn this into a quadratic “mess”. Instead, we will choose two factors of 110 such that they are 1 apart and will multiply to give 110. We must conclude that n must be 10. If the numbers are simple to deal with, you may want to use your fingers to get the answer rather than apply the formula. Example 2: What is the smallest integer bigger than 1 that is at once a perfect square and a perfect cube? The smallest number bigger than 1 is 2, and an integer that is at once a perfect square and a perfect cube must be (22)3 = 26 = 64.

TRIVIA: The smallest number bigger than 1 and is at once a perfect square, a perfect cube and a perfect fifth is ((22)3)5 = 230 = 645




Example 3: “IF 10-9 + 10-8 + 10-7 + 10-6 + 10-5 = N•10-9, What is N?” We notice that N•10-9 is the same as N/109

Let us multiply both sides by 109 so that we can get rid of the 109 in the denominator of the fraction on the right side of the equation.

We get: 109•10-9 + 109•10-8 + 109•10-7 + 109•10-6 + 109•10-5 = N 109-9 + 109-8 + 109-7 + 109-6 + 109-5 = N 100 + 101 + 102 + 103 + 104 = N 1 + 10 + 100 + 1,000 + 10,000 = N 11,111 = N

or N = 11,111 (Answer) Example 4: If 22x + 3 = 1/128, What is x? We notice that we have a decimal number on the right-side of the equation, and an exponential value on the left side. Our objective is to bring values on both sides of the equation to a common basis. We notice that 128 can be expressed as 27. (If you are wondering how to figure this out, start dividing 128 by 2 and see how many 2’s you will get as you go down to 2. 128 = 2•64 = 2•2•32 = 2•2•2•16 = 2•2•2•2•8 and so on) The reciprocal of an exponential value is the same as the base raised to negative exponent. 1/27 = 2-7

So, we have: 22x+3 = 2-7 We apply the rule 5 and get the equation: 2x + 3 = -7 or,

x = -5 (Answer)

Example 5:

If 43x-7 = 1/512, what is x? Our first instinct will be to see whether we can write 512 as an exponent with a base of 4. If we start dividing 512 by 4, we get a 2 as a factor in the end. That is not a good situation. We must get a 4 throughout so that we can express the given number as an exponential value. Let us do the next best thing, or carry out the plan B. Plan B is to express both 4 and 512 as an exponent in terms of a base value of 2. We have: (22)3x-7 = ½9 = 2-9 (Rule 4) Or, 26x-14 = 2-9 (Rule 3) We apply the rule 5, and get the equation: 6x – 14 = -9 or

x = 5/6 (ANSWER) Example 6: How many digits, including repetition of the same digit, are required to express 10100 in decimal notation? We know that 101 = 10 or 2 digits 102 = 100 or 3 digits 103 = 1,000 or 4 digits and so on.

We can see that the number of digits is always one more than the exponent value, if the base is 10. Therefore, we will have 101 digits when 10100 is expressed as a decimal number. In fact, the exponent on top of a base of 10 specifies how many zeros there will be in the decimal number, and there is a 1 in front of the zeros.

TRIVIA The number of digits required to express 100100 is obtained when we express 100100 as (102)100 or 10200. We can determine that we will require 200+1 or 201 digits to express 100100 as a decimal number, including repetitions of the same digit,0. The number of digits required to express 10001000 (which is the same as 103000) will be 3,001, including repetitions of 0. You get the drift, don’t you?




Example 7: Data Sufficiency If y = 22x + 1, what is y?

1. 22x = 1/64 2. y = 2x-2

We need to know the value of x so that we can determine the value of y. Statement 1 lets us decide the value for x. We notice that 1/64 is the same as 2-6 and we can set up an equation connecting 2x and –6. (2x = -6 or x = -3. If x is –3, then y = 22x+1 = 2-5 = 1/32) We will conclude that statement 1 alone is sufficient. We must not move on to examine statement 2, and our choices are either choice1 or choice4. Statement 2 gives us a value for x in terms of x but then we have another value for y in terms of x in the stem. If we look at the statement 2 information in the light of the stem information, we can conclude that 2x+1 = x – 2 or x = -3. Once again, y = 2-5 or 1/32. Statement 2 alone is also sufficient to answer the question in a unique fashion. Either statement alone is sufficient to get the answer. We must pick choice 4. Note: You must remember to use the statement information along with any information that is provided at the outset in the stem as part of the question itself.

MANIPULATE THE BASES We should be able to look at the bases of exponential values, and express them as factors so that we can decide in what convenient manner we can rewrite the given value. For example, if we have 65, we should be able to see that we can write this as (2•3)5 or as 25•35. Flexibility with concepts and values is what the test attempts to get the handle on. So, get flexible. Consider the following problem: IF 521

x 411 = 2x10N, what is N? Notice that on the right side, we have 2 and 10 as the ‘base’ values whereas on the left, we have 5 and 4. We need to get the same ‘bases’ on both sides of the equation in order to relate apples to apples. We can write 4 in terms of 2 as 4 = 22. We get the following picture: 521 x (22)11 = 2x10N

521 x 222 = 2 x 10N 521 X 221 X 21 = 21 X 10N (notice that we split up 222 as 221 times 21 so that we can combine 521 and 221 to get (5x2)21 = 1021. We can thus get 10 as base values on both sides of the equation. Manipulation is the name of the game here0. We have now the following, more workable situation: (5x 2)21 x 2 = 2 x 10N We get: 1021 x 2 = 2x 10 N Cancel the factor 2 on both sides, and we get, 1021 = 10N or N = 21. (answer) NOW, PLAY WITH THE FOLLOWING PROBLEM AND SEE WHETHER YOU GET THE ANSWER SPECIFIED: “IF 36X = 8100, WHAT IS (3X-1)3 ?” (ANSWER: 10/3) HINT: WRITE 36X AS (33X)2 = 8100 = 902 AND GET 33X = 90. Then, write (3X-1)3 AS 33X TIMES 3-1 = 1/33

TIMES 33X. Then, substitute 90 for 33x to get 1/27 times 90 or 90/27 or 10/3. (all fractions must be reduced to their simplest forms)




Multiplication is repeated addition.

Multiplication is a process that involves repeated addition. If we are multiplying 2 by 5, we get the result of 10 by adding 5 two times or by adding 2 five times. 2•5 = 2 + 2 + 2 + 2 + 2 2•5 = 5 + 5 You will be required to understand this principle and process when you deal with the following concept in exponents. Let us say that we have 22 We can write 22 = 2•21

= 21 + 21 We can write 23 = 2•22

= 22 + 22

If we had 37, we can write: 37 = 3•36 = 36 36 +36

If we had 59, we can write it as: 59 = 5•58

= 58 + 58 + 58 + 58 + 58 As you can see, any exponential value can be written as the sum of as many values as the base number on the left, and each of the values in addition set-up on the right

will have the same base and an exponent that is one less than that on the left. For example, if you take a look at the set-up for 59, you will notice that there are 5 values in an addition setup on the right and each of those values has the same exponent value of 8, and that each of those exponents on the right is 1 less than the value of the exponent on the left. Example: IF 2a = 2b + 2c, then which of the following must be true? I. a > b > c II. b = c III. a = b + 1 = c + 1 Using the above analysis, we can see that there are two values in the addition setup on the right, and that a must be 1 more than b and c, and that b and c must be equal to each other. We can see that condition I is not valid, and II and III are valid. We will pick a choice that says: II and III only.




FACTORIAL Factorial is a concept that is used to express the product of consecutive positive integers from 1 all the way to the number sitting in front of the factorial sign. Conventionally, we will go down in steps of 1 starting from the integer in front of the factorial sign all the way down to 1. All the consecutive integers from 1 to the value in front of the factorial sign are factors of given number factorial For example, 7! is the product of 7•6•5•4•3•2•1. Consecutive integers from 1 through 7 are factors of 7! 5! is the product of 5•4•3•2•1. Consecutive integers from 1 through 5 are factors of 5! You should also be able to see that 7! is the same as 7•6!, and is the same as 7•6•5! and so on. Why do we need to be able to do this? Because, we will have to deal with factorial values as factors when we deal with permutation and combination problems, and it will make our lives a lot easier if we can cancel out factors and deal with simpler values. Example: (x + 1)! = ? x! We can express (x + 1)! as (x+1)•x! We have: (x + 1)! = (x + 1)•x! = (x + 1) x! x! And the answer is a simple (x + 1). Example 2: What is (x - 1)! = ? X! We notice that the expression on top is one “fry short of a happy meal”. Let us top up the meal with the “missing” fries and multiply both numerator and the denominator by x. We get: x•(x - 1)! = x! = 1 (Answer) x• x! x• x! x

Of course, we could have reached the same conclusion by writing the denominator expression as x•(x-1)! and by canceling out the (x-1)! on the denominator and the denominator. Example 3: What is (x – 2)! ? (x – 4)! We notice that the value on the denominator is 2 more than that on the denominator. We can write (x – 2)! as (x-2)•(x-3)•(x-4)! We have: (x – 2)! = (x-2)•(x-3)•(x-4)! (x – 4)! (x –4)! = (x-2)•(x-3) Answer Example 4: If N = 21! + 17, which of the following cannot be a factor of N?

I. 21 II. 17 III. 38

Let us revisit the concept of factors here. A factor must divide N a whole number of times. Any value that does not do so cannot be a factor of N.

Can 21 be a factor? No, it cannot be, because, while 21 will divide 21! a whole number of times, it will leave a remainder of 17 when it divides N. Therefore, N cannot be a factor of N.

Can 17 divide N a whole number of times? Yes, It can and does. We notice that 21! has a 17 in the run-down to 1 and will cancel out the 17 in the denominator, leaving a whole number. The stand-alone 17 will also yield a whole number when divided by 17. Therefore, we conclude that 17 is a factor of N.

Can 38 divide N a whole number of times? No, it cannot. We notice that 38 can be expressed in terms of its factors 2 and 19, and these values will cancel out the 19 and 2 that are factors of 21! But 38 cannot divide the stand-alone 17 a whole number of times, and N, when divided by 38, will yield a remainder of 17, not a whole number. Our conclusion is that only the value in II is a factor of N. Conditions I and III are not.




OPERATORS AND FUNCTIONS Problems involving Function statements are usually presented in the GMAT as “operators” problems. A typical functions problem will go as follows: If F(X) = X2 – 2X + 3, What is F(-3)? We are required to find the value for the function of x when x = -3. We replace the X’s on the right side of the statement with –3, and get the value for F(-3) as F(-3) = (-3)2 – 2(-3)+ 3 Or, F(-3) = 18. In the GMAT, function statements are defined in terms of “operators”. An example of such a problem is shown below: Example: :The operator @ defines the relationship between a and b, for all a’s and b’s:

a@b = ab – b2 +a/b

Here the operator represents in an abridged form a relationship between a and b Exercise: The relationship between a and b is expressed by

the following operation: a@b = ab – b2 +a/b

What is the value of if

a@2 = - 33/2 How do we solve this problem ? Whenever we see the operator flanked by two variables, we can write the expression involving those two variables as specified. In this problem, we know that the value of b is 2. We are required to find the value for a. Let us write a@2 as a.2 – 22+ a/2 What did we do? We simply plugged in the value for b as 2 in the specified expression. We also have a value specified for a@2, which is -33/2 . We can now set up an equation as follows and solve for the value of a:

a.2 - 2.2 + a/2 = -33/2 or 2a - 4 + a/2 = - 33/2 or

5a/2 -4 = - 33/2 or

+ 4 +4 5a/2 = -33/2 + 4 = - 25/2 or 5a = - 25 or a = -5

Exercise 2: If x *y = xy + 2 (x +2y) for all integers x and y, then 4*(-5) = ? The operator * specifies a relationship between the two variables in terms of the expression involving those two variables. In our problem, x is 4 and y is -5. We simply plug in these values into the expression xy + 2 (x + 2y) to get: 4.(-5) + 2 (4 + 2(-5)) = -20 + 2 (4 - 10) = -20 + 2 ( -6 )

4*(-5) = -20 - 12 = - 32 Exercise 3: For all real numbers v, the operation v* is defined by the equation v* = v - v/3. If (v*)* = 8, then V = ? V* is an abbreviation for v-v/3 or 2v/3. You will understand V* to stand for “function of v” statement. This means that every time we see * next to a v, we simply replace v by 2v/3. Let us work on the expression within parenthesis first to get:

(v*)* = (2v/3)* Now, we have one more operator to get rid of. Let us invoke the original function statement: v* = 2v/3 Instead of a v, we now have 2v/3 in (2v/3)*. This means that we have to replace v with 2v/3 to get rid of the operator. We have: (2v/3)* = 2/3•(2v/3) = 4v/9 We have got (v*)* reduced to 4v/9, and we have a value specified for this expression, 8. Now we set up an equation as follows to get the value for v.

4v/9 = 8, or v = 9.8 / 4 = 18

The answer is: v = 18




Inequalities Inequalities make a statement that the values on either side of the sign are not equal to each other. The inequalities may specify that the value on the one side is bigger than or less than the value on the other side. ( < or >) We can carry out the same arithmetic operations – additions, subtractions, multiplication, and division – on an inequality, but you must remember the following caveat: When you do an operation involving a negative multiplication or a negative division on an inequality on both sides, or if you simultaneously invert the values on both sides, the sign of the inequality will be reversed. You must remember to do it every time such negative multiplication or negative division is performed. Example: If 3 – 2x < -7, find an expression for x. Let us move the number to the right. We get: -2x < -7 – 3 < -10 Let us divide both sides by –2 and flip the sign. We get: x > (-10/-2) > 5 The inequality expression for x is x > 5. Example 2: Find an expression for x, if 5 – 7x > -2. -4 Let us multiply both sides by –4 and flip the sign, because we are carrying out a negative multiplication. We get: 5 – 7x < 8 Let us move the number 5 to the right. The sign remains unchanged. We get: -7x < 3 Let us divide both sides by –7, and remember to flip the sign one more time. We get: x > -3/7

Ranges of values for a variable specified in an

inequality:

We know that if x2 = 4, then x has two roots: +2 or –2. What do we know about the ranges of values for x if x2 > 4? We should know that the values for x will lie in two ranges: x > 2 or x < -2 x x -2 0 +2

Another interesting fact you should know about is that in a small range between 0 and 1, any squared value will be less than the original value. For example, if x is ½, then x2 is ¼ and ¼ < ½. This reality exists only when x is a positive fraction less than 1.

Example: DATA SUFFICIENCY

Is x2 < x?

1. x > 0

2. x < 1

We need to know whether x lies in the following range: 0 < x < 1, because only in this range any squared value will be less than the original value. Statement 1 tells us that x is positive. IF x is ¾, then x2 is 9/16, and x2 < x. But then x could be 2, in which case x2 > x. The statement 1 gives us “all over the map” solutions. Statement 1 alone is not sufficient for a unique solution. Statement 2 tells us that x is less than 1. If x is ½, then x2 is ¼ and x2< x. But if x is –3, then x2 > x because 9 > -3. Once again, statement 2 alone is not good for a unique solution. When we combine the two statements, we notice that 0 < x < 1, and in this range we know that x2 < x. We must pick choice3 in the test.




You should expect to get about 50 percent of GMAT problems in a Word setting. The test is to see whether you can conceptualize verbal descriptions of a mathematical problem and deal with it by setting up simple mathematical expressions or equations. Come to think of it, we all do it in our management careers almost on a daily basis. In real-life, the word problem you may have to deal with will have the following form: “John was hired as a sales representative in July 1999, and his quarterly sales quota was $25,000. He met his quota in the first two quarters after hiring but his sales performance in the last quarters were only 80% of his quota, and he is facing an increase in his quota come January 2001 by 25%. It is unclear whether John will be able to meet his increased quota or not. Therefore, we need to chat with him, and fire him if necessary.” Sounds like a management case study to you? If you cannot understand the numbers presented in this verbal diarrhea, the chances are you are not going to be able to have a meaningful chat with your employee. What is worse, if your employee knows that your quantitative skills are as good as Governor Bush’s ability to pronounce “subliminal”, then he can run circles around you by throwing all kinds of numbers when you meet with him belly to belly. The bottom line is, you better come up to speed with these monstrous word problems. Management career does require some level of quantitative reasoning and understanding because that

is what presentations will be about in meetings and conferences. Now that we have established why we need to be good with word problems, let us see what kinds of word problems you may have to deal with in the GMAT. We have seen earlier in this module how we can deal with some “word problems” dealing with “proportions”, and algebra. Let us get some more practice with word problems in this chapter. The word problems will test your ability to pay attention to details, and to do simple set ups leading to a resolution of the problem. EXAMPLE 1: “During a certain season, a certain team won 75% of its first 100 games, and 40% of the remaining games. If the team won 60% of all games it played for the entire season, how many games did the team play in total?” When we deal with a word problem, we will set as X the quantity that needs to be determined. In this problem, we know that the team played 100 games and some more. Let us call the additional games X games. We get the following picture: INITIAL GAMES PLAYED: 100 ADDITIONAL GAMES PLAYED: X TOTAL GAMES PLAYED: 100 + X Let us see what else we need to do with the information provided in the problem. The team won 75% of the first 100 games, and 40% of the additional games.




Number of games won = 75% of 100 + 40% of X = 75 + 0.4X The problem also tells us that the number games won is equal to 60% of all games played. Number of games won is also = 60% of (100 + X) = 0.6 (100 + X) = 60 + 0.6X We are looking at two different expressions referring to the same information, namely the number of games won in total. Our equation is now ready to emerge: 75 + 0.4X = 60 + 0.6X Let us move the numbers to one side and the variable to the other. We get: 15 = 0.2X or X = 15/0.2 = 150/2 = 75 We now know the number of additional games played. The question asks us to determine the total number of games played. The answer is 100 + 75 = 175. This problem was not a difficult one once we cracked it. Understand the problem clearly and proceed to set it up one step at a time. Let us try one more and see how we can translate wordy dilemma into a manageable mathematical form.

EXAMPLE 2: “Salesperson A’s compensation for any week is $450 plus 5% of the portion of A’s total sales in excess of $1,000 for that week. Salesperson B’s compensation is 7% of B’s total sales for that week. For what amount of total weekly sales, in dollars, would both salespersons earn the same compensation?” Once again, we need to determine the dollar value of sales for which both people will earn the same income for a given week. Let us call that sales dollar value X. A’s compensation

= 450 + 5% of (X – 1000) = 450 + 0.05(X – 1000) = 450 + 0.05X – 50 = 400 + 0.05X We took (X –1000) when we applied 5% because the commission is paid on the excess over $1,000. B’s compensation = 7% of X = 0.07X At a sales level of X dollars, both people make the same income. Our equation is: 400 + 0.05X = 0.07X Let us move the variable to one side of the equation and isolate the number on the other side. 400 = 0.02X or 40000 = 2X (multiply both

sides by 100 to get rid of the decimal places after the decimal

point)

or, X = $20,000 Both people will make the same level of commission at a sales level of $20,000.




Now that we know what a word problem is all about, let us see what other word problem types we may come across in the GMAT.

AGE PROBLEM

Age problem usually involves two people and their ages will be specified in some relationship. Usually, you will start off by defining each person’s age as of today and go from there. Also, be sure to use the first letter of the name as the variable instead of X and Y. EXAMPLE: “Anna’s age today is 11 times Bob’s age today. Three years ago, Anna’s age was 21 times Bob’s age then. How old will Anna be 10 years from now?” Let us start with the ages of the person as of today. AGE TODAY: ANNA’S = A BOB’S = B Today, Anna’s is 11 times Bob’s age. The set up for this information is simple: A = 11•B Three years ago, Anna was A – 3, and Bob was B – 3. Three years ago, Anna’s was 21 times Bob’s age then. The set up for this information is: A – 3 = 21• (B – 3)

We have two equations and two unknowns. Let us write 11 B for A in the second equation, and get: 11B – 3 = 21B – 63 Let us isolate the numbers on one side and B on the other side. We get: 10B = 60 or B = 6 If B is 6, then A is 11 times B or 66. Anna is 66 years of age today. 10 years from now, Anna will be 66 + 10 = 76 years of age. How can the same problem be presented to us without numbers and in a higher difficulty level? Take a look at the one that follows: “The combined age of John and Lisa is y years of age. If John is 12 years older than Lisa, how old will John be, in terms of y, y years from now?” How do we begin to deal with the problem presented above? Let us say that J is John’s age today, and L, Lisa’s age today. We have: J + L = y ……. (1) We are also given that J = L + 12 . (2) We are asked to get John’s age y years from now. In order to do this, we need to find John’s age today. Let us get Lisa’s age in terms of John’s age from the equation (2) above. We get L = J – 12 Let us replace L with J – 12 in (1) to get: J + J – 12 = y or 2J = y + 12

Or, J = ½. y + 6 The above equation gives us John’s age as of today. How do we determine John’s age y years from now? We simply add y to John’s current age.




We get: J + y = ½. y + 6 + y = 3/2. y + 6 John’s age, y years from now, will be

3/2.y + 6 EXAMPLE 3: “Is Sam were twice as old as he is, he would be 40 years older than Jim. If Jim is 10 years younger than Sam, how old is Sam?” Let Sam be S years of age today, and Jim be J years of age today. If Sam were twice S, he would be 40 + J. The set up is: 2S = J + 40 Jim is 10 years younger than Sam. How do we write this information mathematically? J = S –10 In order to find out the value for S, we need to get rid of J from the first equation. Let us replace J with S – 10 in the first set up, and get: 2S = S – 10 + 40 S = 30 Sam is 30 years of age today and Jim is 20 years of age today. If Sam were 60, he would be 40 years older than 20. Our verification tells us that we got the correct answer. Here is a quickie: “If Lisa was X years of age Y years ago, how old was she Z years ago?” Lisa is today X + Y years of age. Z years ago, she was X + Y – Z years of age. Pretty simple, huh?

DISTANCE PROBLEMS We all know that the Distance traversed is the product of AVERAGE SPEED and time of travel. If we traveled at an average speed of 60 miles per hour for 4 hours on the Interstate 90, we would have traveled 240 miles. DISTANCE = AVERAGE SPEED• TIME Average speed = Distance / Time Time of travel = Distance / Average speed These are the three different re-arrangements of the same equation you should remember. It will also be useful to know the concept of relative speed, and use it when you deal with distance problems. You will find that relative speed concept helps you breeze through some Distance problems, and get to the answer in short order. THEOREM OF RELATIVE SPEED: “If A and B are moving in diametrically opposite directions either away from each other or towards each other, then the relative speed when both A and B are moving is the sum of their individual speeds.”




Let us say you are proceeding southbound on I-95 and traveling at 65 m.p.h, and another driver goes up North on I-95 and doing 70 m.p.h, when you pass each other for a brief moment, it is as if you stood still and the other “jerk” whisked past at 135 miles per hour. A more gruesome way to remember this concept is by understanding what happens in a head-on collision. If two cars collide head-on, and if one car was doing 65 miles per hour and the other was doing 70 miles per hour, the impact of the collision will be the same as if either driver slammed into a brick wall at 135 miles per hour. Now you know what those crash test dummies go through. You will have to apply this concept when you deal with a problem that specifies that A and B are some distance apart and are moving towards each other along a straight line. The question will be: How far does one person travel before he or she meets the other person traveling in the opposite direction towards him or her? S1 m.p.h S2 m.p.h A B D miles Distance traveled by A before A meets B = S1/(S1 + S2) • D Distance Traveled by B before B meets A = S2/(S1 + S2) • D Distance traveled by each person will be in proportion to the relative speed (S1 + S2). If people are moving in the same direction, then the relative speed is the difference of the two speeds. If A and B are moving in the same direction on I-95, and the speeds are 65 m.p.h and 70 m.p.h respectively, when they pass each other, it is as if one person stood still and the other moved at 5 miles per hour.

Let us see how we can apply this concept when we deal with some distance problems. EXAMPLE 1: “The Distance between Boston and Toronto is 600 miles. Car A leaves Toronto 8 A.M. Eastern Standard Time and Car B leaves Boston at the same time. If Car A travels at an average speed of 60 miles per hour and Car B at an average speed of 50 miles per hour, how far would Car A have traveled before it meets Car B?” Let us do this problem the conventional way, and then see how much faster we can do it by using the relative speed concept. We begin by attempting to construct a mathematical model of this word problem: Meeting Point x 600-x Toronto Boston

Car A travels at 60 mph and B at 50 mph average speed. Let X be the distance from Toronto where the two cars meet. By default, the car from Boston would have traveled the difference between 600 and X miles or 600-X miles. Since the two cars leave at the same time, they would have traveled for the same length of time when they meet. However, they would have traveled different distances because they were traveling at different rates of speed. Time of travel for Car A: X/60 Hours Time of Travel for Car B; (600-X)/50 Hr Time of travel is the same for both cars. ∴ x /60 = (600 - x)/50 We have one unknown and we need just one equation. We have got one equation and we couldn’t possibly ask for more. Let us cross multiply to get: 50x = 60(600-x)




50x = 36000 - 60x Or 110x = 36000, or x = 36000/110 We solve for x and we get x = 327.2 miles. Car A would have traveled 327.2 miles before it meets Car B. Car B would have traveled 272.8 miles before it meets Car A. Let us do it faster with the relative speed concept. 60 M.P.H 50 M.P.H TORONTO BOSTON 600 MILES Because the trains are moving in diametrically opposite directions, the relative speed is the sum of the two speeds or 110 miles per hour. Distance traveled by Car A Before it meets Car B = Car A’s speed•Distance RELATIVE SPEED = 60•600/110 =3600/11 = 327.2 Miles. Distance traveled by Car B before it Met Car A = Car B’s speed•Distance RELATIVE SPEED = 50•600/110 = 3000/11 = 272.8 MILES Can you see that Relative speed way of getting the distance traveled by each car may be a faster way than the algebraic way? But regardless of what approach you use, you will get the same answer. Now. Let us add a small wrinkle to the problem and state that Car A leaves an hour ahead of Car B. How far from Toronto do the cars meet under this changed scenario?

What is different about this problem? We notice that Car A travels an hour longer than Car B does before it meets car B. Time of travel for Car A = Time of Travel for B + 1 hour. The distance from Toronto where the cars meet is still x miles. Car B still travels 600-x miles. Time of travel for car A = x/60 Time of travel for Car B = (600-x)/50 How do we factor this information into the new equation? Time for Car A = Time for Car B + 1 i.e. x/60 = (600-x)/50 + 1 = (650-x)50 Let us cross multiply to get: 50x = 60(650-x) Or 110 x = 39,000 Or, x = 39000/110 = 354 miles. The cars will meet at a distance that is 354 miles from Toronto. Let us do it the Relative speed way. When only one car is traveling, the relative speed is the speed of that one car. Car A gets a head start and it travels 60 miles closer to Boston in the first hour. When Car B starts traveling, the distance between Toronto and Boston is no longer 600 miles but only 600-60 = 540 miles. When both cars are moving at the same time towards each other, the distance traveled by each will be computed using this new distance separating the two cars. Relative speed will still be the same as 110 miles per hour. Therefore, distance traveled by Toronto Car = Distance traveled in Hour 1 + Proportional distance traveled when both cars are moving Distance traveled by car A = 60 + 60•540/110 = 60 + 294

= 354 Miles. The same answer as by the other method.




Remember: For “Distance/Speed/Time” problems, your starting point will always be the relationship:

Distance = speed X time EXAMPLE 2: “Sam and Sylvia are 50 miles apart and start walking towards each other along a straight line path. Sam’s average speed is 5 m.p.h and Sylvia’s is 4 miles per hour. If Sylvia leaves an hour before Sam starts walking towards Sylvia, how far, in miles, from where Sylvia started does she meet Sam?” Let us do it the Relative Speed way. 4 miles per hour 5 miles per hr Sylvia Sam 50 miles Because the two folks are walking in the diametrically opposite directions, the relative speed when both are walking is 4 + 5 = 9 miles per hour. Initially, Only Sylvia is walking. She would have walked 4 miles closer to Sam in the First hour, before Sam begins to walk. The effective distance between the two is now 46 miles and not 50 miles. We will use 46 when we compute the proportional distance Sylvia walked. Therefore, distance walked by Sylvia before she met Sam = Distance walked by Sylvia in the first hour + Proportional distance walked by her when both were walking at the same time = 4 + 4•46/9 = 4 + 184/9 = 4 + 204/9 = 244/9 Miles Therefore, Sylvia walked 244/9 Miles before she bumped into Sam. Sam would

have walked the difference of 50 and 244/9 Miles or 255/9 miles. Can you see how the relative speed concept helps you deal with distance problems a lot faster when two people are walking towards each other. What happens if people are moving in the same direction? We said that the relative speed is the absolute difference of the two speeds. Let us see how we can put this information to use in the following problem. EXAMPLE 3: “Car A and Car B leave Kalamazoo at 9 a.m., and travel towards Detroit along I-94 East. Car A’s average speed is 60 miles per hour and Car B’s average speed is 70 miles per hour. An hour after starting, Car B pulls into a rest area, while Car A kept traveling at the same average speed without stopping towards Detroit. If Car B leaves the rest area 15 minutes after it stopped at the rest area, and continued to travel at the same average speed as before, how far from the rest area will Car B catch up with Car A?” In one hour, Car B would have gone 70 miles from start and Car B would be 10 miles behind Car A because car A’s average speed is only 60 miles per hour. Therefore, at the end of 1 hour, car A is ahead 10 miles of Car B. Car B stops at the rest area for 15 minutes and Car A keeps traveling at the same average speed. In 15 minutes, Car A would have gone an additional 15 miles (1/4th of 60 miles) from where it was at the end of Hour 1. This means that Car A will be 5 miles ahead of Car B when it leaves the rest area 15 minutes later. When Car A




leaves the rest area and begins traveling at the same average speed as before, the relative speed between Car A and Car B is 10 miles per hour (difference between 70 miles per hour and 60 miles per hour). How long does Car B take to travel 5 miles at a relative speed of 10 miles per hour? (It is as if Car A stopped and Car B traveled at a relative speed of 10 miles per hour). 30 minutes. In 30 minutes, Car B would travel 35 miles at 70 miles per hour speed, and the answer is: Car B would catch up with Car A at 35 miles from the rest area. Can you see how the concept of relative speed and some reasoning were used in solving this problem? Get used to both processes. Also, remember that the relative speed concept can be used ONLY if two objects are moving. If there is just one object, we will go back to the basic concept of Distance = average speed • time

TRAVEL OVER THE SAME DISTANCE IN

TWO DIFFERENT LEGS OF THE JOURNEY

If the journey involves two legs at two different average speeds S1 and S2, and if the two legs are identical in distance, then the average speed for the total trip involving the two legs is obtained by using the formula: Average Speed = 2S1S2 / (S1 + S2) Note that the average speed cannot be computed as a straight average of the two speeds.

Example: “A salesperson travels from London to Liverpool at an average rate of speed of 40 miles per hour, and returns immediately on reaching Liverpool to London, traveling during this leg of the journey at 50 miles per hour. What is the average speed for the whole trip?” We can tell that the journey is over two legs at different average speeds, and the distance per leg is the same. We need to use the formula to obtain the average speed. In this problem, S1 = 40 miles per hour and S2 = 50 miles per hour.

Average Speed = 2S1S2 / (S1 + S2) = 2•40•50/(40 + 50) = 2•40•50/90 = 400/9 = 444/9 miles per hour Notice that the average speed is not 45 miles per hour, which is what we would have obtained had we taken the straight average. EXAMPLE 2: “Tiffany travels from A to B at an average speed of 30 miles per hour and returns immediately to A on reaching B, traveling back at an average speed of 50 miles per hour. If she took 3 hours for the round trip, what is the distance between A and B?” We notice that the journey involves two legs of identical length and two different average speeds. Average speed for the round trip = 2•30•50/(30+50) = 300/8 miles per hour.




How do we compute the distance? We know that the average speed for the round trip is 300/8 miles per hour, and the round trip time is 3 hours. Therefore, round trip distance = round trip average speed• round trip time = 300•3/8 = 900/8 miles. One way distance between A and B is one half the round trip distance. Distance between A and B = ½ •900/8 = 900/16 = 56 ¼ miles In some other distance problems, we have to go back to the basics and apply the rule: Distance = average speed • time Example “The distance between Chicago and Omaha, Nebraska can be traveled can be traveled 2 hours sooner if the average speed of travel were 10 miles per hour faster. If car is traveling at an average speed of 50 m.p.h from Chicago to Omaha, what is the distance between the two cities?”

In this problem, we are dealing with just one average speed and some “time” information. Let us go back to the basics and work our way up. If the car takes T hours to go from Chicago to Omaha traveling at 50 miles per hour, then the same car would take (T – 2) hours if it traveled at 60 miles per hour. The distance traveled over T hours at 50 miles per hour is the same as that traveled over (T – 2) hours at 60 miles per hour. Our set up is: 50•T = 60•(T – 2) Or, 50•T = 60•T – 120 Or, 10•T = 120 Or, T = 12 hours. The car would have taken 12 hours to travel from Chicago to Omaha at 50 miles per hour. Therefore, the distance between the two cities is: 50 times 12 = 600 miles. Let us see how the same problem can be presented as a high difficulty level question in the following page.




Example 2: “A train traveling at “m” mph arrives at its destination 100 miles away 2 hours behind schedule. At what rate of speed, in terms of m, should the train have traveled to arrive on time?” How do we deal with this problem? What is average speed? It is Total distance

traveled/ total time Let S be the speed at which the train should have traveled to arrive at its destination on time. Let T be the scheduled time of travel. Traveling at m miles per hour, the train takes 2 hours longer or T +2 hours. We get 100 = m (T+2) .....................(1) If the train had traveled at S mph, it would have arrived on time, T. We get 100 = S . T ......................... (2) or S = 100/ T ........................... (3) We are asked to express S in terms of m. Can we express T in terms of m from(1) so that we can plug in that information in (3) and get S in terms of m ? Let us play around with (1). We have 100 = mT + 2m Or m T = 100 - 2m or T = (100-2m)/m .................. (4) Let us plug in this value for T in terms of m in (3) to get S = 100/T = 100 / (100-2m)/m = 100m/(100-2m) We see a common factor of 2 in both the numerator and the denominator. Let us take it out. We get S = 50m/(50-m).

Example 3: “John travels for 6 hours at an average speed of 50 mph, and then travels T hours at an average speed that is 10 mph more than that during the first 6 hours. If the average speed for the whole journey was 58 mph, how long, in hours, did John travel?” Once again, our starting point is the relationship: DISTANCE = SPEED x TIME What are we required to find out in this problem? We are required to find out the value for (6 + T) in hours. How do we read, in mathematical terms, the information in the problem ? • John traveled 50 X 6 = 300 miles in the

first hour. • John traveled (50 + 10) .T = 60T

miles in the next leg of the journey lasting T hours.

• Total distance traveled = 300 + 60.T • Total time traveled: T + 6 • Average speed = Distance/ Time • Average Speed = (300 + 60T)/(T+6) • We also know that the average speed is

58 mph. • We are ready to set up an equation as

follows: (300 + 60T)/)T+6) = 58 Or 300 + 60. T = 58 ( T+6) Or 300 + 60. T = 58. T + 348 Or 2 T = 48 Or T = 24 hours.

John traveled for 6 + T = 6+24 hours = 30 hours.

As you can see, we have to proceed step by step, one step at a time, and then use the value specified to set up an equation to solve the problem.




Example 4: “ A salesperson travels for 3 hours at “m” miles per hour, and travels another 60 miles during the next 4 hours. What is the average speed for the whole trip?”

Average speed = Total distance/ total time

• Distance traveled in the first 3 hours =

3.m miles (Notice that we have to compute the distance from the time/speed information specified)

• Distance traveled in the next 4 hours = 60 miles

• Total distance traveled = (3m + 60) • Total Time traveled = 3 + 4 = 7 hours • Average speed = Total Distance ÷ Total time = (3m+60) ÷ 7= (3m+60)/7 A messy fraction. But that is answer.

WORK PROBLEMS Work Problems deal with the rates at which certain persons or machinery work alone or together and you will be required to calculate the rate at which they work together or alone. Let us illustrate a typical problem dealing with work. “John and David, working together, can complete the job in 6 hours. If John, working alone, can complete the job in 10 hours, in how many hours will David, working alone, complete the job?” The good news about this type of problem is that there is just one way to do it: You calculate what fraction of the work will be completed per unit of time for each person working alone and for the combined efforts.

In our example, if John and David take 6 hours to complete a job, then John and David, working together, will finish 1/6th of the work per unit of time (in one hour). John, working alone, takes 10 hours to complete the job, and will have completed 1/10th of the work in the same unit of time. Let d be the length of time David will require to complete the work, acting on his own. Per same unit of time (in one hour), David would have completed (1/d)th of the work.

Fraction of work completed by John per hour + Fraction of work completed by David per hour = Fraction of work completed by John and David together.

The above information translates into the following mathematical equation: (1/10) + (1/d) = (1/6) or (d+10)/10d = 1/6 Or 6(d+10) = 10d Or 4d = 60 hours or d = 15 hours Solving for “d”, we get 15 hours, meaning that David will require 15 hours to complete the work acting alone. RULE: If T1 is the length of time taken by worker 1 to complete a job working alone, and if T2 is the length of time taken by worker 2 complete the same job working alone, at their respective constant rates, then if worker 1 and 2 work together, they will complete (T1 + T2)/ (T1.T2) of the job in 1 hour. Or, Workers 1 and 2, working together, will take (T1T2) / (T1 + T2) hours to do the job, working together at their respective constant rates. RULE: If X machines, each working at the same constant rate, take N hours to finish a job, then one machine, working alone at the same constant rate, will require X times N hours to finish the job. For example, if 5 machines, working at the same constant rate, take 4 hours to finish the job, then one machine, working alone, will require 5 times 4 or 20 hours to complete the job. If we use 3 machines to finish the same job, then the 3 machines will take 20/3 = 6 2/3 hours to do the job, working together.




Example 2: If 8 Large hoses can fill a pool in 6 hours, and 6 small ones can fill the same pool in 10 hours, working at their respective constant rates, how long would it take for 6 large hoses and 6 small hoses to fill the pool, working together? You will notice that we have information pertaining to how long it takes for 8 Large hoses to fill the tank, but we are not going to use all 8. We will use only 6 Large hoses to fill the pool with 6 small ones. If it takes 6 hours to fill the pool with 8 large hoses, then it would take 6 large hoses 8/6 X 6 = 8 hours to fill the same pool. This value (8 hours) is our T1 value. What is our T2 value? 10 hours corresponding to 6 small hoses. So. How long will it take to fill the pool for 6 small and 6 large hoses working together at their respective constant rates? Time to fill the pool = T1 T2 (T1 + T2) = 8.10 / 18 = 4 4/9 hours 6 small and 6 large hoses, working together at their respective constant rates, will fill the pool in 4 4/9 hours. Example 3: “The machine shop is required to produce 10,000 nuts in 2 hours. The foreman has at his disposal two machines, X and Y. Machine X can produce 10,000 nuts in 5 hours, working alone. At what rate should the machine Y be operating so that 10,000 nuts will be produced in 2 hours?” The Number Of Nuts required to be produced in one hour = 5,000 nuts. Machine X, acting alone, can produce 2,000 nuts in an hour. Therefore, the

machine Y should be set to produce 3,000 nuts in an hour, or 10,000 nuts in 3.333 hours so that X and Y, operating together, can produce 10,000 nuts in 2 hours. Example 4: “A machine working at a constant rate had produced 1/4th of the total job by 10:20 a.m. and 1/3rd of the total job by 10:40 a.m.. If the machine continue to work at the same constant rate, at what time will the machine finish the job?” When we deal with work problems, we need to compute what fraction of work is accomplished per unit of time. In this problem, the machine had done 1/3 – ¼ = 1/12th of the total job in 20 minutes between 10:20 a.m. and 10:40 a.m. This means that the machine does 1/12th of the job in 20 minute increments. Working at this rate, the machine will take 12 times 20 minutes from start to finish, or 240 minutes or 4 hours. The machine had done 1/3rd of the job by 10:40 a.m. and it will take 2/3rd of 4 hours to do the remaining 2/3rd of the job or 160 minutes, which translate into 2 hours and 40 minutes beyond 10:40 a.m. That will put us at 01:20 p.m. for the finish time. Example 5: “A machine started on July 2nd at 10:00 a.m. and ran continuously at a constant rate for 263 hours to finish a production lot. At what time and on what date did the machine finish the operation?” 263 hours are 10 days and 23 hours, or 11 days less an hour. We need to go forward 11 days from 10 a.m. on July 2nd and move back an hour. That will put us at 9 a.m. on July 13th. That is when the machine will complete the operation.




MIXTURE PROBLEM A mixture problem is a weighted-average problem, and weighted average is a concept that you will deal with a lot in GMAT. If you recall, Distance-speed-time problem is also a weighted-average problem when you are asked to determine the speed as a ratio of total distance and total time traveled. “How many gallons of a solution that is 20% alcohol must be added to 8 gallons of a solution that is 10% alcohol so that the resulting solution is 16% alcohol?” Let “x” represent the number of gallons of the 20% solution. The amount of alcohol in “x” gallons of 20% solution will be (0.2x) gallons The amount of alcohol in 8 gallons of 10% solution will be 8(0.1) = 0.8 gallons. The amount of alcohol in the resulting solution of (x + 8) gallons will be: (0.2x + 0.8) gallons We want this amount to be equal to 16% of the total solution volume of (x+8) gallons. The equation we get for this information is:

0.2x + 0.8 = 16%(x+8) = 0.16x + 1.28 Solving for “x” , we get 0.04x = 0.48 or X = 12 gallons. Remember: You must calculate the total solution volume, and the amount of alcohol in that solution, and set up an equation that expresses the alcohol content as a percent of the total solution volume.

Example 2: How many gallons of a 80% alcohol solution must be added to a 75% alcohol solution so that the resulting solution will be 20 gallons of 77% solution? Let us say that we must mix X gallons of 80% solution with 20-X gallons of 75% solution so that we get a resulting solution of 20 gallons. What is the amount of alcohol in X gallons of 80% solution? 0.8 X gallons What is the amount of alcohol in (20-X) gallons of 75% solution? 0.75(20-X) gallons. What is the total amount of alcohol in the mixture ? 0.8X + 0.75(20-X) = 15 + 0.05X We want this amount of alcohol to be equal to 77% of the mixture volume of 20 gallons. This requirement translates into the following equation: 15 + 0.05X = 0.77 (20) = 15.4 Or 0.05X = 0.4 or X = 0.4/ 0.05

X = 8 gallons We must mix 8.0 gallons of 80% alcohol solution with 12.0 gallons of 75% solution to get 20 gallons of 77% alcohol solution. If we want to truly understand why a “mixture problem” is nothing but a “weighted average” problem, we should consider the following problem.




Example 3:

“If one group of 10 students with an average GMAT total score of 610 is mixed with another group of 12 students with an average GMAT total score of 590, what is the average GMAT total score of the new group (mixture)?” The total “total” score of the first group of 10 students is 6,100. The total “total” score of the second group of 12 students is 7,080. The size of the new group (mixture) is 22. The sum of the two totals is: 13,180 The new average score of the mixture is: (13,180 ÷22) =599 Example 4: Let us see what happens if we mix sales figures for two months at a donut shop: “Dunkin’ Donuts sold twice as many dozens of donuts in the Month of August 98 as in the month of July 98. If the average price per dozen of donuts in the month of July 98 was $6.95 and the average price per dozen of donuts in the month of August 98 was $7.25, what was the average price per dozen of the donuts for the two months?” Solution: If the quantity sold (in dozens) during July 98 was “N” dozens, then the quantity sold in August 98 was “2N” dozens. Revenue from the sale of “N” dozen donuts in July 98 = 6.95N dollars

Revenue from the sale of “2N” dozen donuts in August 98 = 2N. (7.25) = 14.5N dollars. Total revenue for the months of July and August 98 = 6.95N + 14.5N = 21.45 N dollars. Total quantity sold during the two months = N + 2N = 3N dozen donuts. Average price per dozen of donuts for the two-month period = Total Revenue / Total Quantity = 21.45N/3N = 7.15 dollars.




INTEREST PROBLEMS: There are two types of interest: Simple Interest, and Compound Interest. With Simple Interest, interest is computed on the principal amount only. For example, 6% simple interest on a principal amount of $100.00 is $6.00. With compound interest, interest is calculated on “principal plus accrued or earned interest”. With 6% compound interest, compounded annually, the $100.00 principal amount will earn $6.00 interest in the first year and $6.36 interest in the second year, the compound interest in the third year will be $6.74 Example: “If $10,000 is invested at 12% annual interest, compounded monthly, what is the balance in the account after 3 months of starting the investment?” 12% annual rate translates to 1% monthly rate of interest. The Account Balance after Month 1: $10,000 + 1% of $10,000 = $10,100 The Account Balance after Month 2 : $10,100 + 1% of $10,100 = $10,201 The Account Balance after Month 3 : $10,201 + 1% of $10,201 = $10,303 The formula to compute the Principal Plus compounded interest over a period of “3” months or years at a rate of interest per same period of “1%” is: P (1+1/100)³ , where P is the principal amount.

Example 2: An amount of money, N, was invested in an account paying interest at 10% per annum compounded annually. If the amount of money in the account at the beginning of year 4 was $14,641, what was the initial value of investment, N? Asset Interest Accrual Value Earned Value N 0.1 N 1.1N 1.1N 0.11N 1.21N 1.21N 0.121N 1.331N 1.331N = $ 14, 641 or N = $11,000 Get used to dealing with compound interest problems by setting up a matrix as shown. You will be using the same approach when you deal with “depreciation” problems, where the depreciation is calculated on the value of the asset at the beginning of the year. In a compound interest situation, interest is calculated on the accrual value, which is the sum of principal amount and the interest earned. In compound depreciation problem, depreciation is computed on the residual value of the asset, which residual value is the asset value at the beginning of the year minus the depreciation during the year.




DEPRECIATION PROBLEMS “Depreciation” Problem is the exact opposite of “Interest” Problem. Depreciation refers to reduction in the value of an asset, and not growth. Let us say that we have an asset - Computerized Milling Machine - that was bought for $100,000. If the tax laws allow the company to take depreciation at a rate of, say, 20%, then the asset will lose $20,000 each year every year until the asset is completely depleted in 5 years. Example 1: “An asset valued at N dollars depreciates at a uniform rate of 15% of its initial value each year. If the residual value (value after depreciation) after 4 years is $12,000, what was the initial value of the asset?” How do we read and set up this problem? The value remaining after 4 depletions- or 60% depreciation - is $12,000. This information can mean only one thing: That $12,000 represents (100-60) = 40% of the value N of the asset. Now we are ready to set up an equation: 40% of N = $12,000 OR 0.4N = $12,000

OR N = 12,000 ÷ 0.4 = $30,000 That was not a difficult problem, was it ? What happens if the information is specified differently as follows ?

Example 2: “An asset of value N depreciates at the rate of 1/4 of its value at the beginning of each year every year. What is the residual value of the asset at the end of year 4?” Notice that the depreciation is not taken on the initial value of the asset, but rather on the value of the asset at the beginning of each year. The best way to deal with this type of problem is to set up a matrix as shown below: Year Value Dep Residual Value 1 N N/4 3N/4 2 3N/4 1/4(3N/4) 3/4(3N/4) 3 3/4(3N/4) 1/4(3/4(3N/4)) 3/4(3/4(3N/4))

4 3/4(3/4(3N/4)) 1/4(3/4(3/4(3N/4) 3/4(3/4(3/4(3N/4)) Do you notice a pattern emerging here ? The residual value at the end of year 4 will be (3.3.3.3)N / (4.4.4.4)= 81N/256 What will be the residual value at the end of year 5 in this problem? It will be 3/4th of 81N/256 = 243N/1024 and so on.......




DISCOUNT PROBLEMS: If P is the original price which is discounted by d%, the new price becomes (100-d)% of P. Note that 10% discount on a price represents 90% of the original price. Let us consider an example: “Mary bought a mink coat for $750.00. If that price represented a 25% discount on the original price, which included a 50% mark-up on the cost, what was the cost to the retailer of the mink coat?” Mary paid $750.00 which was 75% of the original sticker price (which is the same thing as saying 25% discount on the original price). The original sticker price was, therefore, $750.00 ÷0.75 equal to $1,000.00 This sticker price of $1,000 was 150% of the cost (50% markup or profit). The cost to the retailer was, therefore, $1,000 ÷ 1.5 = $666.67. A simple variation on the Discount Problem becomes a Profit Problem. Profit is Selling Price minus Cost. “ A used car dealer buys a vehicle for $2,500.00. 1. At what price should the dealer sell the vehicle to realize a gross profit of 50% of the cost of the vehicle? 2. At what price should the dealer sell the vehicle so that the profit represents 50% of the selling price?” Part 1 of the problem is simple. The dealer wants 50% profit on the cost. So

he must sell the vehicle for 150% of the cost or 1.5 X $2,500.00 = $3,750.00 Part 2 of the problem is a little more complex. The dealer wants the profit to be 50% of the selling price. Let the selling price be “s”. The Desired Profit is: 0.5s. Therefore, the Cost Plus the Desired Profit must equal the selling price: We get the following equation: $2,500.00 + (0.5)s = s Solving for “s”, we get $5,000 as the selling price required to realize a profit of 50% of the selling price. Now, try this one for size: “An item was bought by a retailer for $1,000 and sold for Price S, which represented a 25% discount on the sticker price. If the item had been sold at the sticker price, the profit would have been 40% higher than what the retailer netted from the sale. The profit on the sale of the item was what percent of the item’s selling price ?” Hint: Use the following relationships: COST + MARK UP = STICKER PRICE STICKER(1-D/100) = SELLING PRICE where D is the discount percentage SELLING PRICE - COST = PROFIT Profit on the selling price is (S – 1000) Because the selling price is 75% of sticker, we have: Sticker ( 1-25/100) = S (selling Price) Or ¾. Sticker = S Or Sticker = 4/3 . S What is the hypothetical profit on the sticker price? Sticker – 1000 = 4/3.S – 1000 We are told that the profit on the sticker price is 40% more than the actual profit. We get the equation: 4/3S – 1000 = 1.4(S-1000) or 4S – 3000 = 4.2S – 4,200 Or 0.2S = 1,200 or S = 6,000 Profit on the selling price = 6000 – 1000 = 5000 Profit as a percent of selling price = 5000/6000 X 100 = 82%




STATISTICS - PROBABILITY If there are N equally likely outcomes, and R out of these outcomes are favorable to event A, then the probability of event A is R/N. In the GMAT, you have 5 choices, and the probability of your picking a correct response by “winging” it is 1/5. What is the probability that you will get two answers in a row correct by picking choices randomly? It is 1/5 times 1/5 or 1/25. What is the probability that you will get 6 answers in a row correct by picking choices randomly? It is (1/5)6 or, 1/15,625. What is the probability that you will get at least one of the 6 questions correctly answered by “winging it”? This is a little tricky question. The probability of getting at least one out of 6 questions correct is the same as 1 minus the probability of the WORST CASE SCENARIO of getting them all wrong. The probability of getting them all wrong is (4/5)6 or 4096/15625. Therefore, the probability of getting at least one of the six questions correctly answered by selecting the choices randomly is 1 - 4096/15625 Or, 11529/15625 or 0.737. Not bad at all. Let us get a solid perspective on the probability of getting different numbers of correct answers by randomly selecting choices in the GMAT, especially when you are about to run out of time. Let us say that we have 4 questions remaining, and we decide to “shoot craps”. The probability of getting at least 1 correct response is about 0.6 (1 minus the probability of getting all wrong responses), and the probability of getting at least 2 correct responses is about 0.18, and the probability of getting at least 3 questions correct is about 0.027, and the probability of getting all correct responses is 1/625, or 0.0016. How did we get these probabilities? Take a look at example 4. When you are running out of time in the GMAT, and have, let’s say, 6 questions remaining, you are better off selecting a choice for the remaining questions because the unanswered questions will be counted as incorrect responses. As you can see, you have a

minuscule chance of getting all of the answers correct, but the odds are better than winning a lottery by picking 6 numbers out of the first 49 consecutive integers; the probability of winning a 649 lottery is 1 in 14 million. Remember: When you deal with probability (and permutation), you will multiply when the question is about the probability of doing x AND y; you will add when the problem asks you to find the probability of doing x OR y. The probability of getting question 1 and question 2 correct by “winging it” in the GMAT is the product of 1/5 and 1/5, or 1/25. If there is an assortment of balls in a box: Red : 15, Blue: 20, and Green : 10 And we would like to select a ball randomly, what is the probability that the ball selected will be a red one or a blue one? The probability of a red ball is 15 out of 45, and the probability of a blue ball is 20 out of 45. Therefore, the probability that the ball selected will be a red or a blue ball is 35 out of 45, or 7/9. Example 2: If there is an assortment of balls in a box: Red : 15, Blue: 20, and Green : 10 And we would like to select two balls randomly without replacement, what is the probability that the balls selected will be a red one and a blue one? Here, the order of selection is important. We need to consider that the first ball picked is Red and the next blue or the other way around. Because the probability of picking Red first followed by Blue next is the same, we can compute the probability of picking one red ball and one blue ball, and then multiply the result by 2. Let us say that we need to compute the probability of getting a red ball when the first ball is drawn. The probability is 15 out of 45 or 1/3. After we have drawn a ball, we have only 44 remaining in the box when the second ball is drawn. The probability that the second ball will be a blue is 20 out of 44, or 5/11. The probability that the second ball drawn will be a blue, given that the first one was a red is the product of the two probabilities: 1/3 times 5/11. Or 5/33. Because the selection could be Red

Probability is the same as Proportionality. If 40% of a group is made up of women, then the probability of picking a woman in the group by random selection process is 4/10th or 2/5th. Ratios specify probability information too: If the ratio of men to women in a group is 2 to 3, then we know that men are 2 parts out of a total of 5 parts, and women are 3 parts out of a total of 5 parts. Therefore, the probability of selecting a man randomly is 2/5th and the probability of picking a woman randomly is 3/5th. In GMAT, if 20% of the answer choices are correct, then the probability of picking a correct answer through random selection is 20% or 1/5th.




followed by Blue or vice versa, the required probability is 2 times 5/33 = 10/33. Example 3: Joe Hitter has a batting average of 0.3. (He is likely to hit 3 ball out of 10 balls.) What is the probability that he will hit 4 times in a row when he comes to bat? The probability that he will hit 4 times in a row when he comes to bat is (0.3)4 or 0.081. The probability that he will “fail to hit” 4 times in a row when he comes to bat is (0.7)4 , or 0.2401. Finding the probability of an event though “exceptions”. Sometimes, the easiest way to find the probability of an event is by subtracting the probability of a restriction or an exception from the total probability value of 1. We will do this every time we see the phrase “at least” mentioned in the problem. Example 4: Sam took a test in which he had to select responses from 5 choices to each question. Towards the end of the test, Sam had 4 questions remaining and he decided to randomly select answers because he did not have the time to set up the problems and work them out. I. What is the probability that Sam will

get at least one question correctly answered by taking recourse to random selection of choices for the remaining four questions?

The probability of getting at least 1 correct response is obtained by subtracting the probability of getting all answers incorrect. In a multiple choice test with 5 choices per question, the probability of a correct response is 1/5 and the probability of an incorrect response is 4/5 when the choices are made randomly. Therefore, the probability of all-wrong responses is (4/5)4

Probability of getting at least 1 correct response by randomly selecting answers for the remaining 4 questions is 1 – (4/5)4 = 1 – 256/625 = 369/625 ϕ 0.6 How did we determine that the probability of getting 4 answers in a row wrong as (4/5)4? We know that the probability of wrong answer for each question with 5 choices is 4/5. The probability of getting the first answer wrong AND the second one wrong AND the third one wrong AND the fourth one wrong is the product of individual probabilities.

In the above problem, what is the probability that Sam will get at least 2 correct answers by randomly picking choices for the 4 remaining questions? The probability of 2 or more correct responses is the same as 1 minus the probability of all wrong responses minus the probability of just 1 correct response. We have determined that the probability of getting all 4 wrong responses is (4/5)4. What is the probability of just 1 correct response? There are 4 different scenarios with just 1 correct response.

Q1 Q2 Q3 Q4

Response C W W W

Response W C W W Response W W C W Response W W W C

The probability for the scenario 1 as presented in row 2 in the table above is 1/5•4/5•4/5•4/5. [(43/54)] The probability for the scenario 2 as presented in row 3 in the table above is 4/5•1/5•4/5•4/5. [(43/54)]




The probability for the scenario 3 as presented in row 4 in the table above is 4/5•4/5•1/5•4/5. [(43/54)] The probability for the scenario 4 as presented in row 5 in the table above is 4/5•4/5•4/5•1/5. [(43/54)] The above four scenarios have each one correct response, and three incorrect responses. The probability of getting scenario 1 or scenario 2 or scenario 3 or scenario 4 is the sum of the probabilities for each scenario, and the probability for each scenario is the same. Therefore, the probability of getting just 1 correct response is: 4 • [(43/54)] = [(44/54)] = (4/5)4 Now we are ready to compute the probability for 2 or more correct responses. Probability for 2+ correct responses = 1 – (Prob. of all wrong responses) – (Prob. of just 1 correct response) = 1 - (4/5)4 - (4/5)4 = 1 – 256/625 - 256/625 = 1 – 512/625 = 113/625 ϕ 2/11 ϕ 0.18 As you can see, the probability of getting 2 or more responses by “crap shooting 4 questions” in the test is about 1/3

rd the probability of getting at least 1 correct response.

Coin Toss and Dice throw When we toss a coin, there are two possible outcomes: Heads or tails. The probability of getting a Heads or a Tails is the same and is equal to ½. What happens when we toss the same coin twice or toss two coins simultaneously?

There are four different outcome scenarios: HT, HH, TH, TT As a rule, if a coin is tossed N times or if N coins are tossed, the number of different outcome scenarios will be 2N. For example, if 4 coins are tossed simultaneously or if the same coin is tossed 4 times, then there are 24 or 16 different outcome scenarios. What happens when we throw a die, which is a cube-shaped object with 6 square faces and each face is marked with different consecutive integers from 1 through 6? The probability that we will get any of the values from 1 through 6 on the face showing up will be 1/6. The probability of getting a 4 is 1/6. The probability of getting a 5 is 1/6 and so on. Let us say that we throw the same die 300 times. Given that the probability of getting a 5 is 1/6, how many 5’s should we expect in this exercise? We should expect 300 times 1/6 or fifty 5’s to show up. The probability does not mean that we will get a 5 exactly 50 times. The concept of probability tells us that it is reasonable to expect that the value 5 will show up 50 times. The value may show up 70 times or only 20 times in an actual exercise. RULE: If the events can occur in different order, we need to consider the probability of obtaining the desired outcomes for each of the different orders. If we need to pick answers to two remaining problems(each with 5 choices, one being correct) on the GMAT or GRE, the probability of getting one correct and one wrong answer in a random selection is 2 times 1/5 times 4/5 because we can pick the correct answer to the first and the wrong answer to the second, or vice versa.




Example 1: “If a coin is tossed 4 times, what is the probability that at least 1 heads will turn up?” “At least 1 heads” means 1 or more heads. This means that the “undesired” outcome scenario is one in which it will be all tails. When we toss a coin 4 times, we will have 24 or 16 different outcome scenarios. Out of these 16 different outcome scenarios, there will be just 1 scenario in which it will be all heads or it will be all tails. Probability of “all tails” or “all heads” is 1/16. Therefore, the probability of getting at least 1 heads is = 1 – probability of all tails. = 1 – 1/16 = 15/16. TRIVIA: In a 5-coin toss situation (or, if the same coin is tossed 5 times), the number of different outcome scenarios is 25 or 32, and the probability of getting at least 1 heads is 1 – 1/32 or 31/32. In this exercise, the probability of “all heads” or “all tails” is 1/32. TRIVIA: IF a coin is tossed N times or if N coins are tossed, there will be exactly N outcome scenarios in which there will be just 1 heads and there will be exactly N outcome scenarios in which there will be just 1 tails. Probability of getting just 1 heads in the above exercise is N/2N. For example, if a coin is tossed 4 times, the probability of getting just 1 heads or the probability of just 1 tails is 4/24 = 4/16 or ¼. Example 2: If a coin is tossed 5 times, what is the probability that at least 2 heads will show up? We are looking at the probability of getting 2 or more heads. Probability of getting 2 or more heads = 1 – probability of 0 heads(all tails)

- probability of just 1 heads

= 1 – 1/32 – 5/32 = 26/32 = 13/16 You will see that 13/16 is also the probability for getting at least 2 tails. Example 3: If two dice are thrown at the same time, what is the probability that the sum of the values on the faces turning up will be a 7 or a 9? As a rule, if a die is thrown N times, or if N dice are thrown at the same time, the number of outcome scenarios will be 6N, where 6 is the number of outcome scenarios per die throw. In this exercise, the number of different outcome scenarios will be 62 or 36. The desired outcome scenarios are the following:

Die 1 Die 2 Sum Face value 1 6 7 2 5 7 3 4 7 4 3 7 5 2 7 6 1 7 3 6 9 4 5 9 5 4 9 6 3 9

We notice that there are 10 different outcome scenarios out of a total of 36 that will produce the desired sum of 7 or 9. Therefore, the probability of getting a 7 or a 9 in the above exercise is 10/36 or 5/18. We will see later on in this module that probability question can be asked in any setting. Just remember that probability is the ratio of the number of outcomes favorable to a given event to the total number of all outcomes.




Example 4: A club has M men and W women. 3 more men join the club and 7 women resign their membership. If one person is to be randomly selected for an assignment, what is the probability that a woman will be selected from the members in the club? Let us set up the information in the form of a table so that we can get a clear perspective:

Before the change

After the Change

Men M M + 3

Women W W - 7

Total M + W M + W - 4

Probability that a woman will be selected

No of women/Total membership = (W – 7) / (M + W - 4)

The probability that a man will be selected is given by the expression: (M + 3) / (M + W – 4)

Problems in probability can be asked in any setting, and occasionally you will get a probability question that is more like a decreasing balance depreciation problem. Take a look at the next example on the following page.

Example 5: “In a certain zoo, the probability that a new-born rhino will die is 10% for each month during a six month period. If the zoo had a new-born rhino population of 300 on June 1, 90, approximately how many rhinos can be expected to be alive at the end of month 3?” In month 1, 10% of 300, or 30 are expected to die. Number expected to live at the end of month 1 is 300 – 30 = 270. In month 2, 10% of 270 can be expected to die. Number expected to live at the end of month 2 is 270 – 27 = 243. In month 3, 10% of 243, or 24 can be expected to die, and approximately 243 – 24 = 219 can be expected to survive. And that is the answer.




Probability questions in Geometry setting:

Example 1: “A circular-shaped garden is surrounded by a circular path of uniform width 3 feet. If the radius of the circular garden is 10 feet, and if a point is to be selected randomly from within the circular-shaped region including the path, what is the probability that the point chosen will actually lie in the path?” (Hint: The area of a circular region is π•(radius)2). In order to answer this question, we need to find the area of the circular path and take that area as the ratio to the total area of the circular region including the path. The area shown in blue is the garden and the area around it and shown in red is the path. The circular region includes the width of the path. The radius of the circular region is equal to the radius of the garden plus the radius of the path.

The radius of the circular region = 10 + 3 = 13 feet Area of the circular region = π•132 = 1696 The radius of the garden = 10 feet. Area of the garden = π•102 = 100π Area of the path = 169π - 100π = 69π

The probability of selecting a point that will lie in the path from within the circular region including the path is: 69π / 169π = 69/169 And that is the answer. Notice that we cannot simplify the above fraction further because 69 and 169 do not have any common factors. RULE: ALL FRACTIONS MUST BE REDUCED TO A FORM IN WHICH THERE ARE NO MORE COMMON FACTORS FOR THE VALUES IN THE NUMERATOR AND IN THE DENOMINATOR. PROBABILITY REFRESHER: PROBABILITY IS ANOTHER WAY OF DEFINING THE PROPORTION. If, in a population of N values, there are R items of the desired kind, then the probability of picking R in a random selection is R/N. This is also the proportion of R in the population. If the probability or proportion of event A is P(A) and that of event B is P(B), then the probability that A and B will both occur is the product of two probabilities. If A and B can occur in different order, then we need to factor in the different order in which events A and B can occur when computing the probability. EXAMPLE: There are 10 boys and 12 girls in a group. In a random selection of two children, what is the probability that one boy and one girl were picked? Notice that the first selection could be a boy and the next a girl, or the first a girl and the next a boy. Unless the problem stipulates the order of selection, we need to consider the probability that the events may occur in any of the two different possible orders. The probability that a boy is picked first is 10/22 or 5/11. Given that a boy was picked first, the probability that a girl is picked next is 5/11 times 12/21 or 5/11 times 4/7 = 20/77. Because the selection could also be done in a different order – girl first followed by boy- we need to multiply this probability by 2 to get the required probability: 2 times 20/77 or 40/77 We cannot reduce this fraction any further, and will leave it as is.




Example 2: 2X X X X X 2X Three squares are drawn as shown above. The sides of the smallest square shown in green are X and X, and the sides of the middle square (in red) are 2X and 2X, and those of the outer square are 4X and 4X. If we want to select a point randomly from within the larger square shaped region, what is the probability that the point will actually lie in the L-shaped region shown in red? The area of the middle square is (2X)2 = 4X2. The area of the smallest inner square that lies within the middle square = X2 Area of the L-shaped region

Shown in red = 4X2 – X2 = 3X2 We need to find the area of the larger square shaped region so that we can compute the probability. The larger square has sides 4X and 4X, and

the area of the larger square is (4X)2 = 16X2. Therefore, the probability that a point randomly picked from within the larger square shaped region will lie in the L-shaped region shown in red is 3/16. (The ratio of 3X2 to 16X2). EXAMPLE:

“If X is a multiple of 5, and also a multiple of 6, what is the probability that X will be a multiple of 20”?

We have seen in our discussions on multiples that if X is a multiple of 5 and a multiple of 6, then X must be a multiple of 30, the least common multiple of 5 and 6. NOTE: “MULTIPLE OF” MEANS ‘ANY INTEGER TIMES’. “X is a multiple of 5” means that X is ‘any integer times 5’. This means that X must go in steps of 30. The likely values are: X : 30, 60, 90, 120, 150, 180, and so on. If X is equal to 30, it is not divisible by 20. If X is equal to 60, then it is. We can see that for every two values, X is divisible by 20 for just one of them. Therefore, the probability that X will be a multiple of 20, given that it is a multiple of 5 and a multiple of 6, is ½ .

EXAMPLE: DATA SUFFICIENCY “If a box contains some Red Balls, Blue Balls, and Green balls, and if one ball is drawn randomly from the box, what is the probability that a Red ball was NOT drawn?”

1. The box contains 25 red balls. 2. There are 65 balls in all in the box.

Our N.T.K for this problem is that we need to know how many balls in all there are, and how many of the non-red kind there are so that we can computer the probability. Statement 1 gives us nothing of what we need to know. We must kill choices A and D and move on to examine statement 2. We have just 3 choices remaining: B, C, and E. Statement2 also gives us exactly one-half of what we need to know. WE still do not know




how many of the non-RED balls we have in the box. We must kill choice B now and proceed to combine the two statements. When we combine the two statements, we notice that we know that there are 65 balls in all, and there are 40 of non-Red kind, and we can estimate the probability of drawing a non-Red ball. We must pick Choice C.




Permutations and Combinations Permutation is about selecting or arranging things in an ordered or role-defined way. Each item that is ordered or selected is given a UNIQUE Identity. If we have to seat 5 people in 5 row seats, then in what seat any person, say person 4, is seated along with who is to his right and who is to his left will determine the unique identify for this person. That is why 12345 is a different seating arrangement from 12543. Notice that in the first seating arrangement, person 4 is in position 4 but had 3 to his right and 5 to his left. In the latter, he had 5 to his right and 3 to his left, even though he still occupies position 4. Another example of a permutation problem is when we need to make teams and give ‘titles’ or ‘roles’ to each member of a team. If the problem requires that we make a team of 3 using 5 available people, and that we give titles to the chosen three (one person is the captain, the other vice-captain, and the third navigator), then we are dealing with a PERMUTATION PROBLEM. On the other hand, if the problem did not require that we assign titles to the chosen three, then it becomes a COMBINATION PROBLEM. Combination is about selection at a time and in a non-ordered and non-role-defined fashion. Each item chosen does not receive any unique I.D. If we are required to make a team of 3 using 5 available people, and if the problem does not specify any roles or titles for the chosen, then we are looking at a COMBINATION PROBLEM. Think of COMBINATION as PERMUTATION adjusted for the illogical replication. For example, if we need to make a team of 3 using A, B, and C, then permutation problem would treat ABC, ACB, BAC, BCA, CAB, CBA as 6 different teams. Because A, B, and C cannot be in 6 different teams simultaneously (unless each has multiple personalities), we must conclude that all 6 permutations are mere variant expressions of the same team ABC. NUMBER OF COMBINATIONS = (NUMBER OF PERMUTATIONS)/(NUMBER CHOSEN)!

The simplest of permutation problems involves seating N number of people in N different seats, say, for a photo-shoot.

COUNTING METHODS THEOREM

“If We can do one thing in X ways, and another thing in Y ways, we can do both things in X times Y different ways." Let us say that we have 4 people to seat in 4 different seats. For the first seat, we have 4 choices. Having seated one person, we can choose from the remaining 3 persons any one to be seated in the second seat. Having seated 2 people, we can choose from the remaining 2 people for the third seat. Having seated 3 people, we have just 1 person remaining for the last seat. Therefore, the number of different ways we can seat 4 people in 4 different seats is the product of 4, 3, 2, and 1 or 4 factorial ways. FORMULAS

1. If we need to arrange N items in R positions or row-seats, then the number of different PERMUTATIONS NPR = N!/(N – R)!

2. If we need to arrange 7 people in 3 different row seats, then we can determine the number of seating arrangements using the above formula or by saying that the first seat can be filled in 7 ways, the second in 6 ways, and the third in 5 ways. According to the Counting methods rule, the number of different seating arrangements is 7 times 6 times 5 = 210.

IF we need to make teams of R people using N available people, then the number of different COMBINATIONS, NCR = N!/[R! (N – R)!].




Example 1: “Chairman Chris and four directors are to be seated for a meeting. Chris must be at the head of the table. How many different seating arrangements can be made subject to this requirement?” We notice that for all practical purposes, the problem boils down to one of seating 4 different people in 4 different seats because Chris is immovable. We can do the seating arrangements in 4! Or 24 different ways.

As a rule, If there are N people to be seated in N chairs, we can do it in N! different ways. Example 2: “How many different ways can we seat Anna, Bob, and Chris in a row subject to the restriction that Bob and Chris cannot be seated next to each other in any order?” We notice that since Bob and Chris cannot be seated next to each other, we must put Anna smack in the middle. This configuration makes Anna “immovable”. The problem boils down to seating 2 people in 2 different seats. We can do it in 2!, or 2 different ways. BOB ANNA CHRIS CHRIS ANNA BOB Example 3: “How many different ways can we seat 4 different people A, B, C, and D in 4 different row seats subject to the restriction that A and B cannot be together in any order?” Without any restriction, we can seat A, B, C, and D in 4!, or 24 different ways. Let us see how many seating arrangements we lose because of the restriction imposed. A and B can be seated in seat 1 and seat 2 in two different ways next to each other, and for each of these arrangements, D and C can be seated in seat 3 and seat 4 in two different ways.

We will, therefore, lose 4 seating arrangements in which A and B will be together in seat 1 and seat 2 in different order. Similarly, A and B can be seated in seat 2 and seat 3 in two different ways next to each other, and for each of these arrangements, D and C can be seated in seat 1 and seat 4 in two different ways. Once again, we will lose 4 seating arrangements in which A and B will be next to each other in seats 2 and 3 in different order. Lastly, A and B can be seated in seat 3 and seat 4 in two different ways next to each other, and for each of these arrangements, C and D can be seated in seats 1 and 2 in two different ways. We will lose 4 arrangements in which A and B will be seated in seated in seats 3 and 4 next to each other in different order. Altogether, we will lose 12 seating arrangements in which A and B will be next to each other at various seating configurations. If we subtract the “verboten” arrangements from the total of 24, we get left with 24 – 12 = 12 permissible seating arrangements in which A and B will not be next to each other in any order. In the GMAT, we have not so far come across a problem that imposes restriction on 2 people when 4 people are to be seated. But if you come across a problem of this type, you know what to do. If the problem specifies seating arrangement for 5 people and specifies restriction that any two people out of the 5 cannot be seated next to each other, just remember that we can have 72 permissible seating arrangements. How did we get this? As a rule, If there are N people to be seated in a row, and a restriction is imposed about seating specified TWO entities next to each other in any order, then the number of permissible seating arrangements subject to this restriction is obtained by using the formula:

N! – 2•(N-1)! Let us say that we have 6 people to be seated in a row, and there is a restriction about seating




two specific people next to each other. The number of permissible seating arrangements subject to this restriction is: 6! – 2•(6-1)! = 720 – 2•5! = 720 – 240 = 480 WE can seat 6! In 480 different ways subject to the restriction that a specified entity cannot be seated next to another specified entity. If there is no restriction imposed, then we can seat 6 people in 6!, or 720 different ways.

SEATING OF PEOPLE AROUND A TABLE

Instead of “row” seating, let us say that we need to seat N people around a table that seats exactly N people, the number of seating arrangements we can make is (N-1)!. This is because we need to fix one person in one seat and keep that person as a reference for seating the remaining persons or objects. However, if the problem specifies a restriction about seating any two people in any order around the table, then the number of seating arrangements subject to this restriction is:

(N-1)! - 2•(N-2)! Example: “If 6 people are to be seated around a table that seats exactly six people, and if any two specified people cannot be seated next to each other in any order, how many seating arrangements can be made?” Number of seating arrangements consistent with the restriction imposed = 5! - 2•4! = 120 – 48 = 72 ways Let us consider another problem that imposes a different type of restriction. We will have to use simple reasoning and get the answer. Let us see how we can do it.

EXAMPLE: “If 3 boys and 3 girls are to be seated in a row in six seats such that no two people of the same gender can be next to each other in any order, how many seating arrangements can be made?” The restriction is that boys cannot be seated together and girls cannot be right next to each other. This means that we have to put the girls and boys in alternate seats Boys could be in 1 – 3 – 5 or in 2 – 4 – 6. Likewise, girls could be in 2 – 4 – 6, or in 1 – 3 – 5. Boys could be seated in 1-3-5 configuration in 3 factorial or 6 ways, and the girls can be in 2-4-6 configuration in 3 factorial or 6 ways. We can seat the boys in 1-3-5 and girls in 2-4-6 in 6 times 6 or 36 different ways. (Remember the counting methods theorem we stated at the beginning of this chapter?) But then the boys could be in configuration 2-4-6 and the girls could be in 1-3-5 configuration. These two configurations give us another 6 times 6 = 36 different seating arrangements. Therefore, total number of seating arrangements in two different seating configurations for boys and girls is 36 + 36 = 72 different ways.




Permutation of digits of a

number Let us say that we have four digits 1, 2, 3, and 4. If we were asked how many different four digit numbers can be formed by using these digits? We need to know whether we are allowed to repeat the digits or not. For example, we can go 1111, 2222, 3333, 4444, 1112,1122, and so on. If repetition is allowed, then for the first digit of the number, we have four different choices among the digits 1 through 4. For the second digit, we have another four choices. For the third digit, we have yet another four choices. And for the last digit, we have still another four choices. Therefore, the number of ways of forming four digit numbers by being able to repeat digits is 4•4•4•4 = 256 or 256 different four digit integers. What happens if we are not allowed to repeat digits? For the first digit, we have four choices. For the second digit, we have three choices. For the third digit, we have two choices. For the fourth digit, we have just one choice. According to the Counting methods theorem, we can form four digit integers without repeating digits in 4•3•2•1 or 4 factorial = 24 ways. We can put together 24 four-digit numbers by using four digits without repetition.

REMEMBER: 1. The sum of all 24 four-digit integers

formed by permutation of the digits 1, 2, 3, and 4 without repetition is 66,660.

2. The sum of all 6 three-digit integers formed by permutation of the digits 1,2, and 3 without repetition is 1,332.

EXAMPLE: “What is the sum of all the three digit integers that can be formed by permutation of the digits 2, 3, and 5 without repetition?” We know that there are 3 factorial or 6 different 3 digit integers because repetition of digits is not allowed. In each decimal place, each of the three digits is going to occur twice exactly. In our case, the numbers are easy to set up and add: 235 253 325 352 523 532 We can see that each digit occurs in each decimal place exactly twice. The sum of the six three-digit numbers formed by permutation of the digits 2, 3, and 5 without repetition is 2,220.




PERMUTATION INVOLVING ORDERED

SELECTION If we are asked to select R items out of N items in an ordered or role-defined way, we can use the Permutation formula:

NPR = N! / (N-R)! Example 1: “How many different 3-member teams can be formed from 4 men and 5 women so that one person in the team will be the captain, the other chef, and the third janitor?” Do we see a role-definition here? Yes. We must conclude that it is a permutation problem. The problem is one of selecting 3 people out of 4+5 = 9 people in a role-defined manner. We use the permutation formula: 9P3 = 9! / (9-3)! = 9.8.7.6! / 6! = 9.8.7 = 504 ways. We can select 3 member-teams in a role-defined manner out of 9 people in 504 different ways. Example 2:

Let us say that the problem specified the following: “How many different 3-member teams can be formed out of 4 men and 5 women so that a woman will be the captain, and the second person selected will be the chef and the third janitor?” WE can choose any one of the 5 women to be the captain, and for each selection of a woman as captain, we have to select 2 others in a role-defined manner from the remaining 8 people. The number of different ways we can select the 3-member team subject to the specification is:

5• 8P2 = 5• 8!/(8-2)! = 5•8•7 = 280 ways Permutation problems also give us an insight into women’s psychology. The next time your female significant-other wants to go shopping for yet another new pair of shoes, show her the following permutation problem.




Example 3: “Lisa’s wardrobe consists of the following items: 10 dresses; 12 handbags; 15 pairs of shoes; IF Lisa wears one dress, one pair of shoes and carries one handbag , how many different dress-shoes-handbag combos can Lisa make using the items in her wardrobe?” This type of problem is best dealt with as a tree.

B1 S1 B2 S2 • • D1 • • • • B12 S15 D2 • • • D10 As we can see, Lisa can choose to wear any one of the 10 dresses on a given day. For each dress selection, she can choose to carry any one of the 12 handbags, and for each dress-handbag selection, she can choose any one of the 15 pairs of shoes. This means that for each dress Lisa chooses to wear, she can make 12 times 15 or 180 combinations of shoes and handbags. Given that Lisa has a choice of 10 dresses, she can make 10 times 180 or 1800 different dress-shoes-handbag combinations. Enough to make Lisa not wear the same combination two days in a row for close to 5 years. (Use this information to suggest to your lady that buying another pair of shoes may not be such a good idea. Just kidding.) You will notice that the number of different assortments is simply the product of the numbers of each of the items in the combination. (That is what the counting methods theorem is all about) A similar problem can be faced by McDonald’s manager. Let us say that McDonald’s has 5 different burgers, 4 different fries and 3 different choices of drinks, and let us also say that a combo meal consists of one burger, one set of fries, and one drink. The number of different

combo meals that McDonald’s can offer is the product of 5, 4 and 3, or 60. Let us say that we place a restriction on any two items in the available list. Let us say that drink 3 and fries 2 cannot be offered together in any one combo meal. How many combos do we lose because of this restriction? For each burger, we will lose one combination of Fries2-Drink3. For 5 burgers, we will lose 5 combos in all. We still have 60-5 = 55 permissible combo meal packages.

The same logic applies when we put a restriction on what Lisa can wear. Let us say that Lisa does not like to wear D1 and Shoes 12 together. (Colors don’t match, may be?). How many combinations does she lose in this process? She will lose one combination of D1 and S12 for each handbag she can carry. There are 12 handbags and she will lose 12 combinations in all. She can still make 1800 – 12 = 1788 combinations, after taking into account her dislike for D1-S12 coordination.




The “combo” problem can be presented pictorially as an inverted triangular maze. Take a look at the following problem: Example:

A B C D E F G H K

Let us say that we have a really famished monkey at node A of the inverted-triangular maze shown, and that it wants to get to the banana at node K. If the monkey can move along any of the paths connected by straight line segments, paths such as ABFK, ABCGK, ABCGFK, ABFGK, ABFGHK, and so on, how many different paths can the monkey take to get to the banana from node A ? (The monkey must take the shortest length possible for any path.) If (s)he is a smart monkey, (s)he can go down the path AFK and get to the banana in no time. If (s)he is not so smart, (s)he can take any of the other alternative paths: AFK --- Shortest route AFGK AFGHK ABFK ABFGK ABFGHK ABCGK ABCGFK ABCGHK ABCDHK ABCDHGK ABCDHGFK ABCDEHK ABCDEHGK ABCDEHGFK As you Can see, the monkey can move along 15 paths to get to node K from node A. If you pay attention to the bold letters in each path, you will notice that from node A, there are three paths: AFK, AFGK, AFGHK Similarly, from node B, there are 3 paths: ABFK, ABFGK, ABFGHK

You will notice that other nodes C, D, and E have similarly 3 paths leading to node K along the three middle nodes F, G, and H. We conclude that there are 5 X 3 = 15 paths that will take the monkey from A to K. Is there a simpler way to figure this out? A B C D E F G H K Simply multiply the number of nodes on top by the number of nodal points before the destination node K. We have 5 nodes A, B, C, D, and E on top We have 3 nodes F, G, and H at the next level before the destination node K. So, how may different paths do we have from A to K ? 5 X 3 = 15. Let us create some additional intermediate level nodes before the “banana” node and see what happens . A B L C M D E

F S G T H N P R K Now we have 7 nodal points on the top line: A, B, L, C, M, D, & E We have 5 nodal points next to the top level F, S, G, T, & H We have 3 nodal points before the destination. N, P & R So. How many different paths can the mad monkey take from node A to the banana node K? Simply multiply the number of nodes in each level: 7 X 5 X 3 = 105 paths. What if we turned this problem around and made it a “Probability” question?




Suppose that I asked you: “What is the probability that the monkey will move along the path AFNK? We recognize that AFNK is just one path out of 105 possible paths. Therefore the probability that the monkey will move along path AFNK is simply 1 out of 105 or 1/105. Let us take a look at another type of maze, although it is not related to permutation concept.




MAZE PROBLEMS Imagine you are looking at a street map of Manhattan with North-South Avenues and East-West Numbered streets. B A If you are required to walk from intersection A to intersection B along a route that is confined to the square grid of four avenues and three streets shown in the map above, how many routes from A to B can you take that have the minimum possible length ? The best way to understand this problem is to number the intersections and determine how many different combinations of streets and avenues exist from A to B that have the same minimum length . 9 10 B 6 7 8 3 4 5 A 1 2 You can take any of the following routes to go from A to B: A-1-2-5-8-B A-1-4-5-8-B A-1-4-7-8-B A-1-4-7-8-10-B A-3-4-5-8-B A3-4-7-8-B A - 3 - 4- 7-10-B A-3-6-7-8-B A-3-6-7-10-B A-3-6-9-10-B You can take any of these 10 routes and travel the same distance.

Is there a simpler way to figure this out? When you have a grid with three vertical lines, start writing consecutive integers from 1 from the top horizontal line, and simply add up the numbers from top to bottom as shown below: 1 B 2 3 4 A 1+2+3+4= 10 If you had one more horizontal line, you will write 5 and add 1 through 5 to get 15. If you had three vertical lines and 6 horizontal lines, you will write 1 through 6 and add them up to get 21 routes. If you had seven horizontal lines and three vertical lines, you will add 1 through 7 to get 28 as the value for the number of possible routes with the same length . What if you had 4 vertical lines and a number of horizontal lines ? Let us say you had 4 vertical lines and 4 horizontal lines, you will add the 4 consecutive ODD integers from 1:

1+3+5+7 = 16 What if you had 4 vertical lines and 5 horizontal lines ? You add up 5 consecutive ODD numbers from 1 through 9:

1+3+5+7+9 = 25 What if you had 4 vertical lines and 6 horizontal lines ? You will add the 6 consecutive ODD integers from 1: 1+3+5+7+9+11 = 36 You get the drift, don’t you?

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Combination problems Combination is selection at a time and order of selection is immaterial. We said earlier in this section that you will make a distinction between a permutation problem and a combination problem simply by paying attention to whether any role-definition or ordering is specified in the problem or not. Example of a permutation problem: “How many different 3 member teams can be selected from 5 people so that one person chosen will be the president, the next secretary, and the third treasurer?” You can clearly see that the chosen people will have roles to play. If the role definition is missing or ordering is missing you must treat the problem as a combination problem. Example of a combination problem: “How many different r member teams can be selected out of n persons?” As a rule, r people can be selected at a time out of n people in nCr = n! r!(n-r)! Example1: “How many different 3 member teams can be selected out of 5 people?” The problem is one of finding a number value for the expression 5C3. 5C3 = 5! 3!•(5-3)! = 5.4.3! = 10 3!•2! We can make 10 different 3-member teams out of 5 people. In the GMAT, you should expect combination problems specified with a restriction or an

exclusion. Take a look at the following examples. Example 2: “ How many different 2-element subsets can be formed out of the set {1,2,3,4,5,6,7} subject to the restriction that (2,5) is not a valid subset?” We have to find out how many different 2-element subsets we can form out of 7 numbers and exclude 1 for the invalid subset. The problem is one of finding a number value for the expression: 7C2 - 1. The number value for 7C2 is: 7! = 7•6•5! = 21 2!•5! 2•1•5! Therefore, the number of valid 2-element subsets is 21 – 1 = 20. And that is the answer. When you see the phrase “at least” specified in a combination problem, we must deal with this aspect of the problem by eliminating the undesired combinations from the total of all combinations. Let us see how this works in the following example. Example 3: “How many 3-member rowing teams can be selected from the pool of 4 men and 5 women so that each team will have at least 1 woman in it?” We notice that the restriction is with respect to “all-men” teams. Let us find out how many 3-member teams can be formed out of 4+5 = 9 people without any restriction. No restriction number of teams = 9C3. The number value for 9C3 is 9!/3!•6! = (9•8•7)/3•2 = 84 teams. Now, let us deal with the restriction: “all-men” teams are forbidden. The total of 84 teams includes 4C3 or 4 all-men teams. If we subtract 4 from 84, we get 80 as a number for 3-member teams with at least one woman in them.

Combination involves SELECTION AT A TIME. In a permutation situation, if we are to seat 3 people, we will seat them in 3 different seats. In the combination situation, all three will occupy the same seat or the same box. PERMUTATION: COMBINATION In the permutation situation, the order of selection is important. For example, we can choose to seat A first, B next, and C last, or B first , A next, C last and so on giving rise to 6 different permutations. In the Combination story, it is not material who was selected first as long as A, B, and C are together in a box. How many games must be scheduled if 5 teams play in a tournament and each game is played between two teams? This is a combination problem because there are 2 teams on the field at a given time. It does not matter whether Yankees were chosen to Diamondbacks or the other way around, as long as Yankees play the D-backs.

A B

C

A B C

Think of Combination as a problem that corrects that duplication involved in permutation. Number of combinations = Number of permutations divided by the factorial value of the number of people chosen at a time. Let us say that the problem is one of choosing a group of 3 from 7 people, and the question is how many different ‘teams of 3’ can be made? If we started out as a permutation problem, the first person can be chosen in 7 ways, the second in 6, and the third in 5 ways for a total of 7•6•5 = 210 teams of 3. But this number involves duplication because it considers ABC to be a different team from ACB or from BAC. In order to make the correction required to annul the duplication, we need to divide 210 by 3! To get the number of Non-duplicated teams of 3 as 210/6 = 35. Remember: Combination is non-duplicated permutation or permutation adjusted for the multiplicity or for the ordered arrangements.




Example 4: “If a 3-member flight crew is to be chosen from a pool of 6 experienced and 5 amateur astronauts, what is the probability that an all-amateur team will be selected if each team that can be formed is as likely to make it to the mission as any other?” We have to find out the total number of 3-member teams that can be formed from 6+5 = 11 astronauts, and also find out the number of all-amateur teams included in the total, and take the ratio. Number of 3-member teams = 11C3 = 165 Number of all-amateur 3-member teams

= 5C3 = 10 Probability of an all-amateur team being chosen = 10/165 = 2/33. (Answer) Example 5: “In the above problem, how many different 3-member crews can be selected subject to the restriction that each team have at least one experienced astronaut in it?” The problem tells us that we have to exclude “all amateur” teams from the total number of 3-member teams that can be formed. We have to get a number value for 11C3 – 5C3 = 165 – 10 = 155. 155 teams will have at least one experienced astronaut in them. Example 6: “How many different ways can we choose 4 distinct integers from (1,2,3,4,5,6, and 7) so that their sum will be even?” Notice that there are 4 odd integers and 3 even integers in the set. You can get an even number as the value for the sum of integers by combining • all 4 odd integers; or by combining • 2 odd and 2 even integers; We cannot combine 3 odd and 1 even or 3 even and 1 odd integers, because the sum will not be even.

We can combine 4 odd integers out of 4 odd integers in just one way: We pick all of them. We can choose 2 odd integers out of 4 odd integers in 4C2 or 6 different ways. And, for each of these 2-value selections of odd integers, we can choose 2 more even integers in 3C2 or 3 different ways, for a total of 6•3 = 18 ways. Therefore, the total number of ways we can combine 4 integers so that the sum will be even is 1 + 18 = 19 ways.

See if you can compute the number of different ways we can combine 4 integers from {1,2,3,4,5,6,7} so that the sum will be an odd value as 16 ways. Example 7. “There are six people in a room and each person shakes hands with each of the others exactly once. How many handshakes are exchanged?” The above problem is the same as asking you how many different ways can we put together a team of 2 people out of 6 people because when 2 people shake hands, we are dealing with 2 people at a time out of the six people in the group. We also notice that the order of handshakes is not important or material here. For example, A shakes hands with B or the other way around will still count as one handshake. We must treat this as a combination problem. We are choosing 2 at a time out of 6 people. No of handshakes is 6C2 = 6! /2!•4! = 15.




If there are 10 people in a room and each one shakes hands with each of the others exactly once, there will be 10C2 or 45 handshakes. You get the idea here, don’t you? The same problem can be presented as scheduling of games problem. Take a look at the one below: “10 teams are to play in a tournament, and each team must play in a game each of the other teams exactly once. How many games need to be scheduled?” We notice that this is a problem of choosing 2 teams at a time out of 10 teams. Number of games to be scheduled

= 10C2 = 10!/2!•8! = 10•9/2 = 45 games. IF the problem specified that each team must play each of the other teams exactly twice, then we double 45 and get 90 as the number of games that we need to schedule. If the teams played each other three times, we will multiply 45 by 3 to get 135 games, and so on. Take a look at another COMBINATION problem. LEAGUE RESULTS

TEAM NUMBER OF GAMES WON

A 4 B 7 C 9 D 2

E 2

X ? According to the incomplete table above, if each of the 6 teams in the league played each of the other teams exactly twice and there were no ties, how many games did team X win? (Only 2 teams play in a game) (A) 4 (B) 5 (C) 6 (D) 8 (E) 10 6 teams to choose from and we need to choose two teams at a time to play in a game. And there is a “home” game and an “away” game. Total number of games played: 2• 6C2 = 2• 6!/2!4! = 2•(6.5)/2 = 30 games There are a total of 30 games played. This means 30 wins and 30 losses, because in each game one team wins and the other loses. The total number of games won by teams A through E is 24. Team X must have won the remaining six games. We must select choice C. COMBINATION is PERMUTATION adjusted for the illogical repetition of the same combination in different order. We have seen how to use a formula. WE can also use an approach that treats the NUMBER OF COMBINATIONS as equal to (NUMBER OF PERMUTATIONS) DIVIDIDED BY THE (NUMBER CHOSEN)!. If the problem requires that we compute the number of different teams of 3 that can be made using 6 people, then the number of permutations of 6 in 3 positions is 6 X 5 X 4. Because permutation involves repetition of the same group in different order, we need to divide this by 3! To adjust for this illogical repetition. Therefore, the number of teams of 3 using 6 people is (6X5X4)/(3X2X1) = 20 teams.




MORE PROBLEMS IN COMBINATION AND PERMUTATION PROBLEM 1: “A Pizza Shop allows its customers to choose a pizza with the same number of toppings as the number of cheese types chosen. If the shop has 4 different types of toppings and 2 different types of cheese, how many different pizza combinations can be made subject to the above restriction?” The problem tells us that a customer can choose a pizza with ONE topping and ONE type of cheese, OR one with TWO toppings and TWO types of cheese. OPTION 1 CHOOSE A PIZZA WITH ONE TOPPING AND ONE TYPE OF CHEESE. A customer can choose 1 topping out of 4 available toppings in 4 ways; 1 cheese out of 2 in two different ways. A customer can choose 1 topping AND 1 cheese type in 4 times 2 = 8 different ways. This means that a customer can order 8 different pizzas each containing a different combination of one topping and one cheese. OPTION 2 CHOOSE A PIZZA WITH TWO TOPPINGS AND TWO CHEESES A customer can choose 2 toppings out of 4 toppings in 4•3/2•1 = 6 ways . Notice that this is a combination because we made a ‘team’ of two toppings on the same pizza. Likewise, the customer can choose 2 cheeses out of the available 2 types in 1 way. A customer can choose 2 toppings and 2 cheeses in 6 times 1 ways or 6 ways. A customer can order 8 different combinations of 1 topping and 1 cheese; and additionally order 6 different combinations of 2 cheese and 2 toppings for a total of 8+6=14 different pizza combinations.

PROBLEM 2: PROBLEM 2: “There are 10 qualified tennis players in a group. If the coach wants to schedule ‘doubles’ games in which a team of 2 players plays another of 2 players, how many different games can be scheduled?” . You should conceptualize the problem before you begin to attack it. We need to do the following: 1. We need to decide how many different teams of

2 people can be made using 10 available people. 2. Then, decide how many different ‘games’

involving two teams of two players each must be scheduled. We must then adjust for the ‘double count’ of the same game by dividing the total by 2.

Let us take care of the first step first. We can make teams of 2 people using the available pool of 10 people in 10•9/2•1 = 45 different ways. Now we need to proceed to the second step. We can choose any ONE of the 45 different teams of 2 players to be one of the two teams. Because the people chosen to be on one team cannot be in an opposing team, we must decide how many different teams of 2 can be made using the remaining 8 people. We can make teams of 2 using the remaining 8 people in 8•7/2•1 = 28 different ways. Therefore, we can pick ONE team from the original pool of 45 teams in 45 different ways. And Any one of these chosen teams can play any of the 28 teams of 2 ‘other’ players in 45 times 28 = 1,260 different ways. But then, we must recognize that this number of 1,260 involves double counts of a game. For example, AB playing CD is counted as a different game from CD playing AB. Therefore, we must divide the value of 1,260 by 2 to get 630 as the number of different games that could be scheduled We could have also obtained the same result by doing the following: Because a game involves 4 players, 2 on each side, we can compute the number of different selections of 4 players from 10 as 10C4 = (10!)/[4! 6!]= 210. Within a group of 4, we can create 3 different games: For example, if the group comprises [A, B, C, D], we can create three different games as follows: AB PLAYS CD; AC PLAYS BD; AD plays BC. Therefore, the total number of different ‘doubles’ games that can be scheduled by using 10 players is 3 times 210 = 630.




PROBLEMS IN CONDITIONAL PROBABILITY USING

PERMUTATION OR COMBINATION APPROACH.

Conditional probability is unaffected whether the problem is treated as one involving ORDER or as one not involving order. If the problem specifies that we need to pick two machines out of a group of 6 machines in which 2 are defective, and requires us to compute the probability that we picked ONE good machine and ONE bad machine, then the computed probability will be unaffected whether consider the selection of two machines as SIMULTANEOUS or as SEQUENTIAL involving order. LET US EXPLAIN HOW THIS IS SO. APPROACH 1: WE TREAT THE PROBLEM AS ONE IN WHICH TWO MACHINES ARE PICKED AT A SINGLE TIME. (COMBINATION APPROACH). Number of selections of 2 machines out of 6 is 6! / 2!(6-2)! = 6!/2! 4! = (6 X 5 X 4!)/2! 4! = 15 ‘teams’ of 2 Number of selections of 2 machines in which one is good and the other bad is 4C1 X 2C1 = 4 X 2 = 8 teams of two machines in which one is good and the other bad. Therefore, the probability that a ‘team’ of 2 machines in which one machine is good and the other bad is selected out of 15 different ‘teams’ of 2 machines is 8/15. APPROACH 2: WE TREAT THE PROBLEM AS ONE IN WHICH MACHINES ARE PULLED IN DIFFERENT ORDER – BAD MACHINE FOLLOWED BY GOOD MACHINE or GOOD MACHINE FOLLOWED BY BAD MACHINE. We need to compute the probability of BAD followed by a good machine, or Good one followed by a bad machine selection. The probability that we will pick a bad machine first and then a good machine is 2/6 times 4/5 = 4/15. The probability that we will pick a good machine first and then a bad machine is 4/6 times 2/5 = 4/5. Because we can do the selection of two machines in two different order, we need to add the probability of the two events and get 4/15 + 4/15 = 8/15.

As you can see, either approach gives us the same result. Let us take a look at a few more conditional probability problems in which either approach is valid and will produce the same result. PROBLEM 2: A roller-coaster has 3 cars – A, B, and C. If Alisha rides the roller-coaster on three different occasions and if she does not have specific preference for any of the three cars, what is the probability that she will ride in each of the three cars during her three different rides? APPROACH 1: On her first visit, Alisha must ride any one of the three cars. Therefore, it is a certainty that she will use one of the cars on her first visit. The probability that she will use one of the three cars on her first visit is, therefore, 1. On her second trip back to the park, she must not ride in the car that she rode the first time. The probability that she will ride in one of the two remaining cars is 2 out of 3 or 2/3. On her last trip, she must ride in the one car she has nor ridden in on her previous two trips. The probability that she will ride in the sole remaining car that was not used on the prior two trips is 1/3. Therefore, the probability that she will ride in each of the three cars on her three different trips is 1 times 2/3 times 1/3 or 2/9. APPROACH 2: We can also deal with the problem by considering the different ORDER in which she can ride the three cars. She can ride them in any of the following 6 different ordered ways: ABC, ACB, BAC, BCA, CAB, or CBA. The probability that she will ride Car A first, B next, and C third is 1/3 times 1/3 times 1/3 = 1/27. This is the same probability for each of the remaining 5 different ordered ways in which she can ride the three cars. Therefore, the probability that she will ride in each of the three cars during her three different trips is 6 times 1/27 or 6/27 or 2/9. As you can see, both approaches will produce the same result.




Consider another problem in which we can use either a combination approach or a permutation approach to arrive at the same result. PROBLEM 3: A box contains 15 Red balls, 20 blue balls, and 15 green balls. If two balls are drawn from the box without replacement, what is the probability that exactly one blue ball and one green ball were picked? APPROACH 1: We consider that the two balls were selected ‘at a time’. This is how we will deal with a combination problem – selection AT A TIME. WE can see that there are 50 balls in the box, and the number of 2 ball ‘teams’ chosen from 50 is (50 X 49) divided by 2! Or 1225 different selections of 2 balls out of 50 balls. Number of 2-ball ‘teams’ in which there is exactly one BLUE ball and one GREEN ball is 15 times 20 or 300 selections of 2 balls of specific color as required. Therefore, the probability that the set containing one green and one blue ball was picked is 300/1225 or 12/49. APPROACH 2: We can pull the two balls in two different order: BLUE FIRST AND THEN GREEN Or GREEN FIRST AND THEN BLUE. If we pull green first and then blue, the probability of this occurrence is 15/50 times 20/49 = 6/49. If we pull BLUE first and then green, the probability of this occurrence is 20/50 times 15/49 = 6/49. Because we can select two balls in two different order, we need to add the probabilities of the two different ordered scenarios to compute the total probability of the desired event. Therefore, the probability that one blue ball and one green ball will be picked in any order is 6/49 + 6/49 = 12/49. The approach is not critical as long as we engage in sound reasoning. CONSIDER ANOTHER PROBLEM and solidify your understanding of Conditional probability. PROBLEM 4:

A group consists of 7 boys and 6 girls. If a team of 2 is to be randomly selected for an expedition, what is the probability that a team comprising one boy and one girl will be chosen, if all the boys and girls are qualified to be on the team? APPROACH 1: COMBINATION APPROACH The number of teams of 2 people chosen from 7 + 6= 13 people is (13 X 12)/(2X1) = 78 teams of 2 The number of teams of 2 people in which there is exactly one boy and one girl is 7C1 times 6C1 = 7X6 = 42. Therefore, the probability that the randomly selected team will have exactly one boy and one girl in it is 42/78 or 21/39 or 7/13. APPROACH 2: ORDERED SELECTION The first member of the team could be a boy and the next a girl, or vice versa. The probability that we will choose a boy first and then a girl to be on the team is 7/13 times 6/12 = 7/26. We can also reverse the order of selection and choose a girl first followed by a boy. The probability that we will pick a girl first followed by a boy is 6/13 times 7/12 = 7/26. Therefore, the probability of picking a team in which there is exactly one boy and one girl is the sum of the two probabilities of the two different ordered selections. The answer is: 7/26 + 7/26 = 14/26 = 7/13. Two different approaches producing the same result. On the following page, we will take a look at a few more conditional probability problems that are typical of the ones that appeared on the recent GMAT/GRE tests.




PROBLEM 5 A gardener is planting two red rose bushes and two white rose bushes in a row. If the gardener is to select each of the bushes randomly, and one at a time, what is the probability that he will plant both red rose bushes in the middle of the row? The desired order of planting is W – R – R – W. The probability that the first bush is White is 2/4. The probability that the first bush is white and the second a Red is 2/4 times 2/3 = 1/3. The probability that the first is white, the second red, and the third red is 2/4 times 2/3 times ½ = 1/6. The probability that the first bush is white, the second a red, the third also red, and the last a white is 2/4 times 2/3 tmes ½ times 1 = 1/6. (If we have planted one white and two red already, then the probability that the last one will be a white is a certainty having a probability of 1). The required probability is 1/6. PROBLEM 6:

Peter goes on a camping trip on Friday evening and will return from the camp-ground at the end of the day it rains on the camp ground. If the probability that it will rain on the camp ground on any day is 1/5, what is the probability that Peter will not return from the camp-ground until the END OF the day on the following Monday? According to the problem, Peter should not return on Saturday or on Sunday but must return on Monday. Accordingly, it must not rain on Saturday or on Sunday but must rain on Monday on the camp-ground. The probability that it will not rain on the camp-ground is 1/5. Therefore, the probability that it will rain on the camp-ground on any day is 4/5. The required probability is obtained by the following operation: PROBABILITY THAT IT WILL NOT RAIN ON SATURDAY TIMES PROBABILITY THAT IT WILL NOT RAIN ON SUNDAY TIMES PROBABILITY THAT IT WILL RAIN ON MONDAY = 4/5 x 4/5 x 1/5 = 16/125 = 0.128




We are going to be dealing with mutually exclusive events, and independent events when we deal with statistical events.. What are mutually exclusive events? Let us say we toss a coin. We have two possible events: heads and tails. When one coin is tossed, heads and tails are mutually exclusive because if “heads” occur, then the “tails” do not. Two events are said to be mutually exclusive when if one occurs, then the other does not. What are independent events?

Independent events are those that do not depend on the occurrence or non-occurrence of the other events for their own occurrence. Let us say that we toss two coins. Then the heads and tails are no longer mutually exclusive, but are independent. This means that we can have heads occurring together and tails occurring together on the two faces showing up. We will represent mutually exclusive events by two circles that do not intersect.

What are some of the other mutually exclusive events? Males and Females, Smokers and non-smokers, Rented and not-rented rooms in a

hotel, Yes and No answers to a question, Pass and Fail a test, and so on.

We will represent the independent events by intersecting circles. In the set of independent events shown in the venn diagram above, a person can take Math only, or Physics only, or Chemistry only, or can take any two subjects, or all three subjects. Probability of mutually exclusive events adds up to 1.0. For example, if the Probability of Heads is 0.5, and that of Tails is 0.5, then the total probability of these mutually exclusive events must add up to 1.0 (Later on in this section, we will see how we can deal with two sets of mutually exclusive events by setting up a table. For example, let us say that we have two mutually exclusive items: Males and Females, and each of these items can be one of the two mutually exclusive items: Smokers and Non-smokers. We will deal with the above situation by setting up a table along the following lines:

Males Females Total

Smokers

Non-smokers

Total

Heads Tails

Math

Physics Chemistry

The sum of the probabilities of independent events will be greater than 1.0 if they overlap.




Sets and Venn Diagrams

If is a set of numbers 1,2,3,4,5,.....,n, then we define the set S as follows: S ={1,2,3,4,.....n} Venn diagrams describe the relationship among the members of a set by circles. Let us consider a simple problem dealing with a set of numbers: “ Each of 30 students in a graduating class were enrolled in marketing, finance or both. If 22 were enrolled in marketing, 18 in finance, how many were enrolled in both subjects?” There are two approaches to solving a problem of this type: Approach 1: we will try to solve this problem without the Venn Diagrams. Let “n” represent the number of people enrolled in both Marketing and Finance. Marketing enrollment of 22 includes “n”. Finance Enrollment of 18 includes “n” as well. Since we know that the total size of the class is 30, we construct the following equation: (22-n) + (18-n) +n = 30 Solving for “n”, we get n=10.

Approach 2: Venn Diagrams Some wise guy said that a picture is worth a thousand words. We would concur with that aphorism and go one step further and state that “one good circle deserves another.” Two intermeshed circles, dividing the graduating class into three sets, become a Venn Diagram, after the guy who invented it. The Venn Diagram for our problem will look like this: Marketing Both Finance \ The number of students enrolled in marketing alone was 22-n. The figure for finance was 18-n. The figure for “both” was n. Since the Total cannot be more than 30, 22-n + 18-n + n = 30, or n =10.

Marix approach to sets problems. Let us say that we have two genders: Males and Females, and two sub-classifications: Smokers and Non-Smokers. Or that we have two questions in a survey: Q1 and Q2, and two possible responses to each question: Yes and NO. Or that we have two types of rooms: air-conditioned and non-air-conditioned rooms, and each type of room could be Rented or Vacant. You will deal with problems involving two sets and two sub-sets using a table or matrix approach. An example of such a problem is discussed next. You will see that these “matrix” problems involve “plugging” in the values provided, and computing the values for the remaining cells through a process of simple addition and subtraction: The values in the cells along a row must add up to give the row totals, and those in the cells along the columns must add up to give the column totals.

22-n n 18-n




Let us illustrate what we mean by that with an example. “One night a certain motel rented 3/4 of its rooms, including 2/3 of its air-conditioned rooms. If 3/5 of its rooms were air-conditioned, what percent of the rooms that were not rented were air-conditioned?” The best way (and probably the only way) to deal with this problem is as follows: Let us say that the motel had 100 rooms. The motel rented 3/4th of its rooms, or 75 rooms. By inference, we can conclude that the Motel did not rent 1/4 of its rooms, or 25 rooms. 3/5th of the motel rooms are air-conditioned, or 60 are air-conditioned. By inference, 2/5th of its rooms, or 40, are not air-conditioned. Sets A/C Non A/c Total Rented 75 Vacant 25 Total 60 40 100 The problem states that 2/3rd of its air-conditioned rooms were rented. This means that 2/3rd of 60, or 40 of the motel’s rooms were rented. That leaves 1/3rd or 60, or 20 rooms Air-conditioned and not rented. We complete the rest of the cells in the table as follows: Sets A/C Non A/c Total Rented 40 35 75 Vacant 20 5 25 Total 60 40 100 Let us take a look at the question now: “What percent of the rooms that were not rented were air-conditioned?” The question is: What percent is 20 of 25? The answer is 20/25 •100% or 80%. That is, 80% of the not-rented rooms were air-conditioned. And that is the answer. Let us take a look at the answer choices: (A) 20% (B) 331/3% (C) 35% (D) 40% (E) 80%

We pick E as the correct answer choice. Example 2: “A survey was conducted among a sample size of N persons. 1/4 answered Yes to Question 1 and 1/3 of those who answered Yes to Question 1 answered Yes to Question 2. Which of the following expressions represents the number of people interviewed who did not answer “Yes” to both questions?” (A) N/7 (B) 6N/7 (C) 3N/12 (D) 7N/12 (E)11N/12 Although you could answer this question without setting up a table/matrix, you may want to do so because you will get a better perspective on the information specified in the problem. Later on in the assignment, you will come across an extension of this problem with additional information, and you will have to set up a matrix before you can make sense of all the details provided. We recognize that there are two sets of mutually exclusive categories of information: Yes and No answers to questions 1 and 2. Remember that Q1 and Q2 are not mutually exclusive because the respondent is required to answer both. We will set up a matrix with the mutually exclusive values Q1Yes and Q1No as row headings and with Q2Yes and Q2No as column headings. (We can set them up the other way around too, as long as we set mutually exclusive values along the rows and columns.)

Q2Yes Q2No Total

Q1Yes 1/3•¼N =1/12 N

¼ N

Q1No ¾ N

Total N

We are given that ¼th of N people said Yes to Q1. By inference, ¾th of N people said No to Q1. We are also told that of the ¼ N people answering Yes to Q1, 1/3rd also said Yes to Q2. That gives us 1/12 of the people interviewed answering yes to both Q1 and Q2. The question is: How many people did not answer yes to both questions? # not answering yes to both questions = N - # answering yes to both q’s. = N – 1/12 N = 11/12 N. Choice E.




Practice Problems involving Sets

The figure above shows enrollment in the three clubs at a local college. How many students are enrolled in just one club? We notice that there are 10 students enrolled in all three clubs. How do we get the number for Math only? We notice that 17 students are enrolled in math and Drama; and 14 in Math and Science; and 10 in all three clubs for a total of 17 + 10 + 14 = 41. We subtract this number from 68, the total number of students taking math, to get the number for “math only”. We get : 68-41 = 27 students for math only. Similarly, How do we get the number for Drama only? We notice that 17 are taking drama and Math; 21 are taking drama and science; and 10 are taking all three for a total of 17+10+21=48. We subtract this number from 79, the total number taking drama to get: 79 - 48 =31 for the number taking drama only. Similarly, we can determine that the number taking Science only is 87 - (21+10+14) = 42. We have 27 students taking Math only; 31 taking Drama only; and 42 taking Science only for a total of 100 students participating in just one activity. Example 2: “40% of all members in a club have seen movie X; 75% of all members have seen movie Y. At least what percent of the members must have seen both movies?”

We recognize that this is a “table” problem because we have two sets of variables, and proceed to set up the matrix as below: Seen X Not seen

X Total

Seen Y ???? 75% Not Seen Y

25% 25%

Total 40% 60% 100% Since 75% of the members have seen the movie Y, we can compute that 25% must have “not seen Y”. Minimum value for “Seen X-Seen Y” is obtained if the entire 25% who have not seen Y have seen X. We conclude that 40% - 25% = 15% of the members must have seen both movies, at least. In fact, the percentage of members who have seen both movies lies in the range 15% to 40%, with 15% as the minimum value. Note: We can get the answer by determining by what percentage the total of X-seen and Y-seen exceeds 100%. The total of X-seen and Y-seen is 115% or 15% more than 100%. This tells us that at least 15% must have seen both movies. Remember: When we have two independent sets, the percentage value by which the total of the sets exceeds 100% must give us the value for the overlapping area. Example: “In a town of 5300 residents, 3900 people own pets. The number of people owning cats is 2,970, and the number of people owning dogs is 1,530. How many people own a cat and a dog?” We notice that the total of people owning dogs and cats is 4,500 and it exceeds the number of pet owners by 4500 – 3900 = 600. We conclude that 600 folks have both dog and cat for pets.




EXAMPLE: “If x@y denotes the number of people owning x only or y only but not both, what is the value of x@y if the number of people owning x is 450, the number of people owning y is 560 and the total people in town is 900? (People in town own either x or y or both.)” We notice that x@y denotes the excess over the total population of x and y. x and y add up to 1010, and the excess over 900 is 110. Therefore, the value of x@y is the value for X only + the value for Y only = (450 – 110)+ + (560 – 110) = 340 + 450 = 790. (X only is computed by subtracting from the population of X the value that belongs to both X and Y; ditto for Y only) EXAMPLE: “Out of 65 cars in a lot, 25 have air-conditioning and 45 have power windows. If 20 cars have both power windows AND air-conditioning, how many cars in the lot have neither air-conditioning nor power windows?” We notice that there are two sets of mutually exclusive categories, and we can deal with this problem by setting up a matrix.

A/C YES A/C NO TOTAL

P/W YES

20 25 45

P/W NO

5 15 20

TOTAL 25 40 65

The values in the gray band are the values derived from the information specified in the problem. We can see that 15 cars have neither A/C nor Power windows.

EXAMPLE: Of the 32 applicants for a job, 15 had at least 5 years of experience, and 18 had graduate degrees. If 4 applicants had neither the 5 year experience nor a degree, how many had both 5 years of experience or better and a degree? Once again, we can deal with this problem by using a matrix because we see two mutually exclusive sets of categories:

5+ Exp. YES

5+ Exp. NO

TOTAL

DEGREE YES

5 13 18

DEGREE NO

10 4 14

TOTAL 15 17 32

The values in the gray band are derived on the basis of the other values in white cells, values specified in the problem. Therefore, the number of people having BOTH experience and a degree is 5 among the applicants. EXAMPLE If A@B denotes the number of items that are in A only or in B only but not in both, and A contains 130 items, and B contains 75 items, and if there are 25 items common to both A and B, how many items are in A@B? A@B = Items in A only + Items in B only If A contains 130 items and 25 are common to both A and B, items in A only = 130- 25 = 105 If B contains 75 items, and 25 are common to both A and B, items in B only = 75 – 25 = 50

Therefore, A@B = Items in A only + Items in B only = 105 + 50 = 155.




Word Problems Involving Geometry A B The Figure to the left is a C D piece of land. All angles h=200 E F are right angles. What is perimeter of the land? H G L=200 Can you recognize that the piece of land will have the same perimeter as if it were a square piece of land with sides 200 m? Since the angles are all right angles, AB + CD + EF = GH BC + DE + FG = AH Perimeter = 2(AH + GH) = 2(200+200) = 800 meters.

DATA INTERPRETATION GMAT Problem Solving section might include questions involving interpreting data based on a table, graph or a chart.

1stQtr

2ndQtr

3rdQtr

4thQtr

0102030405060708090

1stQtr

2ndQtr

3rdQtr

4thQtr

EastWestNorth

The performance of three regional divisions of a company is shown for the 4 quarters of 1995. The figures in Y axis represent percentage achievement of sales targets. Based on the information presented in the graph, answer the following questions:

1. Northern Region’s performance in Quarter 3 fell short of its target by what percent? 2. Which region shows the most fluctuation over 4 quarters in performance (target achievement)? 3. Which region shows the most stable performance in terms of target achievement? You will find that questions of this type test your ability to read a chart, or a graph, or a table and obtain the relevant information. Piece of Cake.




Must-be-true problems that you will encounter in the GMAT problem-solving section require you to make the most sense of the given information, consider an exception, and then answer the question. A “must-be-true” condition is valid under all circumstances and if you can think up an exception, you will be ready to test the conditions. “Must-be-True” problem takes the following form:

“ If y ≠ 3 and (3x)/y is a prime integer greater than 2, which of the following must be true?” I. x = y II. Y = 1 III. x and y are prime integers. (A) None. (B) I only (C) II only (D) III only (E) I and III The best and probably the only way to approach this type of problem is to test each of the conditions against “real-life” situations. Also note the word “must”. It will not be highlighted or appear in bold in the test. 3x/y is a prime integer greater than 2. Which means that it could be 3, 5, 7, 11, 13, 17, etc. Let us pick a prime number: 7. The expression 3x/y can produce 7 for the following values of x = 21 and y=9. Therefore x does not have to be equal to y. So we conclude that the condition I is

false. (Read Condition I to say “x must be equal to y”). We can quickly conclude that condition II is false as well because, as we have shown in the previous paragraph, y does not have to be equal to 1 to satisfy the requirement that 3x/y be a prime number greater than 2. (Read condition II to say “y must be equal to 1”). We can eliminate condition III just as easily because 3x/y can be a prime number greater than 2 for values of x=21 and y = 9. None of these values represents a prime number. (Read condition III to say: “x and y must be prime numbers”). We have concluded that conditions I, II and III need not be true. What is the answer choice? None of the above. We pick and mark (A) as the correct answer. Let us do another question to be more comfortable with “must be true” questions: “If x and y are integers and xy² is a positive, odd integer, which of the following must be true?” I. xy is positive II. xy is odd III. x+y is even. (A) I only (B) II only (C) III only (D) I and II (E) II and III This question tests our knowledge of properties of integers - particularly odd and even integers. (Discussed at the outset of this module). Given: xy² is a positive, odd integer.




What do we know about odd integers? For xy² to be a positive, odd integer, the following conditions must be met: 1. x must be positive and odd. 2. y² must be (and will be) positive and must be, in addition, odd. This condition requires that y be an odd integer - positive or negative so that the square of the integer will be a positive and odd integer. For example, if y is -5, then y² will be 25, an odd, positive integer. If y is 5, then y² will still be a positive and odd. Now Let us look at the “must be” statements: 1. We conclude that the condition I is false because xy² will be a positive, odd integer even for negative odd values of y. xy can be negative and still the requirement that xy² be a positive and odd integer can be met. 2. xy must be odd is a valid and true requirement. Both x and y must be odd to satisfy the requirement that xy² is odd and positive. xy can be odd and negative and still satisfy the requirement of the question. II is true. 3. Condition III must be true as well because if x and y are both odd , then their sum is even. (Remember odd plus odd gives even; odd plus even gives odd; etc.) So. What do we have now? Conditions II and III must be true to satisfy the requirement that xy² be odd and positive. So we pick (E) as the answer choice. Let us deal with just one more question before we move on the basics of Geometry.

“If u>t, r>q, s>t, and t>r, which of the following conditions must be true?” I. u > s II. s > q III. u > r

(A) I only (B) II only (C) III only (D) I and II (E) II and III From the relative positions of q, r, s, t, and u, we can construct the following line: q r t s? u s? The relative positions of q through u are a given and they will look like those as shown along the continuum. Now, let us look at the “must be” statements. 1. The only given information about s is that it is greater than t. Which means that it occurs on the continuum after t but does it occur before u or after u? We do not know from the available information. If s occurs before u, which will still satisfy the condition that s be greater than t, then condition I will not be met. Therefore the condition I fails. 2. Conditions II and III must be true as can be seen from the positions of q, r, s and u from the illustration. We pick (E) as the answer choice. Note: The only “trick” condition in this question is the statement s is greater than t but there is nothing given about s and u to define their relative positions. 1. Make as much sense as possible of the given information before you go to deal with the conditions stipulated. 2. Remember: Must-be-true condition must be valid for all values of the variables defined. Therefore, if you can think of an exception - just one - you know that the condition stipulated is not valid. 3. Do not assume that a number is a whole number, and positive. Unless the numbers are specified as such, they could be anything. Go to Assignment on next page.




ASSIGNMENT 1. If x+y is a prime number, then which of the following “must be” true?

(A) x = 2 True / False

Why?_____________________________________________________________________________________________

(B) Either x or y is 2. True / False

Why?_____________________________________________________________________________________________

(C) Both x and y are prime numbers.

True/ False

Why?_____________________________________________________________________________________________

(D) x.y is not a prime number.

True / False

Why?_____________________________________________________________________________________________

(E) x-y is a prime number.

True / False

Why?_____________________________________________________________________________________________

(F) x.y is a positive integer. True / False

Why?_____________________________________________________________________________________________

2. If x²y³ is a negative odd integer, then which of the following must be true?

(A) y is a negative integer. True/ False.

Why?______________________________________________________________________________________________

(B) x can be a fraction. True / False

Why?______________________________________________________________________________________________

(C) x can be positive or negative, integer or a Fraction. True / False

Why?______________________________________________________________________________________________

(D) Both x and y are odd numbers - fractions or integers. True / False

Why?______________________________________________________________________________________________

(E) x² + y³ is a positive even integer. True / False

Why?_____________________________________________________________




Assignment Continued

3. Thames Toyota Dealership has 120 cars in their lot. The cars come in two colors - Red or Blue and two sizes - mid-size and large. There are exactly 10 large sized Red cars in the lot. If there are three times as many mid-sized cars as there are large ones and the red to blue cars ratio in the lot is 3:2, How many cars in the lot are :

(A) Red and Mid-size?

(B) Blue and Large?

(C) Blue and Mid-size?

Directions: You are required to construct a table as discussed and solve the problem.

4. Machine A produces bolts at a uniform rate of 120 every 40 seconds, and machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 bolts?

5. A survey involving two questions Q1 and Q2 was answered by N persons. 1/3 of those participating answered YES to Q1, and of these 1/5 also answered YES to Q2. If the number of those answering YES to Q1 but NO to Q2 is twice that answering NO to Q1 and YES to Q2, what is the number answering NO to both questions Q1 and Q2?

Hint: Plug in the values given and compute the remaining values to agree with the totals specified.

YES TO Q2 NO TO Q2 TOTAL

YES TO Q1

NO TO Q1

TOTAL




GEOMETRY

1. LINES:

A Line is a Straight Line. It is a no-brainer.

A B

AB is called a line segment and A and B are the end points of the segment. Both the segment and the length of the segment are denoted by the notation AB.

2. INTERSECTING LINES AND ANGLES:

Yº

Yº

The opposite angles are referred to as the Vertical angles and will have the same measure. S

P xº yº R

Q

PQR is a straight line or straight angle (180º). Angle PQS and SQR are adjacent angles and add up to 180º. (xº + yº = 180º).

Two angles whose measures add up to 90º are referred to as complementary angles. If the two measures add to 180º, they are known as supplementary angles.

Two straight lines intersecting each other at 90º are referred to as perpendicular lines. A right angle symbol at the intersection denotes perpendicular lines.

Parallel Lines are two lines in the same plane that do not intersect. The lines are supposed to meet at infinity. A third line intersecting two parallel lines produces the following relationship among the angles of intersection:

xº yº

yº xº

xº yº

yº xº

xº+ yº =180º

A Polygon is a closed plane figure formed by 3 or more line segments, referred to as the sides of the polygon. Each side intersects exactly two other sides at their end points called the vertices. Each interior angle of a convex polygon has a measure of less than 180º.

A Polygon with 3 sides is a triangle; with four sides is a quadrilateral; 5 sides is a pentagon and six sides is a hexagon.




A triangle has 3 angles and the sum of the angle measures is 180º.

The general rule for the sum of the angle measures of a polygon with “n” sides is: (n-2)180º

1. What is the sum of the angle measures of a triangle? A triangle is a polygon with 3 sides. ∴n=3 The sum of its angle measures is: (3-2)•180º = 180º

2. What is the sum of the angle measures of a Quadrilateral?

A quadrilateral is a polygon with 4 sides.

∴ n = 4. The Sum of its angle measures is:

(4-2)•180º = 360º

3. What is the sum of the angle measures of a pentagon?

A pentagon is a polygon with 5 sides. ∴N=5.

The sum of its angle measures is:

(5-2)•180º = 540º

4. What is the sum of the angle measures of a hexagon?

A hexagon is a polygon with 6 sides. ∴N=6

The sum of its angle measures is:

(6-2)180º = 720º

What is the sum of the angle measures of a polygon with 9 sides?

The Perimeter of a polygon is the sum of the lengths of its sides.

A triangle with sides 3, 4 and 5 will have a perimeter of 3+4+5=12 units.

The area of a figure lying in a plane will mean the area of the region enclosed by that figure.

Q

5 5

P xº yº R

The above triangle is referred to as Triangle PQR. PQ, PR and QR are the sides of the triangle. In our illustration, PQ =QR.

If none of the sides in the triangle is the same length as any other, then we are dealing with a scalene triangle.

If two sides of a triangle are equal, then the triangle is called an “isosceles” triangle. If all three sides are equal , i.e. PQ=QR=PR, then the triangle is called an equilateral triangle.

In an isosceles triangle, the two angles opposite the two sides of equal length will have the same measure. In our illustration xº = yº. The third angle will be equal to 180º-(xº+yº) = 180º-2xº = 180º - 2yº since xº and yº are equal. Also, the vertical line drawn from the vertex where the two line segments of equal length meet will divide the opposite side (base) in two equal halves. Such a vertical line will also divide the vertex angle in two equal halves.

In an equilateral triangle, all three angles are equal to 60º each. All sides will be the same length. A vertical line drawn from any vertex of




an equilateral triangle to the opposite side (base) will divide the base in two equal halves. Such a vertical line will also divide the vertex angle in two equal halves.

30o 30o

30o 30o 30o 30o

The sides are divided in two equal halves, and the line segments shown are equal length. The vertex angles are all equal and the vertical lines drawn to the opposite “bases” will divide the vertex angles in equal halves. In fact, the vertex angles are each 60 degrees in an equilateral triangle, and the vertical lines will divide the vertex angles to two 30 degree angles.

Triangles Trivia:

Did you know that if you have two parallel lines, and if you take any two points on one of the lines and draw triangles by picking points on the other parallel line, such triangles will all have equal areas?

L1 C D E F

L2

A B

L1 and L2 are parallel lines, and we choose points A and B on L2. If we use the line segment AB as the base for all triangles ABC, ABE, ABE, ABF, the area of these triangles must be equal. Why ? The area of a polygon is determined by the vertical height. Since these triangles share the same base, and have the same vertical heights (shown in dotted lines), it follows that these triangles must have equal areas as well.

The only set of information we need to know to be able to compute the area of a triangle is information pertaining to the base value and the vertical height.

As long as we know that the triangles share the same base and are drawn between two parallel lines, we will know that these triangles have the same area.




PRACTICE EXERCISE 1

C

3x-2 x 2x+6

A B

3x+4

An isosceles triangle ABC shown above has its sides as marked. IF AC = BC, and if the vertical height is x units, what is the area of the triangle?

Answer: An isosceles triangle has two sides of equal value. In this problem AC and BC have equal values. We can, therefore, write: 3x-2 = 2x+6 or x = 8

Knowing the value for x = 8, we can determine that the base must measure (3 X 8 + 4) = 28 units.

The base is 28 units and the vertical height is 8 units. We have all the information that we require to find the area.

Area of isosceles triangle ABC = 1/2. 28.8 = 112 square units.




A right triangle is a triangle with one right angle (90º). The side opposite the right angle is referred to as the hypotenuse; the other two sides are the legs.

An important property of a right triangle is provided by the Pythagorean theorem.

Pythagorean theorem states that, in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

B

5

3

A 4 C

⊇ABC is a right triangle. AB and AC are the legs of the triangle; BC is the hypotenuse.

The legs have lengths 3 and 4; the hypotenuse must have length 5, based on Pythagorean theorem.

AC² +AB² = BC²

3² + 4² = BC² = 9 + 16 =25

BC = √25 = ± 5

Since the length of a figure in a plane cannot be a negative number, BC must be 5.

You should also know that if the triangle has its sides in any of the following standard triples, its must be a right triangle.

Leg 1 Leg 2 hypotenuse

3 x (n) 4 x (n) 5 x (n)

5 12 13

7 24 25

9 40 41

11 60 61

What do you notice? The square of the smallest length of a leg is the same as the sum of the lengths of the other leg and the hypotenuse. This is an important property arising from Pythagorean theorem.

The notation x (n) in the table indicates that if you multiplied the lengths of all sides by the same positive integer, the relationship among the sides still holds.

For instance, 3, 4,5 is a right triangle. So is 6, 8, 10. So is 9, 12, 15.

5, 12, 13 is a right triangle. So is 10,24,26. So is 15, 36, 39.

You should also remember that 45-45-90, and 30-60-90 triangles have their sides in specific proportions as shown below:

Angles Sides

45º -45º - 90º 1 : 1 : √2

30º -60º -90º 1 :√3 : 2

If the angles are 45º, 45º and 90º, then the opposite sides to these angles are in the ratio 1:1:√2.




To illustrate, consider the following right triangle:

3 xº 3√2 √2 = 1.414

xº xº=45º

3 If one angle in a right triangle is given as 45º, the other two angles must be 45º and 90º. If one side in the above right triangle with one angle 45º is given as 3, the other two sides must be 3 and 3√2.

1. If in the above right triangle, one leg length is 5, the other two sides must be 5 and 5√2.

2. If in the above right triangle, one leg length is 6, the other two sides must be ______ and _________.

3. If in the above right triangle, one leg length is 7, the other sides must be ______ and __________.

4. If the hypotenuse in a right triangle with one angle 45º is 12√2, the leg lengths must be _______ and _______.

5. In a right triangle, the sides are as follows: 4, 4, 4√2. The angles of the triangle must be ____, _____, _____.

6. In a right triangle with one angle 45º, the hypotenuse measures 25√2. The angle opposite the hypotenuse must measure ______º.

Answers: (2) 6, 6√2 (3) 7, 7√2 (4 )12, 12 (5) 45º,45º,90º (6) 90º

30º-60º-90º Right Triangle

In 30º-60º-90º right triangles, the lengths of the sides are in the ratio 1:√3:2 (√3 = 1.732)

C

√3 30º 2

90º 60º

A 1 B

The side AB is opposite the 30º angle and if it measures 1 unit, then the side AC opposite the 60º angle must measure √3(=1.732) units and the hypotenuse (always) opposite the 90º angle must measure 2 units.

1. If in the above right triangle, the length of side opposite the 60º angle is 5√3. The other two sides must be _____ and ________.

2. If in the above right triangle, the sides are 4, 4√3 and 8. Identify the angles opposite each side:

The angle opposite side 4 units long is ____º. The Angle opposite side 4√3units long is ___º. And the angle opposite side 8 units long is _____º.

3. If in the above right triangle, the hypotenuse is 14 units long. The other two sides must measure _____ and ____.

Answers: (1) 5, 10 (2) 30º,60º, 90º (3) 7, 7√3.




The area of a triangle is calculated using the formula:

(length of the altitude) X (length of the base)

2

Altitude: Denotes the segment going from the vertex perpendicular to the side opposite that vertex. This side is referred to as the Base.

C

90º E

A 90º B

D

10

In the above triangle, CD is the altitude with respect to the base AB and AE is the altitude with respect to the base CB.

If the base length is 10 units and the altitude is 5 units, the area is :

10 X 5

2

The area is 25 square units.

Exercises:

1. What is the area of an equilateral triangle of side 6 inches?

We know that the equilateral triangle has all 3 sides of equal measure, in this case 6 inches. All 3 angles are equal to - 60º each. The altitude from a vertex to the base will divide the vertex angle equally 30º - 30º and the base equally - 3 inches - 3 inches in length.

C

6 30º 6

A 60º D 60º B

6

From the properties of a 30º - 60º - 90º triangle (Which is what CDB is), and knowing that DB=3 inches because the altitude from vertex C divides the base in two equal halves, we know that the altitude must be 3√3 (Remember the sides are in the ratio of 1:√3:2).

∴ The Area is (CD X AB) ÷ 2 = (3√3 X 6)÷2

The Area is: 9√3

2. An isosceles right triangle has its hypotenuse measuring 5√2. What is its area? We know that in an isosceles right triangle, the 3 angles are 45º, 45º and 90º and the sides opposite these angles are in the ratio 1:1:√2.

In this exercise, the hypotenuse is 5√2. We know that the other two sides must be 5 and 5.

We can compute the area to be:

(5 X 5)÷2 = 12.5

NOTE: In a right triangle, the two legs constitute the altitude and the base and we can compute the area by multiplying the two leg lengths and dividing the product by 2. The area for any polygon is determined by the vertical height .




ASSIGNMENT

1. A triangle has its 3 sides of following lengths: 3, 3, 3. What are the 3 angles?

Answer:______º, ______º, ________º

2. A right triangle has its hypotenuse measuring 8 units. One of its angles is 30º. What are the other two side lengths?

Answer:__________, _____________

3. A triangle has 3 sides of lengths 27, 36, 45. What type of a triangle is this?

Answer:_________________________

4. An equilateral triangle has sides measuring “a”. What is the area of the equilateral triangle in terms of its side “a”?

Area is:__________________________

5. A right triangle has one leg measuring 4 and the other 6. What is the hypotenuse length? (use Pythagorean theorem and show steps).

The Hypotenuse is: __________

6. A triangle has two angles measuring 55º and 43º. What is the measure of the third angle?

Answer: _________________º

7. An isosceles triangle has two sides 10 and 10. The base where the altitude from the vertex meets measures 16. What is the area of the triangle?(Draw triangle, mark sides and angles appropriately, and solve the problem).

Answer: ____________________

8. If AB X CE = 24, What is the area of the triangular shaped region ABD shown below?(AB and CD are parallel lines)

C D

A E B K

Hint: The vertical height DK shown in dotted lines will determine the area of the triangle ABD. You will notice that the altitude for the triangle ABD is the same as CE.

Answer:

nother important rule with regard to the side lengths of a triangle that you should be

aware of is stated here. A




If the triangle is a right triangle, then the relationship between the values of the side measures is specified by the Pythagorean theorem. For example, Can line segments of lengths 4, 7, and 9 be the values for the sides of a right triangle? We know that in a right triangle, the hypotenuse is the longest side (because it is opposite the largest angle). Of these three values - 4, 7, and 9 - 9 must be the hypotenuse value if these line segments can indeed form a right triangle. The only test we will apply is:

Is 42 + 72 = 92 ? We notice that 16 + 49 is not equal to 81, and we conclude that 4, 7, and 9 could not be the values for the sides of a right triangle.

Could these be the values for the sides of a normal triangle (one that is not right angle triangle)? To determine that, you will apply the following test:

If two sides of a triangle are known, then the third side must have its length measure less than the sum of the other two side lengths and greater than the difference of the other two side lengths.

In other words, if a, b and c are the three sides of a triangle and if a and b are known quantities, then, for a,b and c to be the sides of a triangle, c must satisfy the following condition:

(a+b) > c > (+ve difference of a and b)

Let us consider an example from a recent GMAT:

“If 3 and 8 are the lengths of two sides of a triangular region, which of the following can be the length of the third side? (Notice the “can be” condition here because the third

side can have more than one value and still satisfy the requirement that it is part of a triangle.)

I. 5 II. 8 III. 11

Answer Choices: (A) II only (B) III only (C) I and II only (D) II and III only (E) I,II and III.

We know the values of two sides to be 3 and 8. For the third side to be part of this triangular region described, it must lie within the following values:

(3+8) > the third side > (8-3) which means that the third side can have values greater than 5 and less than 11 and be a part of the triangular region.

From the answer choices, we notice that there is just one value - 8 - that satisfies this requirement and condition and we choose (A) as the correct answer choice.

Note: Greater than 5 means that 5 cannot be a value. Similarly, less than 11 means that 11 cannot be a value.

Let us do one more exercise to clarify this rule for a normal triangle.




“A skating rink is in the shape of a triangle. The two sides of the rink measure 80 ft and 90 ft.

1. What is the possible range of values for the third side?

Answer: The third side can have values in the following range: 170> third side <10. The third side can have any value greater than (but not including)10 ft and less than (but not including) 170 feet.

Quadrilaterals We know that a quadrilateral is a polygon with 4 sides.

What do you call a quadrilateral in which both pairs of opposite sides are parallel? Simple: A parallelogram. What else do we know about parallelograms? The opposite sides of a parallelogram have equal lengths.

E D

F

A B C

In the above parallelogram ABDE,

AB || ED and AB = ED; ∠BED=∠ABE

AE || BD and AE = BD:∠BAD=∠ADE

If AB=6, then DE =6; If BD =5, then AE=5.

The diagonals of a parallelogram bisect each other. i.e. AF=FD & EF=BF

The area of a parallelogram is the product of its altitude (CD) and its base (AB).

A parallelogram is composed of two triangles, in our example ABD and AED. ABD and AED will have equal areas. If the area of ABD is 24, AED will have an area of 24 and the parallelogram will have an area of 48.

Do the Diagonals of a parallelogram intersect each other at right angles? What do you think?

Answer: The diagonals of a simple parallelogram do not bisect each other at right angles. The diagonals of a rhombus do.

What is a Rhombus? A rhombus is a parallelogram with all sides of equal lengths. In our illustration,

If AB=BD=DE=EA, then we have a rhombus for a parallelogram. In a rhombus, all four sides are considered to be “congruent” and the diagonals bisect the vertex angles and each other at right angles.

IN other words, ∠AEF = ∠DEF; ∠EAF = ∠BAF; ∠EDF = ∠BDF.

∠EFD = 90º = ∠BFD

EF=BF=AF=FD

A parallelogram with right angles is a rectangle and if all sides of a rectangle are of equal length, it becomes a square. Is a square a rhombus with all angles equal to 90º. You could say that.




The area of a rectangle is the product of its altitude and its side. In a rectangle, the altitude happens to be another side The area of a square is the square of its side. If one side of a square is 6, then all sides are 6 and the area is 6² = 36.

If a rectangle has two sides of value 4 each and the other two sides of value 6, then its area is 6 times 4 = 24.

The perimeter of a parallelogram is the sum of the lengths of its sides. A square of side 6 will have its perimeter equal to 24. A rectangle with sides 4,4,6,6 will have it perimeter equal to 20.

What else do we need to know about quadrilaterals?

We said that a rectangle has 4 right angles. Can a quadrilateral have 2 right angles and not be a rectangle? You bet it can be. Consider the following shape of a quadrilateral:

90º 90º

We can see that the above shape is not that of a rectangle.

Can a quadrilateral have 3 right angles and not be a rectangle? NO. Because, the internal angles add up to 360º (Remember: (n-2)180º). If the three angles are 90º each, then the fourth angle of a quadrilateral must be 90º as well. It must, therefore, be a rectangle or a square. (If all sides are given as equal

or congruent, then the figure is a square.)

Can a quadrilateral have two parallel sides and the other two sides not parallel to each other. Yes. Consider the following figure: It is called a trapezoid. A trapezoid has two parallel sides and two non-parallel sides.

C 6 ft D

4’ 4’

A 3’ E 6 ‘ F 3’ B

How do you calculate the area of the trapezoid?

If you take a look at the trapezoid again, you will notice that it has:

1. A rectangular shaped region.

2. Two triangular shaped regions.

The area of a trapezoid will be the sum of the areas of the rectangle and the two triangles.

In our illustration, the area of the trapezoid will be:

1. Area of the rectangle = 24 Plus

2. Areas of two triangles = ½.. (3)(4) + ½.(3)(4) = 12

Equal to :36 (See below for another way to get this)

Note: The two triangular regions may not be congruent or have the same area in a trapezoid.




“A soccer field is in the shape of trapezoid with a rectangular portion in the middle with sides 120 yards and 60 yards; the rectangular area is flanked by two triangular regions of total area 450 sq.yards. The total area of the trapezoidal soccer field will be 7,650 sq.yds. (120X60 + 450 = 7,650 sq.yds).

The area of a trapezoid can also be calculated using the formula: ½ . (sum of bases)(height).

In our illustration, the area will be:

½. (12 +6)(4) = ½.(72) = 36 which is the answer we obtained by computing and adding individual areas of rectangle and the two triangular regions.




ASSIGNMENT

1. If ∠ABC in the figure below is a right angle, what is the value of external xo?

xº C

B 55º D

A ∠ABD = 55º

Answer: xº = ______º

2. If a triangular region has two sides of value 17 and 25, which of the following cannot be a value for the third side?

(A) 38 (B) 45 (C) 11 (D) 17 (E) 41

Answer: ________

3. Which of the following groups of numbers could be the lengths of the sides of a triangle?

I. 1, 4. √17 II. 4, 7, √11 III. 4, 9,6

Choices: (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II and III. (Explain your answers)

Answer: __________

4. If the length and the width of a rectangular garden plot were each increased by 25%, what would be the percent increase in the area of the plot?

(A) 25% (B) 33% (C) 45% (D) 56.25% (E) cannot estimate from the information

Answer: ______________

5. The trapezoid shown below is the cross section of a rudder of a ship. If the distance from A to C is 25 feet, what is the area of the cross section of the rudder in square feet? C

D

A B

The lengths are: AD= 4’ BC=7’.

(A)168 (B) 175 (C) 100 (D) 132 (E) 28

(Draw diagram here and show your working)

Answer: ________________




Assignment Continued... 6. If a rectangular photograph that is 10 inches wide by 15 inches long is to be enlarged so that the width will be 22 inches and the ratio of length to width will be unchanged, then the length in inches of the enlarged photograph will be

(A) 33 (B) 32 (C) 30 (D) 27 (E) 25

Answer: ________

7. If L and W are the dimensions of a rectangular region that has area 42 and if L and W are integers such that L > W, What is the total number of possible values of L?

(A) Two (B) Three (C) Four (D) Five (E) Six

Answer: _______

8. In the above question, if the area were 45, and L and W are integers such that L >W, how many possible values for W exist?


Answer: __________

9. The shaded portion ABC of a rectangular lot shown below is a flower bed. If the area of the flower bed is 24 sq.yds and BC=AB + 2, then AC equals:

(A)√13 (B)2√13 (C) 6 (D) 8 (E) 10

C

A B

Answer: _________

10. A rectangular region has the same area as a square region : 64 sq. yards. If one length of one side of the rectangle is half the length of the side of the square and the other side of the rectangle is twice the length of square side, what are the dimensions of the rectangle and the square?(rectangle length, rectangle width, square side)

(A) 8,8,8 (B) 4,16,8 (C) 8,12,12 (D) 4,8,8 (E) 16,8,8

Answer: ____________

Assignment Continued...




11. Which of the following inequalities is equivalent to 10-2x > 18?

(A) x > -14 (B) x > -4 (C) x >4 (D) x < 4 (E) x < -4

Answer: ___________

12. Which of the following inequalities is equivalent to 5x-10y < -20

(A) 2y < -(4+x) (B) x < 2y+4 (C) x >2y+4 (D) 2y>4+x (E) None of the above.

Answer: ____________

13. If n is an integer and

n = (2.3.5.7.11.13) ÷ 91p

Which of the following could be the value of p?

(A) 22 (B) 26 (C) 35 (D) 54 (E) 60

Answer: ________

14. Mary and Jenny live 13 miles apart. They meet at a cafe that is directly north of Mary’s house and due east of Jenny’s

house. If the cafe is 7 miles closer to Mary’s house than to Jenny’s house, how many miles is the cafe from Jenny’s house? (Hint: Use Pythagorean theorem and right triangle properties discussed.) Draw diagram and show steps.

(A) 12 (B) 20 (C) 6 (D) 10 (E) 10√3

Answer: ______________

15. Which of the following equations has one root in common with x²-6x+5=0

(A) x² + 1=0 (B) x²-x-2 = 0 (C)x²-10x-5=0 (D) 2x² - 2=0 (E) x²-2x-3=0

Answer:________________

16. How many positive integers n are there such that 100n is a factor of:

(2³)(5)(5³)

(A) None (B) Five (C) Seven (D) Six (E) Eleven

Answer: ______________________




Circles and Solids

Imagine, if you will, a set of points in a plane that are all located equidistant from a fixed point and you have just imagined up a circle.

A chord of a circle is a line segment that has its end points on the circle. When the chord passes through the center of the circle, it becomes the diameter. A radius is a line segment going from the center of the circle to a point on the circle.

A radius is ½. of the diameter. r is a common notation for the radius and d for the diameter.

The circumference of a circle is the distance around the circle. The circumference is (2πr) or (πd) where π is a constant and has a value of 22÷7 or 3.14

P Q R

The line segment shown in red is the chord of the circle. The blue line is the diameter. Q is the mid point of the line PR and denotes the center of the circle.

PQ =QR = r = radius of the circle

If PR = 10 cm, then PQ=QR=5 cm

The circumference of the circle is 2πr =10π or 31.4

The area of the circle is expressed by the formula: πr² or (πd²÷4)

In our illustration, the area of the circle of diameter 10 cm will be:

π.5² = 25π = π. (10²÷4)

If you traveled around the circumference of the circle, you will have traversed a complete revolution with the number of degrees equal to 360º. In other words, the number of degrees of arc in a circle is 360º.

The length of arc of a circle is calculated by determining the angle the arc forms at the center of the circle.

T

xº S

R

If the angle that the arc RST forms at the center of the circle is 60º, then the arc is 1/6 of the circumference of the circle. i.e. (60º÷ 360º) . circumference

If each vertex of a polygon is located on a circle as in Figure A, then the polygon is inscribed in the circle and the circle is circumscribed about the polygon.




If each side of a polygon is tangential to the circle (meets the circle at just one point) as in Fig.B, then the circle is said to be inscribed in the polygon and the polygon is circumscribed about the circle.

If a triangle is inscribed in a circle such that one of its sides is the diameter of the circle, then the triangle is a right triangle.

C 90º

A B

⊇ABC is a right triangle with ∠ACB =90º. AB is also the diameter of the circle.

Exercise:

“A triangle is inscribed in a circle such that its hypotenuse is the diameter of the circle. If the legs of the triangle are 3, and 4 inches, (1) What is the diameter of the circle? (2) What is the area of the circle? (3)What is the circumference of the circle? (4) What is the perimeter of the triangle?”

From the properties of right triangle and the Pythagorean theorem, we know that if the two legs of a right triangle are 3, the other two sides are 4 and 5 and the and 4, the hypotenuse is 5 inches.

When you find values of 3 and 4 or any multiple thereof for the measure of the legs, you know that the hypotenuse must be 5 or multiple thereof.

∴The diameter of the circle is 5. (Answer to Q1)

∴ The radius of the circle is 2.5.

We are now ready to answer the rest of the questions.

2. Area of the circle: πd²/4 = 6.25π

3. Circumference of the circle: πd=5π

4. Perimeter of the Triangle =3+4+5=12

As you can see, what is required is your ability to quickly determine that the inscribed triangle is a right triangle and from the Pythagorean theorem , to see that the triangle must be 3-4-5 and apply the formula to calculate the required information.

Try this one out, and see if you can get the answer:

“Within a circle of radius 6.5 inches is inscribed a right triangle. If one of the other sides is 5 inches, then (1) What is the area of the inscribed triangle? (2) What is the area of the circle? (3) What is the circumference of the circle? (4) What is the ratio of the circumference of the circle to the perimeter of the triangle?”

(Answers: (1) 30 (2) (169π÷4) (3) 13π (4) 13π:30 )

Did you know that the circumference of a semi-circular shaped region is ½ the circumference of the full circle PLUS the diameter? You need to traverse across the diameter to complete the loop.




3

4 8

Rectangular Solid Cylinder

Pyramid Cone Sphere

A rectangular BOX has 6 faces and its surface area is the sum of the surface areas of all its faces. In our illustration, the rectangular BOX has dimensions 3, 4 and 8. Therefore the surface area of the BOX is 2(3.4)+2(3.8)+2(4.8) = 136.

Notice that there are 3 sets of parallel faces and, therefore, you need to calculate the surface areas of 3 faces and multiply the result by 2 to get the total surface area for the solid.

To calculate the volume of the rectangular solid, multiply length , width and height. (3.4.8 = 96)

Also, if L, W, and H are the three dimensions of the rectangular box, then the maximum length of a diagonal connecting two points on the box is given by the formula:

SQRT(L2 + W2 + H2)

If the dimensions of a rectangular box are 5, 7, and 9 inches, then the maximum length of a diagonal is SQRT(52 + 72 + 92) = SQRT(25 + 49 + 81) = SQRT(155)

CUBE

A CUBE IS A RECTANGULAR BOX THAT HAS ALL SIDES MADE OF SQUARES. The surface area of a cube is 6 times (side)2

The capacity or the maximum volume of a cube is (side)3.

REFER TO GEOMETRY FORMULAS FILE FOR A DETAILED DISCUSSION OF THE PROPERTIES OF A CUBE.




CYLINDERS

In a right circular cylinder shown above, there are two circular bases of equal size connected by a curved surface.

The area of the cylinder is the area of the two circular bases plus the area of the curved surface. The curved surface is nothing but a rectangular sheet metal that has been rolled along its width. What was once width of a rectangular sheet metal has become now the circumference of the curved surface. The area of the curved surface is, therefore, 2.pi.r.h

The area of the circular bases are each πr². That gives 2πr². The area of the curved surface is 2πrh.

The Surface area of a circular cylinder is:

2πr² + 2πrh

In our illustration, if, say, the radius, r=5 and the height, h=10, the surface area will be:

2.( π5²) + 2π.5.10 = 50π+ 100π = 150π

The volume of the cylinder is πr²h which is the same as the product of base area and the height.

In our illustration, the volume of the cylinder is : π (5²).10 = 250π

Exercise: A circular cylinder of height 20 has a base area of 64π. What is the (1) Surface area of the cylinder? (2) Volume of the cylinder?

The base area is 64π. ∴ πr² = 64π; r²=64. r =8 The height “h” is 20.

The surface area of the cylinder is twice the base area plus 2πrh =2(64π)+2π.8.20

Surface Area = 128π+320π=448π

Volume of cylinder =πr²h =π.8².20 = 1280π

CONE A Cone is a

truncated cylinder. The surface area of a cone is ½ that of a cylinder of identical radius

and height, and the capacity or maximum volume is 1/3 that of a cylinder of identical radius and height.




SPHERE

A SPHERE is a ball whose capacity is given by the formula:

4/3 • π•RADIUS3

You are not likely to get questions about Spheres in the GMAT, but you should expect questions about Cylinders, cubes and rectangular boxes.

EXERCISE:

“What is the diameter of a cylinder, which will have the maximum capacity, and which can be placed inside a rectangular box of dimensions 8 inches by 10 inches by 12 inches?”

We can place the cylinder on any of the following three sets of faces:

• 8 inches by 10 inches. The maximum diameter of a cylinder that can be placed on this side is 8 inches, and the height of this cylinder could be the third dimension 12 inches. The capacity of such a cylinder will be π•4•4•12 = 192π

• 8 inches by 12 inches. The maximum diameter is only 8 inches and the height is 10 inches. We know that this cylinder is going to have less capacity than the one discussed earlier.

• 10 inches by 12 inches. The maximum diameter of a cylinder is 10 inches, and the height is the third dimension of the

box, which is 8 inches. The capacity of such a cylinder is: π•5•5•8 = 200π

We can see that the cylinder of 10 inches diameter will have the maximum capacity. We will discuss more practice problems in geometry later on in this chapter.




Pyramids

A pyramid is a polyhedron that has only one base. (The base is the "bottom" of the Egyptian pyramids.) The other faces are all congruent triangles, and they share a common vertex, which is the top point. The base can be any type of polygon. If the base is a triangle, then the pyramid has a total of four faces. The Egyptian pyramids have square bases

and four triangles as faces

These are two illustrations of pyramids, one with a triangular base and triangular faces, and the other with a rectangular base and triangular faces. The volume of a pyramid is a measure of how much it would take to fill the shape. For a pyramid, the formula is:

Where B is the area of the base figure, and h is the height from the

base to the vertex. The volume is expressed in measurement units, cubed, like cubic inches. See if you can imagine little cubes filling up the interior space of the shape.

This formula is true for pyramids of any shape base. As long as you can find the area of the base and you know the height, you can calculate the volume.

What is the volume of a pyramid with a square base with sides of 5 cm, and a height of 3 cm?

V = (5cm x 5cm) x 3 cm Volume = 25 cm3




The surface area is calculated by computing the area of all faces that constitute a pyramid. Since all faces except possibly the base are triangles, you will find Pythagorean theorem handy in solving pyramid problems.

A

S 12 R

B C

P 10 Q

PQRS is the base of this pyramid and is a square. ASP, APQ, AQR and ASR are the four triangular faces. How do we find (1) the measure of AQ=AP=AS=AR?; and (2)the area of this pyramid?

We are given that the altitude AB measures 12 and the sides of the square base are 10 (therefore, BC=5). Using Pythagorean Theorem, let us compute the value for AC.

AC= √AB²+BC² = √144+25=√169=13.

AC is the altitude of the triangular face AQR. We know that QC = 5. (Base is 10 on each side).

AQ = √QC²+AC² =√13²+5²=√194..(1)

Now, we need to calculate the surface area of the pyramid.

To recap, the surface area of the pyramid is the surface area of the base plus the surface areas of the 4 triangular surfaces.

The Area of the square base is 10²=100

The surface areas of the triangular faces are the same.

Surface area of 4 triangular faces= 4.1/2. .bh

In this exercise, b is 10; the height is 13.

Surface Area of triangular faces=260

∴The Surface of the pyramid =100+200 =360 Exercise: Try this one out on your own. In the subject illustration, change the value of the base length to 14 and the height AB to 24. What are the new values for (1) AQ; (2) Surface area of the pyramid?

(Answer: (1)√674 or 26 (approx.) Note: In the test, you will not be required to find the square root of 674. In all probability, the answer choice will be √674.

(2) 896




Practice Problems in Geometry

1. An equilateral triangle of side b and a square of side s have the same area. The ratio of s2 to b2 is: (A) 1 (B) \/2/3 (C) \/3 (D) \/3/2 (E) \/3 / 4 We know that the area of the square is s2 , where s is the side of the square. The area of a triangle is 1/2. Base . height. How do we express the height of an equilateral triangle in terms of its side value? 30o 30o b b 60o 60o b/2 b/2 As we can see, the altitude divides the base in two equal halves: b/2 and b/2 We also notice that the altitude converts the equilateral triangle into two 30-60-90 degrees triangles, with a value of b for the side opposite 90o angle, and a value of b/2 for the side opposite 30o angle. What should be the value for the altitude, the side opposite the 60o angle? \/3. b/2 (Remember: the sides of a 30-60-90 degree triangle are in proportion 1: \/3 : 2 ) We now know the value for the altitude in terms of b, the value for the sides of the equilateral triangle.

Area of equilateral triangle = 1/2. Base . height = 1/2. b. \/3.b/2 =

\/3.b2/4 Since the area of the triangle is the same as that of the square, we are ready to set up the following equation: \/3.b2/4 = s2 or s2 / b2 = \/3 /

4. Choice E.

2. A square of side s is inscribed in a circle of diameter d. The area of the circle in terms of s is: (A) π.s2 (B) π. s2 /2 (C) π.s2 / 4 (D) π. s2 /8 (E) Cannot be determined Solution: We know that when a square is inscribed in a circle, then the diameter of the circle is the diagonal of the square. Is s is the side of the square, then the diagonal is \/2.s because the diagonal divides the square into two 45-45-90 triangles, and the sides are in proportion 1:1:\/2. If the sides are s and s, the diagonal must be \/2.s Therefore, diameter of the circle must be \/2.s as well. Area of the circle in terms of diameter = π d2 /4 = π (\/2.S)2 /4 = π.4.s2 /2 = π.s2 /2. Choice B Note: if the problem asks you for the ratio of the area of square to that of circle, then you would set it up as follows: Area of the square = s2 Area of the circle = pi.d2 /4 = pi.s2/2 Ratio of area of square to that of circle = 2/pi (Take the ratio of s2 and π.s2/2 and we have s2

canceling out leaving 2 on the numerator and π on the denominator).




3. A cube’s volume is 3 times its surface area. What is the volume of this cube? Solution: If we know the relationship of a cube’s volume to its surface area, we can determine all the properties of the cube. If a is the side of the cube, we have: Volume = a.a.a = a3 And Surface area = 6. a2

If the volume is 3 times the surface area, we get: Volume = 3 X Surface Area i.e. a3 = 3. 6a2 or a = 18 Let us plug in this value for a into the “formula” for volume, a3 to get: Volume = 183 = 5,832 cubic units. 4. The radius and the height of Cylinder X are each twice the values of those of Cylinder Y. What is the ratio of volume of X to that of Y ? Solution: Cylinder X Cylinder Y Radius 2r r Height 2h h Volume π.(2r)2.2h π.r2.h (pi.(rad)2.(height)) Volume 8. πi.r2.h π.r2.h We notice that cylinder X has 8 times the capacity of cylinder Y. The ratio of volume of X to that of Y is: 8. π.r2.h/ π.r2.h = 8:1 = 8 Note: You can also choose Radius =1 and Height = 1 for one cylinder, and R=2, and H=2 for the other, and compute the same ratio, without having to mess with stuff like π and R.

5. Cylinder X has twice the capacity of cylinder Y. If cylinder X is 1/3 filled with oil, and cylinder Y is 3/4 filled with the same oil, and if all the oil in Y is poured into X, to what capacity is X filled after this transfer? Solution: We have: X = 2.Y or Y = 1/2. X If Y is 3/4 filled, then 3/4 Y is the volume of oil in Y. And 3/4 . Y = 3/4. 1/2. X = 3X/8 If all the liquid from Y is transfered to X, we are, in effect, adding 3/8 more to the volume of liquid already in X (=1/3. X). Volume of liquid in X after the transfer = 1/3. X + 3/8. X = 17/24. X Cylinder X will be filled to 17/24 of its capacity after all the liquid in Y is poured into X. 6. The radius of circular rim X is twice that of circular rim Y. If the circular rims X and Y move the same distance per second, what is the ratio of the number of revolutions per second that X makes to the number of revolutions per second that Y makes? Solution: Circular Rim X Y Radius 2R R Circumference 2.pi(2R) 2.pi.(R) Number of revs per second N1 N2 Distance Moved

per second 4.pi.R..N1 2.pi.R.N2 (Circumference X No. Of revs/sec) Since both X and Y move the same distance per second by making different number of revs per second, we have 4.pi.R. N1 = 2.pi.R. N2

Or N1 / N2 = 2/4 = 1/2




7. “What is the greatest length of a line segment that can be placed inside a rectangular box of dimensions 4 inches by 6 inches by 8 inches?” The greatest length of a line segment that can be placed inside a rectangular box is the same as the greatest length of the box’s diagonal. The greatest diagonal of a rectangular box is given by the formula: SQRT(L2 + W2 + H2) = SQRT(42 + 62 + 82) = SQRT(16 + 36 + 64) =SQRT(116) =SQRT(4.29) =2•SQRT(29) The greatest length of a line segment that can be placed inside the rectangular box of the specified dimensions is 2•⊕29 8. “What is the greatest length of a line

segment that can be placed inside a circle of radius 5 inches?”

The greatest length of a line segment that can be placed inside a circle is the same as the length of the diameter. The diameter of the specified circle is 10 inches, and that is the greatest length of a line segment we can place inside the circle. The diameter is also called the longest chord of a circle. 9. “Samantha wants to put up a fence around

the rectangular backyard of length 35 feet and width 20 feet, and she wants to leave one side of width 20 feet unfenced. How many feet of fencing does she need to erect?”

The length of the fence is the length of the perimeter minus 20 feet. Or, it is twice length plus one width for a total of 90 feet. That is how many feet of fencing she needs to put up.

10. “If the ratio of surface area of cube X to that of cube Y is 1/16, what is the ratio of the capacity of X to that of cube Y?”

IF X is the side of cube X and Y that of cube Y, then the ratio of surface areas is given by the following set up: 6X2 /6Y2 = 1/16 or X2 / Y2 = 1/16 Let us take square root on both sides to get: X / Y = ¼ Capacity is side cubed, and the ratio of capacity of X to that of Y is: X3 / Y3 = (X/Y)3 = (1/4)3 = 1/64 The ratio of capacity of X to that of cube Y is 1/64. Notice how this problem combined geometry and Exponents in one go. 11. “A Rectangular frame of size 18 inches in

length and 12 inches in width encloses a rectangular picture whose area is ½ the total area of the frame on which it is mounted. Also, the length and width of the picture are in the same proportion as the overall length and width of the frame. What is the length of the picture in inches?”

The “picture” painted by this word problem will look like this: 12” 18” The pink outline shows the picture inside tha frame. The green band shows the area of the frame surrounding the picture. Let L be the length of the picture, and W its width. We have two pieces of information, which translate into the following two equations: L•W = ½ •18•12 = 108 And L/W = 18/12 = 3/2, or W = 2L/3 Let us write 2L/3 for W in the first equation and get: L•2L/3 = 108 Or, L2= 162 or L = √162. = √81•2 = 9√2




12. What is the shaded area shown within a

circle of radius 6 inches? We notice that the shaded area represents 100 degrees out of a total of 360 internal degrees in a circle. Therefore, the shaded area will be 100/360th of the total circular area. Area of the circle = pi.radius2 = pi.62 = 36pi Area of the shaded part = (100/360) • 36pi = 10pi 13. “A rectangular table top consists of a piece

of laminated wood bordered by a thin metal strip along its four edges. The surface area of the table top is X square feet, and the total length of the strip before it was attached was also X feet. IF the table top is 3 feet wide, what is the length of the table top?”

L feet 3 ft The Blue line around the rectangular table top shows the strip that was attached to the table top. Let L be the length of the table top. We also notice that the length of the strip corresponds to the perimeter of the rectangular table top. Perimeter of rectangle = 2•(L + W) Because W = 3, Perimeter = 2•(L + 3) = X We also know that the area of the table top is X. Area is L•W = 3L = X

We can see that X = 2(L + 3) And X = 3L Therefore, we can set up 2(L + 3) = 3L Or, 2L + 6 = 3L Or, L = 6 feet. The length of the table top is 6 feet. 14. If a square region has area X square

feet, what is the length of its diagonal in feet?

We know that the area of a square in terms of its diagonal is ½ •(diagonal)2 = X Or, diagonal2 = 2X Or, diagonal = SQRT(2X) Side of the square is SQRT(X) 15. The size of a television screen is

specified by the length of its diagonal. For example, a 15 inches television has diagonal measuring 15 inches. How many square inches more is the area of a 19 inch square, flat television screen than that of a 21 inch square, flat television screen?

The area of a square region = ½ •(diag)2 Area of 19” diagonal square = ½ •192 Area of 21” diagonal square = ½ •212 Difference in area = ½ •212 – ½ •192 = ½ (212 – 192) = ½ •(21+19)(21-19) (We used the algebraic formula here. Can you notice it? A2-B2 = (A+B)(A-B) ) Difference in area = ½ •40•2 = 40 sq.inches

50o




FORMULAS IN GEOMETRY REFER TO THE GEOMETRY FORMULAS FILE FOR A DETAILED DISCUSSION ON THE FORMULAS AND RULES RELEVANT TO GEOMETRY




CCCoooooorrrdddiiinnnaaattteee GGGeeeooommmeeetttrrryyy

You can expect with reasonable certainty at least one question from this chapter in each of the two problem solving sections and in the Data Sufficiency section of the GMAT.

The coordinate geometry deals with information in the coordinate plane. You have a horizontal line called the x-axis and a vertical line called the y-axis. The point at which these axes meet is called the origin, usually designated O. The axes divide the plane into 4 Quadrants, designated I, II, III and IV.

y-axis 3

II A 2 P I

1 x-axis

-4 -3 -2 -1 0 1 2 3 4

III -1 IV

C -2 B

-3

Each point in the plane has an x-coordinate and a y-coordinate. A point in the plane is denoted and identified by an ordered pair of numbers, conventionally denoted (x,y), where the x-coordinate value is the first number and the y-coordinate value is the second number.

The point P in the Quadrant I is denoted by (2,2). Point B is specified by the coordinates (3, -2). Point C is specified by the coordinates (-1,-2).

In Quadrant I, both x and y values are positive. In Quadrant II, x is negative and y is positive. In Quadrant III, x and y are both negative. In Quadrant IV, x is positive and y is negative.

To find the distance AB, use Pythagorean theorem by completing a triangle as shown by the dotted lines. Coordinate geometry is conducive to easy measurement: We can see that AC=CB are both 4 units. Applying the Pythagorean theorem, we can conclude:

AB² = AC² + CB² = 4²+4²=32

AB = √32 =4√2

Exercise:

II I

III IV

In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x-3y ≤ -6 ?

(A) None (B) I (C) II (D) III (E) IV

We are required to find a quadrant, if one exists, such that no point in that plane will satisfy the inequality 2x-3y≤ -6.

Working with -6 on the right side of the inequality is a little clumsy. So we multiply both sides by -1 and reverse the order of inequality as follows: 3y-2x≥6.

To get a handle on a problem of this type, we have to use real-life numbers




into the inequality and see if the condition is met.

In Quadrant I, both x and y are positive. We can always choose a value of x and y such that 3y-2x will be greater than 6. For instance, if we chose y=4 and x=1, 3y-2x has a value of 10, which is greater than 6. Therefore, there does exist at least one point in Quadrant I that can satisfy the inequality. Therefore the answer choice (B) is not valid. We move on.

In Quadrant II, x is negative and y is positive. If we chose a point in Quadrant II : (-1,3), then we have for 3y-2x a value of 11, which is greater than 6. Therefore answer choice (C) is not valid.

In Quadrant III, both x and y are negative. If we chose a point (-5,-1), then we have for 3y-2x a value of 7, which is greater than 6. Therefore, we can find at least one point in Quadrant III such that the inequality is true. We can eliminate choice (D). We move on.

In Quadrant IV, x is positive and y is negative. No point exists in this quadrant such that the inequality is satisfied, because there is a negative sign in front of a positive value x and a positive sign in front of a negative value y in the inequality being tested: 3y-2x ≥ 6. For all values of x and y, the expression on the left side of the inequality will always be negative in this Quadrant and the inequality cannot be satisfied.

Therefore, we pick Quadrant IV as the one containing no point to satisfy the inequality, the correct answer. The correct answer choice, therefore, is (E).

Now, try to do the following on your own and figure out which Quadrant will not have a point such that the inequality -5x + 3y -8 will be satisfied.

II y I

x

III IV

The Correct answer is Quadrant II. Use the same argument that we have discussed earlier to arrive at this conclusion.

If you had difficulty understanding this concept, let us know.

The fastest way to deal with such problems is to identify the quadrant where the signs attached to x and y values are such that the inequality will not be satisfied. In our problem, If -5x + 3y is required to be less than or equal to -8, then look for a quadrant where x values will be negative and y values positive so that the inequality expression -5x + 3y will always be positive. Only quadrant II has -ve xvalues and +ve y values.




COORDINATE GEOMETRY(Contd.) The equation of a straight line in an x-y plane is defined by the equation : y = mx + c Where m is the slope of the line, and c is the y intercept or the value of y when x is 0. In other words, c is the y coordinate value of the point on the y axis where the straight line meets. How do we find the equation of a straight line in an x-y plane? We require to be given information pertaining to the slope and the y intercept. In the alternative, we may simply want to know the x and y values of at least 2 points on the straight line. Let us say that we have two points (5,27) and (20,18) in an x-y plane and want to find out the equation for a straight line connecting these two points. How do we proceed? We start at the starting point, and say that “Let the equation for the straight line be y = mx + C We will find out the values for m and c by using the information about the points that lie on this line. If (5,27) is a point on this line, then we can write 5 for x and 27 for y. We get 27 = m.5 + c ……………… (1) Similarly, if (20,18) is another point on the same line, then we can write 20 for x and 18 for y. We get 18 = m.20 + c……………….(2) What do we see with this picture ? We have two equations and two variables. Can we get the values for m and c ? You bet. Let us subtract equation 2 from 1, so that we can get rid of c and get an equation in terms of m only. We get = m (5-20) Or 9 = -15.m Or m = -9/15 = -3/5 ……(3) Now that we have determined the value for m, let us get the value for c by plugging in the value for m in either equation – 1 or 2. We get : 27 = -3/5 . 5 + c = -3 + c Or, c = 30 ……………………… (4)

What is the equation for the straight line in the x-y plane connecting the points (5,27) and (20,18) ? Y = -3/5 . x + 30 or 3x + 5y = 150 How will this line look in an x –y plane ? +y (0,30) -x + x (50,0) - y As you can see, the straight line will pass through quadrants I, II and IV, but will not pass through quadrant III. How did we get the points (0,30) and (50,0) ? We get (0,30) by writing 0 for x in the equation. A point that lies on the y axis will have x value equal to 0. Similarly, we got (50,0) by writing 0 for y in the equation, because a point that lies on the x axis will have y value equal to 0. If in the GMAT, you got a question that asks: “The straight line 3x + 5y = 150 in an x-y plane will pass through all of the following quadrants except: (A) QI (B) QII (C) QIII (D) QIV (E) None” How will you answer this question ? We can answer this question by constructing a straight line in an x-y plane as shown or by reasoning that the x and y values in QIII are both negative so that 3x + 5y will be a negative value for all points in quadrant III. Therefore, the straight line defined by the subject equation will NOT pass through quadrant III. The slope of a straight line can also be computed from the following formula: “If (x1, y1) and (x2, y2) are two points on a straight line in an x-y plane, then the slope of the line is: (y1 - y2) (x1 - x2) Remember to use the same order for x and y values: If you start with y1 , then you should start with x1 for the denominator. SEE ADDITIONAL DISCUSSION OF THE RULES IN GEOMETRY FORMULAS FILE.

SLOPE defines the constant ratio of the vertical height to the horizontal distance of any point on the straight line. The ratio of the vertical height of any point on the straight line to the horizontal distance of that point from the end of the line where the line meets X axis is a constant. A B C D E F G AD/DG = BE/EG = CF/FG = SLOPE (m)

By setting up an equation for a straight line connecting two specified points, we are able to get the handle on many more points on the straight line.




Now, try this question for size: “What is the equation for a straight line in an x-y plane connecting the points (-3, 45) and (–12,30) ?” Also, determine what quadrant, if any, the line will NOT pass through. Answer:(The equation is: y = 5x/3 + 50 and the line will NOT pass through QIV) You should also expect to get problems in GMAT testing your ability to combine concepts from more than one area, say triangles and co-ordinate geometry. Let us take a look at such a problem: y D -x +x A(-6,0) B C (24,0) -y BD = BC The above figure shows an isosceles triangle ADC such that AD = DC. The triangle lies in an x-y plane. What are the co-ordinates of point D? What do we know about the properties of an isosceles triangle? We know that the vertical line drawn from the vertex where the two sides of equal value meet to the opposite side divides the opposite side (base) in two equal halves. In our problem, AB = BC = ½.. AC Since A lies 6 units to the left of the center of the x-y plane, and C lies 24 units to the right of the center, we determine that AC must be 30, and that AB = BC = ½.. 30 = 15 IF AB is 15 units, and knowing that A lies 6 units to the left of 0, we conclude that B MUST lie 15-6 = 9 units to the right of 0. Which means that the co-ordinates of B are (9,0) where 9 is the distance from the center along the x axis and 0 is the y coordinate value of point B. The problem also specifies that BD = BC = 15.

Which means that the (x,y) values of D must be (9,15). Notice how this problem combined the properties of triangles and those of co-ordinate geometry, and forced us to use a fair bit of reasoning to arrive at the (x,y) values of the point D ? Another problem frequently encountered in the GMAT goes like this: “If (a, 2b) and (a + 4, 2b +k) are two points on a straight line defined by the equation x = -2y + 7, what is the value of k ?” You can solve this problem in two different ways.

APPROACH 1 What is the slope of the line defined by the equation x = -2y + 7 ? We need to express the equation in the form y = mx + c. We get -2y = x - 7 Or -y = ½.. X - 7/2 Or y = - 1/2 .x + 7/2 We conclude that the slope is - ½ and the y intercept is 7/2. Now, let us take a look at the points on the straight line: (a, 2b) and (a+4, 2b + k). We can see that for every 4 units movement of x value, the y value moves by k. The slope of the line in terms of k must be the ratio of the y increment to that of x increment. Slope of the line = k/4 = - ½ or k = -2

APPROACH 2 If (a,2b) is a point on the straight line x = -2y+7, then we can write a for x and 2b for y. We get a = -2. 2b + 7 .................. (1) Or a = - 4b + 7 ................ (2) Similarly, if (a+4, 2b+k) is another point on the same line, we can write a+4 for x and 2b +k for y. We get a + 4 = -2 (2b + k) + 7 ............. (3) Or a + 4 = -4b - 2k + 7 ................. (4) Subtract equation (2) from (4) to get a+ 4 - a = -4b - 2k + 7 -(-4b + 7) 4 = -4b - 2k + 7 + 4b - 7 Or 4 = -2k or K = -2 You can use either approach, but if you are not sure about determining the slope of the equation correctly, then use the approach 2 suggested.




PROBLEM: B(0,30) A(50,0) How many points exist on the line AB between A and B, inclusive, such that the x and y co-ordinate values of the points are integers? We recognize that the slope of the line is -3/5, (how do we get this? Use the formula for computing slope using two points) which means that you can go down y axis in increments of 3 and along x axis in increments of 5. If you went down along y-axis in steps of 3, and drew a line to the sloping line, you will find a point on the sloping line that has both x and y values integers. You can go down along the y axis in steps of 30,27,24,21,18,15,12,9,6,3,0 or there are 11 points along AB with integer values for (x, y).. Or, you can move along the X-axis in steps of 5: 0, 5,10,15,20,25,30,35,40,45,50 . 11 points. The other approach is to set up the equation for the straight line. We notice that the y intercept is 30, and the slope is –3/5. The equation is: y = -3/5. X + 30 If x and y are required to be integers, then we must find x values that are multiples of 5 between 0 and 50. The number of integers that are multiples of 50 between 0 and 50, inclusive, is 11.

ASSIGNMENT:

A is a point in an x-y plane with (x,y) values (-3, 56) and B is another point in the same plane with coordinate values (24,-7). The straightline connecting A and B intersects the Y axis at C and the X axis at D. 1. What is the length of the straightline connecting points A and B? 2. What is the equation for the straight line connecting A and B? 3. What is the slope of the line connecting A and B? 4. What are the coordinates of C and D? 5. How many points exist along the line segment CD such that the x,y values of each of these points will be integers? 6. If (6,35) and (12, 35+k) are two points on the line AB, what is the value for k ? A(-3,56) C(?,?) D(?,?) (24,-7) B




COORDINATE GEOMETRY ASSIGNMENT EXPLAINED

A is a point in an x-y plane with (x, y) values (-3, 56) and B is another point in the same plane with coordinate values (24,-7). The straight-line connecting A and B intersect the Y-axis at C and the X-axis at D. 1. What is the length of the straight-line connecting points and A and B ? 2. What is the equation for the straight- line connecting A and B ? 3. What is the slope of the line connecting A and B ? 4. What are the coordinates of C and D ? 5. How many points exist along the line segment CD such that the x,y values of each of these points will be integers ? 6. If (6, 35) and (12, 35+k) are two points on the line AB, what is the value for k ? A(-3,56) C(?,?) D(?,?) M (24,-7) B 1.The length of the straight-line connecting points A and B is obtained by using the Pythagorean theorem:

AB2 = AM2 + BM2 (1) How do we get the values for AM and

BM? The value for AM is obtained by subtracting the y coordinate values of A and B: (Notice that the y value of B is the same as the y value of M)

Similarly, we get the distance BM by subtracting the x values of A and B. (Notice that the x value of M is the same as the x value of A.)

We get AM = 56 – (-7) = 63 We get BM = 24 – (-3) = 27 Let us plug these values in the equation

(1) above, and get the value for the length of AB:

AB = √63.63 + 27.27 = 9√58 Answer

2. The equation for the straight line connecting A and B is given by the standard form: y = mx + b where m is the slope and b is the y-intercept.

The slope of the line is given by the ratio: (y1-y2)/(x1-x2) Where x1 and y1 values are the coordinates of any of the points, A or B, and x2 and y2 values are the coordinates of the other point.

Slope of the line connecting A and B is : (56 – (-7)) / (-3 – 24) = -63/27 = -7/3 The equation for the straight-line can now be written with this slope value as: Y = - 7/3 x + b How do we get the value for b, the y-intercept ? We can use any of the x, y values of point A or B to get the value for b. Let us write –3 for x and 56 for y to get: 56 = - 7/3. (-3) + b = 7 + b Or b = 49 Therefore, the equation for the straight-line connecting A and B is:

Y = - 7/3 X + 49 3. The slope of the line is -7/3. 4. The coordinates of C are (0,49).

Notice that the y-value of C is the y-intercept, b, value determined above.

5. To get the coordinates of D, we have to write y = 0 and find out the value for the x-coordinate. Why do we do this? Because D lies on the X-axis, where the y -value is zero.

0 = -7/3. X + 49 or X = 21. Therefore, the coordinates of D are (21,0)

6. We notice that the slope of the line is –7/3. Which means that we can go down y axis from C in steps of 7 or move along x-axis in steps of 3 from 0 to D, and that is how many points we will find on the line between C and D with (x,y) values whole number values. We can find (0,49), (3,42), (6,35), (9,28), (12,21), (15,14), (18,7), and (21,0) on the line segment between C and D – a total of 8 points.




7. If (6,35), and (12, 35+k) are two points on

the straight-line AB, the slope in terms of k is given by the expression: (35-35-k) / (6-12) = k/6 But we know that the actual slope is - 7/3, and we can write: k/6 = -7/3 or

k = -14 QUADRATIC

FUNCTION IN AN X-Y PLANE

We have seen that a linear equation of the type Y=mX + b gives rise to a straight line. IF we have a quadratic function of the type Y = X2 – 4, we will get a parabola if we plot X and Y points for various values of X and Y. If Y = X2 – 4, then the curve will be a parabola that looks as follows:

Notice that Y has been replaced with F(X) because Y is in face a function of X even in a linear situation. Instead of plotting Y values in the Y axis, we plot the F(X) values.

If we are given a linear equation of the type F(X1)= mX+b and a quadratic function of the type F(X2) = X2- 4, we

can determine what common points exist for the two curves. Consider the following problem:

“If F(X,F(X)) is a point common to both F(X1) = X1

2 + 8, and F(X2) = X2 + 10, then F(X,F(X)) could be what?”

The common point must satisfy both equations. Therefore, the common point, when substituted into each of the two equations, will yield:

F(X) = X2 + 8 AND

F(X) = X + 10

As we can see, both equations are for the common F(X), and we can write: X2 + 8 = X + 10

Or, X2 – X – 2 = 0 or (X-2)(X+1) = 0

X = 2 or X = -1

The corresponding values for F(X) are:

If X = 2, then F(X) = X+10 = 12

IF X = -1, then F(X) = X+10 = 9

Therefore, F(X,F(X)) could be (2,12) or (-1, 9). The parabola and the straight line will intersect at these two points. In the final analysis, a problem of this type is simple algebra asking you to set a quadratic expression equal to a linear expression, and then to solve for X. Once we get X, find the corresponding value for the other variable such as F(X).

-4 -3 -2 -1 0 1 2 3 4X VALUES

-5-4-3-2-10123456

F(X)

PARABOLAQUADRATIC FUNCTION

IF X=0, F(X) = 2 OR -2




The Data Sufficiency section of the GMAT does not require you to solve a problem presented; but, instead, to determine at what stage or point do you have sufficient information pertaining to the problem so that the problem can be solved.

The Data Sufficiency section will require all the information that you have learned in this module thus far; the only difference is that, in data sufficiency, you are going to be given information in a piece-meal fashion, and you will be required to do the following:

Be sure to make a note of any standard properties pertaining to the problem. For instance, if the problem is about a 30-60-90 triangle, you should know that the sides are in a standard proportion, and you will be required to use this knowledge in the context of determining whether you can answer the question.

Be sure to make a note of any information that is furnished along with the question itself. You are required to combine this information with the additional information in statements 1 and 2 independently, and determine whether a unique solution is possible.

In this Section of the test, you will be given a question with two accompanying

statements: 1 and 2. Your task is to determine if you can answer the question asked based on the information presented in statement 1 alone or statement 2 alone or using the information presented in both the statements.

We will illustrate the appropriate strategy for this section with a couple of examples.

1. If √x / y = n, what is the value of x?

(1) y•n=10 (2) y=40 and n=1/4

Let us rearrange the given information as follows: √x = n•y = y•n (commutative rule of algebra)

Let us look at the statement 1 now. It says that yn=10 which is equal to √x. We have sufficient information to determine the value of x. The answer choice will be either A or D. Let us make a check mark against (1) and move on to statement (2).

When you read statement (2), do not let the information presented in (1) influence you. In other words, put your blinkers on with regard to statement 1 while reading statement 2.

Statement 2 gives values for y and n from which we can determine the value of x. Statement 2 alone is sufficient too.

Therefore, we have a situation where Each statement ALONE is sufficient to answer the question. We pick answer choice D as the correct one.

Remember: Even though we reached a point where we could answer the question based on information in statement 1 alone, the test directions




require you to analyze statement 2 before you decide if you must pick A or D.

Let us do one more to clarify this section:

“Is the prime number p equal to 37 ?”

(1) p = n² + 1 where n is an integer.

(2) p² is greater than 200.

Let us look at statement 1 alone. Since n is an integer capable of taking on any value, p may be a prime number if n=6 but if n=5, p is not only not a prime number but is nowhere close to 37.

We conclude that statement 1 alone is not sufficient and we make a cross mark against statement 1.

Let us look at statement 2 and consider it solely on its own merits.

The fact that p² is greater than 200 indicates that p is greater than 14. Which means that p could be 17 or 19 or 23 or 29 or 31 or 37 or 41. Since there is no unique value for p we can determine based on the information provided, we conclude that statement 2 alone is not sufficient. We put a cross mark against 2.

Can we combine the information in 1 and 2 and then be in a position to determine if p is 37 ?

p could be a prime number with a value of n=4 in the first statement. Such a value will satisfy the requirement in statement 2 as well. Therefore we cannot conclude on the basis of combined information in both statements that p is 37.

We must pick E as the answer choice, meaning that statements (1) and (2) together are not sufficient.

Let us try one more for the road:

“Is x < 0 ?”

(1) -2x > 0 (2) x³ < 0”

Let us redefine the question. The question is: Is x a negative number?

Consider statement (1) alone. For -2x to be a positive number, x must be a negative number. We can, therefore, conclude on the basis of statement 1 alone that x is less than 0. We put a check mark against (1). The answer choice is A or D.

Now we consider statement 2 alone. That x³ is a negative number (less than 0) indicates that x must be a negative number.(Cube of a positive number is positive and a negative number is negative). Therefore statement 2 alone is sufficient to answer the question. We place a check mark against statement 2 also.

What do we have now? We have a situation where either statement alone is sufficient and we pick D as the answer choice.

We will deal with specific Data Sufficiency reasoning techniques later on in Module II supplement, which follows after the Problem Solving Exercises.

Now, go to the assignment involving problem solving exercises from recent GMAT test sections.




ou are going to be constantly tested with regard to your ability to reduce word descriptions of

problems to mathematician-speak, a.k.a equations, and solve for as yet unknown quantities. This Section of exercises will test your ability to do precisely that.

Example: Of the three varieties of lettuces, the Romaine variety has 10 times the nutrients content of the Iceberg variety and 5 times the combined nutrients content of the Iceberg and Boston varieties. If the Boston variety has the nutrients content of 20, what is the nutrients content of the Romaine lettuce?

Step 1: Reduce the verbosity into equations: Let R, I and B represent the nutrients content respectively of the Romaine, Iceberg and the Boston varieties of Lettuce.

R = 10. I --------- (1)

R = 5. (I + B) ------(2) & B = 20 --(3)

Step 2: Reduce the number of equations by substituting known values into equations. When we substitute 20 for B in (2), we get R = 5(I + 20) -------(4)

We have 2 expressions for R (1 and 4) and they are equivalent. We get: 10.I = 5 (I+20) = 5.I + 100 or I = 20.

From (1), we calculate R = 10.20 = 200

You are going to need this skill to do well in the GMAT. Get used to it.

Please complete the following assignment.

Solving with Equations: Assignment

1. X is twice the value of Y and three times the difference of Z and Y (Z > Y). If Z is 4, what are X and Y?

(Answer: x=24/5 and y = 12/5)

2. In a weight-lifting competition, King Kong had the weights of his first three lifts total 900 pounds. If the combined weight of his second and third lifts is 50 pounds less than three times the weight of his first lift and the weight of his third lift is 50 pounds more than his first lift, what was the weight of King Kong’s second lift?

<answer: x=237 ½ y=375 z=287 ½ >

3. The average (arithmetic mean) of x, y and z is 6. The product of the three

Y




positive numbers is 192. If x is 4, what is yz? What is y+z?

<Answer: yz=48; y+z = 14>




4. The equation N = d.q + r is used to denote the relationship between the dividend(N),divisor(d), the quotient(q) and the remainder(r).

Express the following numbers as an equation with the relationship shown above.

(A) N = 78 d = 9

<78 = 9•8 + 6>

(B) N = 108 d = 7

<108 = 7•15 + 3>

(C) N = a d = 6

< a = 6•q + r>

5. A jar contains candies in three colors: Red, Blue and Green. There are exactly 120 candies in the jar. If there are twice as many red candies as green ones and three times as many blue candies as green ones, how many red candies are there in the jar?

<Red=40; Green=20; Blue = 60>

6. If John’s age n years ago was m, how old will John be p years from now? (Express John’s Age, J = )

<J = m+n+p>

MODULE II SUPPLEMENT

1. A recreation club has 40 members, How many male members does the club have? (1) Exactly 1/4 of its male members take swimming lessons. (2) Exactly 1/5 of its female members take swimming lessons. In a data sufficiency problem, you are required to determine at what point you have sufficient information to answer the question with a unique value. Which means that each statement must be good for a unique value for the information sought: How many male members are there in the club? Let us first examine the problem statement itself before we move on to examine the statements (1) and (2). What information do we have from the stem? Simply this: There are 40 members in the club. We have no other information given. Let us move on to examine statement (1). Statement (1) tells us that exactly 1/4 of the club’s members take swimming lessons. What do we understand this statement to mean? Simply this: The number of male members in the club is a multiple of 4. Which means that the possible numbers are 4, 8, 12, 16, 20, 24, 28, 32, and 36. Why is 40 not a possible number for male members? Let us be chivalrous and leave some room for the female members, OK? Do we have a unique value generated from information in statement (1) ? Nope. What do we do at this stage of the game? We simply eliminate statement (1) as sufficient by itself and must move on to examine statement (2). What possible answer choices do we have at this juncture? B, C, or E.




Let us make the following notations against the question so that we will stay focused as to our possible choices: 1. A recreation club has 40 members, How many male members does the club have?

X (1) Exactly 1/4 of its male members take swimming lessons.

B,C,E (2) Exactly 1/5 of its female members take swimming lessons. What did we just do? We put a cross mark against statement (1) meaning that we have eliminated that as a possible choice and even as we begin to examine statement (2), we want to keep in focus that our possible choices at this point are B,C, or E. So we write B, C,or E against statement (2). Let us examine statement (2). Statement (2) tells us that exactly 1/5 of the female members are taking swimming lessons. What do we understand this statement to mean? The number of female members in the club is a multiple of 5. Which means that the possible numbers are 5, 10, 15, 20, 25 30, and 35. As a corollary, we understand that this statement means that the possible values for male members are 35, 30, 25, 20, 15, 10, or 5. Do we get a unique value for the male membership from this statement? Hardly. We have a range of values for male membership which is not good enough. So what do we conclude? We conclude that statement (2) alone is not sufficient to answer the question: How many male members are there in the club? Now, what possible answer choices do we have? C or E. Let us cross out the B against Statement (2) to indicate that we have eliminated B is no longer a viable answer choice.

1. A recreation club has 40 members, How many male members does the club have?


B,C,E (2) Exactly 1/5 of its female members take swimming lessons. When we combine the information (1) and (2), what do we see? Statement (1) tells us that the possible values for male membership are 4, 8, 12, 16, 20, 24, 28, 32, and 36. Statement (2) tells us that the possible values for male membership are 35, 30, 25, 20, 15, 10, and 5. Do we see a common number in both statements? Yes, we do. 20. What does that mean? Simply this: When we combine statement (1) and (2) we can determine that the number of male members in the club is 20, a unique value. Therefore the combined information is good enough to answer the question with precise information and we pick C as the answer choice. 1. A recreation club has 40 members, How many male members does the club have?


B,C,E (2) Exactly 1/5 of its female members take swimming lessons. Do you understand the reasoning process involved with data sufficiency problems and the procedure involved in picking the correct answer choice?




Remember: Each statement alone or together must be good for a UNIQUE value for the information sought. If you get a range of values from the statements, either alone or together, then you must choose E. Remember the following drill: You pick Answer A if statement (1) alone generates a unique value for the information sought but statement (2) alone does not. You pick Answer B if statement (2) alone is good for a unique value for the information sought but statement (1) alone is not. You pick Answer D if statement (1) and (2), each considered alone, are good for a unique value for the information sought. You pick Answer C if statement (1) and (2) together generate a unique value for the information sought. You pick Answer E if none of the above is true. That means that the statements (1) and (2), either alone or together, cannot generate a unique value for required information.




Let us illustrate the process and the procedure with another data sufficiency problem. 2. A jar contains 30 marbles, of which 20 are red and 10 are blue. If 9 of the marbles are removed, how many of the marbles left in the jar are red? (1) Of the marbles removed, the ratio of the number of red ones to the number of blue ones is 2 : 1 (2) Of the first 6 marbles removed, 4 are red. What is the information we are seeking? How many marbles left in the jar are red after removing 9 marbles ? Before we go to examine the statements, remember that each statement alone or together must be good for a unique value for the information sought. Let us examine statement (1) The ratio of red ones to the blue ones is 2:1 of the 9 marbles removed. Which means that 2/3 of the marbles removed were red and 1/3 of the marbles removed were blue. (Or 6 red ones and 3 blue ones). Remember: We do not have to bother to compute how many red ones were removed etc. Suffice it to know that the information given is good for a determination of the value. Once we can determine from the information how many red ones were removed, we can determine how many were left behind in the jar. So what do we conclude? That statement (1) is sufficient for the purposes of answering the question with a unique value. What are the answer choices we have at this point in time? A or D. Remember: We still have to examine statement (2) to determine if it can generate a unique value for the information sought too, in which case the choice will be D. So we proceed to make the following markings on the test booklet against the statements.

2. A jar contains 30 marbles, of which 20 are red and 10 are blue. If 9 of the marbles are removed, how many of the marbles left in the jar are red? A , D (1) Of the marbles removed, the ratio of

√ the number of red ones to the number of blue ones is 2 : 1 (2) Of the first 6 marbles removed, 4 are red. Let us proceed to examine statement (2) now. From this statement we know that the first 6 marbles removed had 4 red ones. We do not know what the next 3 were. They could have been all blue or all red or any combination of the two color. Since we cannot precisely estimate the value for the red marbles removed - and, therefore, the value for the red ones remaining in the jar - we eliminate Statement (2) as sufficient information. In the process of eliminating statement (2) as sufficient, we have eliminated D as an answer choice. We therefore pick A and mark it on the answer sheet or use the mouse to click the answer choice (Only for GMAT-CAT) 2. A jar contains 30 marbles, of which 20 are red and 10 are blue. If 9 of the marbles are removed, how many of the marbles left in the jar are red? A , D (1) Of the marbles removed, the ratio of

√ the number of red ones to the number of blue ones is 2 : 1 X (2) Of the first 6 marbles removed, 4 are red. Let us take another question to get familiar with the reasoning process and the elimination procedure for the answer choices.




3. If w + z = 28, what is the value of wz ? (1) w and z are positive integers. (2) w and z are consecutive odd integers. What is the information being sought? The value of wz. Remember: Each statement alone or together must be good for a unique value for wz. If you get a range of possible values, then what do we do to that statement? We deep-six it; we say “good riddance to bad rubbish”; we say adios; we say sayonara and move on. Now. Let us start with the statement (1). We understand from statement (1) that w and z can have any range of positive values. For instance these are the possible values for w and z: w 1 2 ........... 25 26 27 z 27 26 3 2 1 Do you see what is wrong with this picture here? Can we generate a unique value for wz from this statement? No way, Jose. So what do we do? We conclude that statement (1) is not sufficient to answer the question precisely. We, therefore, put an X against statement (1) and write B, C, E against statement (2) to indicate that those are the possible choices we are faced with at this juncture. 3. If w + z = 28, what is the value of wz ? X (1) w and z are positive integers. B, C, E (2) w and z are consecutive odd integers. Let us move on to examine statement (2).

If w and z are consecutive odd integers and if they add up to 28, then the only set of values for w and z are 13 and 15. (If w is 13, then z is 15 or vice versa). Can we now determine the value of wz precisely? You bet. So what do we conclude? We conclude that statement (2) ALONE is sufficient to answer the questions precisely and we cross out C and E to pick B as the answer choice. We go to the answer sheet and blacken the corresponding oval, or move the mouse to click on the answer choice (GMAT-CAT). 3. If w + z = 28, what is the value of wz ? X (1) w and z are positive integers. B, C, E (2) w and z are consecutive odd integers. Let us consider another question from GMAT. 4. Last Friday a certain shop sold 3/4 of the sweaters in its inventory. Each sweater sold for $20. What was the total revenue from the sale of these sweaters? (1) When the shop opened for business last Friday, there were 160 sweaters in its inventory. (2) All but 40 sweaters in the shop’s inventory were sold last Friday. What is the information we are seeking? Revenue. What is revenue? Number of sweaters sold times the price of each sweater. What do we know from the information given to us on a silver platter? We know the selling price: $20.00. What information do we need now? We need to know how many sweaters were sold in precise numbers. You get the drift here? We have to determine ahead of time what is the nature of “sufficient” information we are seeking before we move on to examine the statements. Here the “sufficient” information is the quantity of sweaters sold and




that will help you calculate revenue using the selling price of each sweater specified. Let us examine statement (1) Statement (1) tells us that the shop had 160 sweaters when it opened its doors for business that day. Combine this with the fact that 3/4 of the sweaters in the inventory were sold, we can determine how many were sold. (3/4 of 160). Is statement (1) sufficient information? YES. So what do we do now? We write A, D against statement (1) in the test booklet to indicate that those are the choices we have at this point in time and move on to examine statement (2) 4. Last Friday a certain shop sold 3/4 of the sweaters in its inventory. Each sweater sold for $20. What was the total revenue from the sale of these sweaters? (1) When the shop opened for business A, D last Friday, there were 160 sweaters in its inventory. (2) All but 40 sweaters in the shop’s inventory were sold last Friday. What does statement (2) tell us? Remember we have to examine statement (2) independently of our knowledge of statement (1) information. Statement (2) tells us that 40 sweaters were left unsold. From the information specified in the question, what fraction of sweaters were unsold? 1/4 . How do we know this? Because the problem states that the shop sold 3/4 of its inventory leaving 1/4 unsold. If 40 represents 1/4 of the inventory, we can determine how many were sold. (3 times as many). Is statement (2) good for the information we are seeking. You bet. So what do we have now? We have a situation in which both statements standing alone are good for the information we are seeking. In this situation what is our answer choice? D.

4. Last Friday a certain shop sold 3/4 of the sweaters in its inventory. Each sweater sold for $20. What was the total revenue from the sale of these sweaters? (1) When the shop opened for business A, D last Friday, there were 160 sweaters in its inventory. √ (2) All but 40 sweaters in the shop’s inventory were sold last Friday. Let us consider one more to drive home the reasoning process before we move on to assignments. 5. What is the value of x? (1) x is a prime number (2) 31 x 37 What is the information we are seeking? Value of x . A precise value of x . No less. Let us examine statement (1) Statement (1) tells us that x is a prime number. Is there just one prime number? No. How many prime number values we can think of for possible values of x ? Endless. Since we cannot get a precise or unique value for X from statement (10, what do we do? We conclude that it is not sufficient and move on to examine statement (2). What are the possible answer choices we have at his stage of the game? B, C, or E. Let us do the drill to make sure that we stay focused on our options.




5. What is the value of x? X (1) x is a prime number B, C, E (2) 31 x 37 Statement (2) specifies a narrow range of possible values for x. But then it is a range and not a unique value. So we conclude that statement (2) alone is not sufficient to answer the question. We eliminate B as a possible answer choice and move on to examine the combined information in statements (1) and (2). The answer choices are limited to C or E at this stage. 5. What is the value of x? X (1) x is a prime number B, C, E (2) 31 x 37 When we combine statements (1) and (2), we understand that x is a prime number and in the range specified in statement (2), we have at least two prime numbers: 31 and 37. Once again, the combined information at best gives us 2 possible values for x. Is that good enough. No way. We need a single, precise, unique value for the variable from either or the combined statements. Since we do not get that, we have to conclude that E is the best choice. 5. What is the value of x? X (1) x is a prime number B, C, E (2) 31 x 37 We go and mark (or click with the mouse) answer choice E on the answer sheet (computer screen).

The reasoning process and the procedures involved in choosing and eliminating answer choices need to be mastered by you until it becomes second nature. How long does it take to do that? 30 minutes for some and 60 minutes for most others. The following assignment is intended to give you the much-needed practice with the reasoning process, and the procedure involved in choosing and eliminating answer choices.




In this assignment, explain your reasoning and the procedures for eliminating and choosing answer choices as discussed earlier in this module for each question. 1. What is the capacity, in liters, of a certain aquarium? (1) Three liters is 1/2 of the capacity of the aquarium. (2) One-half liter is 1/12 of the capacity of the aquarium. 2. In Triangle PQR, what is the measure of angle P? (1) Angle Q is a right angle. (2) The measure of angle R is 17 degrees.




3. What amount did Jean earn from the commission on her sales in the first half of 1988? (1) In 1988 Jean’s commission was 5 percent of the total amount of her sales. (2) The amount of Jean’s sales in the second half of 1988 averaged $10,000 per month more than in the first half. 4. A certain car went from one town to another without stopping. What was the car’s average speed for the trip? (1) The car traveled the 90-mile trip in 2 hours. (2) The car traveled the first 40 miles of the trip in 1 hour.

5. What is the value of x? (1) 2x + 3y = 12 (2) 5x + 7y = 29 6. Does x = y? (1) | x | = | y | (2) x2 = y2




7. Dan took 20-question multiple-choice test in psychology. If Dan answered every question, did he answer at least 12 questions correctly? (1) Dan answered fewer than 40 percent of the questions incorrectly. (2) Dan answered at least 25 percent of the questions incorrectly. 8. If a < x < b and c < y < d, Is x < y? (1) a < c (2) b < c 9. Is (3x + 8)/ (x + 2 ) an integer? (1) x is an integer (2) x = 0

10. How many people did Apex company employ in 1990? (1) The company employed 538 more people in 1991 than in 1990. (2) The company employed 20 percent more people in 1991 than in 1990. 11. Of the four numbers represented on the number line below, is r closest to zero? q r s t (1) q = -s (2) -t < q




12. Is the integer n divisible by 20? (1) n is divisible by 5 (2) n is divisible by 6 13. If Mark saved an average (arithmetic mean) of $80 per week for 3 consecutive weeks, how much did he save the second week? (1) The average amount that Mark saved per week for the first 2 weeks was $60. (2) The amount that Mark saved the first week was 1/2 the amount he saved the second week and 1/3 the amount he saved the third week.

14. If x is not equal to - 1 , which is greater, 1/(x + 1) or x / 2? (1) x ≥ 0 (2) x < 3 15. In a certain two-digit integer, the ratio of the units digit to the tens digit is 2 to 3. What is the integer? (1) The tens digit is 3 more than the units digit. (2) The product of the two digits is 54.




1. What is the capacity, in liters, of a certain aquarium? (1) Three liters is 1/2 of the capacity of the aquarium. (2) One-half liter is 1/12 of the capacity of the aquarium. Statement 1 tells us that ½ (Capacity) = 3 liters. We can determine that the capacity of the aquarium is 6 liters. Statement 1 alone is sufficient to answer the question. We still need to examine statement 2 independently, and determine whether we can answer the question using statement 2 alone also. Our choices narrow to A or D. Statement 2 tells us that 1/12 (Capacity) = ½ liter. We can determine that the capacity of the aquarium is 12 liters. Statement 2 is also independently sufficient to answer the question. We must pick Choice D. 2. In Triangle PQR, what is the measure of angle P ? (1) Angle Q is a right angle. (2) The measure of angle R is 17 degrees. We need to use our knowledge of the properties of a triangle while examining the statements. If we have information about 2 angles, we can determine the third angle value. Statement 1 tells us that Q is a right angle. The other two angles must add up to 90 degrees as well, but we have no means of knowing what P must be. All that we know is that Angle P + Angle R = 900. This information is not enough to let us determine what P must be. We conclude that Statement 1 alone is not sufficient. We must move on to examine statement 2. Our choices are B, C and E. Statement 2 tells us that angle R is 170. We can conclude that P + Q = 1630. This information is not sufficient to answer the question because P and Q can have any number of combinations of values adding up to 1630. We must conclude

that statement 2 alone is also not sufficient to answer the question. We have eliminated B as a viable choice at this juncture. We must now combine the two statements, and see if we can make a determination of what P must be. When we combine the two statements, we notice that P+Q+R = 180o and Q+R = 107o. We must conclude that P must be 63o. A unique solution. We must pick C as the answer. 3. What amount did Jean earn from the commission on her sales in the first half of 1988? (1) In 1988 Jean’s commission was 5 percent of the total amount of her sales. (2) The amount of Jean’s sales in the second half of 1988 averaged $10,000 per month more than in the first half. We must pre-determine what information will help us answer the question here. We require the following information to be able to answer the question: ◊ Dollar value of sales in the first half of 1988 ◊ Jean’s commission rate Statement 1 tells us that the commission rate is 5%. We have no information about the dollar value of sales. We must conclude that statement 1 alone is not sufficient. Our choices are B,C or E. Statement 2 tells us that Sales in 2nd half = Sales in 1st half + $60,000 This information alone is not sufficient to determine what was the dollar value of sales in the first half. Also, we do not know from this statement at what rate Jean was paid commission. We must find statement 2 also wanting. Our choices are C or E. Even when we combine the two statements, we do not have any information that helps us determine the dollar value of sales in the first half. We cannot answer the question using the statements independently or combined. We must pick E.




4. A certain car went from one town to another without stopping. What was the car’s average speed for the trip? (1) The car traveled the 90-mile trip in 2 hours. (2) The car traveled the first 40 miles of the trip in 1 hour. 5. What is the value of x?

(1) 2x + 3y = 12 (2) 5x + 7y = 29

6. Does x = y? (1) | x | = | y | (2) x2 = y2 Statement 1 tells us that the absolute values of x and y are equal. We must remember that the absolute value signifies the distance of the value on a number line from the center, and that the absolute values of 3 and –3 are the same. WE cannot conclude with any certainty that x and y are equal to each other. Our choices are B,C or E. Statement 2 tells us that x and y have the same absolute values. For example, x could be +2 and y could be –2, and x2 = y2. Or, x and y could both be +2 and +2. We have no means of knowing whether the former scenario is valid or the latter. Statement 2 is also not good for a unique determination. We have eliminated B as a viable choice. We are left with C and E. When we combine the two statements, we notice that we are dealing with the same information in two different forms, and that we cannot answer the question with any certainty. We must pick choice E. 7. Dan took 20-question multiple-choice test in psychology. If Dan answered every question, did he answer at least 12 questions correctly? (1) Dan answered fewer than 40 percent of the questions incorrectly. (2) Dan answered at least 25 percent of the questions incorrectly. The question is: Did Dan answer at least 60% of the questions correctly. Statement 1 tells us that Dan answered less than 40% incorrectly. This means that Dan answered more than 60% correctly. Statement 1 is good for a unique determination of the answer to the question. Our choices are A or E. Statement 2 is not good for a unique solution. The fact that Dan answered more than 25% incorrectly could mean that he answered 30% incorrectly (70% correctly) or all 100% incorrectly. Statement 2 gives us “all over the map” solution. We must pick choice A.

Statement 2 gives us an equation with two variables, x and y. We require at least 2 sets of independent information to determine the value for x. Our choice are B, C or E. Statement 2 also gives us a two variable information, and we cannot determine what x must be using statement 2 alone. We have eliminated B as a viable option now. When we combine the two statements, we notice that we have two INDEPENDENT sets of information about x and y, and we can solve for x and y. We must pick choice C

We need to know the total distance traveled, and the total time it took to travel the distance. Average speed = Distance/Time Statement 1 tells us that the total distance traveled was 90 miles, and the total time of travel was 2 hours. We can determine the average speed using these two items of information. Our choices are A or D. Statement 2 is not so precise. This statement tells us that a distance of 40 miles was traveled in the first hour. We do not know how many more miles or how much longer the car traveled. Statement 2 is not good for a unique solution. We must pick choice A.




8. If a < x < b and c < y < d, Is x < y? (1) a < c (2) b < c 9. Is (3x + 8)/(x + 2) an integer? (1) x is an integer (2) x = 0

10. How many people did Apex company employ in 1990? (1) The company employed 538 more people in 1991 than in 1990. (2) The company employed 20 percent more people in 1991 than in 1990. Statement 1 tells us that Number in 1991 = Number in 1990 + 538 We cannot determine a unique value for the number of people employed in 1990 using this information alone. Our choices are B, C or E. Statement 2 tells us that Number in 1991 = Number in 1990+0.2 (Number in 1990) WE cannot determine the number employed in 1990 using this information alone. Our choices narrow to C or E. When we combine the two statements, we notice that 0.2 X (Number employed in 1990) = 538 We can conclude that the number employed in 538 must be 5 times 538. The combined information is good for a unique determination of the number of people employed in 1990. We must pick C as the choice.

Statement 1 is not good for a unique determination because a could be 1, x 2, c 5 and y 6. (In this case x < y). Or, a could be 5, x 10, c 6 and y 7. (in which case, x > y). Statement 1 gives us “may be, but not necessarily” solution to the question. We must move on. Our choices are B, C or E. Statement 2 is right on the money. IF b < c, we can combine the two inequalities into one whole: a<x<b<c<y<d. This picture tells us that x < y. Statement 2 is good for a unique answer. We must pick B.

We are looking for a value for x so that we can answer the question definitively. Statement 1 tells us that x is a whole number.If x = 0, then the expression is an integer. If x=1, another integer, the expression is not a whole number. We are beginning to see conflicting answers right here. We must conclude that statement 1 is not good for a unique answer. We must move on to examine statement 2. Our choices are B, C or E. Statement 2 tells us that x = 0. Using this information, we can determine one way or the other whether the given expression is an integer or not. We can answer the question uniquely using this information alone. We must pick B. (Statement 1 alone is not sufficient, but statement 2 alone is sufficient.)




11. Of the four numbers represented on the number line below, is r closest to zero? q r s t (1) q = -s (2) -t < q Statement 1 tells us that q and s have the same absolute values, and that q and s lie the same distance from 0 on either side of 0. This means that 0 must occur smack in the middle of q and s, and the only value that is between q and s is r, and r must be closest to 0. Statement 1 alone is sufficient to answer the question definitively. We must now examine statement 2 to determine whether we can answer the question also using the statement 2. Our choices are A or D. Statement 2 tells that the mirror image of t occurs to the left of q on the number line shown. We know that 0 occurs midway between t and –t, but we have three values – q, r, and s – between t and –t. Any one of these three values could be close to 0. Statement 2 is not good for a unique determination as to whether r is the closet value to 0. We must pick choice A. (Statement 1 alone is sufficient but statement 2 alone is not sufficient).

12. Is the integer n divisible by 20? (1) n is divisible by 5 (2) n is divisible by 6 The question is: Is n a multiple of 20? Statement 1 tells us that n is a multiple of 5. What are the possible values for n? N could be 5, 10, 15, 20, 25, 30, 35, 40,…… If n is 5, then it is not a multiple of 20. If n is 20 or 40 or 60 (all multiples of 20), then it is. We are looking at a “may be, but not necessarily” type of an answer. Not good. Our choices are B, C or E. Statement 2 tells that n could be 6,12,18,24,30,… 60,….. If n is 12, it is not a multiple of 20. If n were 60, it is. Once again, “may be, but not necessarily” kind of an answer. We must now combine the two statements and see if we can make any sense of the statements. When we combine the two statements, we determine that n is a multiple of 30 (least common multiple of 5 and 6). If n is a multiple of 30, n could be 30, 60, 90, 120, and so on. If n is 30, it is not a multiple of 20 If n were 60, then it is. Once again, we are looking at a “may be, but not necessarily” answer here. We must give up, and pick E.




13. If Mark saved an average (arithmetic mean) of $80 per week for 3 consecutive weeks, how much did he save the second week? (1) The average amount that Mark saved per week for the first 2 weeks was $60. (2) The amount that Mark saved the first week was 1/2 the amount he saved the second week and 1/3 the amount he saved the third week. When you are dealing with a data sufficiency question, be sure to make a note of any information that is provided in the stem. You will be required to use this information in addition to the information in the statements 1 and 2, and to decide whether you can make a unique decision about the question posed. Let us say that W1, W2, and W3 are the savings in the weeks 1,2, and 3 respectively. We have: W1 + W2 + W3 = 3X80 =$240 (An average of $80 per week for 3 weeks gives us a total of $240) We notice that we have a 3-variable equation here. We will require 3 independent sets of information to be able to answer the question about W2. Let us now examine the statements one by one. Statement 1 tells us that W1 + W2 = 120 This statement alone is not sufficient, even in the light of the stem information, because Week 1 savings could be 0 and week 2 savings could have been all 120, or the other way around (not to mention other scenarios in which W1 and W2 can have any combinations of values adding up to $120). We conclude that statement 1 alone is not sufficient. Our choices narrow to B, C or E. Statement 2 gives us two additional sets of independent information: W1 = ½ . W2 and W1 = 1/3 . W3 We have to use these independent equations along with the original equation in the stem. We have 3 equations and 3 variables. We can find the values for Week 1, Week 2, and Week 3 savings, and answer the question in a definitive way. We conclude that statement 2 alone is sufficient, but statement 1 alone is NOT. We must pick B as the answer here.

14. If x is not equal to - 1, which is greater, 1/(x + 1), or x/2? (1) x ≥ 0 (2) x < 3 We need to know the actual value of x so that we can answer the question. Statement 1 tells us that x is greater than, or equal to, 0. If x is 0, then 1/(x+1) > x/2 IF x is 1, then the two expressions are equal If x is 2, then x/2 > 1/(x+1) We are getting “all over the map” answers here. Not unique. We must conclude that statement 1 alone is not sufficient. Our choices narrow to B, C or E. Statement 2 tells us that x < 3. If x is 2, then x/2 > 1/(x+1) IF x =1, then the expressions are equal. IF x = 0, then 1/(x+1) > x/2. Another “all over the map” information. Statement 2 alone is also not sufficient. We have choices C or E to contend with. When we combine the two statements, we notice that we are examining the same scenarios that we did under statements 1 and 2 independently. Even the combined information is not sufficient to answer the question in a unique fashion. We must pick E. 15. In a certain two-digit integer, the ratio of the units digit to the tens digit is 2 to 3. What is the integer? (1) The tens digit is 3 more than the units digit. (2) The product of the two digits is 54. The information in the stem tells us that we are dealing with one of the following three 2-digit numbers: 32, 64, or 96. Only these numbers have the digits in the specified ratio. (units’ to tens’ is 2 to 3). Statement 1 tells us that the tens digit is 3 more than the units’. The only two digit number that satisfies this condition and the ratio specified is 96. Our choices are A or D. Statement 2 tells us that the number must be 96 because the product of the digits cannot be 54 if the number were 32 or 64. Either statement is sufficient alone. We must pick Choice D.




MMMoooddduuullleee IIIIII aaassssssiiigggnnnmmmeeennntttsss EEExxxppplllaaaiiinnneeeddd::: If X is a value such that A and B are the two factors, we can write X = A.B In algebraic setting, A and B could stand for algebraic expressions. For example, A could be (n-3) and B could be (3n + 7), in which case X could stand for (n-3) (3n+7) = 3n2 - 2n - 21 When you come across an algebraic expression, your first task is to determine if the expression can be simplified and reduced to a more manageable form. Let us take an example of a “clumsy” algebraic expression: 4x(x-2)-2x+4 = ?? (4x-2) We notice that -2x+4 in the numerator can be written -2(x-2), and we recognize that (x-2) is sitting as a factor in the first part of the numerator expression. We can write the numerator as: 4x(x-2)-2(x-2) = (x-2) (4x-2) What did we do here? We noticed that (x-2) was a common factor, and took it out. We were left with 4x from the first part and -2 from the second part, and we grouped them together to get (x-2)(4x-2) Let us rewrite the expression in terms of these factors as: (x-2)(4x-2) = (x-2) (4x-2) This is the summary of the drill we performed: 1. We recognized common factors in parts of an expression, and re-wrote a part of the expression in terms of its factors. 2. We then took out the common factor to both parts of the expression, and grouped the remaining values to write an expression in terms of two factors. 3. We found to our pleasant surprise that one of the factors in the numerator expression was the same as the denominator expression. These values canceled each other out, leaving behind a simpler, more manageable expression.

Now try the following on your own and see if you can simplify the expression to a more manageable form: 1. 6x3+3x2+6x+3 = 6x3+6x+3x2+3 (x2+1) (x2+1) = 6x(x2+1) + 3(x2+1) (x2+1) = (x2+1) (6x+3) (x2+1) = 6x+3 = 3(2x+1) 2. 5x3+7x2-25x-35 = x2(5x+7) –5(5x+7) (x2-5) (x2-5) = (5x+7)(x2-5) (x2-5) = (5x+7) 3. 7x2(6x+5)-18x-15 = 7x2(6x+5) – 3(6x+5) (7x2-3) (7x2-3) = (6x+5)(7x2-3) (7x2-3) =(6x+5)

4. 3x(x2+2)-7x2-14 = 3x(x2+2) –7(x2+2) (3x-7) (3x-7) = (x2+2)(3x-7) (3x-7) = (x2+2)




ASSIGNMENT 1. If x+y is a prime number, then which of the following “must be” true?

Before we begin to take a look at the conditions specified, let us set up some scenarios, and test possible real life values for x and y. Remember: “Must be true” questions should be answered on the basis of an exception and not on the basis of a solitary agreement. But if we can find one exception, then we are ready to say “no” to the condition specified.

We notice that x and y are not specified as whole numbers. We must be sure to check positive values, negative values, whole number values and non-whole number values for x and y such that the sum will be a prime integer.

x 1 -1 ½ +7¼ -25

y 2 +4 ½ -2 ¼ +36

x+y 3 3 5 11

Notice that the sum row is full of prime integers but the x’s and y’s could be anything we choose them to be.

Now, let us take a look at the conditions and see whether they stack up against the scenarios specified above. If we find one exception, we will say “no” to the condition stipulated.

(A) x = 2 True / False

x does not have to be equal to 2

(B) Either x or y is 2. True / False

x and y could be negative fractional values

(C) Both x and y are prime numbers.

True/ False

X and y do not have to be prime integers. See the scenarios we have created at the outset.

(D) x.y is not a prime number.

True / False

x.y could be a prime number because if x is 1 and y is 2, then the sum is 3, a prime number and the product is 2, a prime number. We have an exception here, may be the only exception we will ever have, but an exceptional scenario is sufficient to say “no” to the condition specified.

(E) x-y is a prime number. True / False

x-y does not have to be prime. See the scenarios we set up at the outset.

(F) x.y is a positive integer. True / False x or y could be negative and fractional, and the product does not have to be an integer, let alone a positive one.




2. If x²y³ is a negative odd integer, then which of the following must be true?

Once again, we notice that x and y are not specified as whole numbers. We should be sure to check out fractional values as possible values for x and y consistent with the outcome specified.

Because x2y3 is specified as a negative odd integer, we must conclude that y must be a negative number but x could be positive or negative because of the “power of 2” attached to it.

x -1 1/3 -9

x2 1 1/9 81

y -3 -3 -1/3

y3 -27 -27 -1/27

x2y3 -27 -3 -3

We notice that x and y could be fractional values as well as whole number values.

Now, let us take a look at the conditions specified, and check how they stack up.

(A) y is a negative integer. True/ False.

y is negative but does not have to be an integer.

(B) x can be a fraction. True / False

see scenarios above

(C) x can be positive or negative, integer or a Fraction. True / False

see scenarios above

(D) Both x and y are odd numbers - fractions or integers. True / False

see scenarios we set up.

(E) x² + y³ is a positive even integer. True / False

x2+y3 does not have to be an integer, let alone a positive even integer, because either x or y could be a fraction.




Assignment Continued

3. Thames Toyota Dealership has 120 cars in their lot. The cars come in two colors - Red or Blue and two sizes - mid-size and large. There are exactly 10 large sized Red cars in the lot. If there are three times as many mid-sized cars as there are large ones and the red to blue cars ratio in the lot is 3:2, How many cars in the lot are :

(A) Red and Mid-size?

(B) Blue and Large?

(C) Blue and Mid-size?

Values indicated in red are specified in the problem. The rest of the values are obtained by subtracting from the column and row totals.

LARGE MID TOTAL

RED 10 62 72

BLUE 20 28 48

TOTAL 30 90 120

4. Machine A produces bolts at a uniform rate of 120 every 40 seconds, and machine B produces bolts at a uniform rate of 100 every 20 seconds. If the two machines run simultaneously, how many seconds will it take for them to produce a total of 200 bolts?

Machine A produces 3 bolts per second. Machine B produces 5 bolts per second. Together, A and B produce 8 bolts per second. To produce 200 bolts, working together, A and B will take

200/8 =25 seconds.

5. A survey involving two questions Q1 and Q2 was answered by N persons. 1/3 of those participating answered YES to Q1, and of these 1/5 also answered YES to Q2. If the number of those answering YES to Q1 but NO to Q2 is twice that answering NO to Q1 and YES to Q2, what is the number answering NO to both questions Q1 and Q2?

.

YES TO Q2 NO TO Q2 TOTAL

YES TO Q1 1/5(N/3) 4N/15 N/3

NO TO Q1 2N/15 2N/3

TOTAL N

The values in the red cells are computed from the information in the problem.(Yes to Q1 and No to Q2 is TWICE Yes to Q2 and No to Q1)




ASSIGNMENT

1. A triangle has its 3 sides of following lengths: 3, 3, 3. What are the 3 angles ?

Answer:__60_º, _60__º, ____60_º

2. A right triangle has its hypotenuse measuring 8 units. One of its angles is 30º. What are the other two side lengths?

Answer:___4___, ____4\/3___

3. A triangle has 3 sides of lengths 27, 36, 45. What type of a triangle is this?

Answer:____Right Triangle (3:4:5 proportions)

4. An equilateral triangle has sides measuring “a”. What is the area of the equilateral triangle in terms of its side “a”? 30o 30o

a \/3.a/2 a

60o a/2 a/2 60o a

Area is: \/3/4. (side)2 = \/3/4. a2

5. A right triangle has one leg measuring 4 and the other 6. What is the hypotenuse length?(use Pythagorean theorem and show steps).

\/42+62 = \/52 = 2\/13

6. A triangle has two angles measuring 55º and 43º. What is the measure of the third angle?

Answer: _82º

7. An isosceles triangle has two sides 10 and 10. The base where the altitude from the vertex meets measures 16. What is the area of the triangle?(Draw triangle, mark sides and angles appropriately, and solve the problem).

10 10

8 8

16

We notice that the altitude divides the base into two equal halves, and the whole triangle into two 3:4:5 triangles, with the hypotenuse measuring 10 and one side measuring 8. We must conclude that the altitude must measure 6 corresponding to proportion 3. Knowing that the height is 6 and the base is 16, we can calculate the area: ½ . (6)(16) = 48 sq.units.

Answer: _____48_______

8. If AB X CE = 24, What is the area of the triangular shaped region ABD shown below?(AB and CD are parallel lines)

C D

A E B K

Hint: The vertical height DK shown in dotted lines will determine the area of the triangle ABD. You will notice that the altitude for the triangle ABD is the same as CE. Area= ½ (24) = 12 sq.units

The sides of a 30-60-90 triangle are in proportion 1:\/3:2




ASSIGNMENT

1. If ∠ABC in the figure below is a right angle, what is the value of external xº?

xº C

B 55º D

A ∠ABD = 55º

Angle CBd is 90-55=35o. The external angle x is the supplementary angle, and must be 180-35 = 145o

Answer: xº = __145º

2. If a triangular region has two sides of value 17 and 25, which of the following cannot be a value for the third side?

(A) 38 (B) 45 (C) 11 (D) 17 (E) 41

The third side must lie in the following range: 8 < Third Side<42. The only value that is not in the range is 45.

Answer: ___B__

3. Which of the following groups of numbers could be the lengths of the sides of a triangle?

I. 1, 4. √17 II. 4, 7, √11 III. 4, 9,6

Choices: (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II and III. (Explain your answers)

We apply the range rule here and notice that all sets of values will satisfy the rule. For example, the set of values in I are consistent with the range rule requirements: 3<\/17 <5

Similarly, the sets of values in II and III will check out ok too. (Try that). You may notice that the values in I are the values for a right triangle as well.

Answer: ___E_____

4. If the length and the width of a rectangular garden plot were each increased by 25%, what would be the percent increase in the area of the plot?

(A) 25% (B) 33% (C) 45% (D) 56.25% (E) cannot estimate from the information

The area of a rectangle is the product of the two sides. IF L becomes 5/4.L and W becomes 5/4 .W, then the new area is: 5/4.L.5/4.W = 25/16.LW = 1.5625LW.

We notice that the new area is 56.25% more than in the original scenario.

Answer: _____D_____

5. The trapezoid shown below is the cross section of a rudder of a ship. If the distance from A to C is 25 feet, what is the area of the cross section of the rudder in square feet? C

D 25

A 4 7 B

The lengths are: AD= 4’ BC=7’.

(A)168 (B) 175 (C) 100 (D) 132 (E) 28

In the right triangle ABC, AC, hypotenuse, is 25 and one of the sides BC is 7. We must recognize the classical triple 7:24:25 proportion, and conclude that AB must be 24.




The area of a trapezoid is ½ (sum of two parallel sides).(height separating the two parallel sides)

Area of trapezoid = ½ .24.(4+7) = 132 Choice D

Assignment Continued... 6. If a rectangular photograph that is 10 inches wide by 15 inches long is to be enlarged so that the width will be 22 inches and the ratio of length to width will be unchanged, then the length in inches of the enlarged photograph will be

(A) 33 (B) 32 (C) 30 (D) 27 (E) 25

This is a simple proportion problem:

Let X be the new Length.

10/15 = 22/X or X = 33 inches.

Choice A

7. If L and W are the dimensions of a rectangular region that has area 42 and if L and W are integers such that L > W, What is the total number of possible values of L?


The question is: How many different ways can we set up 42 as the product of two integers? We can go: 42X1, 21X2, 14X3, 7X6 (the first value is that of L and the second that of W). We have 4 possible choices here. C is the answer.

8. In the above question, if the area were 45, and L and W are integers such that L >W, how many possible values for W exist?


When the area is 45, we can go: 45X1, 15X3, and 9X5. Three combinations consistent with the specification that L>W and L.W=45.

B is the answer.

9. The shaded portion ABC of a rectangular lot shown below is a flower bed. If the area of the flower bed is 24 sq.yds and BC=AB + 2, then AC equals:

(A)√13 (B)2√13 (C) 6 (D) 8 (E) 10

C

(X+2)

A X B

We have: ½ .X.(X+2) = 24, or X.(X+2)=48

Knowing that X is positive, can we think of two values that are 2 apart and will multiply to give 48? Yes, 6 and 8. AB must be 6 and BC must be 8. We recognize the 3:4:5 proportion here, and must conclude that AC must be 10 (6,8,10 triangle in 3:4:5 proportion).

Answer: E




10. A rectangular region has the same area as a square region : 64 sq. yards. If one length of one side of the rectangle is half the length of the side of the square and the other side of the rectangle is twice the length of square side, what are the dimensions of the rectangle and the square?(rectangle length, rectangle width, square side)

(A) 8,8,8 (B) 4,16,8 (C) 8,12,12 (D) 4,8,8 (E) 16,8,8

It takes more time to read this problem than answer it. If the area of a square is 64, we know that the side of the square must be 8 units. If the rectangle has the same area and has one side that is twice the side of the square, we know that the length must be 16 and the width MUST be 4.

Choice B.

11. Which of the following inequalities is equivalent to 10-2x > 18?

(A) x > -14 (B) x > -4 (C) x >4 (D) x < 4 (E) x < -4

10-2x > 18 or –2X > 8 or -X > 4

Now let us divide both sides by –1 and flip the sign of the inequality.

We get X < -4. Choice E

12. Which of the following inequalities is equivalent to 5x-10y < -20

(A) 2y < -(4+x) (B) x < 2y+4 (C) x >2y+4 (D) 2y>4+x (E) None of the above.

We notice that the answer choices are in terms of both X and Y. We need to express the inequality in terms of both and decide which one matches the answer.

5x-10y < -20 or x –2y <-4

or, x < 2y-4 ………………. (1)

or, -2y < -4-x or 2y > 4+x ……. (2)

We notice that (2) matches choice D.

13. If n is an integer and

n = (2.3.5.7.11.13) ÷ 91p

Which of the following could be the value of p?

(A) 22 (B) 26 (C) 35 (D) 54 (E) 60

This is a problem testing your knowledge of factors. Let us try to simplify the expression to the right of = sign: We notice that 91 cancels out the 7 and 13 on the numerator leaving us with a simpler n = (2.3.5.11)/p

Knowing that n is an integer, we conclude that p must be such that n must continue to be an integer. The only value for p that will produce an integer value for n is 22. Choice A.




14. Mary and Jenny live 13 miles apart. They meet at a cafe that is directly north of Mary’s house and due east of Jenny’s house. If the cafe is 7 miles closer to Mary’s house than to Jenny’s house, how many miles is the cafe from Jenny’s house? (Hint: Use Pythagorean theorem and right triangle properties discussed.) Draw diagram and show steps.

(A) 12 (B) 20 (C) 6 (D) 10 (E) 10√3

We must set up Mary and Jenny on the hypotenuse of a right triangle and the configuration will appear as follows:

Jenny X Cafe

13 (X-7)

Mary

We see a 13 on the hypotenuse of a right triangle and must check to see whether the other two sides are consistent with the classical triple, 5:12:13. If one side is 7 more than the other, then the sides must be 5 and 12. We conclude that the Café is 5 miles from Mary’s house and 12 miles from Jenny’s house.

Remember: You must learn to recognize the triples so that you do not waste time by setting up quadratic equations and finding the roots.

Choice A

15. Which of the following equations has one root in common with x²-6x+5=0

(A) x² + 1=0 (B) x²-x-2 = 0 (C)x²-10x-5=0 (D) 2x² - 2=0 (E) x²-2x-3=0

We notice that the roots of the given equation are 1 and 5.

(x2-6x+5) = (x-1)(x-5) = 0 or x =1 or 5.

Now, check to see whether 1 or 5 will satisfy the choice equations. Notice that Choice A is not worth messing with because the x does not have a real root. (x2 = -1, and that cannot be). We must not waste time checking out choice A. We must move on. We will see that choice D has a root of 1 in common with the stem equation. D is the answer.

Answer:________________

16. How many positive integers n are there such that 100n is a factor of:

(2³)(5)(5³)

(A) None (B) Five (C) Seven (D) Six (E) Eleven

When we express the exponential expression as a decimal number, we get 5000. If 100n is a factor of 5000, then

5000/100n = integer or 50/n = integer.

n can have all the positive values that are factors of 50.

Factors of 50 are: 1,50,2,25,5,10.

n can have 6 positive values. D is the answer.




Assignment: Integers Explained (p.44)

1. Integer 9 is not a divisor of which of the following values of y? (A) 63 (B) 108 (C) 36 (D) 200 (E) 126 Your Answer:______D _________ 2. Even integer when added to an Odd integer results in _ODD__ integer. 3. Even integer when multiplied by an even integer results in ___EVEN_____integer. 4. An odd integer when subtracted from an even integer results in _____ODD________ integer. 5. A prime number n is an integer that has exactly _TWO_______ different __POSITIVE_ divisors and the divisors are 1 AND ITSELF. 6. Which of the following numbers cannot be prime number(s) and why? (A)3 (B)169 (C) 32 (D) 83 (E) 59 Your Answer: 32 AND 169 ARE DIVISIBLE BY MORE THAN 2 FACTORS. 7. The only even prime number is___2____. 8. The expression n/n=1 is true for all values of n except ___0_________

9. How many factors or divisors do the following integers have? What are they? (A) 81 : 1,81, 3, 27, 9 (B) 140:1,140;2,70;4,35;5,28;7,20,10,14. (C) 256:1,256; 2,128; 4,64; 8,32, 16 (D) 171: 1, 171; 3,57; 9,19; 10. The difference of two consecutive even integers is ____2_____________ 11. The difference of two consecutive odd integers is ______2____________ 12. What is the next integer in the following sequence of even integers? 2n, 2n+2, 2n+4, 2n+6,___2n+8______ 13. Which of the following are odd, even or indeterminate integers? (n is not 0) Odd Even Can’t Say (A) 2n+3 x O O (B) n+5 O O X (C) 2n+24 O X O (D) 2n-3 X O O (E) 2(n-7) O X O (F) 3(2n-1) x O O (G) (2n+15) - (2n-4) X O O (H) (2n+64) + (3n+5) O O X (I) n(2n-11)÷n X O O (J) (2n+1)(2n+6) O X O Hint: n takes on all values (except 0), positive and negative. 14. If x and y are prime numbers and (x+y) is a also a prime number, then x or y must be equal to _____2______

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