Aptitude Handbook Www.vtulife.com

MANGALORE

APTITUDE HANDBOOK FORMULAS WITH EXAMPLES

Prepared by:SUSHMA SHETTY

[email protected]

This handbook is been circulated on self risk. Nobody can be held responsible if anything is wrong or any improper informationor insufficient information provided in it.

References: www.iutraining.com, www.myethnus.com

Visit www.vtulife.com for all placement materials,vtu notes, syllabus and question papers
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Prepared By: Sushma Shetty

1 www.vtulife.com INDEXTOPICS PAGE NUMHUMAN RELATIONS 2NUMBER SERIES 4ALPHABETIC SERIES AND LETTER CODING 6SET THEORY, VENN DIAGRAMS 8DIRECTION SENSE 11PERCENTAGE 12PROFIT AND LOSS 15INTEREST 16AVERAGES, MIXTURES AND ALLIGATIONS 17PERMUTATIONS AND COMBINATIONS 19FACTORS AND MULTIPLES 22NUMBER GAMES 24LINEAR EQUATIONS 27PROPORTIONS AND VARIATIONS 29PUZZLES 31TIME AND WORK 34SPEED, TIME AND DISTANCE 36PROBABILITY 40DATA SUFFICIENCY 42CUBES 43NON VERBAL REASONING 44DATA INTEGRITYTHE ART OF FACING INTERVIEWS

4650
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2 www.vtulife.com HUMAN RELATIONSMothers or Fathers son BrotherMothers or Fathers daughter SisterMothers or Fathers brother UncleMothers or Fathers sister AuntMothers or Fathers father GrandfatherMothers or Fathers mother GrandmotherMothers or Fathers father GrandfatherGrandsons or grand daughters daughter Great grand daughterMothers mother Maternal grandmotherMothers father Maternal grandfatherFathers mother Paternal grandmotherFathers father Paternal grandfatherSons wife Daughter-in-lawDaughters husband Son-in-lawHusbands or wifes sister Sister-in-lawSisters husband Brother-in-lawBrothers wife Sister-in-lawHusbands or wifes brother Brother-in-lawBrothers son NephewBrothers daughter NeiceUncle or aunts son or daughter CousinHusbands wife or wifes husband SpouseBrother or sister Sibling

1. Pointing to a girl I said to Priyanka, The girl on the stage is the second daughter of the wife ofthe only son of the grandmother of my younger sister. How is the girl on the stage related tome?Solution:

Answer: Sister
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Similar question 1: Pointing to a photograph, Arshadh said he is my fathers mothers daughtershusbands son. How is he related to the boy in the photograph?

a) Cousin b)Uncle c)Brother d) NephewAnswer: (a)

Similar question 2: Anil introduces Rohit as the son of the only brother of his fathers wife. How isRohit related to Anil?

Solution:

Answer: Cousin

2. A and B are brothers. C and D are sisters. As son is Ds brother. How is B related to C?Solution:

Answer: UncleSimilar question 1: A has 3 children. B is the brother of C and C is the sister of D, E who is thewife of A is the mother of D. there is only one daughter of the husband of E. How is D related to
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B?Solution:

Answer: Brother

NUMBER SERIESTo measure your ability to reason without words.

For number series:

Increase or decrease? Patterns in every number, every second number or every third number. (Eg: 2,5,8,11. Or

1,4,7,8,13,12, or 1,2,3,3,4,6,9,6,9, ) Increase or decrease gradual (addition or subtraction), steep (multiplication or division) or

exponential (exponents). Standard patterns.

ARITHMETIC SERIES:The next numbers are obtained by adding or subtracting a fixed number.Ex: 3,5,7,9,11,. (adding 2)

12,10,8,6,4, (subtracting 2)

GEOMETRIC SERIESSuccessive numbers are obtained by multiplying or dividing a fixed number by the previousnumber.Ex: 4,8,16,32,. (multiplied by 2)

15,-30,60,-120,. (multiplied by -2)
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SERIES OF SQUARES, CUBES, ETCFormed by squaring or cubing every successive number.Ex: 2,4,16,256,..

3,9,81,6561,.2,8,518,..

COMPLEX SERIESGenerally the mixed series.

I. Two layer arithmetic series:Ex: 1,2,5,10,17,26,37.

Solution:

It can be also solved as:02+1, 12+1,22+1,32+1,.. denoted as f(x)=x2+1.

II. Three layer arithmetic series:Ex: 336, 210, 120, 60, 24, 6, 0, Solution:

336 210 120 60 24 6 0 0126 90 60 36 18 6 0

36 30 24 18 12 66 6 6 6 6

It can also be solved as:13-1, 23-2, 33-3, 43-4,.

III. Other series:Ex: Fibonacci series (1,1,2,3,5,8,13,.)

Prime numbers (2,3,5,7,11,13,..)

EXAMPLE QUESTIONS:1. 9, 15, 23, 33, ?

a.44 b.36 c.38 d.452. 2, 5, 14, 41, ?

a.86 b.108 c.122 d.1643. 3, 11, 8, 16, 13, ?

a.20 b.21 c.22 d.25
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4. 4, 5, 7, ?, 19a.56 b.16 c.17 d.11

5. Find the odd one out in this series : 14, 7, 28, 21, 36a.14 b.7 c.28 d.36

ANSWERS:1.d 2.c 3.b 4.d 5.d

ALPHABETIC SERIES AND LETTERCODING1 2 3 4 5 6 7 8 9 10 11 12 13A B C D E F G H I J K L M14 15 16 17 18 19 20 21 22 23 24 25 26N O P Q R S T U V W X Y Z

SERIES EXAMPLES:1. Find the next term: BCA, EFD, HIG

a.JLK b.KLJ c.LKM d.MLKsolution: BCA, BCA+333, EFD+333, HIG+333= KLJ

2. Find the missing term: KALC, MBND, ____, QGRFa.NEOF b.ODPE c.NDOE d.OEPFSolution:first letters: K, M, _, Q (difference of 2. So missing letter is O)second letters: A, B,_, G (difference is 1,2,_,7. Ie 5. So missing letter is D)third letters: (difference is 2. So its P)fourth letters: (difference is 1. So E)answer is ODPE.

3. BOLT:TLOB::a.POLICE:ECILOP b.CHILL:LLHIC c.HORSE:SEROHAnswer: (a)

4. BC:EF::a.FG:HI b.JK:NO c.ST:XY d.JK:MNAnswer: (d)
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CODING:

1. WITH LETTERS:

PATTERNS

Forward Sequence:Ex: If ROBUST is coded as QNATRS, then ZXCMP will be ?a.YWBLO b.YYBNO c.AWDLQ d.AUDNQAnswer is (d) (difference of -1)

Backward Sequence:Ex: If LATE is coded as OZGV, what is the code for DOME ?

Solution: L is the 12th letter from forward and O is the 12th letter from backward.Similarly A is the 1st letter from first and Z from last.similarly other letters. ANSWER: WLUV.

Skipped Sequence:Ex: If FACT is coded as IDFV. What is the code for CLAD?Solution: FACT+3333=IDFV, then CLAD+3333=FODG.

REVERSING LETTERSEx: If DOMINOS is SONIMOD, then QWERTY is coded as YTREWQ.

2. WITH NUMBERS:

MATHEMATICAL OPERATORSEx: If LATE is coded as 38, then MAKE is coded as ?

Its the sum of all letters. 13+1+11+5=30.

RANDOM NUMBERINGA shopkeeper uses a code ASCVBGH=28 , where A=Rs.1, S=Rs.2 and so on. What would be the price ofCASH?

Its 3+1+2+7=13.

3. MIXED LETTER CODING:Ex: If nso ptr kli chn stands for sharma gets marriage gift, ptr lmn wop chn for wife gives marriagegift, tti wop nhi stands for he gives nothing, what is the code for gives ?

Answer: wop
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8 www.vtulife.comSET THEORY, VENN DIAGRAMSSET: Well defined collection of objects.

TYPES

SINGLETON SET:A set consisting of single element. Ex: {5}, {ant}.

EMPTY SET/NULL SET/VOID SET:A set containg no element. Ex: {} or

EQUIVALENT SET:Two sets A and B are equivalent sets if number of elements in both sets is same. Ex: {1,2,3,4,5} and{q,w,e,r,t}.

EQUAL SET:Two sets A and B are said to be equal if every element in A is in B and vice versa. Ex: {A,B,C} and{A,B,C,B,C}

UNIVERSAL SET:A set which contains all the sets in the given context. Ex: If A={1,3,5}, B={3,6,7,8}. Then universal setU={1,3,5,6,7,8}

SUBSET AND SUPERSETTwo sets A and B, if all the elements in A are in B, then A is said to be the subset of B and B is thesuperset of A.

If B contains atleast one element more than that of A, then A is the proper subset of B.

OPERATION ON SETS

UNION OF SETSAB
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INTERSECTION OF SETSAB

DIFFERENCE OF SETSA-B is

COMPLEMENT OF A SETA

ALGEBRAIC LAWS OF SETS

IDEMPOTENT LAW:A A =A, A A= A

IDENTITY LAW:A = A, A=

COMMUTATIVE LAW:AB=BA, AB=BA

ASSOCIATIVE LAW:A(BC)=(AB)C, A(BC)=(AB)C
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DISTRIBUTIVE LAW:A(BC)=(AB)(AC), A(BC)=(AB)(AC)

DE-MORGANS LAW:(AB)=AB, (AB)=AB

VENN DIAGRAM PROBLEMS

1) Select the diagram which best suits the relation.

1. Animals, cows, dogs Answer: 52. Doctors, surgeons, musicians Answer: 33. Dogs, pets, rabbits Answer: 44. Automobile, four wheeler, car Answer: 15. Maruti 800, Cars, Zen Answer: 2

2) Given diagram represents 3 classes of population. The triangle represents the school teachers.Square represents the married persons. Circle represents persons living in joint families.

1. Married persons living ib joint families but not school teachers is represented bya.C b.F c.D d.A

2. Person who live in joint families are unmarried and who do not work as school teachers area.C b.B c.E d.D

3. Married teachers living in joint familiesa.C b.B c.D d.A

4. School teachers who are married but do not live in joint familiesa.C b.F c.A d.D

5. School teachers who are neither married nor living in joint familiesa.F b.C c.B d.A
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Answers: 1.c 2.c 3.b 4.a 5.a

DIRECTION SENSE

PYTHAGOREAN THEOREM:

PROBLEMS:1) Anubha walks 12km towards south. From there she walks 8km towards north. Then, she walks

3km towards west. How far is she with reference to her starting point?a)7km b)5 km c)8km d)6km
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solution:

2) Sanoob walks 20m in south direction. He then turns left and moves 15m where he turns right.After he walks 30m in this direction, he turns right and moves 15m. how far is he from hisstarting point and which direction is he facing?a)10m,north b)50m,west c)10m,west d)50m,east

solution:

Answer: 50m,west

PERCENTAGEPercentage is a way of expressing a number as fraction of 100.

Some Important Conversions:=50% 1/3=33.33% =25% 1/5=20% 1/6=16.66%

1/7=14.28% 1/8=12.5% 1/9=11.11% 1/10=10% 2/3 = 21/3 =233.33 = 66.66
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1. If 2 values are respectively x% and y% more than a third value, then thefirst is (100+x)/(100+y) 100% of the second.

Ex: Two numbers are 20% and 40% more than a third value. What % is first of the second?

Ans: 120/140 100 = 85.71%

2. If A is x% of C and B is y% of C, then A is x/y 100% of B.Ex: Two numbers are respectively 20% and 40% of a third value. What % is first of second?

Ans: 20/40 100 =50%.

3. X% of a quantity is taken by the first. Y% of the remaining is taken bythe second and z% of the remaining by the third person. Now, if A is leftin the fund, then there was ( )( )( ) in the beginning.

Ex: Farah loses 12% of her money and after spending 60% of the remainder,she is left with Rs. 210. Howmuch did she have at first?

Ans:= ( )( ) = Rs. 596.64. X% of a quantity is added. Then again y% is added to that and again z%

is added. Now it becomes A. Then the initial quantity is ( )( )( ) .Ex: Vishal had 5000 in his locker 2 years ago. In the first year, he deposited 10% of the amount in hislocker. Again he deposited 20% of the increased amount in the second year. How much amount ispresent in his locker now?

Ans:= ( )( ) = Rs. 3787.875. Percentage increase = ( ) 100

Ex: The exam fees was Rs.560 last year. This year it became Rs. 610. Find the percentage increase.

Ans:= 100 = 8.9%6. Percentage increase = ( ) 1007. If there is a increase in % or quantity of a number.

Ex: The price of a laptop was Rs. 50,000 last year. This year it increased by 15%. What is the new price?

Ans: increase => 1+0.15=1.15. new price= 50,000 1.15 = Rs. 57,500.
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8. If there is a decrease in % or quantity of a number.Ex: The price of a laptop was Rs. 50,000 last year. This year it decreased by 15%. What is the new price?

Ans: increase => 1-0.15=0.85. new price= 50,000 0.85 = Rs. 42,500.

9. When population P increases by x% for one year and then decreases byy% the second year and again increases by z% the third year, then thepopulation after 3 years will be ( )( )( )

.

The population of a city is 50,000. It decreased by 10% in the first year, increased by 20% in second yearand decreased by 30% In the third. What is the population after 3 years?

Ans:

= 37,800.

10. If a number A is x% more than B, then B is ( 100 ) % lessthan A.

11. If a number A is x% less than B, then B is ( 100 ) % morethan A.

Ex: The length of a plot is decreased by 25%. By how much % should the breadth be increased so thatarea remains the same?

Ans: 25/(100-25) 100 = 33.33%

12. If a number is first increased by x% and then decreased by y%,then there is (x-y- )% increase or decreases, according to the positiveor negative sign respectively.

Ex: If the price is increased by 15% and sale is decreased by 5%, then what will be the effect on income?

Ans: 15-5-155/100 = 9.25%

His income increases by 9.25%

13. The pass marks in a exam is x%. if a student secures y% ad fails byz%, then the maximum marks is 100(y+z)/x.

Ex: A student has to score 35% marks to get through. If he gets 35 marks and fails by 35 marks, find themaximum marks set for the examination?

Ans: 100 (35+35)/35 = 200.
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15 www.vtulife.com PROFIT AND LOSSCost price (CP) is the price at which the good is bought.

Selling price (SP) is the price at which the good is sold.

Marked price (MP) is the price on the tag of the product.

Discount (D) is the price reduction on the MP.

For profit For loss

P=SP-CP L=CP-SP

%P=P/CP100 %L=L/CP100

%SP=CP(100+P)/100 %SP=CP(100-L)/100

%CP=(100SP)/(100+P) %CP=(100SP)/(100-L)

1. SP=MP-D

2. %D=D/MP100

3. In case of successive discounts of x% and y%, the effective discount is(x+y-xy/100)%.

4. If value of a certain quantity is first increased by x% and then decreasedby x%, then net change is a decrease of x2/100 %.

PROBLEMS:1. Find the cost price of a mobile which is sold for Rs. 5000 at a loss of 25%.

Ans:SP=CP(100-25)/100=0.75CP 0.75CP=5000 CP=5000/0.75 = Rs. 6,666.67

2. Find the cost price of a mobile which is sold for Rs. 5000 at a profit of 25%.Ans:SP=CP(100+25)/100=1.25CP 1.25CP=5000 CP=5000/1.25 = Rs. 4000
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3. Sowmya sells a good to Supreetha at a profit of 25% and Supreetha sells it to Sahana at a profitof 20%. If Sahana pays Rs.225 for it, what was the cost price for Sowmya?Ans:225 (100/(100+25)) (100/(100+20)) = Rs.150

4. A 20% reduction in the price of rice enables a person to buy 10Kg more for Rs.400.what is thereduction price per Kg? Also find the original price.Ans:There is saving of 20% on Rs.400 ie 4000.2= Rs.80 For this Rs.80, a person buys 10kg of rice. Reduced price per Kg = Rs.80/10 = Rs.8 Original price = 8 (100/(100-20)) = Rs.10

5. What will be the selling price of a toy if successive discounts of 10% and 20% are given onmarked price of Rs.1000?Ans:

10+20- = 28%

INTERESTInterest is a percentage of some amount of money.

The money lent or borrowed is called principal.

Amount= principal+interest.

The interest can be computed in two ways:

Interest is calculated for succeeding periods on the original sum borrowed i.e., on the originalprincipal only. This is simple interest.Simple interest, I= Pnr/100Where P is the principal, n years and r% is the rate of interest per annum.The amount A due after n years= P+I.

Interest is calculated for each year on the amount due till that period. This is compound interest.Compound interest, CI= A-P.Where A=P[ 1+r/100]n

PROBLEMS1. Find the simple interest on Rs.306.25 from march 3rd to july 27th at 3 %
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Ans:Number of days between march 3rd to july 27th = 146Number of days in a year = 365

3 % = 15/4 1/100

Interest= 306 146/365 15/4 1/100

= Rs. 4.592. What is the amount due on a loan of Rs. 1000, taken at 16% p.a. simple interest, at the end of

two years?Ans:P=Rs. 1000, Rate r=16%, n=2 years.I=Pnr/100= (1000216)/100= Rs. 320.

3. The compound interest on a certain sum of money at 10% p.a. in 2 years in Rs. 2100. Find thesum.Ans:Let P= Rs. 100r= 10%, n=2A=P[1+ r/100 ]n= 100[1+10/100]2= Rs. 121CI=A-P = 121-100 = Rs. 21If the CI is Rs. 21, the principal is Rs.100, then if the CI is Rs. 2100, the principal will be Rs. 100.

AVERAGES, MIXTURES ANDALLIGATIONSAVERAGESAverage is termed as the mean of all values.

Average=

Ex: If Namrata scores 24, 21 and 25 in her first, second and third internals respectively, then what is heraverage marks?

Ans: = 23.33

WEIGHTED AVERAGE:This concept is used when working on two or more groups of data.
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Ex: The average weight of 3 sections A,B,C are respectively 49kg, 52kg and 55kg. number of students insection A are 45, in B 56 and in C it is 61. What is the average?

Ans: = 8472/162 = 52.29kgAVERAGE RELATED TO SPEED:

1. If a person travels a distance at a speed of x km/hr and the same distance at a speedof y km/hr, then the average speed during the whole journey is given by km/hr.

For 3 equal distances it is .

Ex: Manjunath drives his car at a speed of 20km/hr from his house to PACE, and from PACE to his houseat a speed of 30km/hr. what is the average speed of whole journey?

Ans: = 24km/hr.ALLIGATION AND MIXTURESA set of values is referred to as a single group. When two or more groups are combined together, then itis called mixture.

Rule of allegation is used to find the ratio of quantities in which the two entities are mixed.

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By the allegation rule: 0.15/0.25 = 15/25 = 3/5

They must be mixed in the ratio 3:5.

2. A mixture of a certain quantity of milk with 16 litres of water is worth 90 p per litre. If pure milkbe worth Rs. 108 per litre how much milk is there in the mixture?

Ans:

90/18 = 5

Quantity of milk in the mixture = 516 = 80 litres.

PERMUTATIONS AND COMBINATIONSPERMUTATION means arrangement of things. When r things are to be selected out of n things(r


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Ex: How many number of 5 digits can be formed with digits 1,2,3,4,5,6,7 and 8 such that no digit is usedtwice?

Ans:

N=8, r=5

8P5= 8!/(8-5)! = 6720.

COMBINATION means selection of things. When r things are to be selected from n things (r


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If clockwise and anticlockwise directions are distinguishable. The number of clockwisepermutations of n different things in n positions is given by (n-1)!

If clockwise and anticlockwise directions are indistinguishable. The number of clockwise

permutations of n different things in n positions is given by ( )!!PROBLEMS

1. Nawaf wants to pick 3 balls, one from the shelf consisting of 10 red color balls and 2 from theshelf consisting of 7 green color balls. In how many ways can he pick?Ans:10C17C2 = 210 ways.

2. There are 4 chairs and 10 people who want to sit. Each chair can be occupied by one person.After the 1st chair is occupied by a person. Then 9 people wait to occupy the remaining 3 chairs.Once the 3rd chair is occupied 8 people wait to occupy the 2 chairs and after that 7 people waitto occupy the one chair. 6 people left out. Find out the number of ways to fill the 4 positions.Ans:10C4 4! = 10P4 = 5040 ways.

3. In how many ways can the letters of the word ENGINEERING be arranged?Ans:There are 11 letters and the letter E is repeated 3 times, N also 3 times, G twice and I also twice.

Number of arrangements= !! ! ! ! = 554400 ways.4. Thaha has 4 pockets and 6 coins. In how many ways can he put the coins in his pockets if each

pocket has enough space for the coins?Ans:444444= 46 = 4096.

5. How many words can be formed using the letters of the word VTULIFE such thata. F and E occupy the 2nd and 3rd place?b. The words V,L and E occupy only odd places.

Ans:

a. _ F E _ _ _ _ or _ E F _ _ _ _.In 5! Ways other words can fill the other places and in 2 ways.So 5! 2! Ways.

b. V,L and E can occupy 1st,3rd,5th and 7th places i.e 4P3 = 4!Rest of the 4 words occupy 4 places in 4! Ways. Number of arrangements = 4! 4! Ways.
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22 www.vtulife.comFACTORS AND MULTIPLESFACTOR of a given number is the number which can exactly divide the given number. 1 is a universalfactor. Ex: 2 is a factor of 4,6,8,10.

MULTIPLE of a number is the number that is obtained by multiplying the given number with anotherinteger. Ex: 24 is a multiple of 2, 9 is a multiple of 3.

DIVISIBILTY RULES:Number isdivisible

byRule Example

2 Last digit of number must be divisible by 2 1278, here 8 is divisible by 2

3 Sum of all digits must be divisible by 3 249, 2+4+9=15 is divisible by 3

4 Last 2 digits must be divisible by 4 5420, 20 is divisible by 4

5 Last digit must be either 0 or 5 5510, 5515

6 Must be divisible by 2 and 3 438

7 ((number of 10s in num)-2last digit) mustbe 0 or divisible by 7

8064, (806-24=798)Number is big, so again check for 798.

(79-28=63). 63 is divisible8 Last 3 digits must be divisible by 8 762920, 920 is divisible by 8

9 Sum of all digits must be a multiple of 9 5247, 5+2+4+7=18 is divisible by 9

11 (sum of digits in even places)-( sum ofdigits in odd places) is 0 or a multiple of 11 1375, (3+5)-(1+7)=0

12 Number divisible by 3 and 4. 852, divisible by 3 and 4

FACTORIZATION: When a number is expressed as a product of its factors, then the number is saidto be factorized.

Ex: Factorization of 24

So 24= 2223
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HIGHEST COMMON FACTOR (HCF) OR GREATEST COMMON DIVISOR(GCD)The HCF of 2 or more numbers is the greatest number which exactly divides these numbers.

Ex:

HCF of 24,90. First factorize them.

24=2223= 2331

90=2335= 213251

Now, the common factors with minimum power 2131=6 is the HCF.

LEAST COMMON MULTIPLE (LCM)LCM of a given set of numbers is the smallest number which is divisible by each number in the given set.

Ex:

LCM of 24,90. First factorize them.

24=2223= 2331

90=2335= 213251

Now, the all the factors with maximum power 233251=360 is the LCM.

Another method:

LCM of 24,90 is 23415=360.
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SOME PROPERTIES:

1. For any 2 given numbers, HCFLCM= Product of the two numbers.

2. HCF of fractions= HCF of numerators of all fractions/LCM ofdenominators of all fractions.

3. LCM of fractions= LCM of numerators of all fractions/HCF ofdenominators of all fractions.

PROBLEMS1. Find the least 3 digit number which when divided by 24,32,48 and 72 leaves a remainder 14 in

each case.Ans:The LCM of 24,32,48 and 72= 2532= 288x = LCM(24,32,48,72)k+14 = 288k+14Substituting k values=0,1,2,For k=1, 288+14=302 is the least 3 digit number obtained.

2. Find the smallest natural number which when divided by the numbers 200,250 and 350 leavesthe remainders 140,190 and 290 respectively.Ans:LCM(200,250,350)k-60 = 7000k-60For k=0, its -60.For k=1, ans is 6940.

3. Find the largest number which on dividing 92,145 leaves remainders 2 and 1 respectively.Ans:HCF(92-2,145-1) = HCF(90,144) =18.

NUMBER GAMESCYCLICITYEx: Consider the series 4,5,6,7,8,4,5,6,7,8,4,5,6,7,8,4,5,

Here we observe that each term is repeating after the 5th term. So the cyclicity is 5. Now the nth term inthis series is the remainder of n/5.

The 16th term is found by 16%5=1, the first term is 4 in the cycle. So 16th term is 4.
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Cyclicty of x: Units digit of xn as N

takes values 1,2,3,4, Cyclicity Example

1 1,1,1,1 1 Always 1

2 2,4,8,6,2,4, 4 Units digit of 27. 7%4=3.

3rd digit is 8.

3 3,9,7,1,3,9, 4 Units digit of 39, 9%4=1.

So 3.

4 4,6,4,6,4, 2 410, 10%2=0, so the last

term in cycle. 65 5,5,5, 1 Always 5

6 6,6,6,6, 1 Always 6

7 7,9,3,1,7,9,3, 4 716, 16%4=0. So lastdigit in cycle. Ie 1

8 8,4,2,6,8,4, 4 813, 13%4=1, so 8

9 9,1,9,1, 2 938, 38%2=0. So 1.

SOME PROPERTIES:

1. If x1+x2 is divided by n, the remainder will be r1+r2.

2. If x1-x2 is divided by n, the remainder will be r1-r2.

3. If x1*x2 is divided by n, the remainder will be r1*r2.Ex:

What is the remainder when 634734 is divided by 5?

Ans:

6%5=1 7%5=2

62%5=1 72%5=4

63%5=1 73%5=3

. 74%5=1

. 75%5=0

Cyclicity is 1 76%5=0

77%5=0

.
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.

.

Rest of the values =0,

So remainder = remainder(634)remainder(734)=10=0.

TYPES OF NUMBERS1. NATURAL: The numbers 1,2,3,4,5,.2. WHOLE: The numbers 0,1,2,3,4,.3. INTEGER: The positive and negative whole numbers and 0 : .,-3,-2,-1,0,1,2,3,.4. RATIONAL: Numbers which can be written in p/q format where p and q are interger but q is not

0.5. PRIME: Divisible evenly only by 1 and themselves. Eg: 5,7,116. COMPOSITE: Numbers that are not prime. Eg: 4,6,97. EVEN: Numbers evenly divisible by 2.8. ODD: Numbers not evenly divisible by 2.

SOME FORMULAES

(a+b)(a-b)=a2-b2

(a+b)2=a2+b2+2ab

(a-b)2= a2+b2-2ab

(a+b)3=a3+b3+3ab(a+b)

(a-b)3= a3-b3-3ab(a+b)

Sum of all first n natural numbers= n(n+1)/2Ex: 1+2+3+4+..+25=2526/2=325

Sum of first n odd numbers=n2

Sum of first n even numbers=n(n+1)

Sum of squares of first n natural numbers= n(n+1)(2n+1)/6Ex: 12+22+32++102=10(11)(21)/6 = 385

Sum of cubes of first n natural numbers= [n(n+1)/2]2Ex: 13+23+..+63= [6(6+1)/2]2= (21)2=441
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When two numbers, after being divided by a certain number of three digits and the remainderis the same in both the cases, then the difference between the two numbers must be perfectlydivisible by the third number.Ex: 12345 and 21705 being divided by a certain number of three digits and the remainder is the same inboth the cases. Find the divisor and the remainder.

21705-12345=9360

(Check among the options which perfectly divides 9360 and gives the same remainder for 21705 and12345)

Else, 9360=222233513

Combination of above divisors which give 3 digit numbers X= {720, 144, 240, 180, 360, 624, 936, 468,234, 177,..}

We get 234 as the answer.

PROBLEMS:1. How many numbers upto 1000 are divisible by 3 and 5 together?

Ans: LCM of 3 and 5= 15.Now, 1000/15=66.66The quotient 66 is the answer.

2. How many numbers between 200 and 600 are divisible(multiples of) by 12?Ans: 600/12=50200/12=16.6615-16=34 numbers.LINEAR EQUATIONS

An equation is an expression of equality of two terms.

Linear Equations are classified based on the highest power of the variables involved. If the highestpower is one, its linear equation.

Ex: 3x+9=25 (linear equation in one variable)

2x+7y=21 (linear equation in two variables, the graph will be a straight line)
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Consider two simultaneous equations: ax+by+c=0 and ux+vy+z=0.Then 3 cases are possible:

Type of solution Criteria to identity Graphical Interpretation

Unique solution Intersection of two lines

Infinite solutions = = Co-incident linesNo solution or Inconsistent = Parallel lines

Solving linear equation in two variables:Ex:

2x+4y=10 and 3x+y=9

2x+4y=10 --------------------------- (1)

3x+y=9 -----------------------------(2)

To eliminate variable x:

(1)*3-(2)*2

Gives: 6x+12y=30

-6x-2y=-18

-----------------------------

0+10y=12 => y=1.2

Substituting in y: 3x+1.2=9 => x=7.8/3= 2.6

PROBLEMS:1. 10 years ago Shrees mother was 4 times older than her daughter. After 10 years, the mother

will be twice older than the daughter, what is the present age of Shree?Ans: Let Shrees present age be x and her mothers be y.10 years ago, Shrees age is x-10 and her mothers age= y-10.4(x-10)=(y-10)4x-40=y-10 => 4x=y+30-------------------- (1)10 years hence:2(x+10)=y+102x+20=y+10 => 2x=y-10----------------------(2)(1)-(2) => 2x=20 => x=10 and y=2x+10 (from (2))y=30.
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2. Omar bought 5 pens, 7 pencils and 4 erasers. Thaha bought 6 pens, 8 erasers and 14 pencils foran amount which was half more than what Omar paid. What percent of the total amount paidby Omar was paid for the pens?Ans:Let x denote pens, y denote pencils and z denote erasers.5x+7y+4z=R -----------------(1) //by Omar6x+14y+8z=1.5R -----------------(2) //by Thaha(1)*2-(2)=> 10x-6x=2R-1.5R=> 4x=0.5R=> x=0.125RFor 5 pens= 5x= 0.625R i.e, 62.5%

3. The numerator of a fraction is 3 less than denominator of the fraction. If number is increased by1, the numerator is increased by 4 while denominator remains the same. Find the number.Ans: let numerator be x.

when increased by 1,+ 1= also given that numerator increases by 4. So (2x+3)-x=4=> x+3=4x=1

and => number is 4. Amrita Kottari buys two types of pens. One type of Rs.2 and other of Rs.5. total she spent Rs.

100. If she buys a total of 44 pens, find the number of Rs.5 pens she bought?Ans:Let number of Rs.2 pens Amrita bought be xAnd Let number of Rs.5 pens Amrita bought be yGiven 2x+5y=100 and x+y=44.Solving them, we get y=4 and x=40. Required answer= 4.

PROPORTIONS AND VARIATIONSRATIO: The number of times one quantity contains another quantity of the same kind is called of thetwo quantities.

Ex: Let A and B weigh 50kg and 30kg respectively. If we are to compare weights of A and B, we can saythat

The ratio of weight of A to that of B= = = 5/3
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Ratio in terms of percentage: for the above example, 5/3 100= 166.6%, for ,its 100= 50%

Compound Ratio: It is obtained by multiplying together the antecedents for a new antecedent, and theconsequents for a new consequent.

Ex: For 2:5, 6:2, 3:9 compound ratio = = 36/90=2/5Inverse Ratio: If 2:3 be the given ratio, then : or 3:2 is the inverse ratio.

If in x litres mixture of milk and water, the ratio of milk and water is a:b, the quantity of water to be

added in order to make this ratio c:d is ( )( ) .PROBLEMS:

1. Naveen can do a piece of work in 8 days. Nawaf is 60% more efficient than Naveen. Find thenumber of days it takes Nawaf to do the same piece of work.Ans: Naveen: NawafEfficiency 100:160Days: 160:100 => 8:5

Number of days taken by Nawaf= 8 / = 5 days.2. In 30 litres mixture of milk and water , the ratio of milk and water is 7:3. Find the quantity of

water to be added in the mixture in order to make this ratio 3:7.Ans: 30 (77-33)/(3(7+3)) = 40 litres.

3. Sum of 3 numbers is 98. If the ratio between the first and second is 2:3 and that between thesecond and third is 5:8, find the first number.Ans:

a:b::2:3c:d::5:8

then, ratio among first, second and third quantities is = ac:bc:bd 10:15:24

So the first number= 10 = 20

NOTE: Similarly in case of 4 numbers and the ratio is given as a:b, c:d, e:f. Then ace:bce:bde:bdf.

4. If Rs. 1000 is divided among A,B,C in such a way that A gets 2/3 of what B gets and B gets ofwhat C gets, then what is As share?Ans:a=2, b=3, c=1, d=4ab:bc:bd6:3:12

Then As share= 6 = Rs. 285.71
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5. The ages of Shaam and Balakrishna are in the ratio 3:1. If twenty years hence, the ratio will be7:5, then what is the present age of Shaam?Ans:

= 7/5

7 + 140 = 15 + 100 x=5 PUZZLES

ARRANGEMENT PUZZLES:Placing persons or things in linear or circular order.

Ex: Six persons A,B,C,D,E,F are sitting in two rows, three in each. E is not at the end of any row. D issecond to the left of F. C is the neighbor of E and is sitting diagonally opposite to D. B is the neighbor ofF.

Ans:

Row 1 __ __ __

Row 2 __ __ __

E is not at the end of any row =>

Row 1 __ E __

Or Row 2 __ E __

D is second to the left of F =>

Row 1 D E F

Or Row 2 D E F

C is the neighbor of E (ie E cannot sit with D and F) and (E) is sitting diagonally opposite to D =>

Row 1 D __ F

Row 2 __ E C

B is the neighbor of F =>

Row 1 D B F
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Row 2 A E C

SELECTION PUZZLES:Ex: The starting line-up for the school volleyball team is chosen from following two groups:

Group A: Manjunath, Suhaim, Fameez, Mohan.

Group B: Thaha, Omar, Jickson, Yajnesh, Vishal.

The following requirements must be met:

1) Three players are chosen from group A and three from group B.2) Jickson plays only if Fameez plays.3) Suhaim and Fameez both dont play.4) If Jickson plays, then Yajnesh does not.5) Exactly 3 of the four spikersManjunath, Fameez, Jickson, Vishal must be chosen.

1. If Jickson plays, which of the following must also play?a. Yajnesh or Manjunath.b. Suhaim or Thaha.c. Suhaim or Manjunath.d. Manjunath or Vishal.e. Yajnesh or Omar.

From condition (2), Jickson-- Fameez. From condition (5), choosing Manjunath or Vishal. Sooption (d)

2. All of the following pairs of players play together EXCEPT:a. Mohan and Suhaim.b. Jickson and Manjunath.c. Omar and Manjunath.d. Manjunath and Fameez.e. Vishal and Yajnesh.

Starting with option (a),

Both are from group A, so other 3 players from group B must be chosen. Also, all 3 from group Bmust be spikers. But neither option (d) nor (e), gives exactly 3 spikers. So the 3rd spiker cannotbe selected. Answer is (a).

TASK ASSIGNMENT PUZZLE:Ex:

Professor Viru, Head of CSE at SP university, is making the time table. Besides himself, there are fourother professors- Walia, Nayar, Dhar, Ehsan.
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Their availability is subject to the following constraints:

1) Walia cannot teach on Monday or Thursday.2) Dhar cannot teach on Wednesday.3) Ehsan cannot teach on Monday or Friday.4) Nayar can teach anytime.5) Viru cannot teach evening classes.6) Walia can teach only evening classes.7) Viru cannot teach on Wednesday if Nayar teaches on Thursday, and Nayar teaches on8) Thursday if Viru cannot teach on Wednesday.9) At any given time there are always three classes being taught.

Ans:

Following table indicates X on the elimation grid (at time when teacher does not work).

M T W T FWalia am X X X X X

pm X XDhar am X

pm XNair am

pmEhsan am X X

pm X XViru am

pm X X X X X

Questions will be of this form:

1. At which time can Walia, Dhar and Ehsan all be teaching?a. Monday morningb. Friday eveningc. Tuesday eveningd. Friday morninge. Wednesday morning.

Ans: from table, (c)
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34 www.vtulife.com TIME AND WORKIf M1 number of men work do W1 works in D1 days, M2 number of men work do W2 works inD2 days, then:

M1D1W2=M2D2W1 or = If we include working hours, then:

M1D1T1W2=M2D2T2W1

If we include efficiency, then:

M1D1T1E1W2=M2D2T2E2W1

Ex:

5 men can prepare 6 toys in 7 days working 8 hours a day. Then in how many days can 10 men prepare11 toys working 12 hours a day?

Ans:

51178=10612D2

D2= = 4.28

If A can do a piece of work in x days and B can do it in y days, then A and B together can dothe same work in days.

Ex: Arvind can do a piece of work in 15 days and Ashwin can do it in 10 days. How long will they take ifthey both work together?

Ans:

Method 1: (1510)/(15+10)= 6 days //using above formula

Method 2:

Work done by Arvind in 1 day = 1/15

Work done by Arvind in 1 day = 1/10.

Work done by Arvind and Ashwin in 1 day = 1/15+1/10 = 1/6

1/6 X = 1
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X= 6 days.

If A can do a piece of work in x days, B can do it in y days,C in z days. Then A, B and C togethercan do the same work in days.

If A and B can do a piece of work in x days and A alone can do it in y days, then B alone can dothe same work in days.

PROBLEMS:1. If Shaheen can build a wall in 20 days, Mithesh can build the same in 30 days. Devi can destroy

the wall in 60 days. If all the 3 work together, how many days is it required to build the wall?Ans:1/20+1/30-1/60=1/15 1/15 X= 1 X = 15 days.

2. 36 men work 10 hours every day and can complete a work in 20 days. 8 days after they startedworking, 18 more men joined them and worked for 10 hours each day. How many more dayswill they take to complete the remaining work?Ans: = + = + 7200 = 2880 + 540 D2 D2 = 8 days.

PIPES AND CISTERNSUsing method 2 from above work and time problem.

Ex:

A backyard swimming pool is being filled by 2 different hoses. One hose can fill the pool in 10 hours,other hose needs 14 hours. How long will it take for the two hoses to fill the pool if they are turned on atthe same time?

Ans:

x/10 + x/14 = 1

7x+ 5x=70 12x=70 x= 5.833 i.e 5 hours, 60*0.833=50 mins. //60 mins per hour.
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36 www.vtulife.comSPEED, TIME AND DISTANCEDistance= SpeedTime

If the speed of a body is changed in the ratio a:b, then the ratio of the timetaken changes in the ratio b:a.

x km/hr=(x )m/sec

x metres/sec=(x )km/hrEx:

Yashaswini covers 50km from Udupi to Mangalore in 2 hours. How mush time will Sahana take to coverthe same distance if shes a slow driver and drivesat 3/5 times the speed of Yashaswini?

Ans: Speed of Yashaswini = D/T = 50/2=25Km/hr.

Speed of Sahana= 3/5 Speed of Yashaswini= 3/5 25 = 15Km/hr

Time taken= D/S = 50/15 = 3.33hrs.

3hours, 60 0.33= 20 mins. //1hour=60mins

If a certain distance is covered at x km/hr and the same distance is covered aty km/hr then the average speed during the whole journey is km/hr.Ex:

1. Sanoob covers a certain distance by car driving at 70km/hr and he returns back on a scooter at55km/hr. Find the average speed of the whole journey.

Ans: = 61.6 km/hr.2. Arshadh walking at a speed of 10km/hr from hostel reaches college 15 mins late. Next time he

increases his speed by 2 km/hr, but still he is late by 5 mins. Find the distance of his college fromhostel.Ans: Difference in time= 15-5=10mins = 10/60 = 1/6 hrs.

Speed during first journey= 10km/hr.His speed during next journey= 10+2=12km/hr.

Distance=ST= (1/6)=10kms.
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3. Vishal does a journey in 10hrs. if he covers his first half of his journey at 21km/hr and the secondhalf at 24km/hr. find the distance.Ans: Let distance be x km.

Then x/2 is travelled at speed of 21km/hr and remaining x/2 at 24km/hr.Then, time taken for whole journey, T= D/S/ + / = 10hrs.=> x= 224kms.

RELATIVE DISTANCE:Distance of one moving body with respect to another moving body at the start and at the end of thejourney is called relative speed.

TYPES:1. When two bodies are travelling in the SAME direction.

Average speed= S1-S2

Time = =

2. When two bodies are travelling in the OPPOSITE direction.Average speed= S1+S2

Time = =

Ex:

1. Car A travels at a speed of 60km/hr and car B at a speed of 80km/hr in the same direction. Car Ais 65km ahead of car B. how much time will Car B take to meet Car A.Ans: Speed= 80-60=20km/hr

Initial distance between car A and car B = 65km.Final distance between car A and car B = 0km.Time= (65)/(20)= 3.25 i.e 3 hours, 60*0.25=15 mins.

2. Car A travels at a speed of 60km/hr and car B at a speed of 80km/hr in opposite direction. Car Ais 840km apart from car B. how much time will Car B take to meet Car A.Ans: Speed= 80+60=140km/hr

Initial distance between car A and car B = 840km.Final distance between car A and car B = 0km.Time= (840)/(140)= 6 hours.

TRAINS:RELATIVE SPEED concept is taken into consideration.
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1. Time taken for a train to cross a pole = distance/speed

= length of the train/speed of the train.

2. Time taken for a train to cross a bridge or a railway station

= (length of the train+length of the bridge)/speed of the train.

PROBLEMS:1. What is the time taken for a train of length of 200 metres travelling at a speed of 10 m/s to

completely cross a pole?Ans: 200/10 = 20sec.

2. What is the time taken for a train of length of 200 metres travelling at a speed of 10 m/s tocompletely cross a platform of 200 metres?Ans: (200+200)/10 = 40sec.

3. Two trains 120 metres and 100 metres in length respectively are running in opposite directions ,at a speed of 40km/hr and 32km/hr respectively. In what time will they completely clear of eachother from the moment they meet?Ans: speed= 40+32=72km/hr = 725/18 = 20m/s

time= (120+100)/20 = 11sec4. Two trains 120 metres and 100 metres in length respectively are running in same directions , at

a speed of 40km/hr and 32km/hr respectively. In what time will they completely clear of eachother from the moment they meet?Ans: speed= 40-32=8km/hr = 85/18 = (20/9)m/s

time= (120+100)/(20/9) = 99sec5. Two trains start at the same time from Bangalore and Mangalore and proceed towards each

other at the rate of 80km and 90km per hour respectively. When they meet, it is found that onetrain has already travelled 160km more than the other. Find the distance between Bangaloreand Mangalore.Ans: Mangalore train moves 10km more in 1 hour = 10km/hr.

160km more travelled in T=D/S= 160/10 = 16hrs.They are moving in opposite direction, so average speed , S= 80+90= 170km/hr.In 16 hrs, D= ST= 170 16 = 2720 km.Distance = 2720kms.

STREAMS: Boat and swimmer problems. Still and moving water of the stream. Upstream and downstream.
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1. If speed of the boat is x and if speed of the boat is y then, UPSTREAM (boat is movingagainst the stream) speed of boat= x-y. DOWNSTREAM (boat is moving with thestream) speed of boat= x+y.

2. If x km/hr be the boat/swimmers rate in still water and y km/hr be the rate of thecurrent. Then

a) Swimmers rate with current= x+y

b) Swimmers rate against current=x-y

3. If x is the speed of the boat downstream and y is the speed of the boat upstream, then

a) Speed of the boat in still water=(x+y)/2

b) Speed of the water/stream= (x-y)/2

Ex: A man can row upstream at 10km/hr and downstream at 16km/hr. find the mans rate in still waterand the rate of the current.

Ans: Rate in still water = (10+16)/2 = 13km/hr.Rate of current = (16-10)/2 = 3km/hr.

RACES:

1. CIRCULAR TRACKS:

L is the length of the track.

Circular Tracks Running in samedirectionRunning in opposite

direction

Speed of A is x and of Bis y.

Time taken to meet forthe first time L/(x-y) L/(x+y)

Time taken to meet forthe first time at the

starting pointLCM of (L/x , L/y) LCM of (L/x , L/y)

Speed of A is x, B is yand of C is z.

Time taken to meet forthe first time

LCM(L/(x-y) , L/(y-z) ,L/(z-x))

LCM(Relative time ofAB,BC,CA)

Time taken to meet forthe first time at the

starting pointLCM of (L/x , L/y) LCM of (L/x , L/y)
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2. LINEAR TRACKS:1. A can give B a 200m startup and C a 300m start in a race of 1 km. How many metres startup can

B give to C in a 1km race?Ans:

So 1km, when A completes 1000m, B completes 800m and C completes 700m.When B completes 100m, ie 25% more, even C completes 25% more, ie 875. Then 1000-

875=125m.

CLOCKS:Angle in a circle = Angle in a clock = 360

Hour space = 360/12 = 30

Minute space = 360/60 = 6 PROBABILITYPROBABILITY:Measurement of uncertainity.

Probability of occurance of a event = P(E) = ( )( ) , N(E) is number of favourable outcomes of theoccurance of the event. N(S) is the total number of outcomes in the sample space.

RANDOM EXPERIMENT:It is an experiment, trial or observation that can be repeated numerous times under the same conditionswhile the outcome of the experiment cannot be predicted with certainity, before the experiment isconducted. Eg; Tossing a coin, Rolling a dice.

SAMPLE SPACE:Set of all possible outcomes of a random experiment.

EVENT:Subset of a sample space.

p=1-q where p=occurance of the event, q= not occurance of that event.

Probability of occurance of 2 events E1 and E2, P(E1E2)= P(E1)+P(E2)-P(E1E2)
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CARD:In a pack of cards, 52 cards. In which 13 spades, 13 hearts, 13 diamonds, 13 clubs. Hearts and diamondsare red in color. The clubs and spades are black in color. In each set of these 13 cards, there are nine innumber- cards: 2 to 10 are four face cards: ace, jack, king and queen.

PROBLEMS:1. When a dice is thrown, what is the probability that (a) number is odd? (b) number is prime or

odd?Ans: (a) dice: 1,2,3,4,5,6

Odd numbers: 1,3,5.

Required probability = ( , , )( , , , , , ) = 3/6= 0.5 = 50%(b) P((2,3,5)(1,3,5))= P(2,3,5)+P(1,3,5)-P((2,3,5)(1,3,5))

= 3/6 + 3/6 2/6 = 2/32. When 2 dice are thrown, what is the probability that (a) sum of numbers appeared is 6 or 7? (b)

sum of numbers appeared is >1 and


42 www.vtulife.com DATA SUFFICIENCYData sufficiency problem consists of a question and two statements, labeled (1) and (2). Each statementhas some data. You must decide whether the data given in the statements is sufficient for answering thequestions. The options will be given as follows:

(a) Statement (1) alone is sufficient to answer to the given question, but (2) alone is not sufficient.(b) Statement (2) alone is sufficient to answer to the given question, but (1) alone is not sufficient.(c) Both (1) and (2) together are suffiecient to answer the question, but neither of them alone are

not sufficient.(d) Each statement (1) and (2) alone are sufficient to answer to the given question.(e) Both (1) and (2) together are not suffiecient to answer the question, additional data required.

Example:

1. How many people are there at the party?(1) The party has between 20 and 30 people.(2) The number of people at the party is a prime number between 24 and 30.

Given options:

a) Statement (1) alone is sufficient to answer to the given question, but (2) alone is notsufficient.

b) Statement (2) alone is sufficient to answer to the given question, but (1) alone is notsufficient.

c) Both (1) and (2) together are suffiecient to answer the question, but neither of them aloneare not sufficient.

d) Each statement (1) and (2) alone are sufficient to answer to the given question.e) Both (1) and (2) together are not suffiecient to answer the question, additional data

required.

Ans: First check option (1). This is not enough to guess the answer. Now check option (2). Primenumbers between 24 and 30 = 29.

So, option (b) is the answer.

2. How many students in a class play football ?(1) Only boys play football.(2) There are forty boys and thirty girls in the class.

Ans: option (e). It is not mentioned whether all the boys or a proportion of them play football.
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43 www.vtulife.com CUBES6 square faces, 12 edges and 8 corners.

1. If a cube is cut into (p+q+r) cuts such that p cuts on x-axis, q cuts on y-axis and rcuts on z-axis, then the total of cubes/cuboids (identical/non-identical) the cube canbe divided into is (p+1)(q+1)(r+1).

2. A cube can be cut in x,y,z axis to get maximum number of identical pieces, we need tocut the cube in all the three axis such that the difference between one axis cut and theother axis cut is minimum ie cuts should be as uniformly distributed as possibleamong the three axes such that number of cuts made in atleast two of the axes isequal.

3. When a cube is dipped in paint and cut into identical smaller cubes. If n is thenumber of pieces on each side.

Total number of cubes n3

One face painted cubes 6(n-2)2

Two face painted cubes 12(n-2)

Three face painted cubes 8

None of the faces coloured (n-2)3

PROBLEMS:1. What is the maximum number of identical pieces we obtain if a large cube is cut 13 times?

Ans: we have to divide the cube in x,y and z axis. Now 13/3=4.33. so cut into 3 axis through4,4,5 cuts. // (4+4+5=13)

Hence number of pieces = (5+1)(4+1)(4+1)=150

2. A large cube is painted green and then cut to obtain 216 small identical cubes.
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(a) The number of small cubes that have one face painted?(b) The number of small cubes that have 3 faces painted?(c) The number of cubes with atmost 2 faces painted?(d) The number of small cubes has 0 faces painted?

Ans: finding n.(n)(n)(n)=216(n)3=216n=6

(a) One face painted = 6(n-2)2= 6(6-2)2= 96(b) Three faces painted = 8(c) At most 2 faces painted = 0 faces painted+1 faces painted+2 faces painted=

(n-2)3+96+12(n-2) = 64+96+48 = 208(d) 64

NON VERBAL REASONINGProblems based on continuation of figures, analytical reasoning.

PROBLEMS:1.

Ans: option (2)

2.
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Ans: option (4)

3. Find the number of triangles in the figure below.

Ans: label the figure

The simplest triangles are AHG, AIG, AIB, JFE, CJE and CED i.e. 6 in number. The trianglescomposed of two components each are ABG, CFE, ACJ and EGI i.e. 4 in number. The trianglescomposed of three components each are ACE, AGE and CFD i.e. 3 in number. There is only onetriangle i.e. AHE composed of four components. Therefore, There are 6 + 4 + 3 + 1 = 14 trianglesin the given figure.

4. Select a figure from amongst the Answer Figures which will continue the same series asestablished by the five Problem Figures.

Ans: (3)

5.

Ans: (2)6. Choose the alternative which is closely resembles the mirror image of the given combination.
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Ans: (4)

DATA INTERPRETATIONTABLE CHARTStudy the following table and answer the questions based on it. The following table gives the percentageof marks obtained by seven students in six different subjects in an examination.

Student

Subject (max marks)

Maths Chemistry Physics Geography History Computer

150 130 120 100 60 40Ayush 90 50 90 60 70 80Aman 100 80 80 40 80 70Sajal 90 60 70 70 90 70Rohit 80 65 80 80 60 60

Muskan 80 65 85 95 50 90Tanvi 70 75 65 85 40 60Tarun 65 35 50 77 80 80

1. What are the average marks obtained by all the seven students in Physics?Ans: Average marks obtained in Physics by all the seven students= 1/7 [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120)+ (50% of 120) ]= 1/7 [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]= 1/7 [ 520% of 120 ]= 89.14

2. The number of students who obtained 60% and above marks in all subjects is?Ans: 2From the table it is clear that Sajal and Rohit have 60% or more marks in each of the six subjects.

3. What was the aggregate of marks obtained by Sajal in all the six subjects?Ans: Aggregate marks obtained by Sajal
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= [(90% of 150)+(60% of 130)+(70% of 120)+(70% of 100)+(90% of 60)+(70% of 40) ]= [ 135 + 78 + 84 + 70 + 54 + 28 ] = 449

4. What is the overall percentage of Tarun?Ans: 60%Aggregate marks obtained by Tarun = [(65% of 150)+(35% of 130)+(50% of 120)+((77% of100)+(80% of 60)+(80% of 40) ]

= [ 97.5 + 45.5 + 60 + 77 + 48 + 32 ] = 360.The maximum marks (of all the six subjects) = (150 + 130 + 120 + 100 + 60 + 40) = 600.Overall percentage of Tarun = (360/600 100)% = 60%

BAR CHARTThe bar graph given below shows the data of the production of paper (in lakh tonnes) by three differentcompanies X, Y and Z over the years.

1. For which of the following years, the percentage rise/fall in production from the previous year isthe maximum for Company Y?Ans: Percengate change (rise/fall) in the production of Company Y in comparison to the previousyear, for different years are:For 1997: ((35-25)/25 100) % = 40%For 1998: ((35-35)/25 100) % = 0For 1999: ((40-35)/35 100) % = 14.29%For 2000: ((50-40)/40 100) % = 25%Hence, the maximum percentage rise/fall in the production of Company Y is for 1997.

2. What is the ratio of the average production of Company X in the period 1998-2000 to theaverage production of Company Y in the same period?Ans: Average production of Company X in the period 1998-2000

= [1/3 (25+50+40) ] 115/3 lakh tons.Average production of Company Y in the period 1998-2000
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= [1/3 (35+40+50) ] 125/3 lakh tons.Required ratio = (115/3) / (125/3) = 115/125 = 23/25.

3. The average production for five years was maximum for which company?Ans: Average production (in lakh tons) in five years for the three companies are:For Company X = [1/5 x (30 + 45 + 25 + 50 + 40)] = 190/5 = 38.For Company Y = [1/5 x (25 + 35 + 35 + 40 + 50)] = 185/5 = 37.For Company Z = [1/5 x (35 + 40 + 45 + 35 + 35)] = 190/5 = 38.Average production of five years is maximum for both the Companies X and Z.

4. In which year was the percentage of production of Company Z to the production of Company Ythe maximum?Ans: The percentages of production of Company Z to the production of Company Z for variousyears are:For 1996: (35/25 100) = 140%For 1997: (40/35 100) = 114.29%For 1997: (45/35 100) = 128.57%For 1997: (35/40 100) = 87.5%For 1997: (35/50 100) = 70%Clearly, this percentage is highest for 1996.

5. What is the percentage increase in the production of Company Y from 1996 to 1999?Ans: Percentage increase in the production of Company Y from 1996 to 1999[ (40-25)/25 100 ] % = 60%

PIE CHARTThe following pie-chart shows the percentage distribution of the expenditure incurred in publishing abook. Study the pie-chart and the answer the questions based on it.

1. If for a certain quantity of books, the publisher has to pay Rs. 30,600 as printing cost, then whatwill be amount of royalty to be paid for these books?Ans: Let the amount of Royalty to be paid for these books be Rs. r.Then, 20 : 15 = 30600 : r
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r = Rs. (3060015 / 20) = Rs. 22,950.2. What is the central angle of the sector corresponding to the expenditure incurred on Royalty?

Ans: Central angle corresponding to Royalty = (15% of 360) = 543. Royalty on the book is less than the printing cost by?

Ans: Printing Cost of book = 20% of C.P.Royalty on book = 15% of C.P.Difference = (20% of C.P.) - (15% of C.P) = 5% of C.P.Percentage difference = [ difference/printing cost 100 ] %

= [ 5% of C.P/printing cost 100 ] % = 25%
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50 www.vtulife.comTHE ART OF FACING INTERVIEWSDo what you love and Love what you do. Asmall and cute saying which you should alwayskeep in mind. Because you can enjoy and growonly if you do a work which you like or youhave interest on.

I have seen many of my friends crying becausethey were eliminated for Placement drives andinterviews. Actually they are all lucky becauseat least they got a chance to attend interviewsand its an experience. Till the 7th Semester myaggregate was 59.8%. I did not even have the

minimum eligibility to attend interviews.

But after completing Engineering in Computer Science with first class. I have attended more than 10interviews till date. Not just in software domain but also for marketing, HR, Sales, management and thelist goes! Because I like attending interviews. The basic thing which you need to attend interview isconfidence and for that you should know English. I know many of them who dont have a good languagebut are street smart and knows the work. But Im very sorry to say that 99% of the interviewers checksyour language. Even if you have strong technical knowledge without language they are not going to hireyou. To improve your English the most effective way is to talk. I also recommend to read THE HINDUnewspaper daily. By reading newspaper you can improve your vocabulary by using the new words youfind in the newspaper in your daily conversation.

Many of you all can clear the aptitude test if you prepare. But when you get into next level, it may beGroup Discussion or interview you may get eliminated as you are not used to it or have experienced it.For tackling this problem many colleges have placement training in which they teach how to ace this.And whenever you get a chance to speak in front of an audience. Go ahead By doing this slowly you cancompletely eliminate your stage fear.

For any face to face interview. There are 5 golden rules that you have to keep in mind

Here it goes

Know Yourself Know your academics Research the position you seek
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Positive Impression Be Honest

KNOW YOURSELF

The first thing which every interviewer sees is whether we have the skills that are considered to becrucial to the position which we have applied for. So in order to convince the interviewer that you domeet the requirements that they have in mind, you have to be able to communicate your skills well. Andfor that you should know about your abilities. Your strengths and Weaknesses.

KNOW YOUR ACADEMICS

Brush your technical knowledge. The more you know says your ability to gain knowledge. Apart from theknowledge on your core stream. You should know what is happening in the world. Current affairs is veryimportant and for that you have to read newspaper daily. Apart from technical knowledge. Make sureyou have a firm grip on whatever you have written in your resume.

RESEARCH THE POSITION YOU SEEK

Know about the position you are seeking for. Its roles duties etc and also about the company. Gothrough the companies Wikipedia page.

CREATE A POSITIVE IMPRESSION

The most important of all is to create a positive impression and this happens within few minutes of theinterview. Most selection decisions are made within five minutes of an interview. The interviewer checkson your body language, appearance and attitude in the face of stress. So dress appropriately, smile, becalm, act mature even if you are not!! And dont lose your control

BE HONEST

Never lie during an interview. Be honest Be natural and above all be yourself. If you dont knowsomething say you dont know, dont try to act smart and conceal your ignorance. Always know thatinterviewers are smarter than you. You may not know all the answers but what the interviewer sees ishow you tackle a problem or your approach to it

FEW GENERAL TIPS

Make sure you know the time the interview will be held. Confirm that you will be attending Arrive at least 15 minutes before the time Always carry a file to carry your required documents Be calm look at everyone and relax
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Be a good listener If you did not understand a question request the interviewer to repeat. Youwont lose your job because of that

Dont always speak about the salary

Keeping all this in mind. Do your homework, be positive and give it your very best shot. Above all thatlearn from your experience and never give up hope.

BECAUSE YOU ARE THE BEST

Best of Luck for your Future

Sanoob Sidiq
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Image Courtesy The Economic Times
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Aptitude Handbook Www.vtulife.com

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