GMAT Problems

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Difficult Problems from the Math Section 1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=? Sol: First term a=2, common difference d=2 since even number therefore sum to first n numbers of Arithmetic progression would be n/2(2a+(n-1)d) = n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80 therefore n=79 which is odd... 2. The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat? (A) $4.50 (B) $5.10 (C) $5.30 (D) $5.50 (E) $5.60 Soln: 320 + 5x = 580 - .41x; x = # of days after price is same; solving for x; x is approximately 48, thus the required price is 320 + 5 * 48 = 560 cents = $5.6 3. How many randomly assembled people do u need to have a better than 50% prob. that at least 1 of them was born in a leap year?

Transcript of GMAT Problems

Page 1: GMAT Problems

Difficult Problems from the Math Section

1. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...

2. The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents

per day while the price of wheat is decreasing at a constant rate of cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50(B) $5.10(C) $5.30(D) $5.50(E) $5.60

Soln: 320 + 5x = 580 - .41x; x = # of days after price is same;

solving for x; x is approximately 48, thus the required price is 320 + 5 * 48 = 560 cents = $5.6

3. How many randomly assembled people do u need to have a better than 50% prob. that at least 1 of them was born in a leap year?

Soln: Prob. of a randomly selected person to have NOT been born in a leap yr = 3/4Take 2 people, probability that none of them was born in a leap = 3/4*3/4 = 9/16. The probability at least one born in leap = 1- 9/16 = 7/16 < 0.5Take 3 people, probability that none born in leap year = 3/4*3/4*3/4 = 27/64. The probability that at least one born = 1 - 27/64 = 37/64 > 0.5Thus min 3 people are needed.

4. In a basketball contest, players must make 10 free throws. Assuming a player has 90% chance of making each of his shots, how likely is it that he will make all of his first 10 shots?

Ans: The probability of making all of his first 10 shots is given by

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(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%

5. AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determinedAns: AB + CD = AAASince AB and CD are two digit numbers, then AAA must be 111 Therefore 1B + CD = 111B can assume any value between 3 and 9If B = 3, then CD = 111-13 = 98 and C = 9If B = 9, then CD = 111-19 = 92 and C = 9So for all B between 3 & 9, C = 9

Therefore the correct answer is D (C = 9)

6. A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?(A) 12(B) 14(C) 16(D) 18(E) 20

Ans: race 1 :- ta = tb-6 ( because A beats B by 6 sec)race 2 :- Ta = tb+2 ( because A looses to B by 2 sec)

By the formula D= S * Twe get two equations480/Sa = 432/Sb -6 ------------1)480/Sa = 336/Sb +2------------2)Equating these two equations we get Sb = 12ta,Sa stand for time taken by A and speed of A resp.

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7. A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?(A) 1/4(B) 1/3(C) 1/2(D) 2/3(E) ¾

Ans: Let X be the fraction of solution that is replaced.

Then X*25% + (1-X)*40% = 35%

Solving, you get X = 1/3

8. A person buys a share for $ 50 and sells it for $ 52 after a year. What is the total profit made by him from the share?(I) A company pays annual dividend(II) The rate of dividend is 25%(A) Statement (I) ALONE is sufficient, but statement (II) alone is not sufficient(B) Statement (II) ALONE is sufficient, but statement (I) is not sufficient(C) BOTH statements TOGETHER are sufficient, but NEITHER statement alone is sufficient(D) Each statement ALONE is sufficient(E) Statements (I) and (II) TOGETHER are NOT sufficient

9. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?(A) 2/27(B) 1/9(C) 1/3(D) 4/27(E) 2/9

Ans: Case I: Red ball first and then white ball

P1 = 3/9*2/9= 2/27

Case 2: White ball first and then red ball

P2 = 2/9*3/9 = 2/27

Therefore total probability: p1 + p2 = 4/27

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10. What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4B) sqrt (2)C) 1.7D) sqrt (3)E) 2.0

Ans: The equation of the line will be 3x - 4y - 12 = 0.This crosses the x and y axis at (0,-3) and (4,0)

The circle has the origin at the center and has a radius of 1 unit.

So it is closest to the given line when, a perpendicular is drawn to the line, which passes through the origin.

This distance of the line from the origin is 12 / sqrt (9 + 16) which is 2.4 [Length of perpendicular from origin to line ax +by + c = 0 is

mod (c / sqrt (a^2 + b^2))]

The radius is 1 unit.

So the shortest distance is 2.4 - 1 unit = 1.4 units

11.

In the square above, 12w = 3x = 4y. What fractional part of the square is shaded?

A) 2/3

B) 14/25

C) 5/9

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D) 11/25

E) 3/7

Sol: Since 12w=3x=4y,

w:x=3:12=1:4 and x:y=4:3

so, w = 1x = 4y = 3

the fractional part of the square is shaded:{(w+x)^2 - [(1/2)wx + (1/2)wx +(1/2)xy + (1/2)w(2w)]}/(w+x)^2

= {(w+x)^2 - [wx + (1/2)xy + w^2)]}/[(w+x)^2]

=[(5^2) -(4+6+1)]/(5^2)

= (25 - 11)/25

= 14/25

12. The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

20 25 40 45 75 Ans: The answer 25 doesn't refer to a temperature, but rather to a range of temperatures.

The average of the 5 temps is: (a + b + c + d + e) / 5 = 50One of these temps is 45: (a + b + c + d + 45) / 5 = 50Solving for the variables: a + b + c + d = 205In order to find the greatest range of temps, we minimize all temps but one. Remember, though, that 45 is the lowest temp possible, so: 45 + 45 + 45 + d = 205Solving for the variable: d = 7070 - 45 = 25

13. If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being

divisible by 8? 25% 50% 62.5% 72.5% 75%

Soln: E = n*(n+1)*(n+2)

E is divisible by 8, if n is even.No of even numbers (between 1 and 96) = 48

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E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)Such numbers total = 12(7,15,....)

Favorable cases = 48+12 = 60.Total cases = 96P = 60/96 = 62.5

Method 2:From 1 to 10, there are 5 sets, which are divisible by 8.(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)

So till 96, there will be 12 * 5 such sets = 60 sets

so probability will be 60/96 = 62.5

14. Kurt, a painter, has 9 jars of paint: 4 are yellow2 are redrest are brownKurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellowBrun X if the paint contains 3 jars of brown paintJaune X if the paint contains at least 2 jars of yellowJaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42b) 37/42c) 1/21d) 4/9e) 5/9

Sol: 1. This has at least 2 yellow meaning..

a> there can be all three Y => 4c3 OR b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

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Total 40Total possibilities = (9!/3!6!) = 84Adding the two probabilities: probability = 74/84 = 37/42

15. TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??Soln:Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)Ways for second couple = 2*4!These cases include an extra case of both couples sitting togetherWays in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways)Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!Total ways to arrange the 5 ppl = 5!Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5Thus prob of none seated together = 1 - 3/5 = 2/5

16. An express train traveled at an average speed of 100 kilometers per hour, stopping for 3 minutes after every 75 kilometers. A local train traveled at an average speed of 50 kilometers, stopping for 1 minute after every 25 kilometers. If the trains began traveling at the same time, how many kilometers did the local train travel in the time it took the express train to travel 600 kilometers?

a. 300b. 305c. 307.5d. 1200e. 1236

Sol: the answer is C: 307.5 km Express Train: 600 km --> 6 hours (since 100 km/h) + stops * 3 min.stops = 600 / 75 = 8, but as it is an integer number, the last stop in km 600 is not a real stop, so it would be 7 stopsso, time= 6 hours + 7 * 3 min. = 6 hours 21 min

Local Train:in 6 hours, it will make 300 km (since its speed is 50 km/hour)in 300 km it will have 300 / 25 stops = 12, 12 stops 1 min each = 12 minwe have 6 hours 12 min, but we need to calculate how many km can it make in 6 hours 21 min, so 21-12 = 9 minif 60 min it can make 50 km, 9 min it can make 7.5 kmso, the distance is 300 + 7.5 = 307.5 km

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17. Matt starts a new job, with the goal of doubling his old average commission of $400. He takes a 10% commission, making commissions of $100.00, $200.00, $250.00, $700.00, and $1,000 on his first 5 sales. If Matt made two sales on the last day of the week, how much would Matt have had to sell in order to meet his goal?

Sol: The two sales on Matt's last day of the week must total $33,500.

(100 + 200 + 250 + 700 + 1000 + x) / 7 = 800x = 3350since x is Matt's 10% commission, the sale is $33,500.

18. On how many ways can the letters of the word "COMPUTER" be arranged?

1) Without any restrictions.2) M must always occur at the third place.3) All the vowels are together.4) All the vowels are never together.5) Vowels occupy the even positions.

Sol: 1) 8! = 403202) 7*6*1*5*4*3*2*1=5,040

3) Considering the 3 vowels as 1 letter, there are five other letters which are consonants C, M, P, T, RCMPTR (AUE) = 6 letters which can be arranged in 6p6 or 6! Waysand A, U, E themselves can be arranged in another 3! Ways for a total of 6!*3! Ways

4) Total combinations - all vowels always together= what u found in 1) - what u found in 3)= 8! - 6! *3!

5) I think it should be 4 * 720there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

19. In the infinite sequence A, ,where x is a positive integer constant. For what value of n is the ratio

of to equal to ?

(A) 8(B) 7(C) 6

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(D) 5(E) 4

Sol: The method I followed was to reduce the Q to (x^6/ x ) * ( Y/Y)the eqn An= (x ^ n -1)(1+ x + x^2 + x^3 +.... ) ----------------------------(1)

and the eqn x(1+x(1+x(1+x...)))) which I call Z can be reduced to x( 1+x+x^2+x^3 ..) --------(2)

from (1) and (2) we get An / Z = x^(n-1) / xtherefore for getting answer x^5 (n-1) = 6therefore n=7

Ans: B

20. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?a. (13^4) x 48 x 47b. (13^4) x 27 x 47c. 48C6d. 13^4e. (13^4) x 48C6

Sol: 52 cards in a deck -13 cards per suitFirst card - let us say from suit hearts = 13C1 =13Second card - let us say from suit diamonds = 13C1 =13Third card - let us say from suit spade = 13C1 =13Fourth card - let us say from suit clubs = 13C1 =13Remaining cards in the deck= 52 -4 = 48Fifth card - any card in the deck = 48C1Sixth card - any card in the deck = 47C1

Total number of ways = 13 * 13 * 13 * 13 * 48 * 47 = 13^4 *48*47 ---> choice A

21. Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10? 3 4 5 6 7 *

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Sol:ok consider this

0 + 1+ 2 + 3 + 4 +5 +6..Stop --> Don’t go further. Why? Here’s why.At the worst the order given above is how u could pick the out the slips. Until u add the slip with no. 6 on it, no two slips before that add up to 10 (which is what the Q wants)

the best u can approach is a sum of 9 (slip no. 5 + slip no. 4)

but as soon as u add slip 6. Voila u get your first sum of 10 from two slips and that is indeed the answer = 7 draws 22. Two missiles are launched simultaneously. Missile 1 launches at a speed of x

miles per hour, increasing its speed by a factor of every 10 minutes (so that after

10 minutes its speed is , after 20 minutes its speed is , and so forth. Missile 2 launches at a speed of y miles per hour, doubling its speed every 10 minutes. After 1 hour, is the speed of Missile 1 greater than that of Missile 2?

1) 2)

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.(D) EACH statement ALONE is sufficient to answer the question.(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Sol:

Since Missile 1's rate increases by a factor of every 10 minutes, Missile 1 will be

traveling at a speed of miles per hour after 60 minutes:

minutes0-1010-2020-3030-4040-5050-6060+speed

And since Missile 2's rate doubles every 10 minutes, Missile 2 will be traveling at a

speed of after 60 minutes:

minutes0-1010-2020-3030-4040-5050-6060+speed

The question then becomes: Is ?

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Statement (1) tells us that . Squaring both sides yields . We can

substitute for y: Is ? If we divide both sides by , we get: Is ? We

can further simplify by taking the square root of both sides: Is ? We still cannot answer this, so statement (1) alone is NOT sufficient to answer the question.

Statement (2) tells us that , which tells us nothing about the relationship between x and y. Statement (2) alone is NOT sufficient to answer the question.

Taking the statements together, we know from statement (1) that the question can be

rephrased: Is ? From statement (2) we know certainly that , which is

another way of expressing . So using the information from both statements, we can answer definitively that after 1 hour, Missile 1 is traveling faster than Missile 2.

The correct answer is C: Statements (1) and (2) taken together are sufficient to answer the question, but neither statement alone is sufficient.

23. If , what is the unit’s digit of ?

(A) 0(B) 1(C) 3(D) 5(E) 9Sol:

The unit’s digit of the left side of the equation is equal to the unit’s digit of the right side of the equation (which is what the question asks about). Thus, if we can determine the unit’s digit of the expression on the left side of the equation, we can answer the question.

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Since , we know that 13! Contains a factor of 10, so its

unit’s digit must be 0. Similarly, the unit’s digit of will also have a unit’s digit of 0. If we subtract 1 from this, we will be left with a number ending in 9.

Therefore, the unit’s digit of is 9. The correct answer is E.

24. The dimensions of a rectangular floor are 16 feet by 20 feet. When a rectangular rug is placed on the floor, a strip of floor 3 feet wide is exposed on all sides. What are the dimensions of the rug, in feet?

(A) 10 by 14 (B) 10 by 17 (C) 13 by 14 (E) (D) 13 by 17 (E) 14 by 16

Soln: The rug is placed in the middle of the room. The rug leaves 3m on either side of it both lengthwise and breadth wise. Now, the dimensions of the rug would be the dimensions of the room - the space that it does not occupy. With three 3 on either side, 6 m is not occupied by the rug in both dimensions.

So, rug size = (16-6) X (20-6) = 10 X14

25. How many different subsets of the set {10,14,17,24} are there that contain an odd number of elements?

(a) 3 (b) 6 (c) 8 (d) 10 ( e) 12

Soln: 8 is the answer. The different subsets are

10, 14, 17, 24, 10, 14, 17

14 17 24

17 24 10

24 10 14

26. Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done? a.24 b.6 c.4 d.12 e.3

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Soln: I suggest to first arranging men. This can be done in 6! Ways. Now to satisfy above condition for women, they should sit in spaces between each man. This can be done in 7! Ways (because there will be seven spaces between each man on round table)Total ways = 6! * 7!

27. Find the fourth consecutive even number:(I) the sum of the last two numbers is 30(II) the sum of the first two numbers is 22

the ans is D

please explain why I alone is sufficient?

Soln: I guess, there are 4 consecutive even numbers to start with..Stmt 1) Sum of 3rd even + Sum of 4th even = 30 => 14+16 = 30 => 4th even num = 16.. Sufficient

Stmt 2) Sum of Ist even +Sum of 2nd even = 22 => 10 and 12 are the 2 numbers to begin with, then 3rd num = 14 and 4th num = 16.. Sufficient

Ans – D

28. If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a) A+1b) A+5c) A+25d) 2Ae) 5ASoln: If you divide the sum obtained by adding any 5consecutive numbers by '5', and then you will get the Center number of the sequence itself.

i.e. 1 - 5 = 15/5 = 3 . 1, 2, 3, 4, 5

so, sixth consecutive number will be '3' more than the 'Middle term'i.e. 3+3=6, similarly 3+4=7

Hence going by this. Asked sum would be

[(A/5) + 3]+[(A/5) + 4]+[(A/5) + 5]+[(A/5) + 6]+[(A/5) + 7] = A + 25

29. If P represents the product of the first 15 positive integers, then P is not a multiple of:

a) 99 b) 84 c) 72 d) 65 e) 57

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Solution: If P represents the product of the first 15 integers, P would consist of the prime numbers that are below 15.

2,3,5,7,11,13

Any value that has a prime higher than 13 would not be a value of P.

57 = 3, 19

19 is a prime greater than 13, so the answer is E.

30. 5 girls and 3 boys are arranged randomly in a row. Find the probability that:

A) there is one boy on each end.

B) There is one girl on each end.

Solution: For the first scenario:A) there is one boy on each end.

The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways Total possible outcome = 8!Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

For the second scenario:A) there is one girl on each end.

The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways Total possible outcome = 8!Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14

31. If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

Solution: approach: first fix Jen, and then fix Bob. Then fix the remaining three.

Case 1: When Jen is in the first place.=> Bob can be in any of the other four places. => 4.The remaining 3 can arrange themselves in the remaining 3 places in 3! Ways.Hence total ways = 4*3!

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Case 2: When Jen is in the second place.=> Bob can be in any of three places => 3.The remaining can arrange themselves in 3 places in 3! Ways.

Continuing the approach.Answer = 4*3! +3*3! +2*3! +1*3! = 10*3! = 60 ways.

32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If –1 is in the set, which of the following must also be in the set?I. –3 II. 1 III. 5A. I onlyB. II onlyC. I and II onlyD. II and III onlyE. I, II, and III

Soln: Series property: t => t+2. (Note: for any given number N, ONLY N + 2 is compulsory. N - 2 is not a necessity as N could be the first term...this can be used as a trap.)Given: -1 belongs to the series. => 1 => 3 =>5. DOES NOT imply -3.Hence, II and III (D).

33. A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?

(A) 17/30(B) 2/5 (C) 7/15(D) 4/15 (E) 11/30

Solution: Total no from 1 to 30 = 30total no from 1 to 30 which r multiple of 3 = 10 (eg(3,6,9,12,15,18,21,24,27,30))total no from 1 to 30 which r multiple of 13 = 2 (eg 13,26)P(a or b ) = p(a) + p(b)p(a)= 10/30p(b)=2/30p(a) + p(b) = 10/30+2/30 = 2/5

34. Two numbers are less than a third number by 30% and 37 % respectively. How much percent is the second number less than the first?

a) 10 % b) 7 % c) 4 % d) 3 %

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Solution: .7 & .63

diff in %= (.7 - .63)/.7 * 100= .07/.7 * 100= 10%

35. If y ≠ 3 and 2x/y is a prime integer greater than 2, which of the following must be true? I. x = yII. y = 1III. x and y are prime integers.

(A) None

(B) I only

(C) II only

(D) III only

(E) I and II

37. Someone passed a certain bridge, which needs fee. There are 2 ways for him to choice. A : $13/month+$0.2/time , B: $0.75/time . He passes the bridge twice a day. How many days at least he passes the bridge in a month, it is economic by A way?

A) 11 B) 12 C) 13 D) 14 E) 15

Soln: Let x be the no days where both are equal cost13 + 0.2*2x= 0.75*2x13+.4x=1.5x13=1.1xx=11.81

by plugging in for 12 daysFor A13 +(12*2)*.2=13+4.8=17.8For B24*.75=18

therefore answer is B.

38. Two measure standards R and S. 24 and 30 measured with R are 42 and 60 when they are measured with S, respectively. If 100 is acquired with S, what would its value be measured with R?

39. Every student of a certain school must take one and only one elective course. In last year, 1/2 of the students took biology as an elective, 1/3 of the students took

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chemistry as an elective, and all of the other students took physics. In this year, 1/3 of the students who took biology and 1/4 of the students who took chemistry left school, other students did not leave, and no fresh student come in. What fraction of all students took biology and took chemistry?

A. 7/9 B .6/7 C.5/7 D.4/9 E.2/5

Soln: If total =1Last yearB=1/2C=1/3P=1/6This year B= 1/2 * 2/3 = 1/3C= 1/3 * 3/4= 1/4Tot 1/3+1/4=7/12student left this yearB = 1/2 * 1/3 = 1/6C= 1/3 *1/4 = 1/12tot= 1/6 +1/12 = 1/4So the school has total student this year= Last year student no - total no of student left this year = 1- 1/4=3/4Answer = 7/12 / 3/4 = 7/9

40. If X>0.9, which of the following equals to X?

A. 0.81^1/2 B. 0.9^1/2 C. 0.9^2 D. 1-0.01^1/2

41. There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A.18/35 B16/35 C.14/35 D.13/35 E.12/35

Soln: there has to be equal no of men & women so out of 4 people selected there has to be 2M & 2W.Total ways of selecting 4 out of 8 is 8C4 total ways of selecting 2 men out of 4 is 4C2total ways of selecting 2 women out of 4 is 4C2

so probability is (4C2*4C2)/ 8C2 = 18/35

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42. The area of an equilateral triangle is 9.what is the area of it circumcircle.

A.10PI B.12PI C.14PI D.16PI E.18PI

Soln: area = sqrt(3) / 4 * (side)^2 = 9

so , ( side )^2 = ( 9*4 ) / sqrt(3)

Height = H = sqrt(3)/2 * (side) , so H^2 = 3/4 * Side^2 = 3/4 * (9*4) / sqrt(3)

Radius of CircumCircle = R = 2/3 of ( Height of the Equilateral Triangle )

so , area of Circumcircle = PI * R^2

=> PI * 4/9 * H^2

=> PI * 4/9 * 3/4 * (9*4) / sqrt(3)

=> PI * 4 * sqrt(3)

43.A group of people participate in some curriculums, 20 of them practice Yoga, 10 study cooking, 12 study weaving, 3 of them study cooking only, 4 of them study both the cooking and yoga, 2 of them participate all curriculums. How many people study both cooking and weaving?

A.1 B.2 C.3 D.4 E.5

Soln: We know there are 10 people who do cooking as an activity.

3 -> people who do only cooking4 -> do cooking and Yoga2 -> do all of the activitiesx -> number of people doing cooking and weaving

When you sum all this up, we should have 10. So 3+4+2+x=10 --> x=10-9=1

44. There 3 kinds of books in the library fiction, non-fiction and biology. Ratio of fiction to non-fiction is 3 to 2; ratio of non-fiction to biology is 4 to 3, and the total of the books is more than 1000?which one of following can be the total of the book?

A 1001 B. 1009 C.1008 D.1007 E.1006

Soln: fiction : non-fiction = 3 : 2 = 6 : 4non-fiction : biology = 4 : 3

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fiction : non-fiction : biology = 6 : 4 : 3

6x + 4x + 3x = 1000x = 76 12/13if we add 1, 13 will divide evenly (1001/13 = 77)1000 + 1 = 1001

45. In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

A) 5

B) 10

C) 15

D) 20

E) 25Soln: n(1U2U3) = n(1) + n(2) + n(3) - n(1n2) - n(2n3) - n(1n3) + n (1n2n3)85=50+30+20- [n(1n2) - n(2n3) - n(1n3)] +5[n(1n2) - n(2n3) - n(1n3)] = 20

46. For a certain company, operating costs and commissions totaled $550 million in 1990, representing an increase of 10 percent from the previous year. The sum of operating costs and commissions for both years was(A) $1,000 million (B) $1,050 million (C) $1,100 million(D) $1,150 million (E) $1,155 million

Solution: 1998 = $500 M1999 = $550 M

Sum = $ 1050 MAns:B

47. Fox jeans regularly sell for $15 a pair and Pony jeans regularly sell for $18 a pair. During a sale these regular unit prices are discounted at different rates so that a total of $9 is saved by purchasing 5 pairs of jeans: 3 pairs of Fox jeans and 2 pairs of Pony jeans. If the sum of the two discounts rates is 22 percent, what is the discount rate on Pony jeans?

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(A) 9%(B) 10%(C) 11%(D) 12%(E) 15%Soln: Ans : B

Total discount is 22% = $9

by back solving, in this case ,the discount percent for pony jeans should be less than 11% (22/2)because the price of this product is more.

take choice B10% of 18= 1.8 total discount on 2 pony jeans= $3.622%-10%=12%12% of 15 = $1.8total discount on 3 fox jeans = $5.43fox jeans discount +2 pony jeans discount =$95.4 +3.6=9so answer is B.

48. There are 2 kinds of staff members in a certain company, PART TIME AND FULL TIME. 25 percent of the total members are PART TIME members others are FULL TIME members. The work time of part time members is 3/5 of the full time members. Wage per hour is same. What is the ratio of total wage of part time members to total wage of all members. A.1/4 B.1/5 C 1/6 D 1/7 E 1/8

Soln: What is the ratio of total wage of part time members to total wage of all members.

You have calculated ratio of Part time-to-Full time.

P x/4 3y/5 3xy/4*5 A 3x/4 Y + x/4 3y/5 18xy/20

Ratio= p/a= 3xy/18xy= 1/6

49. If 75% of a class answered the 1st question on a certain test correctly, 55% answered the 2nd question on the test correctly and 20% answered neither of the questions correctly, what percent answered both correctly?10%20%30%

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50%65%Soln: This problem can be easily solved by Venn Diagrams Lets think the total class consists of 100 students

so 75 students answered question 1and 55 students answered question 2

Now 20 students not answered any question correctly

Therefore out of total 100 students only 80 students answered either question 1 or question 2 or both the questions...

So 75+55=130 which implies 130-80=50% are the students who answered both correctly and are counted in both the groups...that’s why the number was 50 more..

Let me know if someone has problems understanding...

50. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9B). 4 and 4C). 3 and 5D). 2 and 6E). 0 and 8

Soln: SD = Sqrt(Sum(X-x)^2/N)Since N is changing from 5 to 7. Value of Sum(X-x)^2 should change from 40(current) to 48. So that SD remains same.

so due to new numbers it adds 8. Choice D only fits here.

51. How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A.14 B.15. C.16 D.17 E.18

Soln: if we arrange this in AP, we get4+7+10+.......+49

so 4+(n-1)3=49: n=16C is my pick

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52. If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Soln: The idea is to find what are the common factors that we get in the answer.

m = 1, k = 30 which is divisible by 1,2,3,5,6,10, 15 and 30m = 2, k = 84 which is divisible by 1,2,3,4,6, ....

As can be seen, the common factors are 1,2,3,6

So answer is 3 and 6

53. If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to(A) 2/3(B) 3/4(C) 4/3(D) 3/2(E) 2Soln: and the answer would be B...here is the explanation...

Let the side of the square be s..then the perimeter of the square is 4sLet the radius of the circle be r..then the perimeter of the circle is 2*pi*r

it is given that both these quantities are equal..therefore

4s=2*pi*r

which is then s/r=pi/2

Now the ratio of area of square to area of circle would be

s^2/pi*r^2

(1/pi)*(s/r)^2

= (1/pi)*(pi/2)^2 from the above equality relation

pi=22/7 or 3.14

the value of the above expression is approximate =0.78 which is near to answer B

54. Two people walked the same distance, one person's speed is between 25 and 45,and if he used 4 hours, the speed of another people is between 45 and 60,and if he

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used 2 hours, how long is the distance? A.116 B.118 C.124 D.136 E.140

Soln: First person-- speed is between 25mph and 45mphso for 4 hrs he can travel 100 miles if he goes at 25mph speedand for 4 hrs he can travel 130 miles if he goes at 45 mph speed

similarly

Second person-- speed is between 45mph and 60mphso for 2 hrs he can travel 90 miles if he goes at 45mph speedand for 2 hrs he can travel 120 miles if he goes at 60 mph speed

So for the first person---distance traveled is greater than 100 and Less than 130

and for the second person---distance traveled is greater than 90 and less than 120

so seeing these conditions we can eliminate C, D, and E answers...

but I didn’t understand how to select between 116 and 118 as both these values are satisfying the conditions...

55. How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even and how many odd?Soln: Odd: fix last as odd, 3 ways __ __ _3_now, left are 5, but again leaving 0, 4 for 1st digit & again 4 for 2nd digit: _4_ _4_ _3_ =48 Odd.

100-48= 52 Even

56. How many 3-digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number?

A) 16 B) 80 c) 160 D) 180 E) 240

Soln: The first digit can be 2, 3, 5, or 7 (4 choices)The second digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 choices)The third digit can be 2, 3, 5, or 7 (4 choices)

4 * 4 * 10 = 160

57. There are three kinds of business A, B and C in a company. 25 percent of the total revenue is from business A; t percent of the total revenue is from B, the others are from C. If B is $150,000 and C is the difference of total revenue and 225,000, what is t?

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A.50 B.70 C.80 D.90 E.100

Soln: Let the total revenue be X.

So X= A + B+ C

which is X= 1/4 X+ 150 + (X-225)

4X= X+ 600 + 4X -900

Solving for X you get X= 300

And 150 is 50% of 300 so the answer is 50 % (A)

58. A business school club, Friends of Foam, is throwing a party at a local bar. Of the business school students at the bar, 40% are first year students and 60% are second year students. Of the first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both. Of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both. A business school student is chosen at random. If the student is drinking beer, what is the probability that he or she isalso drinking mixed drinks?

A. 2/5B. 4/7C. 10/17D. 7/24E. 7/10

Soln: The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

The total number of possible outcomes is the total percent of students drinking beer.

40% of the students are first year students. 40% of those students are drinking beer. Thus, the first years drinking beer make up (40% * 40%) or 16% of the total number of students.

60% of the students are second year students. 30% of those students are drinking beer. Thus, the second years drinking beer make up (60% * 30%) or 18% of the total number of students.

(16% + 18%) or 34% of the group is drinking beer.

The outcomes that result in A is the total percent of students drinking beer and mixed drinks.

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40% of the students are first year students. 20% of those students are drinking both beer and mixed drinks. Thus, the first years drinking both beer and mixed drinks make up (40% * 20%) or 8% of the total number of students.

60% of the students are second year students. 20% of those students are drinking both beer and mixed drinks. Thus, the second years drinking both beer and mixed drinks make up (60% * 20%) or 12% of the total number of students.

(8% + 12%) or 20% of the group is drinking both beer and mixed drinks.

If a student is chosen at random is drinking beer, the probability that they are also drinking mixed drinks is (20/34) or 10/17.

59. A merchant sells an item at a 20% discount, but still makes a gross profit of 20 percent of the cost. What percent of the cost would the gross profit on the item have been if it had been sold without the discount?

A) 20% B) 40% C) 50% D) 60% E) 75%

Soln: Lets suppose original price is 100.

And if it sold at 20% discount then the price would be 80

but this 80 is 120% of the actual original price...so 66.67 is the actual price of the item

now if it sold for 100 when it actually cost 66.67 then the gross profit would be 49.99% i.e. approx 50%

60. If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?

A. 42,500B. 37,500C. 45,000D. 40,000E. 50,000

Soln: This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)

Using the Multiplication Principle:

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= 8 * 10 * 10 * 10 * 5= 40,000

61.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

A. 100B. 25C. 50D. 75E. 3600

Soln: The first step in this problem is to calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. Since order of arrangement does not matter, this is a combination problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r!))

The number of combinations of alcoholic ingredients is

C(5,2) = 5!/(2!(3!))C(5,2) = 120/(2(6))C(5,2) = 10

The number of combinations of non-alcoholic ingredients is

C(5,2) = 5!/(2!(3!))C(5,2) = 120/(2(6))C(5,2) = 10

The number of ways these ingredients can be combined into a drink can be determined by the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

The number of possible drinks is

= 10 * 10= 100

62. The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=?

Soln: The sum of numbers between 1 and n is = (n(n+1))/2

1+2+3+.....+n=(n(n+1))/2 {formula}

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we are looking for the sum of the even numbers between 1 and n, which means:

2+4+6+.....+(n-1) n is ODD=1*2+2*2+2*3+......+2*((n-1)/2)=2*(1+2+3+.....+*((n-1)/2))from the formula we obtain :=2*(((n-1)/2)*((n-1)/2+1))/2=((n-1)/2)*((n+1)/2) =79*80=> (n-1)*(n+1)=158*160=> n=159

63. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510(B) 2620(C) 1404(D) 700(E) 635

Soln: 4W 2M == 5C4.(8C2-1) = 5.(27) = 1353W 3M == 5C3.(8C3-6) = 10.50 = 500total 635

Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.consider it this way.... from my previous reply max possible ways considering we can chose any man = 700

now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...

since they are always chosen together...

For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women

1*5C4 = 5

For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women

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6C1*5C3 = 60

So total number of unfavorable cases = 5 + 60 = 65

Now since we want to exclude these 65 cases... final answer is 700-65 = 635

64. In how many ways can the letters of the word 'MISSISIPPI' be arranged?

a) 1260b) 12000c) 12600d) 14800e) 26800

Soln: Total # of alphabets = 10so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.we have 4 Is3 Sand 2 Ps

Hence # of arrangements = 10!/4!*3!*2!

65. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?(A) 720(B) 360(C) 120(D) 24(E) 21

Soln: two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!

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if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

66. On how many ways can the letters of the word "COMPUTER" be arranged?1. M must always occur at the third place2. Vowels occupy the even positions.

Soln: For 1.7*6*1*5*4*3*2*1=5,040

For 2.) I think It should be 4 * 720there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

67. A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?

Soln: There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV’s and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.

68. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420

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B. 2520 C. 168 D. 90 E. 105

Soln: 1st team could be any of 2 guys... there would be 4 teams (a team of A&B is same as a team of B&A)... possible ways 8C2 / 4.2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B is same as a team of B&A)... possible ways 6C2 / 33rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B is same as a team of B&A). Possible ways 4C2 / 2 4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1

total number of ways...

8C2*6C2*4C2*2C2 -------------------4 * 3 * 2 * 1

=8*7*6*5*4*3*2*1--------------------4*3*2*1*2*2*2*2

= 105 (ANSWER)...

Another method: say you have 8 people ABCDEFGH

now u can pair A with 7 others in 7 ways.Remaining now 6 players.Pick one and u can pair him with the remaining 5 in 5 ways.

Now you have 4 players.Pick one and u can pair him with the remaining in 3 ways.

Now you have 2 players left. You can pair them in 1 way

so total ways is 7*5*3*1 = 105 ways i.e. E

69. In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?Soln: The ways of arranging 5 people in a circle = (5-1)! = 4!For a person seated with 2 neighbors, the number of ways of that happening is 2: AXB or BXA, where X is the person in question.

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So, for each person, we have two such arrangements in 4!. Since we don't want to repeat arrangement, we divide 4!/2 to get 12

70. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!

B) 20!/5(4!)

C) 20!/(4!)^5

D) 20!

E) 5!

Soln: 4 copies each of 5 types.

Total = 20 books.Total ways to arrange = 20!

Taking out repeat combos = 20!/(4! * 4! * 4! * 4! * 4!) – each book will have 4 copies that are duplicate. So we have to divide 20! By the repeated copies.

71. In how many ways can 5 rings be worn on the four fingers of the right hand?Soln: 5 rings, 4 fingers1st ring can be worn on any of the 4 fingers => 4 possibilities2nd ring can be worn on any of the 4 fingers => 4 possibilities3rd ring can be worn on any of the 4 fingers => 4 possibilities4th ring can be worn on any of the 4 fingers => 4 possibilities5th ring can be worn on any of the 4 fingers => 4 possibilities

Total possibilities = 4*4*4*4*4 = 4^5.

72. If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

Soln: Write down n x (2^5) x (6^2) x (7^3) as= n x (2^5) x (3^2) x (2^2) x (7^3), = n x (2^7) x (3^2) x (7^3)

now at a minimum 5^2 and a 3 is missing from this to make it completely divisible by 5^2 x 3^3

Hence answer = 5^2 x 3 = 75

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73. Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

(1) 10050 (2) 5050 (3) 5000 (4) 50000

Soln: Consider 5 15 25 ... 995l = a + (n-1)*d

l = 995 = last terma = 5 = first termd = 10 = difference

995 = 5 + (n-1)*10

thus n = 100 = # of terms

consider 5 10 15 20.... 995

995 = 5 + (n-1)*5

=> n = 199

Another approach...

Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)

Also, series is 5 15 25.... 985 995

# of terms = 100

sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000

74. If the probability of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period?

8/1252/255/168/253/4 Soln: Use binomial theorem to solve the problem....

p = 1/2q = 1/2

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# of favorable cases = 3 = r# of unfavorable cases = 5-3 = 2total cases = 5 = n

P(probability of r out of n) = nCr*p^r*q^(n-r)

75. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336B. 1/120C. 1/56D. 1/720E. 1/1440

The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

There is only one result that results in a win: receiving three aces.

Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r)!)

C(8,3) = 8!/(3!(8-3)!)C(8,3) = 8!/(3!(5!))C(8,3) = 40320/(6(120))C(8,3) = 40320/720C(8,3) = 56

The number of possible outcomes is 56.

Thus, the probability of being dealt 3 aces is 1/56.

76. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336B. 1/120C. 1/56

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D. 1/720E. 1/1440

Soln: Since each draw doesn't replace the cards:Prob. of getting an ace in the first draw = 3/8getting in the second, after first draw is ace = 2/7getting in the third after the first two draws are aces = 1/6thus total probability for these mutually independent events = 3/8*2/7*1/6 = 1/56

77. Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains

A) exactly 1 woman B) at least 1 woman C) at most 1 woman

Soln:

A.) 7C1* 11C3/ 18C4B) 1 - (11C4/18C4)C) (11C4/18C4) + (7C1*11C3/18C4)

78. A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on a particular Saturday morning. Because there are so few mechanics, only 6 can be serviced. (a) If the 6 cars are chosen at random, what is the probability that 3 of the cars selected are domestic and the other 3 are foreign? (b) If the 6 cars are chosen at random, what is the probability that at most one domestic car is selected?

Soln: A) 10C3*15C3/25C6B) Probability of no domestic car + Probability of 1 domestic car = 10C6/25C6 + 15C1 *10C5/25C6

79. How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21?

A. 4,514B. 4,475C. 4,521D. 4,428E. 4,349

Soln: We first determine the number of integers less than 5,000 that are evenly divisible by 15. This can be found by dividing 4,999 by 15:

= 4,999/15= 333 integers

Now we will determine the number of integers evenly divisible by 21:

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= 4,999/21= 238 integers

some numbers will be evenly divisible by BOTH 15 and 21. The least common multiple of 15 and 21 is 105. This means that every number that is evenly divisible by 105 will be divisible by BOTH 15 and 21. Now we will determine the number of integers evenly divisible by 105:

= 4,999/105= 47 integers

Therefore the positive integers less than 5000 that are not evenly divisible by 15 or 21 are 4999-(333+238-47)=4475

80) Find the least positive integer with four different prime factors, each greater than 2.

Soln: 3*5*7*11 = 1155

81) From the even numbers between 1 and 9, two different even numbers are to be chosen at random. What is the probability that their sum will be 8?

Soln: Initially you have 4 even numbers (2,4,6,8)you can get the sum of 8 in two ways => 2 + 6 or 6 + 2so the first time you pick a number you can pick either 2 or 8 - a total of 2 choices out of 8 => 1/2after you have picked your first number and since you have already picked 1 number you are left with only 2 options => either (4,6,8) or (2,4,8) and you have to pick either 6 from the first set or 2 from the second and the probability of this is 1/3. Since these two events have to happen together we multiply them. ½ * 1/3 = 1/6

82) 5 is placed to the right of two – digit number, forming a new three – digit number. The new number is 392 more than the original two-digit number. What was the original two-digit number?

Soln: If the original number has x as the tens digit and y as the ones digit (x and y are integers less than 10) then we can set up the equation:100x + 10y + 5 = 10x + y + 39290x + 9y = 3879(10x+y) = 38710x + y = 43 ==> x = 4, y = 3the original number is 43, the new number is 435

83) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000

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the first one is 8 and the last one is 1000==> ((1000-8)/8) + 1 = 125==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

84) A serial set consists of N bulbs. The serial set lights up only if all the N bulbs are in working condition. Even if one of the bulbs fails then the entire set fails. The probability of a bulb failing is x. What is the probability of the serial set failing?

Soln: Probability of x to fail. Probability of a bulb not failing = 1-xprobability that none of the N bulbs fail, hence serial set not failing = (1-x)^Nprobability of serial set failing = 1-(1-x)^N

85) Brad flips a two-sided coin 8 times. What is the probability that he gets tails on at least 7 of the 8 flips?

1/32

1/16

1/8

7/8

none of the above

Soln: Number of ways 7 tails can turn up = 8C7 the probability of those is 1/2 each Since the question asks for at least 7, we need to find the prob of all 8 tails - the number of ways is 8C8 = 1Add the two probabilities8C7*(1/2)^8 = 8/2^8 -- for getting 7 Prob of getting 8 tails = 1/2^8Total prob = 8/2^8+1/2^8 = 9/2^8

Ans is E.

86. A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller than the person standing in front of him or her. How many such arrangements of the 6 people are possible?

A. 5B. 6C. 9D. 24

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E. 36

Soln: 5 ways

123 124 125 134 135 456 356 346 256 246

87. As a part of a game, four people each much secretly chose an integer between 1 and 4 inclusive. What is the approximate likelihood that all four people will chose different numbers?

Soln: The probability that the first person will pick unique number is 1 (obviously) then the probability for the second is 3/4 since one number is already picked by the first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively. Their product 3/4*1/2*1/4 = 3/32

88. Which of the sets of numbers can be used as the lengths of the sides of a triangle?

I. [5,7,12]II. [2,4,10]III. [5,7,9]

A. I onlyB. III onlyC. I and II onlyD. I and III onlyE. II and III only

Soln: For any side of a triangle. Its length must be greater than the difference between the other two sides, but less than the sum of the other two sides.Answer is B

89. A clothing manufacturer has determined that she can sell 100 suits a week at a selling price of 200$ each. For each rise of 4$ in the selling price she will sell 2 less suits a week. If she sells the suits for x$ each, how many dollars a week will she receive from sales of the suits?

Soln: Let y be the number of $4 increases she makes, and let S be the number of suits she sells. ThenX = 200 + 4y ==> y = x/4 - 50S = 100 - 2y ==> S = 100 - 2[x/4 - 50] = 100 - x/2 + 100 = 200 - x/2

so the answer is that the number of suits she'll sell is 200 - x/2

90. A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the

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price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

A) 20, 35, 70

B) 20, 45, 70

C) 20, 35, 40

D) 35, 40, 70

E) 35, 40, 45

Soln: Add the 5 prices together:20 + 35 + 40 + 45 + 70 = 210

2% of that is 210 x .02 = 4.20

Let x be the stock that rises and y be the stock that falls.

.15x -.35y = 4.20 ==> x = (7/3)y + 27

This tells us that the difference between x and y has to be at least 27. Since the answer choices list the ones that DONT change, we need to look for an answer choice in which the numbers NOT listed have a difference of at least 27.

Thus the answer is (E)

91. if -2=<x=<2 and 3<=y<=8, which of the following represents the range of all possible values of y-x?(A) 5<=y-x<=6(B) 1<=y-x<=5(C) 1<=y-x<=6(D) 1<=y-x<=10 (E) 1<=y-x<=10

Soln: you can easily solve this by subtracting the two inequalities. To do this they need to be in the opposite direction; when you subtract them preserve the sign of the inequality from which you are subtracting.

3 < y < 8multiply the second one by (-1) to reverse the sign2 > x > -2Subtract them to get3 - 2 < y - x < 8 - (-2)1 < y - x < 10

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92. Of a group of 260 people who purchased stocks, 61 purchased A, 88 purchased B, 56 purchased C, 75 purchased D, 60 purchased E. what is the greatest possible number of the people who purchased both B and D?

A:40 B:50 C:60 D:75 E:80

Soln: ANSWER is D (75)

since they have asked us to find out the greatest possible number buying both B as well as D, the answer has to be the smallest no between the two which is 75...as all the guys purchasing D can also buy B and only 75 out of 88 purchasing B can simultaneously purchase D as well....

93. There are 30 people and 3 clubs M, S, and Z in a company. 10 people joined M, 12 people joined S and 5 people joined Z. If the members of M did not join any other club, at most, how many people of the company did not join any club?

A: 4 B: 5 C: 6 D: 7 E: 8

Soln: total no of people = 30no joining M = 10no joining S = 12no joining Z = 5question asked - AT MOST how many people did not join any group?

solution: now since none of the members of M joined any other group, the no of people left = 30-10(for M)=20since the question says at most how many did not join any group, lets assume the all people who join Z also join S. so no of people joining group S and Z are 12 (note that there will be 5 people in group S who have also joined Z)

therefore no of people not joining any group = 20-12=8Hence e

94. Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

1) in a row2) in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats3) around a table

A:1) 4! * 4! * 22) 4! * 4! * 2 - number of ways with John and Susan sitting together= (4! * 4! * 2) - (7 * 3! * 3! *2)

The way that JS arrangements are found is by bracketing J and S and considering it to be a single entity. So a possible arrangement is (B=boy, G=girl)

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(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252(SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252

3) Fix one boy and arrange the other 3 boys in 3! ways. Arrange the 4 girls in 4! ways in the gaps between the boys.

Total arrangements = 3! x 4!

= 6 x 24

= 144

95. From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3

Soln: Total number of ways of selecting 4 children = 6C4 = 15

with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

Hence p = 9/15 = 3/5

96. What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?

(1) 0 (2) 3 (3) 4 (4) 2 (5) None of these

Soln: Remainder of 9*odd /6 is 3remainder of 9*even/6 is 0

9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)

1+9+9^2+.....9^8 is odd.

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Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.

Another way: We get the value as 9*odd/6 = 3*odd/2 Since 3*odd = odd; odd/2 = XXXX.5so something divided by 6, gives XXXX.5, hence remainder is 6*0.5 = 3

97. Find the value of 1.1! + 2.2! + 3.3! + ......+n.n!

(1) n! +1 (2) (n+1)!(3) (n+1)!-1(4) (n+1)!+1(5) None of these

Soln: 1.1! + 2.2! + 3.3! + ......+n.n!=1.1! + (3-1)2! + (4-1)3! +......+ ((n+1)-1) n!=1.1!+3!-2!+4!-3!+.......+(n+1)!-n!

So it is (n+1)! -1 (Answer choice 4)

98. The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x < y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.II. The tens digit of x + y equals 2.III. The hundreds digit of y is at least 5.

A. II only

B. III only

C. I and II

D. I and III

E. II and III

Soln: x= abc y= def

x = a7c y= b5f

x > y and x+y = wxyz.

I. The units digit of x + y is greater than the units digit of either x or y.

It can carryover one digit. False

II. The tens digit of x + y equals 2.

It can be 2 or 3. False

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III. The hundreds digit of y is at least 5.

a+b+1 >= 10a >b so a at least 5. True.

Ans: b

99. Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together? (A) 38 (B) 46 (C) 72 (D) 86 (E) 102Soln: The total number of ways 5 of them can sit is 120when the siblings sit together they can be counted as one entitytherefore the number of ways that they sit together is 4!=24, but sincethe two siblings can sit in two different ways e.g. AB and BA we multiply 24 by 2 to get the total number of ways in which the 5 children can sit together with the siblings sitting together - 48In other words 4P4*2P2the rest is obvious 120-48=72

100. There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

Soln: MuEuG = M + E + G - MnE - MnG - EnG - 2(MnEnG)MnE + MnG + EnG = M + E + G - 2(MnEnG) - MuEuGMnE + MnG+ EnG = 40 + 30 + 35 - 2(15) - 70 = 105 - 30 - 70 = 5

Whenever an intersection occurs between 2 sets, (MnEnG) is counted twice, therefore you deduct one of it. If the intersection occurs between 3 sets, it is counted thrice; therefore you deduct two of it. And so forth.If there are four sets, then the formula is A + B + C + D -(two) -(three)*2 -(four)*3 = total 101. John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

A- 6 B- 10 C- 8 D- 7.5 E- 3.5

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Soln: Together they do 1/20+1/12=4/30 of the task.

Jane and John started the work together, but only John finished the work because Jane gets sick. So let x be the number of days they worked together.

x*3/40+4*1/20=1

x4/30=4/5 and therefore x =6

So in total they worked 6 days on it together and John worked 4 days on it. So total days spent=10, but if the question is asking how many time did they spend working on the project together, then the answer is 6.

102. Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

Soln: Jane gave Karen a 5 m head start means Karen was 5 m ahead of Jane. So after the lead, Karen ran 95m and Jane ran 99.75 m when the race ended.

Let speed of Karen and Jane be K and J respectively and lets say after X minutes Jane overtakes Karen.

1st condition: 95/K = 99.75/J2nd condition JX - KX = 0.25

Solving for JX, we get JX=21/4.Hence Jane needs to run 5.25m more (or total of 105m) to overtake Karen.

103. Out of 20 surveyed students 8 study math and 7 study both math and physics. If 10 students do not study either of these subjects, how many students study physics but not math?

(A) 1 (B) 2 (C) 4 (D) 5 (E) 6Soln: total = gr1 + gr2 - both + neither20 = 8 + P - 7 + 10P = 9, 9 students study both P and M, 7 study both, 9-7 = 2 study only P

104. Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?

(A) 6

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(B) 7 (C) 9 (D) 10 (E) 11

Soln: Let x be the number of days that need to be worked 500x - 1500 > 400x - 550100x > 950x > 9.5 (D)

105. Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?

A. 2/3B. 7/3C.5/2D. 8/3E. 3

Soln: x + y = 4x = 4 - y

x/2 = y/44x = 2y4(4-y) = 2y16 = 6yy = 8/3 (D)

106. If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36

Soln: There are 10C2 ways to pick 2 different people out of 10 people.

10C2 = 10!/2!8! = 45 (D)

107. If operation $ is defined as $X = X + 2 if X is even $X = X - 1 if X is odd,

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what is $(...$($($(15)))...) 99 times?

(A) 120 (B) 180 (C) 210 (D) 225 (E) 250

Soln: $15 gives 14

After that it is an AP with d=2, a=14 and n =99

108. A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

1- Find the probability that the person who tosses first will win the game?

2- What are the odds against A's losing if she goes first?

Soln: look at the conditions; it says that the first person who tosses a head wins.

Let’s say A tosses first.

what is the probability that he wins

H + TTH + TTTTH + TTTTTTH + TTTTTTTTH

i.e. either the first toss is head, or the first time A tosses the coin he gets a tail and B also gets a tail , n in the second throw A gets a head.....

This continues for a max till 5 throws, because the game is for 5 throws only.So, 1. 1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9

2. (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10

109. How many integers less than 1000 have no factors (other than 1) in common with 1000?

(1) 400 (2) 410(3) 411 (4) 412(5) None of the above

Soln: 1000 - multiples of 2 and/or 5

multiples of 2 = 500 (all even #)

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multiples of 5 = (995 -5)/10 + 1 [ Using AP formula]= 100

Answer = 1000 - (500 + 100)= 400

You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The difference in the AP series should be 10 instead of 5 because you're looking for the integers that have 5 as a unit’s digit. Therefore we divide by 10 and not 5.

110. Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

36, 28, 12, 9 or none

Soln: Let the divisor be a.

x = a*n + 11 ---- (1)y = a*m + 21 ----- (2)also given, (x+y) = a*p + 4 ------ (3)adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4)

equate 3 and 4.a*p + 4 = a*(n+m) + 32or a*p + 4 = [a*(n+m) + 28] + 4cancel 4 on both sides.u will end up with.a*p = a*(n+m) + 28.

which implies that 28 should be divisible by a. or in short a = 28 works.

Another method: I think the easiest (not necessarily the shortest), way to solve this is to use given answer choices. Since the remainders are given as 11 and 21, therefore the divisor has to be greater than 21 which leaves with two choices 28 and 36. Try 28 first; let the two numbers be 28+11= 39 and 28+21= 49. Summing them up and dividing by 28 gives (49+39=88), 88/28 remainder is 4, satisfies the given conditions. Check for 36 with same approach, does not work, answer is 28

111. There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

Soln: let’s assume Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.so the total is (1c1.6c3)/8c4 which is 2/7

112. How many 5-digit positive integers exist where no two consecutive digits are the same?

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A.) 9*9*8*7*6 B.) 9*9*8*8*8 C.) 9^5D.) 9*8^4 E.) 10*9^4

Soln: C is correct. The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.Hence 9^5

113. How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (A) 15 (B) 96 (C) 216 (D) 120 (E) 180

Soln: The sum of digits of a multiple of 3 should be div by 3.for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15.For a sum of 12, the digits that can be used : 0,1,2,4,5for a sum of 15: 1,2,3,4,5Number of numbers from the first set = 4.4! (0 cannot be the first digit in the numbers)for the second set : 5!total = 5! +4.4! = 4!(5+4) = 24*9 = 216

Since 0 cannot be the first digit of a number, for the first position, you have 4 choices (all digits except zero). No such constraints exist for the rest of the positions; hence the next choices are 4,3,2,1 - all multiplying up to give a 4!. Had there been no 0 involved, the choices would've been 5! Instead of 4.4!

114. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420 B. 2520 C. 168 D. 90 E. 105

Soln: out of 8 people one team can be formed in 8c2 ways.

8c2*6c2*4c2*2c2= 2520.The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )

115. My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

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a) 5% b) 10% c) 20% d) 25%

Soln: there are actually 20 ways to interchange the letters, namely, the first letter could be one of 5, and the other letter could be one of 4 left. So total pairs by product rule = 20.

Now, there are two cases when it wouldn’t change the name. First, keeping them all the same. Second, interchanging the two EEs together. Thus 2 options would leave the name intact.

Prob = 2/20 = 0.1, or 10%.

116. A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

A. y > ROOT2

B. ROOT3/2 < y < ROOT2

C. ROOT2/3 < y < ROOT3/2

D. ROOT3/4 < y < ROOT2/3

E. y < ROOT3/4

Soln: right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1i.e xy = 2

If x were equal to y, we would have had xy = y^2 = 2. And y = root2

But, x<y and so y>root2.

117. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

Soln: Chance of drawing a blue on the first draw = 2/8, so chance of not drawing a blue on the first draw is 6/8

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similarly chance of not drawing blue on second draw = 5/7

Therefore probability of not drawing blue in 2 draws = 6/8*5/7 = 15/28

118. How many integers between 100 and 150, inclusive can be evenly divided by neither 3 nor 5?

Soln: Number of integers that divide 3:

the range is 100-150Relevant to this case, we take 102 - 150 (since 102 is the first to div 3)102 = 34*3150= 50*3, so we have 50-34+1 = 17 multiples of 3

For multiples of 5,100=5*20150=5*3030-20+1 =11

Now we have a total of 27 integers, but we double counted the ones that divide BOTH 3 AND 5, ie 15.

105 is the first to divide 15.105=15*7150=15*1010-7+1 = 4 integers

So our total is 17+11-4 = 24 integers that can be divided by either 3 or 5 or both.

51 integers - 24 integers = 27 that cannot be evenly divided.

119. Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?

(A) 10% (B) 33 1/3%

(C) 40%

(D) 50%

(E) 66 2/3%

Soln: (x+y)30/100 = x*40/100 + y*25/10030x + 30y = 40x + 25yy = 2x or y/x = 2/1 or y:x = 2:1 hence x = 33 1/3%

120. Sequence A and B. a1=1, b1=k. an=b(n-1)-a(n-1) bn=b(n-1)+a(n-1). What is a4=?

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Soln: a2 = k-1 ; b2 = k+1

a3= (k+1)-(k-1) = 2 ; b3 = (k+1)+(k-1) = 2k

a4 = 2k - 2 = 2(k-1) = 2(b1-a1)

121. A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum? A) 38% B) 42% C) 19% D) 40%

Soln: assume, interest = rso after 3 years total money = 1000*(1+r)^3 = 1000*1.19(1+r)^3 = 1.19so after 6 years total money = 1000*(1+r)^6 = 1000*1.19^2 = 1000*1.42so percentage of interest is 42%

122. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

A. 750 secondsB. 50 secondsC. 250 secondsD. 375 secondsE. 75 seconds

Soln: When two people are running in the same direction the relative speed is a difference in speeds of the two people.

In this case A=3 B=6 C=18

So relative speed of B wrt A is 6-3 = 3m/sRelative speed of A wrt to C is 18-3 =15m/s

Therefore relative distances will be:B wrt A is 750/3 =250C wrt to A 750/15 = 50

So they have to bridge this distance of 250 and 50 between them which is the LCM of 250 and 50 which is 250.

Another Method: Simply put, Runner A's time take to run one lap is 250Runner B's time is 125sand Runner C's time is 41.67s

We can notice that A=2Band thet B=3C

So when 250 s elapse, they will be at their starting point. A will have completed one lap, B 2 laps, and C 6 laps

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123. X percents of the rooms are suits, Y percent of the rooms are painted light blue. Which of the following best represents the least percentage of the light blue painted suits?

1) X-Y2)Y-X +1003)100X-Y4)X+Y-100e)100-XY

Soln: Equation from set theory:

n(AUB)=n(A)+n(B)-n(A^B)

where,

A= % of rooms which are suitesB= % of rooms painted blueA^B means the intersection of the two sets

Now in this case, what we need to find is n(A^B), thereforen(A^B)=n(A)+n(B)-n(AUB)= X + Y - n(AUB)

Now this would be least when n(AUB) is maximum, which would happen if these two kinds of rooms are only two kinds available, making n(AUB)=100

Therefore the answer should be X+Y-100

All Rate Questions

16. In the first half of a certain trip, a car maintained an average speed of 40m/h; in the second half of the trip, the car maintained an average speed of 60m/h. What is the average speed for the whole trip?

avg speed = 2* product of speeds/ sum of speeds2*40*60/(40+60) = 4800/100 = 48 ( avg speed = 2ab/a+b)

31. A ship started off at the rate of a. t hours later, in the same direction, another ship started off at the rate of b. In how many hours the second ship will catch up with the first ship?

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I think there are actually two assumptions that were probably given in the original question, but are left out here:

1) speed b > speed a  (though, this may simply be implied by the wording, anyway)2) ship 'a' and ship 'b' start from the exact same point.

If both of these are true, I would solve it like this:

ship 'a' is moving at a speed of a for t hours (covering some distance) before ship 'b' begins moving toward ship 'a' from the same location that ship 'a' started from.This means that all ship 'b' has to do to catch ship 'a' is make up that distance (essentially, ship 'b' gave ship 'a' a headstart).  To calculate that head start distance, you simply take ship 'a's speed times the hours it traveled at that speed:

head start distance = a * t

Now that we have the distance ship 'b' needs to make up as it chases ship 'a', we just need to calculate the rate at which it will make up this distance. That is, we need to calculate ship 'b's speed RELATIVE TO ship 'a'.  This is simple:

relative speed = b - a

So, the question asks how much time, in hours, it will take ship 'b' to catch up to ship 'a'. This can be calculated using the common formula d = r * t  (distance = rate * time).Rearrange it to solve for TIME, since that's what the question asks for...

             time = distance / rate

Here, distance = "head start distance" = a * t and   rate = "relative speed" = b - a

Plug in, and solve...    time = (a * t) / (b - a)   hours

39. People A and B together can finish 800 products in x hours. If it takes y hours for A alone to finish the 800 products, how many hours will it take for B alone to finish the 800 products?

1/a+1/b = 1/x

1/y+1/b = 1/x

1/b = 1/x - 1/y

1/b = y-x/xy

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b = xy/y-x

where b will be the total hours that B takes to complete 800 products.

40. Tickets for a movie include A, with rate of $4.5, and B, with rate of $6. If 60% of the tickets sold are tickets B, what percent of the total revenue on tickets sales are from B?

Let's assume n\the number of B tickets = 60, so the number of A tickets = 40

Total amount = 60*$6+40*$4.5 = $540,

Percent of the total revenue on tickets B sales = 360/540*100% = 66.6%

41. After every 45 minutes trip, a car will take a 15 minutes break. In highway, the car¡¯s rate is 60m/h, and in country road, the rate is 40m/h. In a 4 hours trip, the car can travel how many more miles in highway than in country road?

53. Three machines can finish a certain work in 36 hours at the same rate. If a 4th machine with the same rate is added in, what is the time to finish the work?

3 machines - 36 hrs, so each machine takes 12 hrs. Fourth machine also has this rate. 4 machines can finish the the work in 36*3/4 = 27 hours.

63. With a speed of 20m/h, a truck will passes through a 4024 feet bridge in approximate how many minutes?

Speed = 20m/hr = (20 * 5280)/60 feet/min (1 mile=5280 feet)

=> Speed = 1760 feet/min

Therefore, time taken = 4024/1760 = 2.28 mins = 2.3 mins

73. A people walked from signpost 1 to signpost 2 at the average rate of 0.125km/m. Is the distance between two signposts greater than 0.8km?1). It cost more than 400 second for the people finish the trip2). It cost less than 450 second for the people finish the trip.

A is the answer.

From the question stem:

People walk 0.125km/m or 1km/8 minutes ot 384 seconds for 0.8 km.

1). It cost more than 400 second for the people finish the trip.

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Is SUFF to answer the question -  the distance between two signposts is greater than 0.8km.

2). It cost less than 450 second for the people finish the trip. Clearly INSUFF.

it can be exactly 384 seconds or 425 seconds.

97. One tank has a capacity of 64800 gallon, one gallon water weights 8.3 kg. A pile can fill the tank with a constant speed in12 hours, what is the weight of the water filled into the tank per min?

747 is the answer.

64800/12 = 5400 gal/hour. 5400/60 = 90gal/minute. 90*8.3 kg = 747.

121. Mary goes from a to b and then from b to c. Speed for a to b is 40 feet/min and speed for b to c is 60feet /min. She took 3 more hrs to go from a to b and the distance between a and b was 90 miles more. What was the time taken for b to c?

1. she took less than 45 mins

2. she took more than 42 mins.

Answer E

i guess the question is:

A - B : Speed : 40ft/min 

Time : 3hrs more than the time taken to travel distance BC  

Distance : 90miles more than BC

B - C : Speed = 60ft/min

144. A car has a 15 gallon-tank. With one gallon gasoline, the car can travel 30 miles when driving in the city, 20miles in the countryside. What expression is the fraction of the full fuel tank of a car uses when driving 60 miles in city and 30 miles in the country?

7/30 it is.

Average mpg=total miles/gallons used = (60/30+30/20)15 = (7/2)/15 = 7/30

All Remainder Questions

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10. Is x divisible by 3?

1). When x is divided by 5, the remainder is 1

2). When x is divided by 15, the remainder is 1

a) The number could be 5 11 16 21 26 31... and varying remainders when divided by 3

b) The number could be 16 31 46 .. any number will give the same remainder when divided by 3

B it is 

58. What is the remainder when 9^845 is divided by 10?

9 it is.

Same explanation as Sangdil's one.

9^2 = 81, 9^3 = 729, patternt will be 1,9,1,9,1,9 ....

And when you divide the digit that ends on 9 by 10, the remainder, of course, will be 9.

64. When integer n is divided by 9, the remainder is 3. Is the integer divisible by 5?

1). When n is divided by 45, the remainder is 30

2). n is divisible by 2

Nos. are 3, 12, 21, 30, 39, 48, 57, 66, 75

1. Divided by 45, remainder is 30. so 30 and 75, divisible by 5

2. 12 and 30 are divisible by 2 but both not by 5. Insufficient.

Hence A

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77. What is the remainder when n is divided by 10?

1). The tens?digit of 11^n is 4

2). The hundreds?digit of 5^n is 6

A

11^4 = 14641 11^14 = xxxxxxxxxxxx241. (it will be 15 digit number u can check using calc.)

similarly 11 ^ 24 = ....841 (a 25 digit number)

we get n = 4 , 14,24,34,44.....   every time dividing by 10 leaves   4..

so stem(1) alone sufficient.

stem(2) after n>2 and n even we get 6 in hundreds place and n odd we get 1 in hundreds place .. it can be better understood by values.

5 ^ 3 = 125 5^4 = 625 5^5 = 3125 5^6 = 15625   so for every n >2 and n is even we get 6 in hundreds place

so n can have 4, 6, 8,10,12... when divided by 10 leaves diffirent values so not sufficient..

clearly answer is (A) ie stem(1) alone sufficient and stem(2) not sufficient.

83. When k^4 is divided by 32, the remainder is 0. Which of the following could be the remainder when k is divided by 32?

2, 4, 6?/P>

Should be 4.

Here is my reasoning. K^4=32n (n is positive integer.) K=2*4sqrt(2n) Because K should be integer, n should be 2^3, 2^7, 2^11, 2^15, etc. Then, K could be 4, 8, 16, 32.. Hence, 4 from choices.

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99. Is x^2+y^2 divisible by 5?

1). When x-y is divided by 5, the remainder is 1

2). When x+y is divided by 5, the remainder is 3

stem 1>> x-y can be 6, 11,16,21,26,31 etc

take 6 for eg, 6 in x-y can be 7-1, 8-2, 9-3,10-4 etc...insufficient, some do get satisfied, some dont..

stem 2>>> x+y can be 8,13,18,23,28 etc

take 8 for eg, 1+7, 2+6, 3+5, 4+4 etc...same case, insufficient...

combining 1 & 2,

x-y/5 with rem=1, x+y/5 with rem 3

7-1/5=5/5==rem 1, 7+1/5, 8/3==rem 3

12-1/5=11/5==rem 1, 12+1/5, 13/5==rem 3

17-1/5=16/5==rem 1, 17+1/5, 18/5 ==rem 3

guess, C is the answer

IMO-C

102. If 56<x<66, where x is an integer. x=?1). When x is divided by 2, the remainder is 12). x+1 is divisible by 3.

E it is.

1). When x is divided by 2, the remainder is 1

x can be 57, 59, 61, 63, 65 INSUFF

2). x+1 is divisible by 3.

x can be 59, 62, 65 INSUFF

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Both INSUFF as well.

107. When a positive integer is divided by 4, the remainder is r; when divided by 9, the remainder is R. What is the greatest possible value of r^2+R? [Uncertain]23, 21, 17, 13, 11

17 it is.

3^2+8=17.

maximum remainder when integer is divided by 4 is 3.

maximum remainder when integer is divided by 9 is 8.

All Sequence questions

4. Sequence s1,s2,s3,..sn, is such that s=1/n-1/n+1 . Is the sum of first n terms greater than 9/10?

1). n>10

2). n<19

The Sum is asked here.

Sn = 1/n - 1/n+1

Sum of first k terms will be:

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/k - 1/(k+1)

= 1 - 1/(k+1)

If k = 9 the Sum would be 1 - 1/10 = 0.9

Therefore, for any k>9 the sum will always be > 0.9.

Stmt 1: Sufficient;

Stmt2: Insufficient

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Reason: if k> 9 sum will be > 9/10

for k= 9 sum will be 9/10

for k < 9 sum will be > 9/10

IMO the answer is A.

8. S(n)=S(n-1)^S(N-2), S1=1,S2=2,S3=2^1....S6/S5=?

S(1) = 1

S(2) = 2

S(3) = 2^1.... (all of the above are given)

Now.............according to the formula

S(4) = S(3)^S(2)

       = (2^1) ^ 2 = 2^(1+1)=2^2..(Substitute S(3) and S(2))

S(5) = S(4)^S(3) = (2^2)^(2^1) = (2^2)^(2) = 2^4

S(6) = S(5)^S(4)= (2^4)^(2^2) = (2^4)^4 = 2^16

Now......

S(6)/S(5) =     2^16/2^4 = 2^12

163 = (42)3 = 46

46 = (22)6 = 212

23. The first term of a geometric sequence is 1/2, and the common ratio is 1/2. The 10th term will fall into which of the following range?

0.1~0.01, 0.01~0.001, 0.001~0.0001,0.0001~0.00001

C

1st term: (1/2)

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2nd term: (1/2) * (1/2) = (1/2) ^2

3rd term: (1/2) * (1/2) * (1/2) = (1/2)^3

....

So, therefore, the 10th term: (1/2)^10 = 1/1024

If you divide 1/1024, so the answer should be .001 (1/1000) ~.0001 (1/10000)

43. If an=3a(n-1)-x, and a5=99, a3=27, x=?A5 = 3A4 - X -----(1)

A4 = 3A3 -X  -----(2)

Let us substitue the value of A4 from (2) in (1)

=> A5 = 3(3A3 - X) -X = 9A3 - 4X

Given A5 = 99 and A3 = 27

=> 99 = 9 * 27 - 4X

=> 4X = 243 - 99

=> X = 144/4 = 36

Or

A4 = 3a(3) -x

a4=3*27-x

a4=81-x

 

a5=3a(4)-x

a5=3(81-x)-x

99=243-3x-x

-144=-4x

36=x

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88. An=An-1 +5, A5=31. A1=?

Answer:11

94. Sequence A consists of 10 consecutive odd numbers and B consists of 5 consecutive even numbers. If the least number in A is 7 greater than the least number in B, the average of the numbers in A is how much greater than average of numbers in B?

Picked numbers B : 2 4 6 8 10   | A : 9 11 13 15 17 19 21 23 25 27

Ans : 12

96. S=1/32 + 1/33 + ... +1/64, S is in which of the following range?

0.5<S<1

128. A person makes a sit-up exercise plan. First day, he will do five sit-up, then 6 in the second day, 7 in the third day...How many sit-up will he do in the following 16 days?

Answer: a1=5, d=1, n=16, S16=na1+n(n-1)/2 d =200

the formula of sum of n terms of an Arithmetic progression (AP) is

S(n) = (n/2)*[ 2a+(n-1)*d ] or n*(a+(n-1)*d/2)

here a = first term

d= 1 (difference between successive terms)

and number of terms = 16

194. A sequence of numbers is 1/2, 1/4, 1/8, 1/16, 1/32, what is the range of the tenth number a10 in the sequence? Answer: 0.0001<a10<0.001

should be:0.0001 < 1/1024 (a10) < 0.001 range.

The tenth number a10 in the sequence will be 1/1024.

All Units Questions

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14. X, y and z are three digits numbers, and x=y+z. Is the hundreds’ digit of x equal to the sum of hundreds’ digits of y and z?

1). the tens’ digit of x equal to the sum of tens’ digits of y and z

2). the units’ digit of x equal to the sum of units’ digits of y and z

yes only A satifies.

Statement 1: 321 + 221 = 542, 952 = 521 + 432 -->satifies

Statement 2: 952 = 521 + 432 but 364 + 240 = 604 ---> does nt satisfy

62. n=?

1). The tens¡¯ digit of 11^n is 4

2). The hundreds¡¯ digit of 5^n is 6

E it is.

From 1, n = 10k + 4 (k>=0, this is the series n lies in)

From 2, n = 2m+3 (where, m>=0)

Thus, there's no value of m and k that can satisfy n, thus the answer is E.

Or

wouldnt the series repeat after 10 numbers?from A: n can be 4,14,24...from B: cannot determine..

Both together also insufficient..Answer E

77. What is the remainder when n is divided by 10?

1). The tens?digit of 11^n is 4

2). The hundreds?digit of 5^n is 6

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stem 1>>

11^0=01 | 11^1=11 | 11^2=121 | 11^3=1331 | 11^4=14641 | 11^5=161051 |

the units digit start from 0,1,2,3,4,5 etc...

take n with units digit =4, 14641...sufficient.....and this applicable for all n values of 14,24,34,etc..

stem 2>>

5^0=1 | 5^1=5 | 5^2=25 | 5^3=125 | 5^4=525 | 5^5=2625 | 5^6=13125

it doent follow a pattern, so - insufficient..

A it is....

Or

11^4 = 14641 11^14 = xxxxxxxxxxxx241. (it will be 15 digit number u can check using calc.)

similarly 11 ^ 24 = ....841 (a 25 digit number)

we get n = 4 , 14,24,34,44.....   every time dividing by 10 leaves   4..

so stem(1) alone sufficient.

stem(2) after n>2 and n even we get 6 in hundreds place and n odd we get 1 in hundreds place .. it can be better understood by values.

5 ^ 3 = 125 5^4 = 625 5^5 = 3125 5^6 = 15625   so for every n >2 and n is even we get 6 in hundreds place

so n can have 4, 6, 8,10,12... when divided by 10 leaves diffirent values so not sufficient..

clearly answer is (A) ie stem(1) alone sufficient and stem(2) not sufficient.

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111. Is number 3 the tens?digit of x?

1). When x is divided by 100, the remainder is 30

2). When x is divided by 110, the remainder is 30

Statement (1). When x is divided by 100, the remainder is 30

This will be true only for 130 ,230,330 ..etc

Ten's digit always 3

SUFF

Statement 2). When x is divided by 110, the remainder is 30

x = 140,250,360  --> ten's digit other than 3

x = 1130 --> ten's digit 3

INSUFF

Answer (A

125. Integers x and y have three digits and z is the sum of x and y. Is the tens?digit of z equal to the sum of the tens?digits of x and y?

1). Both x and y have units?digits greater than 6

2). The sum of tens?digit of x and y is 7

A.

X = ABCY = DEF,Z = (X+Y)

i) Given than C and F are greater than 6, that implies, the tens' digit of Z will NEVER be sum of B and E.  And, hence SUFF.  There will always be carry from sum of units digit of X and Y.

ii) INSUFF, as we do not know about the units' digit. 

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I fell for that GMAT trap.... was looking at B for a moment....... Without the tenths digit we cant really support claims of stmt 2 so it is insufficient.

and as gere79 said stmt is really sufficient if you take a closer look at it.

130. X=10^n*25^2, what is the unit's digit of x?

1). Forget...useless

2). n^2=1

n can be +/-1, so x can be 6250 or 62.5.

Discarding A it is E

Agree, E.

But is given X is an integer, then B.

133. If b is an integer and b<10, x=1+ b/100. b=?

1). 1<=b<=3

2). The thousandth's digit of 10X^2 is equal to the tens's digit of x^2

Ones.......  tenths........hund..............thousand...........ten-thous andths0 .        1           ;           ;    2                     3                      4

.1234 =1/10  + 2/100 +3/1000 + 4/10000

(thousandths’ digit of 10X^2 ) is equal to the tens’ digit of x^2

1000 of 10x^2  = 10 of x^2

take b=1

so x^2 = 1.201 and 10x^2 = 12.01

which is only possible when b=9

so the x^2 value is = 1.09^2 = 1.1881 

so the value of b=9....

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IMO B

134. Of the 24 positive integers, all have the units's digit of 5, 1/3 have tens?digit of 0, 1/3 have tens' digit of 1, 1/3 of tens' digit of 2. What is the tens' digit of sum of 24 numbers?

Pretty straight forward 6 it is

(24 * 5 ) = 120 for the units

8 * 0 = 0 for the 1st 1/3

8 * 1 = 8 for the 2nd 1/3

8 * 2 = 16 for the 3rd 1/3

so the sum of tenths of the 24 numbers = 12(carry over from the units) + 8 + 16 = 36

we are looking for the tenths so answer is 6 .....we carry over 3 to the hundreds side.

140. Is the tens?digit of x greater than that of y?

1). x-y=37

2). The units' digit of x is ... greater than that of y

E it is. But to be sure there must be the indication in the question stem that x and y are positive.

1). x-y=37  INSUFF

For example: x = 64 and y = 27 (6>4) or x = 114 and y = 77 (1<7).

2). The units’ digit of x is … greater than that of y INSUFF

146. If 300<X<400, is the tens?digit of x greater than 5?

1). The units' digit of x is greater than 4

2). When x is added with 237, the hundreds?digit will be equal to 6

IMO B

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hundered's digit cannot be 4; hence the ten's digit got be 7,8,9 in X

to get 600, min 363 must be added..

since 300<X<400, x has to be greater than or equal to 363 or tens digit has to be greater than 5. so B

159. What is the unit's digit of X?1). x/(10^n)=25^22). n^2=1

Let the unit digit be u.

Now1>> x/(10^n) = 625.. Now n can take any value ranging from -ve to +ve.

So u can have 6,2,5,0.

Insuff

2>> n^2=1. ;

n^2=1;

n can be +1,-1.

So u can be 2 or 0.

E is the answer

161. x and y are 2-digt integers. What is the difference between two tens' digit?1). x-y=272). Units' digit of x minus the units' digit of y is greater than 3

C

From 1,

x-y = 27 (consider 2 representative cases x = 97 & y = 70 or x = 86, y=59) .. so the diff of tens digits can be either 2 or 3

From 1 & 2,

The diff between Units' digit of x minus the units' digit of y will be greater than 3 only for cases where the diff between tens digits is 2 (and not for the cases where the diff is 3)

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Coordinate Geometry Questions and Answers

1.) In the xy-plane, does the line with equation y=3x + 2 contain the point (r,s)?

(1) (3r + 2 - s) (4r + 9 - s) = 0(2) (4r - 6 - s) (3r + 2 - s) = 0

Answer: We have to prove, s=3r+2 statement 1:(3r+2-s)(4r+9-s) = 0s=3r+2 substituting in above eqn. satisfies it.s=4r+9 doesn't satisfy the above eqn.So s could be 3r+2 or 4r+9 Means we can't say for sure.Statement 2:(4r-6-s)(3r+2-s) = 0s=4r+6 substituting in above eqn. doesn't satisfy.s=3r+2 satisfies the above eqn.so s could be 3r+2 or 4r+6 so can't say for sure. InsufficientCombining both the statements s=3r+2so satisfies the above eqn.ANS(C)

2.) In the xy-coordinate plane, line l and line k interect at the point (4,3). Is the product of their slopes negative?

1.) The product of the x-intercepts of lines l & k is positive.

2.) The product of the y-intercept of lines l and K is negative.

Answer:Let me try to explain this qn theoretically!let eqn of lines be y1 = mx1 +c1 & and y2 = mx2 + c2Qn: Is m1*m2 negative?

Stmt 1:Product of x intercepts is +ve:x intercept of line l : -c1/m1x intercept of line k : -c2/m2product of x intercepts = c1*c2/m1*m2 is +ve=> c1*c2 & m1*m2 can be +ve or c1*c2 & m1*m2 can be -ve

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Insufficient

Stmt2:Product of y intercepts is -ve:y intercept of line l = c1y intercept of line k = c2product of y intercepts = c1*c2 is -ve.. no info on m1*m2Insufficient

Combining both stmts, since c1*c2 is -ve from stmt 2, m1*m2 has to be -ve so that c1*c2/m1*m2 is +ve (from stmt 1)

Ans C.

3.)

Answer:

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I would say C.

From Stmt 1 we have a+ b = -1 and from 2 we have ab = -6

to find a and bA=> a+b = -1no infoB=> ab=-6no infoA and B togetherSolving these eqns, we get a and b

y=(x+a)(x+b) intersects x-axis at (-a,0) and (-b,0) and not a,0 and b,0

also, ab = -6 , cozsecond stmnt says that graph intersects y-axis at (0,-6)Put x = 0 in y=(x+a)(x+b) and we get -6 = a*b

4.)

Coordinate Geometry Questions and Answers

4.) In the xy-plane, does the line with equation y=3x + 2 contain the point (r,s)?

(1) (3r + 2 - s) (4r + 9 - s) = 0(2) (4r - 6 - s) (3r + 2 - s) = 0

Answer: We have to prove, s=3r+2 statement 1:(3r+2-s)(4r+9-s) = 0s=3r+2 substituting in above eqn. satisfies it.s=4r+9 doesn't satisfy the above eqn.So s could be 3r+2 or 4r+9 Means we can't say for sure.Statement 2:(4r-6-s)(3r+2-s) = 0s=4r+6 substituting in above eqn. doesn't satisfy.s=3r+2 satisfies the above eqn.so s could be 3r+2 or 4r+6 so can't say for sure. InsufficientCombining both the statements s=3r+2so satisfies the above eqn.ANS(C)

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5.) In the xy-coordinate plane, line l and line k interect at the point (4,3). Is the product of their slopes negative?

1.) The product of the x-intercepts of lines l & k is positive.

2.) The product of the y-intercept of lines l and K is negative.

Answer:Let me try to explain this qn theoretically!let eqn of lines be y1 = mx1 +c1 & and y2 = mx2 + c2Qn: Is m1*m2 negative?

Stmt 1:Product of x intercepts is +ve:x intercept of line l : -c1/m1x intercept of line k : -c2/m2product of x intercepts = c1*c2/m1*m2 is +ve=> c1*c2 & m1*m2 can be +ve or c1*c2 & m1*m2 can be -veInsufficient

Stmt2:Product of y intercepts is -ve:y intercept of line l = c1y intercept of line k = c2product of y intercepts = c1*c2 is -ve.. no info on m1*m2Insufficient

Combining both stmts, since c1*c2 is -ve from stmt 2, m1*m2 has to be -ve so that c1*c2/m1*m2 is +ve (from stmt 1)

Ans C.

Introduction

How probable is it to get probability questions on GMAT?

Probability questions are becoming increasingly common. They tend to be bundled among the difficult questions, so high scorers will commonly encounter 1, 2, or 3 of them. If you are a low scorer and are pressed for time, consider skipping most of the material past "Simple Probability." GMAT is a computer-adaptive test, and low scorers aren't likely to encounter the most difficult probability question types.

Do I have to be a genius to solve probability questions?

Absolutely not. Both this brief course and GMAT do not require any math knowledge beyond what you learned in your high school. Just be sure to try solving the problems and get a grip of the solution tools, and you'll crack it. To tell you the secret, the myth of the complexity of the probability theory is simply another way to secure the math instructors' wages.

Page 72: GMAT Problems

What is probability?

Probability is a measure of how likely is an event to happen. It is measured in fractions from 0 to 1 (0 is impossible, 1 is unavoidable or certain). Sometimes it is denoted in percentages, again from 0% to 100%.

What is an event and an outcome?

Event is anything that happens. In probability theory we speak of events having outcomes or results. A coin flip (an event) has two possible outcomes - heads and tails. A die toss has six possible outcomes. When a coin is flipped (an event is tested), one of the outcomes is obtained. Either heads or tails.

How is probability used?

A probability is commonly denoted as p(SomeEvent). So, p(Heads) = 50% means that you have 1 chance in 2 to get heads in a coin flip. This also means that if you flip the coin 100 times, you'll get about 50 heads. But not exactly 50. You may get 49, or 63, or even no heads. But you're most likely to get such a number of heads that will be close to 50. This works for any probability. So, if the probability of getting married after going to the cinema is 3%, out of 1,000 movies you'll be married about 1,000 * 3% = 30 times. Maybe 26 or 34, but the average expectation is 30. That's what you use probability for, apart from cracking GMAT.

 ::outline::

 

Simple Probability: The F/T Rule

In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. This is known as the F/T Rule, and 90% of the problems are solved with this tool. No kidding.

Probability = (# of favorable outcomes) / (# of possible outcomes)

Here're some examples to see how it works:

Example 1 . What is the probability that a card drawn at random from a deck of cards will

be an ace?

Solution

In this case there are four favorable outcomes:

1. the ace of spades,

2. the ace of hearts,

3. the ace of diamonds,

4. the ace of clubs.

Since each of the 52 cards in the deck represents a possible outcome, there are 52

possible outcomes. Therefore, the probability is 4/52 or 1/13. The same principle can be

applied to the problem of determining the probability of obtaining different totals from a

pair of dice.

 

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Example 2. Two fair six-sided dice are rolled; what is the probability of having 5 as the

sum of the numbers?

Solution

There are 36 possible outcomes when a pair of dice is thrown (six outcomes for the first

die times six outcomes for the second one). Since four of the outcomes have a total of 5

[(1,4), (4,1), (2,3), (3,2)], the probability of the two dice adding up to 5 is 4/36 = 1/9.

 

Example 3 . Two six-sided dice are rolled; what is the probability of having 12 as the sum

of the numbers?

Solution

We already know the total number of possible outcomes is 36, and since there is only one

outcome that sums to 12, (6,6 - you need to roll double six), the probability is simply

1/36.

 

Dinosaur example . A blonde girl (G.W. Bush, your boss, or whoever you love too heartily)

was asked once what is the probability of meeting a dinosaur in the street. The answer

was: "50%. I either meet it or not." This is how you DON'T use the F/T rule. When counting

the outcomes, make sure that:

1. all of them are equally l ikely to happen

2. you have not left out any possibil ities when counting T

3. F and T are in the same currency, i.e. if F is combinations and T is permutations,

you'll get an error.

Congratulations! Now you've come through the easy part. If you're fine with moderate GMAT and a modest school in West Virginia or Nevada desert, you may proudly and happily abandon this course right here.

NOTE: The material from here on through the end of the section is dense and intended only for medium to high scorers. Because GMAT is a CAT (computer-adaptive test), it is relatively unlikely that lower scorers will encounter these questions, and, if they are short of time, they are better off putting their time into other sections.

 ::outline::

 

Probability of Multiple Events

For questions involving single events, the F/T rule is sufficient. In fact, it is often sufficient for all other cases too. But, for questions involving multiple events, some other tools may be more appropriate. Even

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when the problem can be solved with F/T, these tools still may provide a more elegant solution. Here're the tools:

1. NOT tool

If you know that the probability of an event (or one of the outcomes) is p, the probability of this event NOT happening (or the probability of it NOT having this given outcome), is (1-p).

p(not A) + p(A) = 1

2. AND tool

If two (or more) independent events are occurring, and you know the probability of each, the probability of BOTH (or ALL) of them occurring together (event A and event B and event C etc) is a multiplication of their probabilities.

p(A and B) = p(A) * p(B)p(A and B and C ... and Z) = p(A) * p(B) * p(C) * ... *

p(Z)

Suppose I will only be happy today if I get an email and win the lottery. I've a 90% chance to get an email and 0.1% chance to win the lottery. What are my chances for happiness? Since email and lottery are independent (getting an email doesn't change my lottery chances, and vice versa), we can use the AND tool: p(email and lottery) = p(email) * p(lottery) = 90% * 0.1% = 0.09%; So I have 9 chances in 10,000... Not bad.

3. OR tool

If two (or more) incompatible events are occurring, the probability of EITHER of them occurring (event A or event B or event C etc) is a sum of their probabilities.

p(A or B) = p(A) + p(B)p(A or B or C ... or Z) = p(A) + p(B) + ... + p(Z)

Incompatible means that they can't happen together, i.e. p(A and B) = 0. In case of two compatible events, the OR tool looks a bit more complicated:

p(A or B) = p(A) + p(B) - p(A and B)

If we know that A and B are independent, we can apply AND tool to rewrite:

p(A or B) = p(A) + p(B) - p(A) * p(B)

Suppose I will now be happy in both cases - either getting an email or winning the lottery. What are my chances to happiness now? p(email or lottery) = p(email) + p(lottery) - p(email) * p(lottery) = 90% + 0.1% - 0.09% = 90.01%; My chances are 9,001 in 10,000 now. I'd rather choose this one.

4. Expressions/Brackets tool

When you're being asked for something complex, try reducing it to events and outcomes, and writing a formula. Use brackets to denote complex events, such as (A and B), or (A and (B or C)), etc. It is common to use AND as if it is multiplication and OR as if it is addition in the order preference, i.e. (A and B or C) = ((A and B) or C), but (A and (B or C)) <> (A and B or C). When you figure out the formula, it'll be easy to reduce it to simple arithmetic operations by using NOT, AND, and OR tools.

5. Elimination tricks

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Given that 0 <= p(A) <= 1, you get the following rules:

1. p(A and B) <= p(A) 2. p(A or B) >= p(A) 3. p(A and B) <= p(A or B)

Thinking of these rules is often an excellent strategy for eliminating certain answer choices.

Example 4. If a fair coin is tossed twice, what is the probability that on the first toss the

coin lands heads and on the second toss the coin lands tails?

1. 1/6

2. 1/3

3. 1/4

4. 1/2

5. 1

Solution . Suppose first toss is A, second is B. We know that p(A_heads) = 50% and that

p(B_tails) = 50%. Also, A and B are independent. So, p(A_heads and B_tails) = p(A_heads)

* p(B_tails) = 50% * 50% = 25% = 1/4. Answer is C.

 

Example 5. If a fair coin is tossed twice what is the probability that it will land either

heads both times or tails both times?

1. 1/8

2. 1/6

3. 1/4

4. 1/2

5. 1

Solution . Let first toss be A, second B.

p(Ah) = p(At) = p(Bh) = p(Bt) = 1/2

p(Ah and Bh) = p(Ah) * p(Bh) = 1/4

p(At and Bt) = p(At) * p(Bt) = 1/4

p((Ah and Bh) or (At and Bt)) = p(Ah and Bh) + p(At and Bt) =

1/4 + 1/4 = 1/2

Note that AND rule works because A and B are independent, and OR rule works because

(Ah and Bh) and (At and Bt) are incompatible.

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Alternatively, you may use F/T rule to solve this. Enumerate outcomes as (HH, HT, TH, TT).

Favorable are HH and TT. So, p = 2/4 = 1/2. Although in this case F/T rule works more

gracefully, the AND/OR approach is sti l l helpful - you can learn it on such easy examples

as this to prepare for the more difficult ones.

 

Example 6 . A bowman hits his target in 1/2 of his shots. What is the probability of him

missing the target at least once in three shots?

Solution . An optimal way to solve this is to think that (missing the target at least once)

= 1 - (hitting it every time). So, p(hitting it every time) = p(shot1_hit and shot2_hit and

shot3_hit) = p(shot1_hit) * p(shot2_hit) * p(shot3_hit) = 1/2 * 1/2 * 1/2 = 1/8; p(missing at

least once) = 1 - p(hitting it every time) = 1 - 1/8 = 7/8.

Alternatively, use the F/T rule. The T are HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM. T

= 8. The F are HHM, HMH, HMM, MHH, MHM, MMH, MMM. F = 7.

In cases like this it is evident that F/T rule soon becomes too hard to apply.

 ::outline::

 

Event Types and Sets Analogy

Compatible vs. Incompatible (Mutually exclusive) Events

Sometimes you have to distinguish compatible and mutually exclusive events. Mutually exclusive are those events that can't happen together. Heads and tails are mutually exclusive events. Formally, two events are mutually exclusive if p(A and B) = 0. Otherwise, they are compatible. Note that mutually exclusive events are independent. (!)

Dependent vs. Independent Events

Most of the events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events if you put the cards back.

An example of a dependent event, one in which the probability of the second event is affected by the first, is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, there are one fewer aces and one fewer spades. This fact affects the F in the F/T rule.

What to do if you encounter dependent events? If possible, try to use F/T rule to the composite event of the two. In the cards example, you may consider counting all 2-card combinations you may draw (T), and then counting those that fit (F). This will be discussed in detail later. But sometimes the events can't be reduced to outcomes that can be counted. In these cases, use the sets analogy.

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Sets Analogy

Remember the familiar problem type about students attending three language classes, say, French, German, and Chinese? There you had to calculate the number of students attending one of the classes, or number of students attending both French and German, but not Chinese, etc? The greatest way to solve such problems is to draw intersecting circles representing the three sets of students, and then to write there their numbers and try to find the answer.

What does it have to do with probability, one might wonder. But this is precisely the way to solve probability problems with dependent events. This charts you may have drawn for simple sets problems are called Venn diagrams in the probability theory. Perhaps to scare you away.

The logic is simple: each event is a language class, and each chance is a student in that class. And the probability of the event is the number of students (chances) attending it divided by the total number of students. Where the classes intersect is where two events happen at once. Mutually exclusive events do not intersect. Finally, independent events intersect in such an interesting way that, supposing French and German classes represent two independent events, the proportion of French students in the German class is the same as the proportion of French students in the school as a whole (100 students, 40 study German, 50 study French, and 20 study both: 20/40 = 50/100).

Conditional Probability

Conditional probability is a simple way to denote proportions you understand with the sets analogy. Simply put, p(A/B) is the probability of event A happening given that event B has already happened, or the number of students attending both A and B classes divided by the number of students attending B class.

So, for any two events, including dependent events, this statement hold:

p(A and B) = p(A) * p(B/A) = p(B) * p(A/B)

This statement, however scary, is self-evident. Look at it. It says that to find the number of students studying French and German you have to either multiply the number of those who study French by the proportion of German scholars in the French class (p(B/A)), or multiply the number of German students by the proportion of French students in the German class (p(A/B)). But that's self-evident, isn't it? So it is with events.

Independent events may, therefore, be defined as such that p(B/A) = p(B), p(A/B) = p(A).

Example 7. What is the probability that a card selected from a deck will be either an ace

or a spade?

1. 2/52

2. 2/13

3. 7/26

4. 4/13

5. 17/52

Solution .Let A stand for a card being an ace, and S for it being a spade. We have to find

p(A or S). Are A and S mutually exclusive? No. Are they independent? Why, yes, because

spades have as many aces as any other suit. Then,

p(A or S) = p(A) + p(S) - p(A) * p(S)

With simple F/T we get:

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p(A) = 4/52 = 1/13

p(B) = 13/52 = 1/4

So,

p(A or S) = 1/13 + 1/4 - 1/52 = 16/52 = 4/13

Sets analogy can help you visualize the formula. Draw two intersecting circles - one for

aces, the other for spades. To get the area (probability) of the figure formed by these two

circles together (all chances that are either aces or spades), you add the areas of aces

and spades and subtract the intersecting area, in order not to count it twice. What we

subtract is the ace of spades that was counted twice.

Another way to think about the question is to just count aces and spades; that is, use the

F/T rule. There are 13 spades in a deck and 3 aces other than the ace of spades already

included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52

possible outcomes, or 16/52 = 4/13.

 

Example 8. If someone draws a card at random from a deck and then, without replacing

the first card, draws a second card, what is the probability that both cards will be aces?

Solution. Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) =

4/52 = 1/13. Given that the first card is an ace, what is the probability that the second

card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B/A) =

3/51 = 1/17, and therefore:

p(A and B) = p(A) * p(B/A) = 1/13 * 1/17 = 1/221

 

Example 9. If there are 30 red and blue marbles in a jar, and the ratio of red to blue

marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles

if you return the marbles after each draw?

Solution. So, there are 12 red and 18 blue marbles. We are asked to draw twice and

return the marble after each draw. Therefore, the first draw does not affect the probability

of the second draw. We return the marble after the draw, and therefore, we return the

situation to the initial conditions before the second draw. Nothing is altered in between

draws; therefore, the events are independent.

p(drawing a red marble) would be 12/30 = 2/5. The same is true for the second draw. Then

p(First_Red and Second_Red) = p(First_Red) * p(Second_Red) = 2/5 * 2/5 = 4/25.

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Example 10. Now consider the same question with the condition that you do not return

the marbles after each draw.

Solution . The probability of drawing a red marble on the first draw remains the same,

12/30 = 2/5. The second draw, however, is different. The initial conditions have been

altered by the first draw. We now have only 29 marbles in the jar and only 11 red. So,

p(Second_Red/First_Red) = 11/29. Using the dependent event formula,

p(First_Red and Second_Red) = p(First_Red) *

p(Second_Red/First_Red) =

2/5 * 11/29 = 22/145

To summarize, if you return every marble you select, the probability of drawing another

marble is unaffected; the events are INDEPENDENT. If you do not return the marbles, the

number of marbles is affected and therefore DEPENDENT.

 ::outline::

 

Learning the Advanced Tools

Detailed discussion of advanced solution tools is out of scope of this lesson, but here're some considerations to get you started:

Combinations. Good understanding of CT formulas (n!, nAk, nCk) is essential to solving complex F/T problems, where both F and T are so large you can't enumerate them manually, but only with a formula. See our Combinations Lesson.

Expectations. Some probability problems deal with money, gains, and bets. Often you have to calculate which bet will be better, or how much it will be worth. The tool that deals with this is Expectation. E = G * p, where G is gain, and p is probability. So, a 10% chance to get $100 is worth (has E) of $100 * 10% = $10. Therefore, it is better than to get $8 for granted, but worse than a 5% chance to get $300 (E = $300 * 5% = $15). Complex expectation works similarly: E1 = E * p, i.e. a 10% chance to get a 25% chance to get $100 is worth 10% * (25% * $100) = $2.5; This is how Expectations work.

Distributions. The three types of distributions are Binominal, Hypergeometric, and Poisson distributions. These are just handy formulas for solving 3 very specific kinds of problems, like these:

If the coin is tossed 5 times, what is the probability that at least 3 out of 5 times it will show heads? (Binominal Distribution)

There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue? (Hypergeometric Distribution)

Each hour an average of ten cars arrive at the parking lot. The lot can handle at most fifteen cars per hour. What is the probability that at a given hour cars will not be accepted? (Poisson Distribution)

As you may have noticed, Poisson and Binominal Distribution problems are alike. In fact, these Distributions are two methods of solving the same kind of problems. The difference is that BD provides accurate but costly (many calculations) method, and PD provides and elegant approximation, and is therefore used only on large numbers.

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While BD and HD are quite likely to appear on GMAT, PD is not. For GMAT Club's members it is an open question whether one can in fact encounter PD on GMAT. In any case, there won't be two questions on PD.

See the Appendix 1 below for an explanation of BD and HD.

S@ifur’s GMAT MAF:28Difficult Problems from the Math Section

[email protected]

10. The sum of the even numbers between 1 and n is 79*80, where n is an odd number, then n=?

Sol: First term a=2, common difference d=2 since even number

therefore sum to first n numbers of Arithmetic progression would be

n/2(2a+(n-1)d)

= n/2(2*2+(n-1)*2)=n(n+1) and this is equal to 79*80

therefore n=79 which is odd...

11. The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents

per day while the price of wheat is decreasing at a constant rate of cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50(B) $5.10(C) $5.30(D) $5.50(E) $5.60

Soln: 320 + 5x = 580 - .41x; x = # of days after price is same;

solving for x; x is approximately 48, thus the required price is 320 + 5 * 48 = 560 cents = $5.6

12. How many randomly assembled people do u need to have a better than 50% prob. that at least 1 of them was born in a leap year?

Soln: Prob. of a randomly selected person to have NOT been born in a leap yr = 3/4Take 2 people, probability that none of them was born in a leap = 3/4*3/4 = 9/16. The probability at least one born in leap = 1- 9/16 = 7/16 < 0.5Take 3 people, probability that none born in leap year = 3/4*3/4*3/4 = 27/64. The probability that at least one born = 1 - 27/64 = 37/64 > 0.5Thus min 3 people are needed.

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13. In a basketball contest, players must make 10 free throws. Assuming a player has 90% chance of making each of his shots, how likely is it that he will make all of his first 10 shots?

Ans: The probability of making all of his first 10 shots is given by

(9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10)* (9/10) = (9/10)^10 = 0.348 => 35%

14. AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1

(B) 3

(C) 7

(D) 9

(E) Cannot be determinedAns: AB + CD = AAASince AB and CD are two digit numbers, then AAA must be 111 Therefore 1B + CD = 111B can assume any value between 3 and 9If B = 3, then CD = 111-13 = 98 and C = 9If B = 9, then CD = 111-19 = 92 and C = 9So for all B between 3 & 9, C = 9

Therefore the correct answer is D (C = 9)

15. A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?(A) 12(B) 14(C) 16(D) 18(E) 20

Ans: race 1 :- ta = tb-6 ( because A beats B by 6 sec)race 2 :- Ta = tb+2 ( because A looses to B by 2 sec)

By the formula D= S * Twe get two equations480/Sa = 432/Sb -6 ------------1)

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480/Sa = 336/Sb +2------------2)Equating these two equations we get Sb = 12ta,Sa stand for time taken by A and speed of A resp.

16. A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?(A) 1/4(B) 1/3(C) 1/2(D) 2/3(E) ¾

Ans: Let X be the fraction of solution that is replaced.

Then X*25% + (1-X)*40% = 35%

Solving, you get X = 1/3

17. A person buys a share for $ 50 and sells it for $ 52 after a year. What is the total profit made by him from the share?(I) A company pays annual dividend(II) The rate of dividend is 25%(A) Statement (I) ALONE is sufficient, but statement (II) alone is not sufficient(B) Statement (II) ALONE is sufficient, but statement (I) is not sufficient(C) BOTH statements TOGETHER are sufficient, but NEITHER statement alone is sufficient(D) Each statement ALONE is sufficient(E) Statements (I) and (II) TOGETHER are NOT sufficient

18. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?(A) 2/27(B) 1/9(C) 1/3(D) 4/27(E) 2/9

Ans: Case I: Red ball first and then white ball

P1 = 3/9*2/9= 2/27

Case 2: White ball first and then red ball

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P2 = 2/9*3/9 = 2/27

Therefore total probability: p1 + p2 = 4/27

10. What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4B) sqrt (2)C) 1.7D) sqrt (3)E) 2.0

Ans: The equation of the line will be 3x - 4y - 12 = 0.This crosses the x and y axis at (0,-3) and (4,0)

The circle has the origin at the center and has a radius of 1 unit.

So it is closest to the given line when, a perpendicular is drawn to the line, which passes through the origin.

This distance of the line from the origin is 12 / sqrt (9 + 16) which is 2.4 [Length of perpendicular from origin to line ax +by + c = 0 is

mod (c / sqrt (a^2 + b^2))]

The radius is 1 unit.

So the shortest distance is 2.4 - 1 unit = 1.4 units

11.

In the square above, 12w = 3x = 4y. What fractional part of the square is shaded?

A) 2/3

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B) 14/25

C) 5/9

D) 11/25

E) 3/7

Sol: Since 12w=3x=4y,

w:x=3:12=1:4 and x:y=4:3

so, w = 1x = 4y = 3

the fractional part of the square is shaded:{(w+x)^2 - [(1/2)wx + (1/2)wx +(1/2)xy + (1/2)w(2w)]}/(w+x)^2

= {(w+x)^2 - [wx + (1/2)xy + w^2)]}/[(w+x)^2]

=[(5^2) -(4+6+1)]/(5^2)

= (25 - 11)/25

= 14/25

12. The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures?

20 25 40 45 75 Ans: The answer 25 doesn't refer to a temperature, but rather to a range of temperatures.

The average of the 5 temps is: (a + b + c + d + e) / 5 = 50One of these temps is 45: (a + b + c + d + 45) / 5 = 50Solving for the variables: a + b + c + d = 205In order to find the greatest range of temps, we minimize all temps but one. Remember, though, that 45 is the lowest temp possible, so: 45 + 45 + 45 + d = 205Solving for the variable: d = 7070 - 45 = 25

13. If n is an integer from 1 to 96, what is the probability for n*(n+1)*(n+2) being

divisible by 8? 25% 50% 62.5% 72.5% 75%

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Soln: E = n*(n+1)*(n+2)

E is divisible by 8, if n is even.No of even numbers (between 1 and 96) = 48

E is divisible by 8, also when n = 8k - 1 (k = 1,2,3,.....)Such numbers total = 12(7,15,....)

Favorable cases = 48+12 = 60.Total cases = 96P = 60/96 = 62.5

Method 2:From 1 to 10, there are 5 sets, which are divisible by 8.(2*3*4) (4*7*6) (6*7*8)(7*8*9)(8*9*10)

So till 96, there will be 12 * 5 such sets = 60 sets

so probability will be 60/96 = 62.5

14. Kurt, a painter, has 9 jars of paint: 4 are yellow2 are redrest are brownKurt will combine 3 jars of paint into a new container to make a new color, which he will name accordingly to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellowBrun X if the paint contains 3 jars of brown paintJaune X if the paint contains at least 2 jars of yellowJaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be Jaune

a) 5/42b) 37/42c) 1/21d) 4/9e) 5/9

Sol: 1. This has at least 2 yellow meaning..

a> there can be all three Y => 4c3 OR b> 2 Y and 1 out of 2 R and 3 B => 4c2 x 5c1

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Total 34

2.This has exactly 1 Y and remaining 2 out of 5 = > 4c1 x 5c2

Total 40Total possibilities = (9!/3!6!) = 84Adding the two probabilities: probability = 74/84 = 37/42

15. TWO couples and a single person are to be seated on 5 chairs such that no couple is seated next to each other. What is the probability of the above??Soln:Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)Ways for second couple = 2*4!These cases include an extra case of both couples sitting togetherWays in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways)Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!Total ways to arrange the 5 ppl = 5!Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5Thus prob of none seated together = 1 - 3/5 = 2/5

16. An express train traveled at an average speed of 100 kilometers per hour, stopping for 3 minutes after every 75 kilometers. A local train traveled at an average speed of 50 kilometers, stopping for 1 minute after every 25 kilometers. If the trains began traveling at the same time, how many kilometers did the local train travel in the time it took the express train to travel 600 kilometers?

a. 300b. 305c. 307.5d. 1200e. 1236

Sol: the answer is C: 307.5 km Express Train: 600 km --> 6 hours (since 100 km/h) + stops * 3 min.stops = 600 / 75 = 8, but as it is an integer number, the last stop in km 600 is not a real stop, so it would be 7 stopsso, time= 6 hours + 7 * 3 min. = 6 hours 21 min

Local Train:in 6 hours, it will make 300 km (since its speed is 50 km/hour)in 300 km it will have 300 / 25 stops = 12, 12 stops 1 min each = 12 minwe have 6 hours 12 min, but we need to calculate how many km can it make in 6 hours

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21 min, so 21-12 = 9 minif 60 min it can make 50 km, 9 min it can make 7.5 kmso, the distance is 300 + 7.5 = 307.5 km

17. Matt starts a new job, with the goal of doubling his old average commission of $400. He takes a 10% commission, making commissions of $100.00, $200.00, $250.00, $700.00, and $1,000 on his first 5 sales. If Matt made two sales on the last day of the week, how much would Matt have had to sell in order to meet his goal?

Sol: The two sales on Matt's last day of the week must total $33,500.

(100 + 200 + 250 + 700 + 1000 + x) / 7 = 800x = 3350since x is Matt's 10% commission, the sale is $33,500.

18. On how many ways can the letters of the word "COMPUTER" be arranged?

1) Without any restrictions.2) M must always occur at the third place.3) All the vowels are together.4) All the vowels are never together.5) Vowels occupy the even positions.

Sol: 3) 8! = 403204) 7*6*1*5*4*3*2*1=5,040

3) Considering the 3 vowels as 1 letter, there are five other letters which are consonants C, M, P, T, RCMPTR (AUE) = 6 letters which can be arranged in 6p6 or 6! Waysand A, U, E themselves can be arranged in another 3! Ways for a total of 6!*3! Ways

5) Total combinations - all vowels always together= what u found in 1) - what u found in 3)= 8! - 6! *3!

5) I think it should be 4 * 720there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

19. In the infinite sequence A, ,where x is a positive integer constant. For what value of n is the ratio

of to equal to ?

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(A) 8(B) 7(C) 6(D) 5(E) 4

Sol: The method I followed was to reduce the Q to (x^6/ x ) * ( Y/Y)the eqn An= (x ^ n -1)(1+ x + x^2 + x^3 +.... ) ----------------------------(1)

and the eqn x(1+x(1+x(1+x...)))) which I call Z can be reduced to x( 1+x+x^2+x^3 ..) --------(2)

from (1) and (2) we get An / Z = x^(n-1) / xtherefore for getting answer x^5 (n-1) = 6therefore n=7

Ans: B

20. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?a. (13^4) x 48 x 47b. (13^4) x 27 x 47c. 48C6d. 13^4e. (13^4) x 48C6

Sol: 52 cards in a deck -13 cards per suitFirst card - let us say from suit hearts = 13C1 =13Second card - let us say from suit diamonds = 13C1 =13Third card - let us say from suit spade = 13C1 =13Fourth card - let us say from suit clubs = 13C1 =13Remaining cards in the deck= 52 -4 = 48Fifth card - any card in the deck = 48C1Sixth card - any card in the deck = 47C1

Total number of ways = 13 * 13 * 13 * 13 * 48 * 47 = 13^4 *48*47 ---> choice A

21. Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into a hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10? 3 4 5

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6 7 *

Sol:ok consider this

0 + 1+ 2 + 3 + 4 +5 +6..Stop --> Don’t go further. Why? Here’s why.At the worst the order given above is how u could pick the out the slips. Until u add the slip with no. 6 on it, no two slips before that add up to 10 (which is what the Q wants)

the best u can approach is a sum of 9 (slip no. 5 + slip no. 4)

but as soon as u add slip 6. Voila u get your first sum of 10 from two slips and that is indeed the answer = 7 draws 22. Two missiles are launched simultaneously. Missile 1 launches at a speed of x

miles per hour, increasing its speed by a factor of every 10 minutes (so that after

10 minutes its speed is , after 20 minutes its speed is , and so forth. Missile 2 launches at a speed of y miles per hour, doubling its speed every 10 minutes. After 1 hour, is the speed of Missile 1 greater than that of Missile 2?

1) 2)

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.(D) EACH statement ALONE is sufficient to answer the question.(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

Sol:

Since Missile 1's rate increases by a factor of every 10 minutes, Missile 1 will be

traveling at a speed of miles per hour after 60 minutes:

minutes0-1010-2020-3030-4040-5050-6060+speed

And since Missile 2's rate doubles every 10 minutes, Missile 2 will be traveling at a

speed of after 60 minutes:

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minutes0-1010-2020-3030-4040-5050-6060+speed

The question then becomes: Is ?

Statement (1) tells us that . Squaring both sides yields . We can

substitute for y: Is ? If we divide both sides by , we get: Is ? We

can further simplify by taking the square root of both sides: Is ? We still cannot answer this, so statement (1) alone is NOT sufficient to answer the question.

Statement (2) tells us that , which tells us nothing about the relationship between x and y. Statement (2) alone is NOT sufficient to answer the question.

Taking the statements together, we know from statement (1) that the question can be

rephrased: Is ? From statement (2) we know certainly that , which is

another way of expressing . So using the information from both statements, we can answer definitively that after 1 hour, Missile 1 is traveling faster than Missile 2.

The correct answer is C: Statements (1) and (2) taken together are sufficient to answer the question, but neither statement alone is sufficient.

23. If , what is the unit’s digit of ?

(A) 0(B) 1(C) 3(D) 5(E) 9Sol:

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The unit’s digit of the left side of the equation is equal to the unit’s digit of the right side of the equation (which is what the question asks about). Thus, if we can determine the unit’s digit of the expression on the left side of the equation, we can answer the question.

Since , we know that 13! Contains a factor of 10, so its

unit’s digit must be 0. Similarly, the unit’s digit of will also have a unit’s digit of 0. If we subtract 1 from this, we will be left with a number ending in 9.

Therefore, the unit’s digit of is 9. The correct answer is E.

24. The dimensions of a rectangular floor are 16 feet by 20 feet. When a rectangular rug is placed on the floor, a strip of floor 3 feet wide is exposed on all sides. What are the dimensions of the rug, in feet?

(A) 10 by 14 (B) 10 by 17 (C) 13 by 14 (E) (D) 13 by 17 (E) 14 by 16

Soln: The rug is placed in the middle of the room. The rug leaves 3m on either side of it both lengthwise and breadth wise. Now, the dimensions of the rug would be the dimensions of the room - the space that it does not occupy. With three 3 on either side, 6 m is not occupied by the rug in both dimensions.

So, rug size = (16-6) X (20-6) = 10 X14

25. How many different subsets of the set {10,14,17,24} are there that contain an odd number of elements?

(a) 3 (b) 6 (c) 8 (d) 10 ( e) 12

Soln: 8 is the answer. The different subsets are

10, 14, 17, 24, 10, 14, 17

14 17 24

17 24 10

24 10 14

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26. Seven men and seven women have to sit around a circular table so that no 2 women are together. In how many different ways can this be done? a.24 b.6 c.4 d.12 e.3

Soln: I suggest to first arranging men. This can be done in 6! Ways. Now to satisfy above condition for women, they should sit in spaces between each man. This can be done in 7! Ways (because there will be seven spaces between each man on round table)Total ways = 6! * 7!

27. Find the fourth consecutive even number:(I) the sum of the last two numbers is 30(II) the sum of the first two numbers is 22

the ans is D

please explain why I alone is sufficient?

Soln: I guess, there are 4 consecutive even numbers to start with..Stmt 1) Sum of 3rd even + Sum of 4th even = 30 => 14+16 = 30 => 4th even num = 16.. Sufficient

Stmt 2) Sum of Ist even +Sum of 2nd even = 22 => 10 and 12 are the 2 numbers to begin with, then 3rd num = 14 and 4th num = 16.. Sufficient

Ans – D

28. If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is:

a) A+1b) A+5c) A+25d) 2Ae) 5ASoln: If you divide the sum obtained by adding any 5consecutive numbers by '5', and then you will get the Center number of the sequence itself.

i.e. 1 - 5 = 15/5 = 3 . 1, 2, 3, 4, 5

so, sixth consecutive number will be '3' more than the 'Middle term'i.e. 3+3=6, similarly 3+4=7

Hence going by this. Asked sum would be

[(A/5) + 3]+[(A/5) + 4]+[(A/5) + 5]+[(A/5) + 6]+[(A/5) + 7] = A + 25

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29. If P represents the product of the first 15 positive integers, then P is not a multiple of:

a) 99 b) 84 c) 72 d) 65 e) 57

Solution: If P represents the product of the first 15 integers, P would consist of the prime numbers that are below 15.

2,3,5,7,11,13

Any value that has a prime higher than 13 would not be a value of P.

57 = 3, 19

19 is a prime greater than 13, so the answer is E.

30. 5 girls and 3 boys are arranged randomly in a row. Find the probability that:

A) there is one boy on each end.

B) There is one girl on each end.

Solution: For the first scenario:A) there is one boy on each end.

The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways Total possible outcome = 8!Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

For the second scenario:A) there is one girl on each end.

The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways Total possible outcome = 8!Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14

31. If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

Solution: approach: first fix Jen, and then fix Bob. Then fix the remaining three.

Case 1: When Jen is in the first place.

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=> Bob can be in any of the other four places. => 4.The remaining 3 can arrange themselves in the remaining 3 places in 3! Ways.Hence total ways = 4*3!

Case 2: When Jen is in the second place.=> Bob can be in any of three places => 3.The remaining can arrange themselves in 3 places in 3! Ways.

Continuing the approach.Answer = 4*3! +3*3! +2*3! +1*3! = 10*3! = 60 ways.

32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If –1 is in the set, which of the following must also be in the set?I. –3 II. 1 III. 5A. I onlyB. II onlyC. I and II onlyD. II and III onlyE. I, II, and III

Soln: Series property: t => t+2. (Note: for any given number N, ONLY N + 2 is compulsory. N - 2 is not a necessity as N could be the first term...this can be used as a trap.)Given: -1 belongs to the series. => 1 => 3 =>5. DOES NOT imply -3.Hence, II and III (D).

33. A number is selected at random from first 30 natural numbers. What is the probability that the number is a multiple of either 3 or 13?

(A) 17/30(B) 2/5 (C) 7/15(D) 4/15 (E) 11/30

Solution: Total no from 1 to 30 = 30total no from 1 to 30 which r multiple of 3 = 10 (eg(3,6,9,12,15,18,21,24,27,30))total no from 1 to 30 which r multiple of 13 = 2 (eg 13,26)P(a or b ) = p(a) + p(b)p(a)= 10/30p(b)=2/30p(a) + p(b) = 10/30+2/30 = 2/5

34. Two numbers are less than a third number by 30% and 37 % respectively. How much percent is the second number less than the first?

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a) 10 % b) 7 % c) 4 % d) 3 %

Solution: .7 & .63

diff in %= (.7 - .63)/.7 * 100= .07/.7 * 100= 10%

35. If y ≠ 3 and 2x/y is a prime integer greater than 2, which of the following must be true? I. x = yII. y = 1III. x and y are prime integers.

(A) None

(B) I only

(C) II only

(D) III only

(E) I and II

37. Someone passed a certain bridge, which needs fee. There are 2 ways for him to choice. A : $13/month+$0.2/time , B: $0.75/time . He passes the bridge twice a day. How many days at least he passes the bridge in a month, it is economic by A way?

A) 11 B) 12 C) 13 D) 14 E) 15

Soln: Let x be the no days where both are equal cost13 + 0.2*2x= 0.75*2x13+.4x=1.5x13=1.1xx=11.81

by plugging in for 12 daysFor A13 +(12*2)*.2=13+4.8=17.8For B24*.75=18

therefore answer is B.

38. Two measure standards R and S. 24 and 30 measured with R are 42 and 60 when they are measured with S, respectively. If 100 is acquired with S, what would its value be measured with R?

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39. Every student of a certain school must take one and only one elective course. In last year, 1/2 of the students took biology as an elective, 1/3 of the students took chemistry as an elective, and all of the other students took physics. In this year, 1/3 of the students who took biology and 1/4 of the students who took chemistry left school, other students did not leave, and no fresh student come in. What fraction of all students took biology and took chemistry?

A. 7/9 B .6/7 C.5/7 D.4/9 E.2/5

Soln: If total =1Last yearB=1/2C=1/3P=1/6This year B= 1/2 * 2/3 = 1/3C= 1/3 * 3/4= 1/4Tot 1/3+1/4=7/12student left this yearB = 1/2 * 1/3 = 1/6C= 1/3 *1/4 = 1/12tot= 1/6 +1/12 = 1/4So the school has total student this year= Last year student no - total no of student left this year = 1- 1/4=3/4Answer = 7/12 / 3/4 = 7/9

40. If X>0.9, which of the following equals to X?

A. 0.81^1/2 B. 0.9^1/2 C. 0.9^2 D. 1-0.01^1/2

41. There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students. What is the probability that the number of men is equal to that of women?

A.18/35 B16/35 C.14/35 D.13/35 E.12/35

Soln: there has to be equal no of men & women so out of 4 people selected there has to be 2M & 2W.Total ways of selecting 4 out of 8 is 8C4 total ways of selecting 2 men out of 4 is 4C2total ways of selecting 2 women out of 4 is 4C2

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so probability is (4C2*4C2)/ 8C2 = 18/35

42. The area of an equilateral triangle is 9.what is the area of it circumcircle.

A.10PI B.12PI C.14PI D.16PI E.18PI

Soln: area = sqrt(3) / 4 * (side)^2 = 9

so , ( side )^2 = ( 9*4 ) / sqrt(3)

Height = H = sqrt(3)/2 * (side) , so H^2 = 3/4 * Side^2 = 3/4 * (9*4) / sqrt(3)

Radius of CircumCircle = R = 2/3 of ( Height of the Equilateral Triangle )

so , area of Circumcircle = PI * R^2

=> PI * 4/9 * H^2

=> PI * 4/9 * 3/4 * (9*4) / sqrt(3)

=> PI * 4 * sqrt(3)

43.A group of people participate in some curriculums, 20 of them practice Yoga, 10 study cooking, 12 study weaving, 3 of them study cooking only, 4 of them study both the cooking and yoga, 2 of them participate all curriculums. How many people study both cooking and weaving?

A.1 B.2 C.3 D.4 E.5

Soln: We know there are 10 people who do cooking as an activity.

3 -> people who do only cooking4 -> do cooking and Yoga2 -> do all of the activitiesx -> number of people doing cooking and weaving

When you sum all this up, we should have 10. So 3+4+2+x=10 --> x=10-9=1

44. There 3 kinds of books in the library fiction, non-fiction and biology. Ratio of fiction to non-fiction is 3 to 2; ratio of non-fiction to biology is 4 to 3, and the total of the books is more than 1000?which one of following can be the total of the book?

A 1001 B. 1009 C.1008 D.1007 E.1006

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Soln: fiction : non-fiction = 3 : 2 = 6 : 4non-fiction : biology = 4 : 3

fiction : non-fiction : biology = 6 : 4 : 3

6x + 4x + 3x = 1000x = 76 12/13if we add 1, 13 will divide evenly (1001/13 = 77)1000 + 1 = 1001

45. In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

A) 5

B) 10

C) 15

D) 20

E) 25Soln: n(1U2U3) = n(1) + n(2) + n(3) - n(1n2) - n(2n3) - n(1n3) + n (1n2n3)85=50+30+20- [n(1n2) - n(2n3) - n(1n3)] +5[n(1n2) - n(2n3) - n(1n3)] = 20

46. For a certain company, operating costs and commissions totaled $550 million in 1990, representing an increase of 10 percent from the previous year. The sum of operating costs and commissions for both years was(A) $1,000 million (B) $1,050 million (C) $1,100 million(D) $1,150 million (E) $1,155 million

Solution: 1998 = $500 M1999 = $550 M

Sum = $ 1050 MAns:B

47. Fox jeans regularly sell for $15 a pair and Pony jeans regularly sell for $18 a pair. During a sale these regular unit prices are discounted at different rates so that

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a total of $9 is saved by purchasing 5 pairs of jeans: 3 pairs of Fox jeans and 2 pairs of Pony jeans. If the sum of the two discounts rates is 22 percent, what is the discount rate on Pony jeans?(A) 9%(B) 10%(C) 11%(D) 12%(E) 15%Soln: Ans : B

Total discount is 22% = $9

by back solving, in this case ,the discount percent for pony jeans should be less than 11% (22/2)because the price of this product is more.

take choice B10% of 18= 1.8 total discount on 2 pony jeans= $3.622%-10%=12%12% of 15 = $1.8total discount on 3 fox jeans = $5.43fox jeans discount +2 pony jeans discount =$95.4 +3.6=9so answer is B.

48. There are 2 kinds of staff members in a certain company, PART TIME AND FULL TIME. 25 percent of the total members are PART TIME members others are FULL TIME members. The work time of part time members is 3/5 of the full time members. Wage per hour is same. What is the ratio of total wage of part time members to total wage of all members. A.1/4 B.1/5 C 1/6 D 1/7 E 1/8

Soln: What is the ratio of total wage of part time members to total wage of all members.

You have calculated ratio of Part time-to-Full time.

P x/4 3y/5 3xy/4*5 A 3x/4 Y + x/4 3y/5 18xy/20

Ratio= p/a= 3xy/18xy= 1/6

49. If 75% of a class answered the 1st question on a certain test correctly, 55% answered the 2nd question on the test correctly and 20% answered neither of the questions correctly, what percent answered both correctly?10%

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20%30%50%65%Soln: This problem can be easily solved by Venn Diagrams Lets think the total class consists of 100 students

so 75 students answered question 1and 55 students answered question 2

Now 20 students not answered any question correctly

Therefore out of total 100 students only 80 students answered either question 1 or question 2 or both the questions...

So 75+55=130 which implies 130-80=50% are the students who answered both correctly and are counted in both the groups...that’s why the number was 50 more..

Let me know if someone has problems understanding...

50. A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9B). 4 and 4C). 3 and 5D). 2 and 6E). 0 and 8

Soln: SD = Sqrt(Sum(X-x)^2/N)Since N is changing from 5 to 7. Value of Sum(X-x)^2 should change from 40(current) to 48. So that SD remains same.

so due to new numbers it adds 8. Choice D only fits here.

51. How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?

A.14 B.15. C.16 D.17 E.18

Soln: if we arrange this in AP, we get4+7+10+.......+49

so 4+(n-1)3=49: n=16C is my pick

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52. If k=m(m+4)(m+5) k and m are positive integers. Which of the following could divide k evenly?

I.3 II.4 III.6

Soln: The idea is to find what are the common factors that we get in the answer.

m = 1, k = 30 which is divisible by 1,2,3,5,6,10, 15 and 30m = 2, k = 84 which is divisible by 1,2,3,4,6, ....

As can be seen, the common factors are 1,2,3,6

So answer is 3 and 6

53. If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to(A) 2/3(B) 3/4(C) 4/3(D) 3/2(E) 2Soln: and the answer would be B...here is the explanation...

Let the side of the square be s..then the perimeter of the square is 4sLet the radius of the circle be r..then the perimeter of the circle is 2*pi*r

it is given that both these quantities are equal..therefore

4s=2*pi*r

which is then s/r=pi/2

Now the ratio of area of square to area of circle would be

s^2/pi*r^2

(1/pi)*(s/r)^2

= (1/pi)*(pi/2)^2 from the above equality relation

pi=22/7 or 3.14

the value of the above expression is approximate =0.78 which is near to answer B

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54. Two people walked the same distance, one person's speed is between 25 and 45,and if he used 4 hours, the speed of another people is between 45 and 60,and if he used 2 hours, how long is the distance? A.116 B.118 C.124 D.136 E.140

Soln: First person-- speed is between 25mph and 45mphso for 4 hrs he can travel 100 miles if he goes at 25mph speedand for 4 hrs he can travel 130 miles if he goes at 45 mph speed

similarly

Second person-- speed is between 45mph and 60mphso for 2 hrs he can travel 90 miles if he goes at 45mph speedand for 2 hrs he can travel 120 miles if he goes at 60 mph speed

So for the first person---distance traveled is greater than 100 and Less than 130

and for the second person---distance traveled is greater than 90 and less than 120

so seeing these conditions we can eliminate C, D, and E answers...

but I didn’t understand how to select between 116 and 118 as both these values are satisfying the conditions...

55. How many number of 3 digit numbers can be formed with the digits 0,1,2,3,4,5 if no digit is repeated in any number? How many of these are even and how many odd?Soln: Odd: fix last as odd, 3 ways __ __ _3_now, left are 5, but again leaving 0, 4 for 1st digit & again 4 for 2nd digit: _4_ _4_ _3_ =48 Odd.

100-48= 52 Even

56. How many 3-digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number?

A) 16 B) 80 c) 160 D) 180 E) 240

Soln: The first digit can be 2, 3, 5, or 7 (4 choices)The second digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 choices)The third digit can be 2, 3, 5, or 7 (4 choices)

4 * 4 * 10 = 160

57. There are three kinds of business A, B and C in a company. 25 percent of the total revenue is from business A; t percent of the total revenue is from B, the others

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are from C. If B is $150,000 and C is the difference of total revenue and 225,000, what is t?

A.50 B.70 C.80 D.90 E.100

Soln: Let the total revenue be X.

So X= A + B+ C

which is X= 1/4 X+ 150 + (X-225)

4X= X+ 600 + 4X -900

Solving for X you get X= 300

And 150 is 50% of 300 so the answer is 50 % (A)

58. A business school club, Friends of Foam, is throwing a party at a local bar. Of the business school students at the bar, 40% are first year students and 60% are second year students. Of the first year students, 40% are drinking beer, 40% are drinking mixed drinks, and 20% are drinking both. Of the second year students, 30% are drinking beer, 30% are drinking mixed drinks, and 20% are drinking both. A business school student is chosen at random. If the student is drinking beer, what is the probability that he or she isalso drinking mixed drinks?

A. 2/5B. 4/7C. 10/17D. 7/24E. 7/10

Soln: The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

The total number of possible outcomes is the total percent of students drinking beer.

40% of the students are first year students. 40% of those students are drinking beer. Thus, the first years drinking beer make up (40% * 40%) or 16% of the total number of students.

60% of the students are second year students. 30% of those students are drinking beer. Thus, the second years drinking beer make up (60% * 30%) or 18% of the total number of students.

(16% + 18%) or 34% of the group is drinking beer.

The outcomes that result in A is the total percent of students drinking beer and

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mixed drinks.

40% of the students are first year students. 20% of those students are drinking both beer and mixed drinks. Thus, the first years drinking both beer and mixed drinks make up (40% * 20%) or 8% of the total number of students.

60% of the students are second year students. 20% of those students are drinking both beer and mixed drinks. Thus, the second years drinking both beer and mixed drinks make up (60% * 20%) or 12% of the total number of students.

(8% + 12%) or 20% of the group is drinking both beer and mixed drinks.

If a student is chosen at random is drinking beer, the probability that they are also drinking mixed drinks is (20/34) or 10/17.

59. A merchant sells an item at a 20% discount, but still makes a gross profit of 20 percent of the cost. What percent of the cost would the gross profit on the item have been if it had been sold without the discount?

A) 20% B) 40% C) 50% D) 60% E) 75%

Soln: Lets suppose original price is 100.

And if it sold at 20% discount then the price would be 80

but this 80 is 120% of the actual original price...so 66.67 is the actual price of the item

now if it sold for 100 when it actually cost 66.67 then the gross profit would be 49.99% i.e. approx 50%

60. If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?

A. 42,500B. 37,500C. 45,000D. 40,000E. 50,000

Soln: This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)

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Using the Multiplication Principle:

= 8 * 10 * 10 * 10 * 5= 40,000

61.A bar is creating a new signature drink. There are five possible alcoholic ingredients in the drink: rum, vodka, gin, peach schnapps, or whiskey. There are five possible non-alcoholic ingredients: cranberry juice, orange juice, pineapple juice, limejuice, or lemon juice. If the bar uses two alcoholic ingredients and two non-alcoholic ingredients, how many different drinks are possible?

A. 100B. 25C. 50D. 75E. 3600

Soln: The first step in this problem is to calculate the number of ways of selecting two alcoholic and two non-alcoholic ingredients. Since order of arrangement does not matter, this is a combination problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r!))

The number of combinations of alcoholic ingredients is

C(5,2) = 5!/(2!(3!))C(5,2) = 120/(2(6))C(5,2) = 10

The number of combinations of non-alcoholic ingredients is

C(5,2) = 5!/(2!(3!))C(5,2) = 120/(2(6))C(5,2) = 10

The number of ways these ingredients can be combined into a drink can be determined by the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

The number of possible drinks is

= 10 * 10= 100

62. The sum of the even numbers between 1 and n is 79*80, where n is an odd number. N=?

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Soln: The sum of numbers between 1 and n is = (n(n+1))/2

1+2+3+.....+n=(n(n+1))/2 {formula}

we are looking for the sum of the even numbers between 1 and n, which means:

2+4+6+.....+(n-1) n is ODD=1*2+2*2+2*3+......+2*((n-1)/2)=2*(1+2+3+.....+*((n-1)/2))from the formula we obtain :=2*(((n-1)/2)*((n-1)/2+1))/2=((n-1)/2)*((n+1)/2) =79*80=> (n-1)*(n+1)=158*160=> n=159

63. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510(B) 2620(C) 1404(D) 700(E) 635

Soln: 4W 2M == 5C4.(8C2-1) = 5.(27) = 1353W 3M == 5C3.(8C3-6) = 10.50 = 500total 635

Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.consider it this way.... from my previous reply max possible ways considering we can chose any man = 700

now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...

since they are always chosen together...

For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women

1*5C4 = 5

For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing

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3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women

6C1*5C3 = 60

So total number of unfavorable cases = 5 + 60 = 65

Now since we want to exclude these 65 cases... final answer is 700-65 = 635

64. In how many ways can the letters of the word 'MISSISIPPI' be arranged?

a) 1260b) 12000c) 12600d) 14800e) 26800

Soln: Total # of alphabets = 10so ways to arrange them = 10!

Then there will be duplicates because 1st S is no different than 2nd S.we have 4 Is3 Sand 2 Ps

Hence # of arrangements = 10!/4!*3!*2!

65. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?(A) 720(B) 360(C) 120(D) 24(E) 21

Soln: two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

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if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

66. On how many ways can the letters of the word "COMPUTER" be arranged?1. M must always occur at the third place2. Vowels occupy the even positions.

Soln: For 1.7*6*1*5*4*3*2*1=5,040

For 2.) I think It should be 4 * 720there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

67. A shipment of 10 TV sets includes 3 that are defective. In how many ways can a hotel purchase 4 of these sets and receive at least two of the defective sets?

Soln: There are 10 TV sets; we have to choose 4 at a time. So we can do that by 10C4 ways. We have 7 good TV’s and 3 defective.

Now we have to choose 4 TV sets with at least 2 defective. We can do that by

2 defective 2 good3 defective 1 good

That stands to 3C2*7C2 + 3C3*7C1 (shows the count)

If they had asked probability for the same question then

3C2*7C2 + 3C3*7C1 / 10C4.

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68. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420 B. 2520 C. 168 D. 90 E. 105

Soln: 1st team could be any of 2 guys... there would be 4 teams (a team of A&B is same as a team of B&A)... possible ways 8C2 / 4.2nd team could be any of remaining 6 guys. There would be 3 teams (a team of A&B is same as a team of B&A)... possible ways 6C2 / 33rd team could be any of remaining 2 guys... there would be 2 teams (a team of A&B is same as a team of B&A). Possible ways 4C2 / 2 4th team could be any of remaining 2 guys... there would be 1 such teams... possible ways 2C2 / 1

total number of ways...

8C2*6C2*4C2*2C2 -------------------4 * 3 * 2 * 1

=8*7*6*5*4*3*2*1--------------------4*3*2*1*2*2*2*2

= 105 (ANSWER)...

Another method: say you have 8 people ABCDEFGH

now u can pair A with 7 others in 7 ways.Remaining now 6 players.Pick one and u can pair him with the remaining 5 in 5 ways.

Now you have 4 players.Pick one and u can pair him with the remaining in 3 ways.

Now you have 2 players left. You can pair them in 1 way

so total ways is 7*5*3*1 = 105 ways i.e. E

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69. In how many ways can 5 people sit around a circular table if one should not have the same neighbors in any two arrangements?Soln: The ways of arranging 5 people in a circle = (5-1)! = 4!For a person seated with 2 neighbors, the number of ways of that happening is 2: AXB or BXA, where X is the person in question.So, for each person, we have two such arrangements in 4!. Since we don't want to repeat arrangement, we divide 4!/2 to get 12

70. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf?

A) 20!/4!

B) 20!/5(4!)

C) 20!/(4!)^5

D) 20!

E) 5!

Soln: 4 copies each of 5 types.

Total = 20 books.Total ways to arrange = 20!

Taking out repeat combos = 20!/(4! * 4! * 4! * 4! * 4!) – each book will have 4 copies that are duplicate. So we have to divide 20! By the repeated copies.

71. In how many ways can 5 rings be worn on the four fingers of the right hand?Soln: 5 rings, 4 fingers1st ring can be worn on any of the 4 fingers => 4 possibilities2nd ring can be worn on any of the 4 fingers => 4 possibilities3rd ring can be worn on any of the 4 fingers => 4 possibilities4th ring can be worn on any of the 4 fingers => 4 possibilities5th ring can be worn on any of the 4 fingers => 4 possibilities

Total possibilities = 4*4*4*4*4 = 4^5.

72. If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

Soln: Write down n x (2^5) x (6^2) x (7^3) as= n x (2^5) x (3^2) x (2^2) x (7^3), = n x (2^7) x (3^2) x (7^3)

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now at a minimum 5^2 and a 3 is missing from this to make it completely divisible by 5^2 x 3^3

Hence answer = 5^2 x 3 = 75

73. Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

(1) 10050 (2) 5050 (3) 5000 (4) 50000

Soln: Consider 5 15 25 ... 995l = a + (n-1)*d

l = 995 = last terma = 5 = first termd = 10 = difference

995 = 5 + (n-1)*10

thus n = 100 = # of terms

consider 5 10 15 20.... 995

995 = 5 + (n-1)*5

=> n = 199

Another approach...

Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)

Also, series is 5 15 25.... 985 995

# of terms = 100

sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000

74. If the probability of rain on any given day in city x is 50% what is the probability it with rain on exactly 3 days in a five day period?

8/1252/255/16

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8/253/4 Soln: Use binomial theorem to solve the problem....

p = 1/2q = 1/2# of favorable cases = 3 = r# of unfavorable cases = 5-3 = 2total cases = 5 = n

P(probability of r out of n) = nCr*p^r*q^(n-r)

75. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?

A. 1/336B. 1/120C. 1/56D. 1/720E. 1/1440

The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.

There is only one result that results in a win: receiving three aces.

Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem.

The number of combinations of n objects taken r at a time is

C(n,r) = n!/(r!(n-r)!)

C(8,3) = 8!/(3!(8-3)!)C(8,3) = 8!/(3!(5!))C(8,3) = 40320/(6(120))C(8,3) = 40320/720C(8,3) = 56

The number of possible outcomes is 56.

Thus, the probability of being dealt 3 aces is 1/56.

76. The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces,

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what is the probability that a player will win a trip?

A. 1/336B. 1/120C. 1/56D. 1/720E. 1/1440

Soln: Since each draw doesn't replace the cards:Prob. of getting an ace in the first draw = 3/8getting in the second, after first draw is ace = 2/7getting in the third after the first two draws are aces = 1/6thus total probability for these mutually independent events = 3/8*2/7*1/6 = 1/56

77. Find the probability that a 4 person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains

A) exactly 1 woman B) at least 1 woman C) at most 1 woman

Soln:

B.) 7C1* 11C3/ 18C4B) 1 - (11C4/18C4)C) (11C4/18C4) + (7C1*11C3/18C4)

78. A rental car service facility has 10 foreign cars and 15 domestic cars waiting to be serviced on a particular Saturday morning. Because there are so few mechanics, only 6 can be serviced. (a) If the 6 cars are chosen at random, what is the probability that 3 of the cars selected are domestic and the other 3 are foreign? (b) If the 6 cars are chosen at random, what is the probability that at most one domestic car is selected?

Soln: A) 10C3*15C3/25C6B) Probability of no domestic car + Probability of 1 domestic car = 10C6/25C6 + 15C1 *10C5/25C6

79. How many positive integers less than 5,000 are evenly divisible by neither 15 nor 21?

A. 4,514B. 4,475C. 4,521D. 4,428E. 4,349

Soln: We first determine the number of integers less than 5,000 that are evenly divisible by 15. This can be found by dividing 4,999 by 15:

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= 4,999/15= 333 integers

Now we will determine the number of integers evenly divisible by 21:

= 4,999/21= 238 integers

some numbers will be evenly divisible by BOTH 15 and 21. The least common multiple of 15 and 21 is 105. This means that every number that is evenly divisible by 105 will be divisible by BOTH 15 and 21. Now we will determine the number of integers evenly divisible by 105:

= 4,999/105= 47 integers

Therefore the positive integers less than 5000 that are not evenly divisible by 15 or 21 are 4999-(333+238-47)=4475

80) Find the least positive integer with four different prime factors, each greater than 2.

Soln: 3*5*7*11 = 1155

81) From the even numbers between 1 and 9, two different even numbers are to be chosen at random. What is the probability that their sum will be 8?

Soln: Initially you have 4 even numbers (2,4,6,8)you can get the sum of 8 in two ways => 2 + 6 or 6 + 2so the first time you pick a number you can pick either 2 or 8 - a total of 2 choices out of 8 => 1/2after you have picked your first number and since you have already picked 1 number you are left with only 2 options => either (4,6,8) or (2,4,8) and you have to pick either 6 from the first set or 2 from the second and the probability of this is 1/3. Since these two events have to happen together we multiply them. ½ * 1/3 = 1/6

82) 5 is placed to the right of two – digit number, forming a new three – digit number. The new number is 392 more than the original two-digit number. What was the original two-digit number?

Soln: If the original number has x as the tens digit and y as the ones digit (x and y are integers less than 10) then we can set up the equation:100x + 10y + 5 = 10x + y + 39290x + 9y = 3879(10x+y) = 38710x + y = 43 ==> x = 4, y = 3the original number is 43, the new number is 435

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83) If one number is chosen at random from the first 1000 positive integers, what is the probability that the number chosen is multiple of both 2 and 8?Soln: Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 0 to 1000the first one is 8 and the last one is 1000==> ((1000-8)/8) + 1 = 125==> p (picking a multiple of 2 & 8) = 125/1000 = 1/8

84) A serial set consists of N bulbs. The serial set lights up only if all the N bulbs are in working condition. Even if one of the bulbs fails then the entire set fails. The probability of a bulb failing is x. What is the probability of the serial set failing?

Soln: Probability of x to fail. Probability of a bulb not failing = 1-xprobability that none of the N bulbs fail, hence serial set not failing = (1-x)^Nprobability of serial set failing = 1-(1-x)^N

85) Brad flips a two-sided coin 8 times. What is the probability that he gets tails on at least 7 of the 8 flips?

1/32

1/16

1/8

7/8

none of the above

Soln: Number of ways 7 tails can turn up = 8C7 the probability of those is 1/2 each Since the question asks for at least 7, we need to find the prob of all 8 tails - the number of ways is 8C8 = 1Add the two probabilities8C7*(1/2)^8 = 8/2^8 -- for getting 7 Prob of getting 8 tails = 1/2^8Total prob = 8/2^8+1/2^8 = 9/2^8

Ans is E.

86. A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller than the person standing in front of him or her. How many such arrangements of the 6 people are possible?

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B. 5B. 6C. 9D. 24E. 36

Soln: 5 ways

123 124 125 134 135 456 356 346 256 246

87. As a part of a game, four people each much secretly chose an integer between 1 and 4 inclusive. What is the approximate likelihood that all four people will chose different numbers?

Soln: The probability that the first person will pick unique number is 1 (obviously) then the probability for the second is 3/4 since one number is already picked by the first, then similarly the probabilities for the 3rd and 4th are 1/2 and 1/4 respectively. Their product 3/4*1/2*1/4 = 3/32

88. Which of the sets of numbers can be used as the lengths of the sides of a triangle?

I. [5,7,12]II. [2,4,10]III. [5,7,9]

A. I onlyB. III onlyC. I and II onlyD. I and III onlyE. II and III only

Soln: For any side of a triangle. Its length must be greater than the difference between the other two sides, but less than the sum of the other two sides.Answer is B

89. A clothing manufacturer has determined that she can sell 100 suits a week at a selling price of 200$ each. For each rise of 4$ in the selling price she will sell 2 less suits a week. If she sells the suits for x$ each, how many dollars a week will she receive from sales of the suits?

Soln: Let y be the number of $4 increases she makes, and let S be the number of suits she sells. ThenX = 200 + 4y ==> y = x/4 - 50S = 100 - 2y ==> S = 100 - 2[x/4 - 50] = 100 - x/2 + 100 = 200 - x/2

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so the answer is that the number of suits she'll sell is 200 - x/2

90. A certain portfolio consisted of 5 stocks, priced at $20, $35, $40, $45 and $70, respectively. On a given day, the price of one stock increased by 15%, while the price of another decreased by 35% and the prices of the remaining three remained constant. If the average price of a stock in the portfolio rose by approximately 2%, which of the following could be the prices of the shares that remained constant?

A) 20, 35, 70

B) 20, 45, 70

C) 20, 35, 40

D) 35, 40, 70

E) 35, 40, 45

Soln: Add the 5 prices together:20 + 35 + 40 + 45 + 70 = 210

2% of that is 210 x .02 = 4.20

Let x be the stock that rises and y be the stock that falls.

.15x -.35y = 4.20 ==> x = (7/3)y + 27

This tells us that the difference between x and y has to be at least 27. Since the answer choices list the ones that DONT change, we need to look for an answer choice in which the numbers NOT listed have a difference of at least 27.

Thus the answer is (E)

91. if -2=<x=<2 and 3<=y<=8, which of the following represents the range of all possible values of y-x?(A) 5<=y-x<=6(B) 1<=y-x<=5(C) 1<=y-x<=6(D) 1<=y-x<=10 (E) 1<=y-x<=10

Soln: you can easily solve this by subtracting the two inequalities. To do this they need to be in the opposite direction; when you subtract them preserve the sign of the inequality from which you are subtracting.

3 < y < 8

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multiply the second one by (-1) to reverse the sign2 > x > -2Subtract them to get3 - 2 < y - x < 8 - (-2)1 < y - x < 10

92. Of a group of 260 people who purchased stocks, 61 purchased A, 88 purchased B, 56 purchased C, 75 purchased D, 60 purchased E. what is the greatest possible number of the people who purchased both B and D?

A:40 B:50 C:60 D:75 E:80

Soln: ANSWER is D (75)

since they have asked us to find out the greatest possible number buying both B as well as D, the answer has to be the smallest no between the two which is 75...as all the guys purchasing D can also buy B and only 75 out of 88 purchasing B can simultaneously purchase D as well....

93. There are 30 people and 3 clubs M, S, and Z in a company. 10 people joined M, 12 people joined S and 5 people joined Z. If the members of M did not join any other club, at most, how many people of the company did not join any club?

A: 4 B: 5 C: 6 D: 7 E: 8

Soln: total no of people = 30no joining M = 10no joining S = 12no joining Z = 5question asked - AT MOST how many people did not join any group?

solution: now since none of the members of M joined any other group, the no of people left = 30-10(for M)=20since the question says at most how many did not join any group, lets assume the all people who join Z also join S. so no of people joining group S and Z are 12 (note that there will be 5 people in group S who have also joined Z)

therefore no of people not joining any group = 20-12=8Hence e

94. Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively.

1) in a row2) in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats3) around a table

A:1) 4! * 4! * 22) 4! * 4! * 2 - number of ways with John and Susan sitting together

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= (4! * 4! * 2) - (7 * 3! * 3! *2)

The way that JS arrangements are found is by bracketing J and S and considering it to be a single entity. So a possible arrangement is (B=boy, G=girl)

(JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252(SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252

3) Fix one boy and arrange the other 3 boys in 3! ways. Arrange the 4 girls in 4! ways in the gaps between the boys.

Total arrangements = 3! x 4!

= 6 x 24

= 144

95. From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/10

B. 4/9

C. 1/2

D. 3/5

E. 2/3

Soln: Total number of ways of selecting 4 children = 6C4 = 15

with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

Hence p = 9/15 = 3/5

96. What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?

(1) 0 (2) 3 (3) 4 (4) 2 (5) None of these

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Soln: Remainder of 9*odd /6 is 3remainder of 9*even/6 is 0

9^1 + 9^2 + 9^3 + ...... + 9^9=9*(1+9+9^2+.....9^8)

1+9+9^2+.....9^8 is odd.

Thus we obtain 3 as a remainder when we divide 9*(1+9+9^2+.....9^8) by 6.

Another way: We get the value as 9*odd/6 = 3*odd/2 Since 3*odd = odd; odd/2 = XXXX.5so something divided by 6, gives XXXX.5, hence remainder is 6*0.5 = 3

97. Find the value of 1.1! + 2.2! + 3.3! + ......+n.n!

(1) n! +1 (2) (n+1)!(3) (n+1)!-1(4) (n+1)!+1(5) None of these

Soln: 1.1! + 2.2! + 3.3! + ......+n.n!=1.1! + (3-1)2! + (4-1)3! +......+ ((n+1)-1) n!=1.1!+3!-2!+4!-3!+.......+(n+1)!-n!

So it is (n+1)! -1 (Answer choice 4)

98. The numbers x and y are three-digit positive integers, and x + y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x < y, which of the following must be true?

I. The units digit of x + y is greater than the units digit of either x or y.II. The tens digit of x + y equals 2.III. The hundreds digit of y is at least 5.

A. II only

B. III only

C. I and II

D. I and III

E. II and III

Soln: x= abc y= def

x = a7c y= b5f

x > y and x+y = wxyz.

I. The units digit of x + y is greater than the units digit of either x or y.

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It can carryover one digit. False

II. The tens digit of x + y equals 2.

It can be 2 or 3. False

III. The hundreds digit of y is at least 5.

a+b+1 >= 10a >b so a at least 5. True.

Ans: b

99. Among 5 children there are 2 siblings. In how many ways can the children be seated in a row so that the siblings do not sit together? (A) 38 (B) 46 (C) 72 (D) 86 (E) 102Soln: The total number of ways 5 of them can sit is 120when the siblings sit together they can be counted as one entitytherefore the number of ways that they sit together is 4!=24, but sincethe two siblings can sit in two different ways e.g. AB and BA we multiply 24 by 2 to get the total number of ways in which the 5 children can sit together with the siblings sitting together - 48In other words 4P4*2P2the rest is obvious 120-48=72

100. There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.

Soln: MuEuG = M + E + G - MnE - MnG - EnG - 2(MnEnG)MnE + MnG + EnG = M + E + G - 2(MnEnG) - MuEuGMnE + MnG+ EnG = 40 + 30 + 35 - 2(15) - 70 = 105 - 30 - 70 = 5

Whenever an intersection occurs between 2 sets, (MnEnG) is counted twice, therefore you deduct one of it. If the intersection occurs between 3 sets, it is counted thrice; therefore you deduct two of it. And so forth.If there are four sets, then the formula is A + B + C + D -(two) -(three)*2 -(four)*3 = total 101. John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work

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on it together?

A- 6 B- 10 C- 8 D- 7.5 E- 3.5

Soln: Together they do 1/20+1/12=4/30 of the task.

Jane and John started the work together, but only John finished the work because Jane gets sick. So let x be the number of days they worked together.

x*3/40+4*1/20=1

x4/30=4/5 and therefore x =6

So in total they worked 6 days on it together and John worked 4 days on it. So total days spent=10, but if the question is asking how many time did they spend working on the project together, then the answer is 6.

102. Jane gave Karen a 5 m head start in a 100 race and Jane was beaten by 0.25m. In how many meters more would Jane have overtaken Karen?

Soln: Jane gave Karen a 5 m head start means Karen was 5 m ahead of Jane. So after the lead, Karen ran 95m and Jane ran 99.75 m when the race ended.

Let speed of Karen and Jane be K and J respectively and lets say after X minutes Jane overtakes Karen.

1st condition: 95/K = 99.75/J2nd condition JX - KX = 0.25

Solving for JX, we get JX=21/4.Hence Jane needs to run 5.25m more (or total of 105m) to overtake Karen.

103. Out of 20 surveyed students 8 study math and 7 study both math and physics. If 10 students do not study either of these subjects, how many students study physics but not math?

(A) 1 (B) 2 (C) 4 (D) 5 (E) 6Soln: total = gr1 + gr2 - both + neither20 = 8 + P - 7 + 10P = 9, 9 students study both P and M, 7 study both, 9-7 = 2 study only P

104. Machine A can produce 50 components a day while machine B only 40. The monthly maintenance cost for machine A is $1500 while that for machine B is $550. If

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each component generates an income of $10 what is the least number of days per month that the plant has to work to justify the usage of machine A instead of machine B?

(A) 6 (B) 7 (C) 9 (D) 10 (E) 11

Soln: Let x be the number of days that need to be worked 500x - 1500 > 400x - 550100x > 950x > 9.5 (D)

105. Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?

A. 2/3B. 7/3C.5/2D. 8/3E. 3

Soln: x + y = 4x = 4 - y

x/2 = y/44x = 2y4(4-y) = 2y16 = 6yy = 8/3 (D)

106. If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!

(B) 10*10

(C) 10*9

(D) 45

(E) 36

Soln: There are 10C2 ways to pick 2 different people out of 10 people.

10C2 = 10!/2!8! = 45 (D)

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107. If operation $ is defined as $X = X + 2 if X is even $X = X - 1 if X is odd, what is $(...$($($(15)))...) 99 times?

(A) 120 (B) 180 (C) 210 (D) 225 (E) 250

Soln: $15 gives 14

After that it is an AP with d=2, a=14 and n =99

108. A and B alternately toss a coin. The first one to turn up a head wins. if no more than five tosses each are allowed for a single game.

1- Find the probability that the person who tosses first will win the game?

2- What are the odds against A's losing if she goes first?

Soln: look at the conditions; it says that the first person who tosses a head wins.

Let’s say A tosses first.

what is the probability that he wins

H + TTH + TTTTH + TTTTTTH + TTTTTTTTH

i.e. either the first toss is head, or the first time A tosses the coin he gets a tail and B also gets a tail , n in the second throw A gets a head.....

This continues for a max till 5 throws, because the game is for 5 throws only.So, 1. 1/2 + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9

2. (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10

109. How many integers less than 1000 have no factors (other than 1) in common with 1000?

(1) 400 (2) 410(3) 411 (4) 412(5) None of the above

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Soln: 1000 - multiples of 2 and/or 5

multiples of 2 = 500 (all even #)multiples of 5 = (995 -5)/10 + 1 [ Using AP formula]= 100

Answer = 1000 - (500 + 100)= 400

You cannot calculate for all multiples of 5 because you have already removed all even integers (including 10, 20, and 30). The difference in the AP series should be 10 instead of 5 because you're looking for the integers that have 5 as a unit’s digit. Therefore we divide by 10 and not 5.

110. Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

36, 28, 12, 9 or none

Soln: Let the divisor be a.

x = a*n + 11 ---- (1)y = a*m + 21 ----- (2)also given, (x+y) = a*p + 4 ------ (3)adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4)

equate 3 and 4.a*p + 4 = a*(n+m) + 32or a*p + 4 = [a*(n+m) + 28] + 4cancel 4 on both sides.u will end up with.a*p = a*(n+m) + 28.

which implies that 28 should be divisible by a. or in short a = 28 works.

Another method: I think the easiest (not necessarily the shortest), way to solve this is to use given answer choices. Since the remainders are given as 11 and 21, therefore the divisor has to be greater than 21 which leaves with two choices 28 and 36. Try 28 first; let the two numbers be 28+11= 39 and 28+21= 49. Summing them up and dividing by 28 gives (49+39=88), 88/28 remainder is 4, satisfies the given conditions. Check for 36 with same approach, does not work, answer is 28

111. There are 8 members; among them are Kelly and Ben. A committee of 4 is to be chosen out of the 8. What is the probability that Ben is chosen to be in the committee and Kelly is not?

Soln: let’s assume Ben has already been chosen. Then I have to choose 3 more people from the remaining, excluding Kelly, that is, three from six people, that’s 6c3.so the total is (1c1.6c3)/8c4 which is 2/7

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112. How many 5-digit positive integers exist where no two consecutive digits are the same?

A.) 9*9*8*7*6 B.) 9*9*8*8*8 C.) 9^5D.) 9*8^4 E.) 10*9^4

Soln: C is correct. The first place has 9 possibilities, since 0 is not to be counted. All others have 9 each, since you cannot have the digit, which is same as the preceding one.Hence 9^5

113. How many five digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits? (A) 15 (B) 96 (C) 216 (D) 120 (E) 180

Soln: The sum of digits of a multiple of 3 should be div by 3.for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15.For a sum of 12, the digits that can be used : 0,1,2,4,5for a sum of 15: 1,2,3,4,5Number of numbers from the first set = 4.4! (0 cannot be the first digit in the numbers)for the second set : 5!total = 5! +4.4! = 4!(5+4) = 24*9 = 216

Since 0 cannot be the first digit of a number, for the first position, you have 4 choices (all digits except zero). No such constraints exist for the rest of the positions; hence the next choices are 4,3,2,1 - all multiplying up to give a 4!. Had there been no 0 involved, the choices would've been 5! Instead of 4.4!

114. A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420 B. 2520 C. 168 D. 90 E. 105

Soln: out of 8 people one team can be formed in 8c2 ways.

8c2*6c2*4c2*2c2= 2520.The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )

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115. My name is AJEET. But my son accidentally types the name by interchanging a pair of letters in my name. What is the probability that despite this interchange, the name remains unchanged?

a) 5% b) 10% c) 20% d) 25%

Soln: there are actually 20 ways to interchange the letters, namely, the first letter could be one of 5, and the other letter could be one of 4 left. So total pairs by product rule = 20.

Now, there are two cases when it wouldn’t change the name. First, keeping them all the same. Second, interchanging the two EEs together. Thus 2 options would leave the name intact.

Prob = 2/20 = 0.1, or 10%.

116. A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

A. y > ROOT2

B. ROOT3/2 < y < ROOT2

C. ROOT2/3 < y < ROOT3/2

D. ROOT3/4 < y < ROOT2/3

E. y < ROOT3/4

Soln: right triangle with sides x<y<z and area of 1 => z = hypotenuse and xy/2 = 1i.e xy = 2

If x were equal to y, we would have had xy = y^2 = 2. And y = root2

But, x<y and so y>root2.

117. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

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Soln: Chance of drawing a blue on the first draw = 2/8, so chance of not drawing a blue on the first draw is 6/8

similarly chance of not drawing blue on second draw = 5/7

Therefore probability of not drawing blue in 2 draws = 6/8*5/7 = 15/28

118. How many integers between 100 and 150, inclusive can be evenly divided by neither 3 nor 5?

Soln: Number of integers that divide 3:

the range is 100-150Relevant to this case, we take 102 - 150 (since 102 is the first to div 3)102 = 34*3150= 50*3, so we have 50-34+1 = 17 multiples of 3

For multiples of 5,100=5*20150=5*3030-20+1 =11

Now we have a total of 27 integers, but we double counted the ones that divide BOTH 3 AND 5, ie 15.

105 is the first to divide 15.105=15*7150=15*1010-7+1 = 4 integers

So our total is 17+11-4 = 24 integers that can be divided by either 3 or 5 or both.

51 integers - 24 integers = 27 that cannot be evenly divided.

119. Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?

(A) 10% (B) 33 1/3%

(C) 40%

(D) 50%

(E) 66 2/3%

Soln: (x+y)30/100 = x*40/100 + y*25/10030x + 30y = 40x + 25yy = 2x or y/x = 2/1 or y:x = 2:1 hence x = 33 1/3%

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120. Sequence A and B. a1=1, b1=k. an=b(n-1)-a(n-1) bn=b(n-1)+a(n-1). What is a4=?

Soln: a2 = k-1 ; b2 = k+1

a3= (k+1)-(k-1) = 2 ; b3 = (k+1)+(k-1) = 2k

a4 = 2k - 2 = 2(k-1) = 2(b1-a1)

121. A person put 1000 dollars in a bank at a compound interest 6 years ago. What percentage of the initial sum is the interest if after the first three years the accrued interest amounted to 19% of the initial sum? A) 38% B) 42% C) 19% D) 40%

Soln: assume, interest = rso after 3 years total money = 1000*(1+r)^3 = 1000*1.19(1+r)^3 = 1.19so after 6 years total money = 1000*(1+r)^6 = 1000*1.19^2 = 1000*1.42so percentage of interest is 42%

122. A, B and C run around a circular track of length 750m at speeds of 3 m/sec, 6 m/sec and 18 m/sec respectively. If all three start from the same point, simultaneously and run in the same direction, when will they meet for the first time after they start the race?

A. 750 secondsB. 50 secondsC. 250 secondsD. 375 secondsE. 75 seconds

Soln: When two people are running in the same direction the relative speed is a difference in speeds of the two people.

In this case A=3 B=6 C=18

So relative speed of B wrt A is 6-3 = 3m/sRelative speed of A wrt to C is 18-3 =15m/s

Therefore relative distances will be:B wrt A is 750/3 =250C wrt to A 750/15 = 50

So they have to bridge this distance of 250 and 50 between them which is the LCM of 250 and 50 which is 250.

Another Method: Simply put, Runner A's time take to run one lap is 250Runner B's time is 125sand Runner C's time is 41.67s

We can notice that A=2Band thet B=3C

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So when 250 s elapse, they will be at their starting point. A will have completed one lap, B 2 laps, and C 6 laps

123. X percents of the rooms are suits, Y percent of the rooms are painted light blue. Which of the following best represents the least percentage of the light blue painted suits?

1) X-Y2)Y-X +1003)100X-Y4)X+Y-100e)100-XY

Soln: Equation from set theory:

n(AUB)=n(A)+n(B)-n(A^B)

where,

A= % of rooms which are suitesB= % of rooms painted blueA^B means the intersection of the two sets

Now in this case, what we need to find is n(A^B), thereforen(A^B)=n(A)+n(B)-n(AUB)= X + Y - n(AUB)

Now this would be least when n(AUB) is maximum, which would happen if these two kinds of rooms are only two kinds available, making n(AUB)=100

Therefore the answer should be X+Y-100

Combinatory and probability

1. In a workshop there are 4 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet?

a) 168.b) 16.c) 80.d) 48.

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e) 56.

2. In a workshop there are 4 kinds of beds, 3 kinds of closets, 2 kinds of shelves and 7 kinds of chairs. In how many ways can a person decorate his room if he wants to buy in the workshop one shelf, one bed and one of the following: a chair or a closet?

a) 168.b) 16.c) 80.d) 48.e) 56.

3. Three people are to be seated on a bench. How many different sitting arrangements are possible if Erik must sit next to Joe?

a) 2.b) 4.c) 6.d) 8.e) 10.

4. How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

a) 648.b) 504.c) 576.d) 810.e) 672.

5. Barbara has 8 shirts and 9 pants. How many clothing combinations does Barbara have, if she doesn’t wear 2 specific shirts with 3 specific pants?

a) 41.b) 66.c) 36.d) 70.e) 56.

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

a) 27.b) 36.c) 72.

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d) 112.e) 422.

7. In jar A there are 3 white balls and 2 green ones, in jar B there is one white ball and three green ones. A jar is randomly picked, what is the probability of picking up a white ball out of jar A?

a) 2/5.b) 3/5.c) 3/10.d) 3/4e) 2/3.

8. Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.b) 1/30.c) 2/5.d) 1/720.e) 3/10.

9. The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black ball out of a glass jar and breaking the jar is 1/Y. What is the probability of breaking the jar?

a) 1/(XY).b) X/Y.c) Y/X.d) 1/(X+Y).e) 1/(X-Y).

10. Danny, Doris and Dolly flipped a coin 5 times and each time the coin landed on “heads”. Dolly bet that on the sixth time the coin will land on “tails”, what is the probability that she’s right?

a) 1.b) ½.c) ¾.d) ¼.e) 1/3.

11. In a deck of cards there are 52 cards numbered from 1 to 13. There are 4 cards of each number in the deck. If you insert 12 more cards with the number 10 on them and you shuffle the deck really good, what is the probability to pull out a card with a number 10 on it?

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a) 1/4.b) 4/17.c) 5/29.d) 4/13.e) 1/3.

12. There are 18 balls in a jar. You take out 3 blue balls without putting them back inside, and now the probability of pulling out a blue ball is 1/5. How many blue balls were there in the beginning?

a) 9.b) 8.c) 7.d) 12.e) 6.

13. In a box there are A green balls, 3A + 6 red balls and 2 yellow ones. If there are no other colors, what is the probability of taking out a green or a yellow ball?

a) 1/5.b) 1/2.c) 1/3.d) 1/4.e) 2/3.

14. The probability of Sam passing the exam is 1/4. The probability of Sam passing the exam and Michael passing the driving test is 1/6.What is the probability of Michael passing his driving test?

a) 1/24.b) 1/2.c) 1/3.d) 2/3.e) 2/5

15. In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

a) 1/5.b) ½.c) 1/3.d) 3/10.e) ¼.

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16. Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary. What is the probability that only girls will be elected?

a) 8/125.b) 2/5.c) 1/30.d) 1/720.e) 13/48.

17. Two dice are rolled. What is the probability the sum will be greater than 10?

a) 1/9.b) 1/12.c) 5/36.d) 1/6.e) 1/5.

18. The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?

a) 1/8.b) 1/6.c) 1/3.d) 1/5.e) ¼.

19. On one side of a coin there is the number 0 and on the other side the number 1. What is the probability that the sum of three coin tosses will be 2?

a) 1/8.b) ½.c) 1/5.d) 3/8.e) 1/3.

20. In a flower shop, there are 5 different types of flowers. Two of the flowers are blue, two are red and one is yellow. In how many different combinations of different colors can a 3-flower garland be made?

a) 4.b) 20.c) 3.d) 5.e) 6.

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21. In a jar there are balls in different colors: blue, red, green and yellow.The probability of drawing a blue ball is 1/8.The probability of drawing a red ball is 1/5.The probability of drawing a green ball is 1/10.If a jar cannot contain more than 50 balls, how many yellow balls are in the Jar?

a) 23.b) 20.c) 24.d) 17.e) 25.

22. In a jar there are 3 red balls and 2 blue balls. What is the probability of drawing at least one red ball when drawing two consecutive balls randomly?

a) 9/10b) 16/20c) 2/5d) 3/5e) ½

23. In Rwanda, the chance for rain on any given day is 50%. What is the probability that it rains on 4 out of 7 consecutive days in Rwanda?

a) 4/7b) 3/7c) 35/128d) 4/28e) 28/135

24. A Four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼b) ½c) ¾d) 15/16e) 1/16

25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits?

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a) 15/16b) 11/16c) 11/12d) ½e) 5/8

26. What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8b) ¼c) ½d) 3/8e) 5/8

27. In how many ways can you sit 8 people on a bench if 3 of them must sit together?

a) 720b) 2,160c) 2,400d) 4,320e) 40,320

28. In how many ways can you sit 7 people on a bench if Suzan won’t sit on the middle seat or on either end?

a) 720b) 1,720c) 2,880d) 5,040e) 10,080

29. In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color?

a) 8b) 23c) 29d) 32e) 53

30. In a jar there are 21 white balls, 24 green balls and 32 blue balls. How many balls must be taken out in order to make sure we have 23 balls of the same color?

a) 23

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b) 46c) 57d) 66e) 67

31. What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

a) 10/216b) 12/216c) 21/216d) 23/216e) 25/216

32. How many diagonals does a polygon with 21 sides have, if one of its vertices does not connect to any diagonal?

a) 21b) 170c) 340d) 357e) 420

33. How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal? Use the formula: number of diagonals: n (n-3)/2 where n is the number of sides. Each vertex sends of n-3 diagonals.

a) 90b) 126c) 210d) 264e) 306

34. What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously? Use the formula: number of diagonals: n (n-3)/2 where n is the number of sides. Each vertex sends of n-3 diagonals.

a) 1/6b) ¼c) ½

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d) 21/216e) 32/216

35. The telephone company wants to add an area code composed of 2 letters to every phone number. In order to do so, the company chose a special sign language containing 124 different signs. If the company used 122 of the signs fully and two remained unused, how many additional area codes can be created if the company uses all 124 signs?

a) 246b) 248c) 492d) 15,128e) 30,256

36. How many 8-letter words can be created using computer language (0/1 only)?

a) 16b) 64c) 128d) 256e) 512

37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?

a) 2520b) 3150c) 3360d) 6000e) 7500

38. A drawer holds 4 red hats and 4 blue hats. What is the probability of getting exactly three red hats or exactly three blue hats when taking out 4 hats randomly out of the drawer and returning each hat before taking out the next one? (Tricky)

a) 1/8b) ¼c) ½d) 3/8e) 7/12

39. Ruth wants to choose 4 books to take with her on a camping trip. If Ruth has a total of 11 books to choose from, how many different book quartets are possible?

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a) 28b) 44c) 110d) 210e) 330

40. A computer game has five difficulty levels. In each level you can choose among four different scenarios except for the first level, where you can choose among three scenarios only. How many different games are possible?

a) 18b) 19c) 20d) 21e) None of the above

41. How many four-digit numbers that do not contain the digits 3 or 6 are there?

a) 2401b) 3584c) 4096d) 5040e) 7200

42. How many five-digit numbers are there, if the two leftmost digits are even, the other digits are odd and the digit 4 cannot appear more than once in the number?

a) 1875b) 2000c) 2375d) 2500e) 3875

43. In a department store prize box, 40% of the notes give the winner a dreamy vacation; the other notes are blank. What is the approximate probability that 3 out of 5 people that draw the notes one after the other, and immediately return their note into the box get a dreamy vacation?

a) 0.12b) 0.23c) 0.35d) 0.45

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e) 0.65

44.The probability of having a girl is identical to the probability of having a boy. In a family with three children, what is the probability that all the children are of the same gender?a) 1/8.

b) 1/6.

c) 1/3.

d) 1/5.

e) E) ¼

45. A buyer buys 3 different items out of the newly introduced 10 different items. If two items were to be selected at random, what is the probability that the buyer does not have both the chosen items?

46. A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?

47. There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?

48. Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, and then drew cards one at a time randomly from the box, without returning the cards he had already drawn to the box. In order to ensure that the sum of all cards he drew was even, how many cards did Jerome have to draw?

49. From (1, 2, 3, 4, 5, 6), one number is picked out and replaced and one number is picked out again. If the sum of the 2 numbers is 8, what is the probability that the 2 numbers included the number 5?

50. Each participant in a certain study was assigned a sequence of 3 different letters from the set {A, B, C, D, E, F, G, H}. If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned, what was the number of participants in the study? (Note, for example, that the sequence A, B, C is different from the sequence C, B, A.)

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51. In how many ways can 11 # signs and 8* signs be arranged in a row so that no two * signs come together?

52. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

A. 21/50

B. 3/13

C. 47/50

D. 14/15

E. 1/5

53. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?A. 15/28

B. 1/4

C. 9/16

D. 1/32

E. 1/16

54. There are 5 red marbles, 3 blue marbles, and 2 green marbles. If a user chooses two marbles, what is the probability that the two marbles will be a different color?

55. A bag contains six marbles: two red, two blue, and two green. If two marbles are drawn at random, what is the probability that they are the same color?

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Explanations:

1. The best answer is C.You must multiply your options to every item. (2 shelves) x (4 beds) x (3 closets + 7 chairs) = 80 possibilities.

2. The best answer is C.You must multiply your options to every item. (2 shelves) x (4 beds) x (3 closets + 7 chairs) = 80 possibilities.

3. The best answer is B.Treat the two who must sit together as one person. You have two possible sitting arrangements. Then remember that the two that sit together can switch places. So you have two times two arrangements and a total of four.

4. The best answer is C.For the first digit you have 9 options (from 1 to 9 with out 0), for the second number you have 9 options as well (0 to 9 minus the first digit that was already used) and for the third digit you have 8 options left.So the number of possibilities is 9 x 9 x 8 = 648.

5. The best answer is D.There are (8 x 9) 72 possibilities of shirts + pants. (2 x 3) 6 Of the combinations are not allowed. Therefore, only (72 – 6) 66 combinations are possible.

6. The best answer is A.First digit is 1, the second is 2, the third can be (7,8,9), the forth can be (3,6,9), the fifth and the sixth are dependent with one another. The fifth one is 3 times bigger than the sixth one, therefore there are only 3 options there: (1,3), (2,6), (3,9). All together there are: 1 x 1 x 3 x 3 x 3 = 27 options.

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7. The best answer is C.The probability of picking the first jar is ½, the probability of picking up a white ball out of jar A Is 3/(3+2) = 3/5. The probability of both events is 1/2 x 3/5 = 3/10.

8. The best answer is B.The probability for the first one to be black is: 4/(4+6) = 2/5.The probability for the second one to be black is: 3/(3+6) = 1/3.The probability for the third one to be black is: 2/(2+6) = 1/4.The probability for all three events is (2/5) x (1/3) x (1/4) = 1/30.

9. The best answer is B.Let Z be the probability of breaking the jar, therefore the probability of both events happening is Z x (1/X) = (1/Y). Z = X/Y.

10. The best answer is B.The probability of the coin is independent on its previous outcomes and therefore the probability for “head” or “tail” is always ½.

11. The best answer is A.The total number of cards in the new deck is 12 +52 = 64.There are (4 + 12 = 16) cards with the number 10.The probability of drawing a 10 numbered card is 16/64 = 1/4.

12. The best answer is E.After taking out 3 balls there are 15 left. 15/5 = 3 blue balls is the number of left after we took out 3 therefore there were 6 in the beginning.

13. The best answer is D.The number of green and yellow balls in the box is A+2.The total number of balls is 4A +8.The probability of taking out a green or a yellow ball is (A+2)/(4A+8)=1/4.

14. The best answer is D.Indicate A as the probability of Michael passing the driving test.

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The probability of Sam passing the test is 1/4, the probability of both events happening together is 1/6 so: 1/4 x A = 1/6 therefore A = 2/3.

15. The best answer is B.Indicate A as the probability of drawing a white ball from the jar.The probability of drawing a red ball is 1/5.The probability of drawing both events is 1/10 so, 1/5 x A = 1/10.Therefore A = ½.

16. The best answer is C.The basic principle of this question is that one person can’t be elected to more than one part, therefore when picking a person for a job the “inventory” of remaining people is growing smaller.The probability of picking a girl for the first job is 4/10 = 2/5.The probability of picking a girl for the second job is (4-1)/(10-1) = 3/9.The probability of picking a girl for the third job is (3-1)/(9-1) = 1/4.The probability of all three events happening is: 2/5 x 3/9 x ¼ = 1/30.

17. The best answer is B.When rolling two dice, there are 36 possible pairs of results (6 x 6).A sum greater than 10 can only be achieved with the following combinations: (6,6), (5,6), (6,5).Therefore the probability is 3/36 = 1/12.

18. The best answer is E.The gender of the first-born is insignificant since we want all children to be of the same gender no matter if they are all boys or girls.The probability for the second child to be of the same gender as the first is: ½. The same probability goes for the third child. Therefore the answer is ½ x ½ = ¼.

19. The best answer is D.The coin is tossed three times therefore there are 8 possible outcomes (2 x 2 x 2). We are interested only in the three following outcomes:(0,1,1), (1,0,1), (1,1,0).The probability requested is 3/8.

20. The best answer is A.We want to make a 3-flower garlands, each should have three colors of flowers in it.There are two different types of blue and two different types of red.The options are (2 blue) x (2 red) x (1 yellow) = 4 options.

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21. The best answer is A.If 1/8 is the probability of drawing a blue ball then there are 40/8 = 5 blue balls in the jar. And with the same principle there are 8 red balls and 4 green ones. 40 – 5 – 8 – 4 = 23 balls (yellow is the only color left).

22. The best answer is A.Since we want to draw at least one red ball we have four different possibilities:

1. Drawing blue-blue.2. Drawing blue-red.3. Drawing red-blue.4. Drawing red-red.

There are two ways to solve this question:One minus the probability of getting no red ball (blue-blue):1-2/5 x ¼ = 1-2/20 = 18/20 = 9/10/Or summing up all three good options:Red-blue --> 3/5 x 2/4 = 6/20.Blue-red --> 2/5 x ¾ = 6/20.Red-red --> 3/5 x 2/4 = 6/20.Together = 18/20 = 9/10.

23. The best answer is C. We have 7!/(4!*3!) = 35 different possibilities for 4 days of rain out of 7 consecutive days (choosing 4 out of seven). Every one of these 35 possibilities has the following probability: every day has the chance of ½ to rain so we have 4 days of ½ that it will rain and 3 days of ½ that it will not rain. We have ½ to the power of 7 = 1/128 as the probability of every single event. The total is 35 x 1/128 = 35/128.

24. The best answer is D.For every digit we can choose out of 8 digits (10 total minus 1 and 4). There are four different options:

5. No even digits6. One even digit.7. Two even digits.8. Three even digits.9. Four even digits.

The probability of choosing an odd (or an even) digit is ½.One minus the option of no even digits: 1- (1/2)4= 15/16. You can also sum up all of the other options (2-5).

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25. The best answer is B.Since 1 appears exactly three times, we can solve for the other four digits only. For every digit we can choose out of 8 digits only (without 1 and 0). Since we have 4 prime digits (2, 3, 5, 7) and 4 non-prime digits (4, 6, 8, 9), the probability of choosing a prime digit is ½. We need at least two prime digits:One minus (the probability of having no prime digits + having one prime digit):There are 4 options of one prime digit, each with a probability of (1/2)4.There is only one option of no prime digit with a probability of (1/2)4.So: [1- ((1/2)4+(1/2)4*4)] = 11/16.

26. The best answer is D.There are three different arrangements of a boy and two girls:(boy, girl, girl), (girl, boy, girl), (girl, girl, boy). Each has a probability of (1/2)3. The total is 3*(1/2)3=3/8.

27. The best answer is D.Treat the three that sit together as one person for the time being. Now, you have only 6 people (5 and the three that act as one) on 6 places: 6!=720. Now, you have to remember that the three that sit together can also change places among themselves: 3! = 6. So, The total number of possibilities is 6!*3!= 4320.

28. The best answer is C.First, check Suzan: she has 4 seats left (7 minus the one in the middle and the two ends), After Suzan sits down, the rest still have 6 places for 6 people or 6! Options to sit. The total is Suzan and the rest: 4*6! = 2880.

29. The best answer is C.The worst case is that we take out seven balls of each color and still do not have 8 of the same color. The next ball we take out will become the eighth ball of some color and our mission is accomplished. Since we have 4 different colors: 4*7(of each) +1=29 balls total.Of course you could take out 8 of the same color immediately, however we need to make sure it happens, and we need to consider the worst-case scenario.

30. The best answer is D.The worst case would be to take out 21 white balls, 22 green and 22 blue balls and still not having 23 of the same color. Take one more ball out and you get 23 of either the green or the blue balls. Notice that you cannot get 23 white balls since there are only 21, however, you must consider them since they might be taken out also.The total is: 21+22+22+1= 66.

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31. The best answer is E. Start checking from the smaller or bigger numbers on the dice. We will check from bigger numbers working downwards: start with 6, it has the following options: (6,5,1), (6,4,2), (6,3,3). Now pass on to 5: (5,5,2), (5,4,3). Now 4: (4,4,4). And that’s it, these are all number combinations that are possible, if you go on to 3, you will notice that you need to use 4, 5 or 6, that you have already considered (the same goes for 2 and 1). Now analyze every option: 6,5,1 has 6 options (6,5,1), (6,1,5), (5,1,6), (5,6,1), (1,6,5), (1,5,6). So do (6,4,2) and (5,4,3). Options (6,3,3) and (5,5,2) have 3 options each: (5,5,2), (5,2,5) and (2,5,5). The same goes for (6,3,3). The last option (4,4,4) has only one option. The total is 3*6+2*3+1=18+6+1 = 25 out of 216 (63) options.

32. The best answer is B. We have 20 vertices linking to 17 others each: that is 17*20=340. We divide that by 2 since every diagonal connects two vertices. 340/2=170. The vertex that does not connect to any diagonal is just not counted.

33. The best answer is A. We have 15 Vertices that send diagonals to 12 each (not to itself and not to the two adjacent vertices). 15*12=180. Divide it by 2 since any diagonal links 2 vertices = 90. The three vertices that do not send a diagonal also do not receive any since the same diagonal is sent and received. Thus they are not counted.

34. The best answer is A.The options for a sum of 14: (6,4,4) has 3 options, (6,5,3) has 6 options, (6,6,2) has 3 options, (5,5,4) has 3 options. We have 15 options to get 14.The options for a sum of 8: (6,1,1) has 3 options, (5,2,1) has 6 options, (4,3,1) has 6 options, (4,2,2) has 3 options, (3,3,2) has 3 options. We have 21 options to get 8.Total: 21+15= 36/216 = 1/6.

35. The best answer is C. The phone company already created 122*122 area codes, now it can create 124*124.1242-1222=(124+122)(124-122) = 246*2 = 492 additional codes.There are other ways to solve this question. However this way is usually the fastest.

36. The best answer is D.Every letter must be chosen from 0 or 1 only. This means we have two options for every word and 28 = 256 words total.

37. The best answer is A.

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The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before). The total is 4*5*3*7*6=2520.

38. The best answer is C.Getting three red out of 4 that are taken out has 4 options (4!/(3!*1!)) each option has a probability of (1/2)4 since drawing a red or blue has a 50% chance. 4*1/16= ¼ to get three red hats. The same goes for three blue hats so ¼+¼ =1/2.

39. The best answer is E.Choosing 4 out of 11 books is: 11!/(4!*7!) = 330 possibilities.

40. The best answer is .On four levels there are 4 scenarios = 16 different games. The first level has 3 different scenarios. The total is 19 scenarios.

41. The best answer is B.The first digit has 7 possibilities (10 – 0,3 and 6). The other three digits have 8 possibilities each. 7*8*8*8= 3584.

42. The best answer is C.Not considering the fact that 4 cannot appear more than once, we have a total of 4*5*5*5*5=2500. Now we deduct the possibilities where 4 does appear more than once (in this case it can appear only twice on the two leftmost even digits). In order to do so, we put 4 in the first and second leftmost digits. The rest of the digits are odd: 5*5*5=125. 2500-125=2375.

43. The best answer is B.The chance of winning is 0.4 and it stays that way for all people since they return their note. The number of different options to choose 3 winners out of 5 is 5!/(3!*2!) = 10. Each option has a chance of 0.4*0.4*0.4*0.6*0.6 = 0.02304 * 10 = 0.2304.

44. The best answer is E.First child could be B or G, similarly 2nd and third could be B or G. Hence total number of ways =2*2*2 =8.

Favorable number of ways = all B or all G (i.e. BBB or GGG)=2hence P(E)= Favorable number of ways/Total number of ways

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=2/8=1/4

45. Number of ways 2 can be selected without taking the three in question. : 7C2Number of ways 2 can be selected out of 10 = 10C2Prob. = 7C2/10C2 = 21/45 = 7/15

46. No. of ways to pick 3 individuals out of 5 couples such that a couple is included is 5C1*8C1 = 5*8 = 40 (order of choice here doesn't matter)

Total no. of ways to pick 3 folks from ten = 10C3 = 10*9*8/(3*2) = 120

Therefore ways to select 3 individuals out of 5 couples such that no couple is included = 120-40 = 80.

47. Total no of ways: 3^41st sec can get a report in 4c1 ways = 42nd sec .........................3c1 way = 33rd sec...........................2c1 way = 2last can be distributed to any of the three = 3

So probability is = 4*3*2*3/ 3^4 = 8/9

48. Lets assume the worst-case scenario.

Jerome draws an odd. Then an even. This gives an odd number. Next he draws another even. Now we have an odd again. Then he draws an even. Again the sum is odd.So, to sum it up,

We have 10 odd and 10 evens. If his first draw is odd, then the next 10 are even, we still have an odd sum. The tie breaker will be the 12th card, which now has to be odd since all evens have been exhausted. So if the first card drawn is odd, then we must DRAW 12 CARDS.

If the first is even, then the second is odd, again we have an odd number. Now we only have 9 evens lefts, we must exhaust all of them to get an odd one. So again, 12 cards.

So the answer is 12. The 12th draw ensures an even sum.

49. It is already given that sum is 8. So total number of events is 5 i.e. (2,6) (6,2) (3,5) (5,3) (4,4)Events in which 5 is included 2.So probability = 2/5.

50. Since order is important, this is a permutation problem, not a combination one# of sequences possible = 8P3 = 336

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36 unassigned => # of participants = 336-36 = 300

51. 1st we place the 11#, now there are 10 places between them and 2 on the extreme left and extreme right of them, total places = 10 + 2 = 12

If * is placed at any of these 12 places, no 2 *'s will be togetherso the number ways will be 12C8 = 12C4 = 495

52. The probability will be 1- probability that no blue marbles are selected1- 3C2 / 10C2 = 14/15Another method: 1 Blue - 1 Red + 1 Red - 1 blue + Both Blue = 7/10*3/9 + 3/10*7/9 + 7/10*6/9= 14/15. Since we are not replacing the marbles, order matters and so 1 Blue- 1 Red is not the same as 1 Red- 1 Blue.

53. 1 st card that is not blue = 6/82nd card that is not blue = 5/ 7

= 30/56 = 15/ 28

Another method: 1 - P(at least 1 blue card)

54. 1 -(probability of same color)

1 -(5C2/10C2 + 3C2/10C2 + 2C2/10C2)=1 - (10+3+1)/45= 1 - 14/45 = 31/45

55. Probability that they are the same color = 1- probability that are NOT the same color.

Probability that they are not the same color = Probability of (1R-1B + 1B-1G + 1G-1R + 1B-1R + 1G-1B + 1R-1G) = 1- 24 /30 = 1/5 (**Order of color matters)

MJJ’s HI Guys:

Thanks for all the shout outs. The GMAT has been a long 3 months journey for me. My big debrief is not quite ready. I'm leaving New York this afternoon and heading back home. I will debrief in a day or two.

In the mean time, I've attached the latest MJJ's(as of October 09) with explanations. Even though I didn't see any MJJ in my exam (I think the United States pool of questions are different from the Indian; Just my feelings), the explanations are excellent and give you a perspective on how to approach GMAT quant questions. I

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have the entire September MJJ's to share with you, but they are on my computer back home. I will do when I get back.

1. Is X>Y?1). (X-Y)/X>0 2). (X^2-Y^2)/X^2>0Stmt 1) (X-Y)/X > 0 => 1-Y/X > 0 => 1 > Y/X => We cant say nothing unless we know the sign of X. INSUFFStmt 2) (X^2-Y^2)/X^2>0 => X^2 > Y^2. Still nothing can be said till we know the relation between X and Y. INSUFFCombining (1) and (2) actually gives us no further info. INSUFF(E)2. As the figure above shows, AB=BC=8, CD=2, angle B and angle C are right angle, and AE=ED=13. What is the area of the ABCDE?Draw a horizontal line DP from D to AB. AP=6, DP=8. Since APD is a right angled triangle => AD (hypotenuse) = 10.So the height of the triangle AED = sqrt(169-25) = sqrt(144) = 12.So area of ABCDE = Area of trapezium ABCD + triangle AED = 1/2*(8+2)*8 + 1/2*10*12 = 40 + 60 = 100 3. A plane traveled k miles in the first 96 minutes of flight time. If it completed the remaining 300 miles of the trip in t minutes, what was its average speed, in mils per hour, for the entire trip?Average speed = Total Distance / Total Time = (k+300)*60/(96+t) [60 is multiplied to convert hrs from minutes]4. Is the positive integer x an even number?1). If x is divided by 3, the remainder is 22). If x is divided by 5, the remainder is 5Stmt 1) x=3k+2 -> 3k could be add or even. So x could be odd or even. INSUFFStmt 2) Doesn’t makes sense! Remainder cannot be 5! (E) 5. What is the tens digit of positive integer x?1). X divided by 100 has a remainder of 302). X divided by 100 has a remainder of 30Stmt 1) x=100k+30 => No matter what the value of k, the tens digit will always be 3. SUFFStmt 2) x=100k+30 => J(D)6. In the xy-plane, are the points (r,s) and (u,v) equidistance from the origin?1). r+s=12). u=1-r and v=1-sDistance of points from origin = sqrt(r^2+s^2) and sqrt(u^2+v^2)Stmt 1) r+s=1, nothing known about values of u and v. INSUFFStmt 2) u^2 + v^2 = 2 - 2(r+s) + (r^2 + s^2) => iff r+s=1, the distances could be equal. Cant say. INSUFFCombining (1) and (2) -> we get the distances equal. SUFF(C)7. Working independently, Tina can do a certain job in 12 hours. Working independently, Ann can do the same job in 9 hours. If Tina works independently at the job for 8 hours and then Ann works independently, how many hours will it take Ann to complete the remainder of the job?Let the job be w.Tina’s rate = w/12Ann’s rate = w/9Tina works for 8 hrs so finishes (w/12)*8 th part of the work. The work left for Ann =

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w – 8*w/12 = 1*w/3Time taken by Ann to finish the job = w/3 / (w/9) = 3 hrs.8. If m and n are positive integers, is m/n a terminal decimal?1). 90<m<1002). n=4Stmt 1) terminality of decimal depends on denominator. INSUFFStmt 2) n=4. Any fraction with 4 as denominator will be terminal. SUFF(B)9. Mary has 11 more nickels than quarters. How many coins does she have if the total value of her coins is $2.65? Quarters = xNickels = x+11x*0.25 + 0.05*(x+11) = 2.65=> x=7 and y=18=> total coins = 2510. A and B are in a line to purchase tickets. How many people are in the line?1). There are 15 people behind A and 15 people in front of B2). There are 5 people between A and B Stmt 1) There is no information about how many people in between A and B.Stmt 2) No info on how many people before A and after BCombining 1 and 2, there is nothing given about the order of A and B [who is infront]. It could be that the when we say “15 people behind A”, B could be one of them. So nothing unique can be said about total number of people. INSUFF(E)11. Working alone, a small pump takes twice as long as a large pump takes to fill an empty tank. Working together at their respective constant rates, the pumps can fill the tank in 6 hours. How many hours will it take the small pump to fill the tank working alone?Let the rate of the smaller pump be r, rate of larger pump is 2r. Combined rate = 2r+r = 3r tank/hr.3r*6 = 1 tank full => r = 1/18 tank/hrSo small pump will fill the tank in 18 hrs. 12. What is the total number of different 5-digit numbers that contain all of the digits 2, 3, 4, 7, and 9 and in which none of the odd digits occur next to each other?The only possible way is to keep 3,7,9 in 1st, 3rd and 5th place. These can be arranged in 3! ways. The remaining two places will be covered by 2 and 4. These can be arranged in 2! WaysTotal = 3!*2! = 12 ways.13. One integer will be selected randomly from the integers 11 to 60, inclusive. What is the probability that the selected integer will be a perfect square or a perfect cube?Probablity = suitable outcomes/ total outcomes = 5/50 = 1/10[5 ways for 16, 25, 36, 49, 27]14. 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total number of different possible seating arrangements for the group?This seems like a straight forward circular permutation problem: (5-1)! = 4!15. The subset of [x, y, z] is x, y, z, xy, xz, yz, xyz. How many subsets of [x, y, z, w] contain w?Number of subsets which contain w = Total number of subsets of [x,y,z,w] – number of subsets of [x,y,z] = 2^4 – 2^3 [including empty set in calculations, it will cancel out anyways] = 8 16. A three-digit code consists of digits from 0 to 9. If the first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot be 0 at the same time, how many codes are possible?

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First digit cannot be 0,1 so total ways of filling first digit= 8 ways2nd digit can be any of the 0 or 1 = 2 waysLast digit can be anything = 10 waysTotal ways = 8*2*10 = 160 waysBut here we have counted some where there are 0 on 2nd and 3rd places, we need to subtract those combinations = 8*1*1 = 8Total ways = 15217. A jar contains 12 balls, including three red and nine blue; another jar contains 20 balls, including eight red and 12 blue. If one ball is selected from each of the two bags, what is the probability that two blue balls will be selected?Probability of getting a blue ball from the first jar = 9/12Probability of getting a blue ball from the second jar = 12/20Total Probability = 9/12 * 12/20 = 9/2018. What is the remainder when positive integer x is divided by 71). X+1 is divisible by 72). X+13 is divisible by 7Stmt 1) x+1 is divisible by 7. => x is one less than the multiple of 7, so it will have remainder of 6. SUFFStmt 2) x+13 = 7k -> x = 7k-13 => 7(k-2)+14-13 = 7(k-2)+1. so the remainder would be 1. SUFF(D)19. A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits 2, 3, 4, 5, 6, 7, 8, 9 and no digit appears more than once in any one code number. How many unassigned code numbers are there?Total number of codes = 8P3 = 336.There are 330 employees, so 6 unassigned codes.20. K=wxyz, where w, x, y, z are prime numbers. Not including 1 and K, how many factors does K have?Using the formula no of factors of K = p^a*q^b… = (1+b)(1+a)…. [includes K and 1]So for K=wxyz, number of factors = 2*2*2*2 = 16Excluding K and 1, we get 16-2 =14.21. Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?Total weight of 30 boxes = 30*18 = 540 pounds.Let there be x 10 pound boxes and y 20 pound boxes => 10x+20y=540 and x+y=3010y = 240 => y=24. x=6.Lets assume we remove ‘z’ 20 pounds boxes, new number of 20 pounds boxes = 24-z20*(24-z) + 60 = 14*(30-z) => 480 – 20z + 60 = 420 – 14z => 540-420 = 6z=> 120=6z => z=20.So we need to take out 20, 20-pound boxes.22. At their respective rate, pump A can fill 1/2 of a tank in 3 hours, and pump B can fill 2/3 of the tank in 6 hours. Working together, how many hours will it take A and B to fill the tank?A’s rate = (1/2)/3 tank/hr = 1/6 tank/hrB’s rate = (2/3)/6 tank/hr = 2/18 = 1/9 tank/hrCombined rates = 1/6 + 1/9 = 5/18 tank/hr=> time taken for them to complete = 18/5 hrs.23. The value of 1/11+1/12+1/13+...+1/20 will be A. Less than 1/3B. Less than 1/2

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C. Greater than 1/2D. Greater than 2/3E. …1/11+1/12+1/13+...+1/20 (10 terms)> 1/20+1/20+1/20…..(10 terms) [replacing each term with 1/20]=> 1/11+1/12+1/13+...+1/20 > 10/20=> 1/11+1/12+1/13+...+1/20 > ½(C)24. Maria invested a certain amount of money at r percent compounded annual interest rate. r=?1). She earned an interest of $200 in the first year. 2). She earned an interest of $220 in the second year.Formula for CI = P(1+r/100)^n [where n is the time]Stmt 1) -> CI for 1 year = 200 = P(1+r/100). Since we P is unknown too, INSUFFStmt 2) -> CI for 1 year = 220 = P(1+r/100)^2. Since P is unknown. INSUFFCombine (1) and (2), 220/200 = (1+r/100) => r can be found. SUFF(C)25. In the xy-plane, line L passes through point (1,p). Does L have a positive slope? 1). L passes through point (-p, 13) 2). L passes through point (0,1)Let the equation of the line L be y=mx+c => p=m+c (since it passes through (1,p))Stmt 1) L passes through point (-p,13) => 13=-pm+c. we get (p-13)/(1+p), could be positive or negative, depending upon p. INSUFFStmt 2) L passes through point (0,1) => 1=c combining with p=m+c => m=p-1. Still depends upon the sign of p. INSUFFCombining them, we still get a quadratic equation for p, giving 2 values of p, one positive other negative. (E)26. In a class of 120 students, 80 can speak English, 70 can speak <ST1:COUNTRY-REGIoN w:st="on">Spain</ST1:COUNTRY-REGIoN>, 60 can speak French. If every student in the class can speak at least one of the languages, at least how many of students in class can speak all the three languages?AUBUC = A + B + C – AnB – BnC – AnC + AnBnC120 = 80 + 70 + 60 – (AnB + BnC + AnC) + AnBnC(AnB + BnC + AnC) – 90 = AnBnCThe least value of AnBnC would be when (AnB + BnC + AnC). So least value of AnBnC could be 0. 27. At the same time, Ken invested x dollars at compounded annual interest rate of 4%, and Marlon invested y dollars at compounded annual interest rate of 6%. x/y=?1). At the end of the second year, Ken and Marlon earned same amount of interest. 2). At the end of the second year, Ken earns an interests of $400, and Marlon eared an interest of $500.CI(Ken) = x(1+4/100)^n and CI(Marlon) = y(1+6/100)^nStmt 1) x(1+4/100)^2 = y(1+6/100)^2 => x/y can be found. SUFFStmt 2) 400= x(1+4/100)^2 and 500= y(1+6/100)^n => x/y can be found . SUFF(D)28. A’s salary for any week is $600 plus 5% of the portion of A’s sales above $2000 for that week. B’s salary for any week is 10% of his sales for that week. For what amount of sales in a certain week would A and B earn the same salary?Lets say A’s sales is X => A’s salary = 600 + 0.05(X-2000)Lets say B’s sales is Y => B’s salary = 0.1YWe need to find what amount of total sales (X+Y) is B’s salary same as A’s.600 + 0.05(X-2000) = 0.1YThis question is not solvable in its current literal form. We have discussed this question before and this was solved [other scoretopians]

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assuming that the question says “A’s salary for any week is $600 plus 5% of the portion of total sales above $2000 for that week” and “B’s salary for any week is 10% of total sales for that week”. But I never agreed to that approach unless the "wording of" question itself is changed [and thus the assumption is discarded]. But as the question stands, I think its incomplete.29. Which of points (4,6) and (-5,5) is closer to the origin?Just find the distance of the points from originDistance (4,6) from origin = sqrt[(4-0)^2 + (6-0)^2] = sqrt(52)Distance (-5,5) from origin = sqrt(25+25) = sqrt(50).(-5,5) is closer.32. A is cycling around a track at a speed of 15 mph. B starts at the same time at a speed of 12 mph. If A and B move at the same direction and the track is 1/2 mile long, how many minutes after they start will A pass B if? 33. What is the tens’ digit of integer X?1). The tens’ digit of X-9 is 12). The hundreds’ digit of 10X is 234. A fish tank is half full of water. If 10 gallons of water were added, the tank would be 7/8 full. What is the capacity of the tank?35. The average of A, B, C, D, and E is C, and C=A+B. In terms of C, D+E=?36. Terry receives 10z coins in addition to what she already had. She now has 5y+1 times as many coins as she had originally. In terms of y and z, how many coins did she have originally?37. The price of a item decline by 67% every 6 months. At the rate, approximately how many years will it take for the $81 item to reach $1?38. Set A has 20 numbers and B has 40 numbers. Is the range of B greater than 45?1). The range of A and B combined exceeds 502). Range of B is greater than range of A39. Sequence 2, 7, 22,…after the first three terms, each term is three times the previous term plus 1, a(n+1)=3an + 1. What is the sum of tens digit and tens digit of the 35 term?

Is the integer N odd? (1) N is divisible by 3 (2)2N is divisible by twice as many positive integer as N.

OA:BFor Condition A, both 6 and 9 are divisible by 3. But, 6 is even and 9 is odd. hence not A

Now to check condition B, let’s pick an odd number. Lets say, N=3, the only numbers that are divisible evenly by 3 are 1 and 3 (total of 2)2N=6 and the only numbers that are divisible evenly by 6 are 1,2,3, 6 (total 4)

Lets pick another odd number….say N=11, only numbers that are divisible evenly by 11 are 11 and 1 (total of 2)

2N= 22 and only numbers divisible evenly by 22 are 1,2,11 and 22 (total of 4)

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Lest pick another odd number, If N=15, only numbers that are divisible evenly by 15 are 1,3, 5, 15 (total of 4 numbers)

2N=30, only numbers that are divisible evenly by 30 are 1,2,3,5,6,10,15,30 (total of 8 numbers).

That is it . Just stop here. Condition B is satisfied for all odd numbers that you have checked so far and therefore answer is B.

If digit h is the hundredths' digit in the decimal d=0.2h6, what is the value of d, rounded to the nearest tenth?1)d<1/42)h<5

Vote D

I: If d < 0.25, then h < 5. SuffYou can't round 0.246 to 0.3. What kind of rounding is that?0.246 = 0.2 to the nearest tenth.Note the word nearest. Is the number 46 nearer to 100 or to 0?

II: Suff

If y is greater than 110% of x, is y greater than 75?

1. x > 752. y-x=10

The set S has following properties:(a)If x in in S, then 1/x is also in S(b) If both x and y are in S, so is (x+y)

Is 3 in S ?(1) 1/3 is in S(2) 1 is in S

If p is a prime number greater than 2, what is the value of p?

(1) There are a total of 100 prime numbers between 1 and p+1(2) There are a total of p prime numbers between 1 and 3,912Could anyone suggest the best way to solve this problem?

Does the integer K have a factor P such that 1< P< K?

1) K> 4!2) 13!+2<=K<=13!+13

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What is the value of x?

1) x2 = 4x

2) x is an even integer.

Is XY<0 ?

1) 1/y > 1/x

2) x>0

Is the number x between 0.2 and 0.7?

(1) 560x < 280(2) 700x > 280

If the units digit of the three-digit positive integer k is nonzero, what is the tens digit of k?(1) The tens digit of k + 9 is 3.(2) The tens digit of k + 4 is 2.

What is the value of X ?............___1. x = \/16

2. x^2 = 16

What is the remainder when the positive integer n is divided by 2? (1) When n is divided by 5, the remainder is an odd integer.(2) When n is divided by 10, the remainder is an odd integer.

x, 3, 1, 12, 8If x is an integer, is the median of the 5 numbers shown greater than the average (arithmetic mean) of the 5 numbers?

1) x > 62) x is greater than the median of the 5 numbers

what is the remainder when postive integer x is divided by 6?

1) when x is divided by 2, the remainder is 1, and x is divided by 3, the remainder is 0.

2) when x is divided by 12, the remainder is 3.

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The numbers 'x' and 'y' are not integers. The value of 'x' is closest to which integer?

1) 4 is the integer that is closest to x+y2) 1 is the integer that is closest to x-y.

) x+y ~4 -> insuff. We know nothing about x it self.2) x-y ~1 -> insuff. Same reason.

Combining:

Let's pick x+y=4.4 and x-y =1.4 ==> x=2.9 the closest integer is 3.Now pick x+y=3.6 and x-y=0.6 ==> x=2.1 the closest integer is 2.So C is out, it is E.

Is N an integer

1. n^2 is an integer2. sq root n is an integer.Data sufficiencyx is an int, is (x^2+1)(x+5) even number?optionsx is odd numberEach prime factor of x^2 is > 7 (this one not able to resolve)

X>0 ?1). x^3>x^2 2). x^4>x^3 A