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[Number System]

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[Number System]


CSAT – II, IAS EDGE

This Documents was prepared under the supervision of Mr. Pramod Singh,

Director, IAS Edge

April 2018

[Number System]


CSAT BOOK - 2

CONTENTS

Chapter 1 – Number Systems

1.1 Introduction to various types of numbers ………………. 1

1.2 Divisibility Rules ………………. 5

1.3 LCM, HCF & their applications ………………. 9

1.4 Cyclicity ………………. 14

1.5 Number of factors of a given number ………………. 16

1.6 Remainders ………………. 20

1.7 Highest power dividing a factorial ………………. 26

Exercise on LCM and HCF ………………. 31

Exercise on Properties of Numbers ………………. 37

Exercise On Divisibility, Sum Of Divisors, Number Of Divisors ………………. 41

Exercise on Number of zeroes ………………. 43

Exercise on Remainder Theorem ………………. 46

Exercise on Unit‘s Digit ………………. 48

Word Based Problems ………………. 50

Past Exam Questions ………………. 52

Chapter 2 – Progressions

Arithmetic Progressions ………………. 67

Exercise on Arithmetic Progressions ………………. 77

Chapter 3- Permutations and Combinations ………………. 79

Chapter 4- Probability ………………. 86

Chapter 5- Set Theory ………………. 94

Chapter 6- Logical Reasoning ………………. 101

Chapter 7- DI based on percentages ………………. 109

Chapter 8- DI based on averages ………………. 112

Chapter 9- DI based on ratio and proportion ………………. 115

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Chapters 1 – Number Systems No one can think of the world without numbers. They are the most primary elements of life and logic that are used everywhere. From the point of view of aptitude based competitive exams it is one of the most important topics in mathematics. The past trends of exams like CAT,XAT, CSAT, SSC, IBPS-PO and other aptitude exams show that while there has always been a variation in the number of questions from chapters like Algebra, Geometry, Permutation & Combination, Probability, Percentages, Time Speed Distance and other topics, questions on Number Systems have always been present. In the following part of the study material we have discussed this chapter under the structure given below. We have also included various models of problems, based on various concepts and problems that have been appearing in various exams.

1.1 Introduction to various types of numbers: Classification of Numbers: - All those numbers that can be represented on the number line are called real Numbers. The number line is a line with an arbitrarily selected point reported as 0 and has positive numbers to its right and negative numbers to its left as shown below

The types of numbers in the number system can be classified as follows:

TYPES OF NUMBERS

Real Complex

Eg 5 + 2i Rational Irrational

Eg 1, 2, 3/2, 2/3 Eg2, 3/4, 7 - - - etc Rational

Integers Fractions

Z = {-, -5, …. 6, 7, . . }

Whole Numbers W = {0, 1, 2, …..}

Natural Numbers N = {1, 2, 3 . . . . } Even Numbers

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Odd Numbers Prime Numbers Composite Numbers Unity Perfect Numbers Triangular Numbers etc. Let us now look at the definitions of each of these types of numbers: Rational Numbers: - A number which can be expressed in the form p/q where p & q

are integers and q 0 is called a rational number. Eg 4, 1/3, 2/5 etc. Any terminating or recurring decimal is a rational number.

Eg 0.333 - - - - - 0.1111 - - - - - - 0.1666 - - - - - - Let y = 0.333 - - - - - -

x = 0. 3 -------------------------- (1)

Multiplying equation (1) by 10 10x = 3.33333 --------- (2) Equation (2) – equation (1)

9x = 3

x = 3/9 = 1/3 Irrational Numbers: Numbers which are not rational but which can be represented by

points on the number line are called irrational numbers. Eg (2, 4 5 etc. , e)

Any non-terminating non-recurring decimal is an irrational number. Integers: A subset of rational numbers of the form p/q, where q = 1. It can be positive or negative. It is further classified under prime numbers& composite numbers. Natural Numbers: All counting numbers like 1, 2, 3 - - - - are called natural numbers. Whole Numbers: When 0 is added to the set of natural numbers, we get the set of whole numbers. Prime Numbers:- A number which doesn‘t have any factor apart from one & itself is called a prime number. Eg. 2, 3, 5, 7, 11, 13 etc. Note: 1 is not a prime number and is said to be neither prime nor composite. Every Prime number greater than 3 can be written in the form of (6k +1) or (6k-1) where k is an integer. Composite Numbers: A number which is not a prime number is called a composite number. Eg 4, 6, 8, 9, 10 etc. In other words, numbers which have at least one more factor apart from 1 and itself are called as composite numbers.

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Relative or co-primes: Two numbers are said to be relative primes or co-prime to

each other if they do not have any common factor other than 1. Example 15 & 16 are co-prime. In fact, all consecutive integers are co-prime to each other.

Fractions: These numbers basically include a part of an integer. Eg 1/4 = 0.25 Types of Fractions: There are six kinds of fractions:

1. Proper fractions: The numerator is less than the denominator. Eg 3/7

2. Improper Fractions: The numerator is more than the denominator. Eg 7/3

3. Mixed Fractions: Numbers which consist of integral as well as fractional part. Eg

(2 + 7

3) =

7

17

4. Compound Fractions: Fractions whose numerator and denominator themselves are fractions.

Eg

8

33

2

= 16/9

5. Complex fractions: Any complicated combination of other type of fractions. Eg 2

4

1 of

4

31

4

6. Continued fractions: it is in form of

3

33

33 or

a

aa

aa

Multiples: If one number is divisible exactly by a second number, the first number is said to be a multiple of the second number and the other number is called as a factor or divisor of the first number.

Eg 42 is a multiple of 7 & 7 is a factor (or divisor) of 42. Even & Odd Numbers: Numbers divisible by 2 are called even whereas numbers that are not divisible by 2 are called odd Numbers. In fact, you should remember that even numbers are often represented in mathematics in the form 2n while odd numbers are represented in the form of 2n+1 or 2n – 1.

Examples Even numbers 2, 4, 6, 8, -12, 200, 1242 etc.

Odd numbers 3, 5, 7, 9, 33,-11 etc. The following operations hold true when you consider even and odd numbers: 1) Even + even = even 2) Odd + odd = even 3) Even + odd = Odd

4) Even any number = even 5) Odd x odd = odd 6) Even/Odd = Even (if divisible)

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7) Odd/odd = odd (if divisible) 8) Odd/Even Never divisible Perfect Numbers: A number is said to be a perfect number if the sum of all its factors excluding itself is equal to the number itself.

Eg 6 has factors of 1, 2, 3 & 6 = 1 + 2 + 3 Other examples are 28, 496, 8128. There are 27 perfect numbers that have been discovered so far. BODMAS Rule :- (Hierarchy of Arithmetic operations)

B Brackets,

O Of

D Division

M Multiplication

A Addition

S Substraction Note: The operations have to be carried out in the order in which they appear in the word BODMAS.

ILLUSTRATIONS SET 1.1: 1. The difference between the number of numbers from 2 to 100 which are not

divisible by any other number except 1 & itself and the numbers which are divisible by at least one more number along with 1 and itself. (a) 25 (b) 50 (c) 49 d) Cannot be determined

Solution: From 1 to 100. The number of numbers which are divisible by 1 & itself only = 25 (these are the prime numbers from 2 to 100) Also the number of numbers which are divisible by at least one more number except 1

& itself (i.e. composite numbers) = 99 – 25 = 74 So, Required difference = 74 – 25 = 49

Option (c)

2. If the sum of (2n +1) prime numbers where nN is an even number, then one of the prime numbers must be (a) 2 (b) 3 (c) 5 (d) 7 (e) 11

Solution: For any n N, 2n + 1 is odd. Also, it is given in the problem that the sum of an odd number of prime numbers = even. Since all prime numbers except 2 are odd, the above condition will only be fulfilled if we have an (Odd + Odd + even) structure of addition. Since, the sum of three prime numbers is said to be even, we have to necessarily include one even prime number. Hence 2 being the only even prime number, it must be included.

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Note: If we add odd number of prime numbers, not including 2 (two), we will always get an odd number, because

number odd timesofnumber oddan - - - - - odd odd odd

option (a) 3. What will be the difference between the largest and smallest four-digit number

made by using distinct single digit prime numbers? (a) 1800 (b) 4499 (c) 4495 (d) 4799 (e) 5175

Solution: Required largest number 7 5 3 2

Required smallest number 2 3 5 7

Difference 5 1 7 5

Option (e)

1.2 Divisibility Rules: The following divisibility rules hold true: Divisibility by 2:-If the last Digit of a number is even ie 0, 2, 4, 6 or 8 then the number is divisible by 2. eg – 842. Divisibility by 3:- If the sum of all digits is divisible by 3, the number is divisible by 3.

Eg – 747 7 + 4 + 7 = 18 is divisible by 3 hence the number is also divisible by 3. (Note an important point here: In case the sum of digits in not divisible by 3, then the remainder of the sum of digits when divided by 3 will also be equal to the remainder of the number itself when divided by 3.) Divisibility by 4 :- If the last two digits of a number is divisible by 4 or is 00, the

number is also divisible by 4. eg 2768964 here 64 is divisible by 4 hence the number is divisible by 4.

Divisibility by 5:- If the last digit is 0 or 5 the number is divisible by 5 eg 125, 525, 220 etc. Divisibility by 6:-A number is divisible by 6 if the number is divisible by both 2 & 3

simultaneously. Eg:- 144 is divisible by 2 & 3 both hence will also be divisible by 6 Divisibility by 8:- If the last three digits of a number is divisible by 8 or have 3 or more

zeroes (eg. 000, 0000 etc), the number is also divisible by 8. eg 632000. Divisibility by 9:-A number is divisible by 9 if the sum of its digits is a multiple of 9.

eg 729 (Note an important point here: In case the sum of digits in not divisible by 9, then the remainder of the sum of digits when divided by 9 will also be equal to the remainder of the number itself when divided by 9. e.g. 241 is not divisible by 9 because 2+4+1=7 is not divisible by 9. The remainder of 241/9 will then be equal to 7 too since 7/9 leaves a remainder of 7.) Divisibility by 11:- If you subtract the sum of digits in the even places from the sum of digits in the odd places (even places and odd places are to be counted from the right),

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the answer of the subtraction should be either 0 or a multiple of 11. In such a case,

the number is said to be divisible by 11. eg 1419 Sum of digits in even places = 1 + 1 = 2 Sum of digits in odd places = 4 + 9 = 13

Required subtraction= 13 – 2= 11, which is a multiple of 11. Note: In case the given difference between the sum of digits in odd and even places is not a multiple of 11, then the remainder this difference leaves when divided by 11 will also be the remainder that the number will leave when it is divided by 11.

e.g. The number 33224512 will give the difference between odd and even places as 12- 10=2. This will be the remainder when the number itself is divided by 11. [Note: This would work only in odd places – even places case. Do not commit the error of even places minus odd places. That would give you a wrong answer for the remainder, though it would predict divisibility correctly.]

ILLUSTRATIONS SET 1.2 1. The difference between two three digit numbers XYZ and ZYX will be equal to

(a) Difference between X and Z ie |X – Z| (b) Sum of X and Zi.e (X + Z)

(c) 9 difference between X and Z

(d) 90 difference between X and Z

(e) 99 difference between X and Z Solution: From the property of numbers, it known that on reversing a three digit number, the difference (of both the numbers) will be divisible by 99. Also, it is known that this difference will be equal to 99 times the difference between the units and

hundreds digits of the three digit number. option (e) 2. When the difference between the number 842 and its reverse is divided by 99, the

remainder will be

(a) 0 (b) 1 (c) 74 (d) 17 (e) None of these Solution: From the property (used in the above question) We can say that the difference will be divisible by 99

remainder = 0 (zero)

option (a) 3. When the difference between the number 783 and its reverse is divided by 99, the

quotient will be? (a) 1 (b) 10 (c) 3 (d) 4 (e) None of these

Solution: The quotient will be the difference between the extreme digits of 783. i.e. 7 – 3 = 4 (This is a property which you should know)

option (d)

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4. A long wood of some length when cut into equal pieces each of 242 cm, leaves a small piece of length 98 cm. If the long wood were cut into equal pieces each of 22 cm, the length of the left wood would be (a) 76 cm (b) 12 cm (c) 11 cm (d) 15 cm (e) 10 cm

Solution: As 242 is divisible by 22. So the required length of left wood will be equal to the remainder when 98 is divided by 22 Hence, 10 [98/22; Remainder 10]

option (e) 5. Find the number of numbers from 1 to 100 which are not divisible by 2.

(a) 51 (b) 50 (c) 49 (d) 48 (e) 52 Solution:The 1st number from 1 to 100, not divisible by 2 is 1 and last number from 1 to 100, not divisible by 2 is 99. Every alternate number (i.e at the gap of 2) will not be divisible by 2 from 1 to 99. (1, 2, 3, - - - - , 95, 97, 99)

So, Required number of nos = 1step)(or gap

no.first - no.last = 501

2

199

Option (b)

Alternate method: - Total number of nos from 1 to 100 = 100 --------------- (i) Now, if we count number of numbers from 1 to 100 which are divisible by 2 and subtract it from total number of numbers from 1 to 100, as a result we will find the number of numbers from 1 to 100 which are not divisible by 2. To count the number of numbers from 1 to 100 which are divisible by 2: The 1st number which is divisible by 2 = 2 The last number which is divisible by 2 = 100 (2, 4, 6, - - - - , 96, 98, 100) Gap/step between two consecutive numbers = 2

So, number of numbers which are divisible by 2 = 1step)(or gap


2

2100

- - - - - (ii) So, from (i) & (ii) Required number of numbers = 100 – 50 = 50

option (b) 6. Find the number of numbers from 1 to 100 which are not divisible by any one of 2

& 3. (a) 16 (b) 17 (c) 18 (d) 19 (e) 33

Solution: From 1 to 100 Number of numbers not divisible by 2 & 3 = Total number of numbers – number of numbers divisible by either 2 or 3. Now, total number of numbers = 100 - - - - (ii) For number of numbers divisible by either 2 or 3:

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Number of numbers divisible by 2 = 1step)(or gap


2

2100

The number of numbers divisible by 3 (but not by 2, as it has already been counted) 1st such no = 3 (3, 9, 15, 21, - - -- - - , 93, 99 gap/step=6 2nd such no = 9 last such no = 99

So, the number of numbers divisible by 3 (but not by 2) = 1step / gap

no.such first - no.such last

= 1716

399

Hence, the number of numbers divisible by either 2 or/and 3 = 50 + 17 = 67 so, from (i), (ii) & (iii) Required number of numbers = 100 – 67 = 33

option (e) 7. Find the number of numbers from 1 to 100 which are not divisible by any one of 2,

3, & 5. (a) 26 (b) 27 (c) 29 (d) 32 (e) 33

Solution: From above question, we have found out that From 1 to 100, number of numbers divisible by 2 = 50 - - - - - (i) Number of numbers divisible by 3 (but not by 2) = 17 - - - - - - - (ii) Now, we have to find out the number of numbers that are divisible by 5 (but not by 2 & 3) Numbers that are divisible by 5 (5) 10 15 20 (25) 30 (35) 40 45 50 (55) 60 (65) 70 75 80 (85) 90 (95) 100 ie. There are 7 such numbers - - - - - - (iii) A better way to find out the number of numbers that are divisible by 5 but not 2 and 3 is to first only consider odd multiples of 5. You will get the series of 10 numbers: 5, 15, 25, 35, 45, 55, 65, 75, 85 and 95 From amongst these we need to exclude multiples of 3. In other words, we need to find the number of common elements between the above series and the series of odd multiples of 3 viz- 3, 9, 15, 21….99. This situation is the same as finding the number of common elements between the two series for which we need to first observe that the first such number is 15. Then the common terms between these two series will themselves form an arithmetic series and this series will have a common difference, which is the LCM of the common differences of the two series. (in this case the Common difference of the two series are 10 and 6 respectively and their LCM being 30, the series of common terms between the two series will be 15,45 and 75.) Thus, there will be 3 terms out of the 10 terms of the series 5,15,25…95 which will be divisible by 3 and hence need to be excluded from the count of numbers which are divisible by 5 but not by 2 or 3. Hence, the required answer would be: 100-50-17-7 = 26

option (a)

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8. Find the number of numbers from 1 to 100, which are not divisible by any one of 2, 3, 5 & 7. (a) 22 (b) 24 (c) 23 (d) 27 (e) 20

Solution: From above question we have seen that from 1 to 100. Number of numbers divisible by 2 = 5 --------------- (i) Number of numbers divisible by 3 but not by 2 = 17 -------------- (ii) Number of numbers divisible by 5 but not by 2 & 3 = 7 -------------- (iii) Number of numbers divisible by 7 but not by 2, 3 & 5; Such numbers are 7, 49, 77, 91 = 4 nos ----------------- (iv) Required number of numbers = Total number of numbers from 1 to 100 – {(i) + (ii) +

(iii) + (iv)} = 100 – (50 + 17 + 7 + 4) = 22

option (a)

1.3 LCM, HCF & their applications: Least Common Multiple (LCM) of two or more numbers is the least number which is divisible by each of these numbers without a remainder. Method to find LCM: (a) LCM by factorization

eg: L.C.M of 2, 4, 8, 12 While finding LCM by this method, one should keep in mind that there should be one common factor for two or more than two numbers.

So, L.C.M = 2 2 1 1 2 3 = 24 (b) LCM by Division: In this process find the prime factors of each number and then the LCM will be got by writing the highest powers of all prime factors. 2 = 2 4 = 22 8 = 23 12 = 22 x 3 LCM = 23 x 3 = 24 Highest Common Factor (HCF): It is the largest factor of two or more given numbers. It is also called GCD (Greatest Common Divisor)

Methods to find HCF: (a) HCF by factorizations

2 2 4 8 12

2 1 2 4 6

1 1 2 3

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eg: HCF of 2, 4, 8, 12 In this case, list the factors of each of the numbers

Factors of 2 1, 2

Factors of 4 1, 2, 4

Factors of 8 1, 2 4, 8

Factors of 12 1, 2, 3, 4, 6, 12 So, Common factor = 2 = HCF The other and of course more common method for finding both the HCF and the LCM involves writing the prime factors of each of the numbers. Then,

For LCM: The product of the highest powers of each of the prime factors present in the prime factorization of all the numbers. e.g. LCM of 17,21 and 245 17=171, 21= 31 x 71, 35= 51 x 72

The LCM will be given by: 17 x 3 x 5 x 72 (taking the highest powers of each of the prime factors present). For HCF: The product of the least powers of all the common prime factors present in the prime factorization of all the numbers. e.g. HCF of 75,125 and 275 75= 3 x 52, 125 = 53 and 275= 52 x 111. The HCF will be given by: 52 (taking the least power of each of the prime factors present). Properties of HCF and LCM:

1. Product of two numbers = LCM of the numbers HCF of the numbers (Note: This is a common property used to solve questions that are of the type where the LCM and HCF of 2 numbers is given & one of the two numbers is given and you have to find the other number. Also note that this property does not hold true for three or more numbers.) 2. Also note that, in order to find the LCM of fractions you need to use: (LCM of Numerators) / (HCF of denominators) 3. Similarly, in order to find the HCF of fractions you need to use: (HCF of Numerators) / (LCM of denominators) 4. The HCF of any set of numbers is always a factor of the LCM of the numbers. 5. For any set of numbers, the factors of the HCF would be common factors of the set of numbers. For instance, if 4 is the HCF of a, b and c then the factors of 4 i.e. 2 and 1 would also divide each of a, b and c. 6. All multiples of the LCM for any set of numbers, would themselves be common multiples of all the numbers of the set. Thus, the LCM of 3 and 4 being 12, all multiples of 12 (viz.24.36.48 and so on to infinite such numbers) would be multiples of both 3 and 4.

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DIFFERENCE BETWEEN HCF AND LCM

HCF of x, y and z LCM of x, y and z

is the Highest Divisor which can exactly divide x, y and z

is the Least number which is exactly divisible by x, y and z

Special Note: How does one spot whether a question is based on LCM or HCF? One important thing to note is that all questions based on LCM would first be based on common multiple of numbers & within those you would be expected to find the least such number. So for instance if one bell tolls every 5 seconds then the times at which it would ring would be multiples of 5. (5, 10, 15, 20, and so on). If another bell tolls at

intervals of 6 seconds, its‘ tolling would be in multiples of 6 (6, 12, 18, 24 and so on). Ringing together would mean a common multiple of 5 and 6 while ringing together for the first time would be the LCM of 5 and 6. Hence, in this case, 30 would be the correct answer. Likewise, if we have to measure 20 feet with a tape such that it is measured exactly, the tape would need to be of a length that is a factor of 20. (So, the tape length possible to be used would be 1, 2, 4, 5, 10 and 20). Similarly, if we have to measure 30 feet with a tape such that it is measured exactly, the tape would need to be of a length that is a factor of 30. (So, the tape length possible to be used would be 1, 2, 3, 5, 6, 10, 15 and 30). If we want to measure both these lengths exactly using the same tape, the tape length would need to be a common factor between 20 and 30. Further, if we are required to find the longest possible tape length to achieve this, the length of the tape would need to be the HCF of 20 and 30. Hence, in this case 10 would be the correct answer.

ILLUSTRATIONS SET 1.3: 1. Three bells chime at intervals of 6 min, 7 min and 16 min respectively. At a certain

time, they begin together. What length of time will elapse before they chime together again? (a) 1 hour 24 min (b) 1 hour 52 min (c) 3 hours 44 min.

(d) 4 hours 40 min (e) 5 hours 36 min Solution: L.C.M. of 6, 7 and 16 = 336

They will chime together again after 336 min. i.e. after 5 hours 36 min

option (e) 2. Three fans can complete 120, 100 and 80 rotations per minute respectively. The

name of the manufacturer company is printed once on each fan. When the fans are in stand by position (off), the printed names for each fan are in the east direction. How much time after switching on the fans, will all these printed names simultaneously be in the east direction again? (a) 2 sec (b) 5 sec (c) 6 sec (d) 3 sec (e) 4 sec

Solution: The first fan completes 120 rotations in 1 minute therefore,

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Time taken to complete 1 rotation = 120

60sec. =

2

1sec,

For 2nd fan, time taken to complete 1 rotation = 100

60sec. =

5

3sec,

For 3rd fan, time taken to complete 1 rotation = 80

60sec. =

4

3sec,

So, the required time will be L.C.M. of the time taken by the three fans to complete one rotation.

Required time = L.C.M of 4

3&

5

3,

2

1sec is:

31

3

4 5, 2, of H.C.F.

3 & 3, 1, of L.C.M. sec

option (d) 3. The lengths of the outer walls of an irregular pentagon building are 165 meters,

243 meters, 405 meters, 288 meters and 372 meters respectively. Find the greatest length of tape by which the five sides may be measured completely? (a) 165 meters (b) 45 meters (c) 9 meters (d) 6 meters (e) 3 meters

Solution: The length of the greatest tape will be equal to the H.C.F. of 165, 243, 405, 283 and 372 which is equal to 3

option (e) 4. A yellow light flashes 3 times per 2 minutes and a blue light flashes 5 times per 3

minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour? (a) 10 (b) 11 (c) 9 (d) 12 (e) 3

Solution: First light flashes after every 40 seconds and second light flashes after every 36 seconds. They will flash together after L.C.M. of 40 sec & 36 sec. = 360 sec = 6 minutes Hence, the number of times they flash together in an hour = 10

option (a) 5. There are 384 males and 432 females. The ratio of minor males to major adult

males is 9:7 and that of minor females to major adult females is 13:5. Find the least possible number of groups of equal size in which all the four categories of people can be divided equally. (a) 24 (b) 36 (c) 16 (d) 20 (e) 34

Solution: Males Females 384 432

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Minor Major Minor Major 9 : 7 13 : 5

216 168 312 120 Size of each group = H.C.F. of 216, 168, 312, 120 = 24

Hence, number of groups of equal size = 345137924

120

24

312

24

168

24

216

option (e) 6. 6. In which of the following cases, will the difference between the G.C.D. (H.C.F) &

L.C.M. of two distinct numbers (say x and y) be minimum: (a)When x and y are co-prime numbers (b)When x and y are perfect squares (c)When one out of x and y is the factor of other (d)When one out of x and y is prime number and another is a composite number (e)Cannot say

Solution: From observation of G.C.D. & L.C.M. of any two number x and y, we observe that, the required difference is minimum when one out of x and y is the factor of other (do it by yourself by taking a few examples)

option (c ) 7. 7. In which of the following cases, will the difference between the G.C.D. (H.C.F) &

L.C.M. of two distinct numbers (say x and y) be maximum: (a)When x and y are prime numbers (b)When x and y are co-prime numbers (c)When one out of x and y is the factor of other (d)When one out of x and y is a prime number and another is a composite number (e)When both are perfect squares

Solution: From various examples, we observe that, when both the numbers x and y are co-prime to each other, then the required difference is maximum

option (b) Note: You might do well to remember both these principles. 8. What will be the maximum value of the H.C.F. of any two distinct co-prime

numbers x and y? (a) 1 (b) minimum {x, y} (c) maximum {x. y}

(d)Average of x and y

2.

yxei (e) xy

Solution: For any two co-prime numbers greatest common factor = 1 and maximum value of H.C.F = 1

option (a) 9. What may be the maximum value of the H.C.F. of any two distinct numbers say x

and y? (a) 1 (b) minimum {x, y}

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(c) Maximum {x. y} (d) average of x and y

2.

yxei (e) xy

Solution: For any two distinct numbers x and y. the maximum value of H.C.F. may be equal to the lesser between x and y. i.e minimum {x, y}

option (b) 10. What will be the difference between the maximum possible HCF &the minimum

possible HCF value of any two co-prime numbers say x and y? (a) 0 (b) 1 (c) {x - y} (d) x + y (e) x/y

(where x > y)

Solution: For any two co-prime numbers. H.C.F = 1 always. There is no question of minimum or maximum.

1 is maximum as well as minimum) So, Required difference = 1 – 1 = 0

option (a) 1.4 Cyclicity: The last Digits of the powers of any number follow a cyclic pattern – i.e. they repeat after certain number of steps. Lets have a look : - Last digit of 21 – 2 Last digit of 22 - 4 Last digit of 23 – 8 Last digit of 24 - 6 Last digit of 25 – 2 So the Last Digit of 25 is the same as the last Digit of 21. Beyond this value it will again start repeating i.e unit‘s digits of 25, 26, 27& 28 will be the same as those of 21, 22, 23& 24. So there is a cyclicity of 4. You would do well to remember the cyclicity of the numbers given below:

When the Last Digit is

Cyclicity is

Unit digit cycle

0 1 Always 0 at any

power

1 1 Always 1 at any power



4 2 4 at 42n+1 6 at 42n

9 2 9 at 92n+1 1 at 92n

2 4 Same as 21 at 24n+1 Same as 22 at 24n+2

Same as 23 at 24n+3

Same as 24

at 24n


Same as 33 at 34n+3

Same as 34

at 34n


Same as 73 at 74n+3

Same as 74

at 74n

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Same as 83 at 84n+3

Same as 84

at 84n

ILLUSTRATIONS SET 1.4: 1. What is the sum of the single digit numbers which cannot be the unit (last) digit of

any perfect square? (a) 18 (b) 9 (c) 20 (d) 12 (e) Can‘t be determined Solution: The digits which cannot be the last digit of any perfect square are 2, 3, 7 and 8

The required sum = 2 + 3 + 7 + 8 = 20

option (c ) 2. What will be the unit‘s digit of

a.

nn

n

n

1

where n is any multiple of 10

(a) 5 (b) 0 (c) any even digit (d) any odd digit (e) either 0 or 5 Solution: For n=10 we will get: 1 + 2 + - - - - - + 10 = 55 (units‘ digit = 5) For n=20 we will get: 1 + 2 + - - - - + 10 + 11 + 12 + - - - - + 20 = 55 + 155 = 210 (units‘ digit = 0) For n=30 we will get: 1 + 2 + - - - - + 10 + 11 + 12 + - - - - + 20 + 21 + 22 + - - - + 30 = 465 (unit digit = 5) For n=40 we will get: 1 + 2 + - - - - + 10 + 11 + 12 + - - - - + 20 + 21 + 22 + - - - + 30 + 31 + - -- - + 40 = 820 (unit digit = 0) Thus we observe that when n is an odd multiple of 10, the units‘ digit is 5 (odd) & when n is an even multiple of 10, the units‘ digit is 0 (even)

So, the units‘ digit will be either 0 or 5

Option ((e)

b.

nn

n

n

1

where n is an odd multiple of 10.

(a) 5 (b) 0 (c) any even digit (d) any odd digit (e) either 0 or 5 Solution: From the above solution we can say that when n is an odd multiple of 10, the units digit of the above expression is 5

option (a)

c.

nn

n

n

1

where n is an even multiple of 10.

(a) 5 (b) 0 (c) any even digit (d) any odd digit (e) either 0 or 5

[Number System]


Solution: From the above solution we can say that when n is an even multiple of 10, the expression will have its unit digit as 0

option (b) 3. What will be the units digit of a number which is equal to the difference between

two numbers, where the first is an odd power of a number with unit digit ‗9‘ and the second is also an odd power of a number with units digit 4. (Assume that the first number is greater in value than the second). (a) 0 (b) 1 (c) 3 (d) 5 (e) 7

Solution: First number = (………….. 9)odd = …………. 9 (always) Second number = (………….. 4)odd = …………. 4 (always) So, difference = ……………… 5 (always) (9 – 4 = 5) (irrespective of the value of the odd power)

option (d) 4. What will be the units digit of the summation of the following series x1 + x2 + x3 + - - - - - x255 + x256 (Where x is a single digit number other than 1 & 6)

(a) 0 (b) 1 (c) 3 (d) 5 (e) 7 Solution: If we put any single digit in the place of x, then x1 + x2 + x3 + x4 = 0 Also, x5 + x6 + x7 + x8 = 0 and so on Sox1 + x2 + x3 + x4 + - - - - - - - - - - - - + x253 + x254 + x255 + x256 Sum = 0 sum = 0

The required unit digit will be zero (0) Option (a)

5. What will be the last two digits of the following expression 67 54 29 28 47 (a) 25 (b) 13 (c) 52 (d) 26 (e) 65

Solution: Last two digits will be equal to the remainder when 67 54 29 28 47 is divided by 100.

i.e = )(13

25

1618

25

2274417

100

47282954 67 remainder

.

So, the required remainder (or, last two digit) is equal to 13 4=52

option (c)

1.5 Number of factors of a given number :

Let N is a composite number such that N = ap. bq. cr where a, b and c are prime factors of N & p, q, r - - - - are positive integers. Then the number of factors of N is given by (p +1) (q + 1) (r + 1) - - - - -

Eg 36 = 32 22

Here 36 has (2 + 1) (2 + 1) i.e. 9 factors. Note that the answer got here by using the above formula also includes 1 & the given number N itself as factors. So if one wants to find out the number of factors that the given number has excluding 1 & itself, find out (p +1) (q + 1) (r + 1) and then subtract 2 from that figure.

[Number System]


Number of ways of expressing a given number as a product of two factors: -

Rule 1: - As a product of two different factors is 1 -......1)........ (r 1) (q 1) (p2

1 ways

{excluding N N}

Rule 2: - As a product of two factors (including N N) is 1 - - - - - 1) (q 1) (p2

1

ways.

ILLUSTRATIONS SET 1.5: 1. The number of factors of a number ‗N‘ is not divisible by 2. What can be said about

the number ‗N‘. (a) N is a prime number (b) N is a composite number (c) N is a composite number but not a perfect square (d) N is a perfect square (e) Cannot say anything

Solution: The number of factors of a number ‗N‘ is not divisible by 2

The number of factors of a number ‗N‘ is not even but odd and we should know that only perfect squares have an odd number of factors.

Hence, N is any perfect square

Option (d) 2. The number of factors of a number ‗N‘ is divisible by 2, but not equal to 2, what

can be said about the number ‗N‘. (a) N is a prime number (b) N is a composite number (c) N is a composite number but not a perfect square (d) N is a perfect square (e) Cannot say anything

Solution: From the property of numbers we know that all numbers except perfect squares have an even number of factors. Further, you should also realize that it is

only for prime numbers that the number of factors is equal to 2. So, the required answer will be any composite number except a perfect square.

option (c) 3. What is the difference between the number of even and odd factors of 1080?

(a) 16 (b) 32 (c) 14 (d) 24 (e) 8

Solution: 1080 = 23 3351 Standard form

Number of even factors = 3 (3 + 1) (1 + 1) = 24

Number of odd factors = 1 (3 + 1) (1 + 1) = 8 So, required difference = 24- 8 = 16

Option (a) Note: In order to find the number of even factors of a number we use the same procedure as that used for the number of factors of the number except for the minor change in the bracket corresponding to the prime number 2 we do not add 1. Hence,

[Number System]


in the above case when 1080 = 23 33 51 gives us (3+1)(3+1)(1+1) as the number of factors, the number of even factors is given by (3) (3+1)(1+1). [Note that in this case all we have done is excluded the count of 20 from the number]. Also, the number of odd factors then is got by replacing the bracket concerning the powers of 2 by the number one. Thus the number of odd factors is given by: (1)(3+1)(1+1). 4. What is the ratio of number of even factors to number of odd factors of 1800?

(a) 5 : 1 (b) 4 : 1 (c) 3 : 1 (d) 2 : 1 (e) 1 : 1

Solution: 1800 = 23 33 52

Number of even factors = (3)(3 + 1) (2 +1) = 36

Number of odd factors = 1 (3 + 1) (2 + 1) = 12 So, the required ratio =3/1

Option (c) 5. What is the difference between the sum of even factors and the sum of odd factors

of 1080? (a) 1092 (b) 78 (c) 1014 (d) 1170 (e) 14

Solution: 1080 = 23 33 51

Sum of even factors = (21 + 22 + 23) (30 + 31 + 32) (50 + 51)

= (2 + 4 + 8) (1 + 3 + 9) (1 + 5)

= 14 13 6 = 1092

Sum of odd factors = 20 (30 + 31 + 32) (50 + 51) = 1 13 6 = 78 So, Required difference = 1092 – 78 = 1014

Option (c) 6. What is the ratio of the sum of even factors to the sum of odd factors of 1800?

(a) 3:31 (b) 31 : 13 (c) 14 : 13 (d) 14 : 1 (e) 1: 13

Solution: 1800 = 23 32 52

Sum of even factors = (21 + 22 + 23) (30 + 31 + 32) (50 + 51 + 52)

= (2 + 4 + 8) (1 + 3 + 9) (1 + 5 + 25)

= 14 13 31= 5642

Sum of odd factors = 20 (30 + 31 + 32) (50 + 51 + 52) = 1 13 31 = 403

So, the required ratio =14:1 option (d) 7. What fraction of the total number of factors of 360 is even?

(a) 4/7th (b) 3/5th part (c) 1/2th part (d) 4/7th part (e) 3/4th part

Solution: 360 = 23 32 51

Number of even factors = (3) (2 + 1) (1 + 1) = 18

Number of odd factors = 1 (2 + 1) (1 + 1) = 6 So, required part = 18/24 = 3/4

Option (e) 9. What part of the total number of factors of 360 is divisible by 10?

(a) 8/11 (b) 3/5 (c) 3/11 (d) 5/8 (e) 3/8

[Number System]


Solution: 360 = 23 32 51

Total number of factors = (3 + 1) (2 + 1) (1 + 1) = 24 For the number of factors divisible by 10, we need to ensure that there is at least one 5 and one 2 in the expression: Hence, we exclude the count of 20 and 50 from our count. Hence the number of factors divisible by 10 would be: 3x (2+1) (1) = 9

So, the required part = 8

3

24

9

360 of factors of no. total

10by divisible are which 360 of factors of no.

Option (e) 10. What part of the total number of factors of 480 are perfect squares?

(a) 1/7 (b) 1/4 (c) 2/3 (d) 1/8 (e) 1/12

Solution: 480 = 25 31 51 Factors of 480 which are perfect squares will be; (i) 20 x30 x50 (ii) 22 x30 x50 (iii) 24 x 30 x50 Except these three numbers, there will not be any other perfect square out of the factors of 480 because any another combination of prime factors(s) with even powers is not possible.

Number of factors which are perfect square = 3

Also, total number of factors = (5 + 1) (1 + 1) (1 + 1) = 24

So, Required part = 8

1

24

3

option (d) 11. What is the ratio of factors of 720 which are perfect squares to the factors of

720 which are not perfect squares? (a) 1 : 1 (b) 1 : 3 (c) 1 : 5 (d) 1 : 4 (e) 1 : 9

Solution: 720 = 24 32 51 ;

Total number of factors = (4 + 1) (2 + 1) (1 + 1) = 30 For number of factor of 720 which are perfect squares: (i) 20 x30 x50 = 1 (by combination of 0th power to prime factors) (ii) 22 x30 x50 (iii) 24 x30 x50 (iv) 20 x 32 x50

(v) 22 32 x50 (vi) 24 32 x50

Number of factors of 720 which are perfect squares = 6 Now, for number of factors of 720 which are not perfect squares: 30 – 6 = 24

So, the required ratio = 4:124

6

option (d) 12. What is the ratio of the sum of the factors of 900 that are divisible by 20 to the

number of factors of 800 that are divisible by 10?

[Number System]


(a) 130 : 1 (b) 130 : 3 (c) 13 : 3 (d) 14 : 7 (e) 7: 14

Solution: 900 = 22 32 5 For sum of factors of 900 which are divisible by 20;

900 = (22 5) (32 5)

The required sum will be 20 times that of the sum of the factors of 32 5

= 20 (30 + 31 + 32) (50 + 51)

= 20 (1 + 3 + 9) (1 + 5)

= 20 13 6 = 1560 For number of factors of 900 which are divisible by 10;

900 = (2 5) (2 32 5)

The required number of factors will be equal to number of factors of 2 32 51

= (1 + 1) (2 + 1) (1 + 1)

= 2 3 2 = 12

So, the required ratio = 1:13012

1560

Option (a) 13. What is the ratio of the sum of factors of 840, which are perfect squares to the

sum of the rest of the factors? (a) 2875 (b) 2885 (c) 2880 (d) 2882 (e) 2878

Solution: 840 = 23 31 51 71 Factors which are perfect square:- (i) 1 (ii) 22 (only two) Their sum = 1 + 22 = 1 + 4 = 5 Now, sum of the rest of the factors = sum of total factors – sum of factors which are perfect squares

= (20 + 21 + 22 + 23) (30 + 31) (50 + 51) (70 + 71) - 5

= (1 + 2 + 4 + 8) (1 + 3) (1 + 5) (1 + 7) – 5

= 15 4 6 8 – 5 = 2880 – 5 = 2875

Option (a)

1.6 Remainders: Remainder is the least positive number which when added to the multiple of a divisor we get the dividend. Divisor ) Dividend ( Quotient . Remainder So, Dividend = Divisor × Quotient + Remainder Features: a) It follows BODMAS Rule. b) In case if a remainder is –ve, it should be added to the divisor to get the positive value of the remainder. [Remember by definition that the remainder always has to be non-negative. Thus for any divisor D it is a rule that the remainder will only be in the range 0<R<D]

[Number System]


The Remainder Theorem: Consider the following question: 17 × 23. Suppose you have to find the remainder of this expression when divided by 12. We can write this as: 17 × 23 = (12 + 5) × (12 + 11) Which when expanded gives us: 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11

When the expression above gets divided by 12, the remainder would only depend on the last term of the expression. Thus, 12 × 12 + 12 × 11 + 5 × 12 + 5 × 11 gives the same remainder as 5 × 11 12 12 Hence, the required remainder is 7. This is the remainder when 17 × 23 is divided by 12. Learning point: In order to find the remainder of 17 × 23 when divided by 12, you need to look at the individual remainders of 17 and 23 when divided by 12. The respective remainders (5 and 11) will give you the remainder of the original expression when divided by 12. Mathematically, this can be written as: The remainder of the expression [A × B × C + D × E]/M, will be the same as the remainder of the expression [AR× BR× CR + DR× ER]/M. Where ARis the remainder when A is divided by M, BRis the remainder when B is divided by M, CRis the remainder when C is divided by M DRis the remainder when D is divided by M and ERis the remainder when E is divided by M, We call this transformation as the remainder theorem transformation and denote it by

the sign ---R Thus, the remainder of 1421× 1423× 1425 when divided by 12 can be given as: (CAT 2002).

1421 × 1423 × 1425 ---R 5 × 7 × 9 = 35 × 9 ---R11 × 9 12 12 12 12

---R gives us a remainder of 3. In the above question, we have used a series of remainder theorem transformations

(denoted by ---R) and equality transformations to transform a difficult looking expression into a simple expression.

[Number System]


Try to solve the following questions on Remainder theorem: Find the remainder in each of the following cases: 1) 17 × 23 × 126 × 38 divided by 8 2) 243 × 245 × 247 × 249 × 251 divided by 12 3) 173 × 261 + 248 × 249 × 250. 13 15 4) 1021 × 2021 × 3021. 14 5) 37 × 43 × 51 + 137 × 143 × 151. 7 9

USING NEGATIVE REMAINDERS Consider the following question: Find the remainder when; 14 × 15 is divided by 8. The obvious approach in this case would be

14 × 15 ---R6 × 7 = 42 ---R 2 (Answer). 8 8 8 However there is another option by which you can solve the same question: When 14 is divided by 8, the remainder is normally seen as +6. However, there might be times when using the negative value of the remainder might give us more convenience. Which is why you should know the following process: Concept Note: Remainders by definition are always non–negative. Hence, even when we divide a number like –27 by 5 we say that the remainder is 3 (and not – 2). However, looking at the negative value of the remainder has its own advantages in Mathematics as it results in reducing calculations. Thus, when a number like 13 is divided by 8, the remainder being 5, the negative remainder is – 3. (Note: It is in this context that we mention numbers like 13, 21, 29 etc as 8n + 5 or 8n – 3 numbers.)

Thus 14 × 15 will give us –2 × –1 R 2. 8 8 Consider the advantage this process will give you in the following question:

51 × 52 ---R– 2 × – 1 ---R 2. 53 53 (The alternative will involve long calculations. Hence, the principle is that you should use negative remainders wherever you can. They can make life much simpler!!!) What if the answer comes out negative?

For instance, 62 × 63 × 64 R–4 × –3 × –2 R – 24 . 66 66 66 But, we know that a remainder of –24, equals a remainder of 42 when divided by 66. Hence, the answer is 42. Of course nothing stops you from using positive and negative remainders at the same time in order to solve the same question:

Thus 17 × 19 R (–1) × (1) R –1 which means the same thing as R 8. 9 9

[Number System]


Note: Divisor + Negative Remainder = Positive Remainder. Dealing with large powers: There are two tools which are effective in order to deal with large powers –: (a) If you can express the expression in the form (ax + 1)n, the remainder will a become 1 directly. In such a case, no matter how large the value of the power n is, the remainder is 1.

For instance, (37/9)12635 ---R (1/9)12635 ---R 1. In such a case the value of the power does not matter. (b) (ax – 1)n. In such a case using –1 as the remainder it will be evident that the aremainder will be +1 if n is even and it will be –1 (Hence a – 1) when n is odd.

e.g. : 31127/8 ---R (–1)127/8 ---R (–1) ---R 7 ANOTHER IMPORTANT POINT: Suppose you were asked to find the remainder of 14 divided by 4. It is clearly visible that the answer should be 2. But consider the following process:

14/4 = 7/2 ---R 1 (The answer has changed!!) What has happened? We have transformed 14/4 into 7/2 by dividing the numerator and the denominator by 2. The result is that the original remainder 2 is also divided by 2 giving us 1 as the remainder. In order to take care of this problem, we need to reverse the effect of the division of the remainder by 2. This is done by multiplying the final remainder by 2 to get the correct answer. Note: In any question on remainder theorem, you should try to cancel out parts of the numerator and denominator as much as you can, since it directly reduces the calculations required.

AN APPLICATION OF REMAINDER THEOREM: Finding the last two digits of an expression: Suppose you had to find the last 2 digits of the expression: 22 × 31 × 44 × 27 × 37 × 43 The remainder the above expression will give when it is divided by 100 is the answer to the above question. Hence, to answer the question above, find the remainder of the expression when it is divided by 100. Solution : 22 × 31 × 44 × 27 × 37 × 43 100 = 22 × 31 × 11 × 27 × 37 × 43 (on dividing by 4) 25

---R 22 × 6 × 11 × 2 × 12 × 18 = 132× 22 × 216 ---R7× 22 × 16

[Number System]


25 25 25

= 154 × 16 ---R 4 × 16 ---R 14 25 25 Thus, the remainder being 14, (after division by 4) the actual remainder should be 56. [Don‘t forget to multiply by 4 !!] Hence, the last 2 digits of the answer will be 56. Note: Similarly finding the last three digits of an expression means finding the remainder when the expression is divided by 1000. Some important algebraic formulae used in Number system: - (a + b)2 = a2 + 2ab + b2 (a – b) 2 = a2 - 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (a + b)3 = a3 + b3 + 3ab (a + b)

(a – b) 3 = a3 – b3 - 3ab (a + b)

a2 – b2 = (a + b) (a – b) a3 + b3 = (a + b) (a2 – ab + b2) a3 - b3 = (a - b) (a2 + ab + b2) a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca) if, a + b + c = 0 then a3 + b3 + c3 – 3abc = 0

ILLUSTRATIONS SET 1.6 1. What will be the remainder when –34 is divided by 5?

(a) 1 (b) 4 (c) 1 (d) –4

Solution: –34 = 5 (-6) + (-4) Remainder = -4, but it is wrong because remainder cannot be negative.

So, -34 = 5 (-7) + 1

option (a) Alternately, when you see a remainder of -4 when the number is divided by 5, the

required remainder will be equal to 5-4=1. 2. What will be the remainder when –24.8 is divided by 6?

(a) 0.8 (b) 5.2 (c) –0.8 (d) –5.2

Solution: –24.8 = 6 (-4) + (-0.8)

So, negative remainder, so not correct –24.8 = 6 (-5) + 5.2 Positive value of remainder, so correct

option (b) 3. If p is divided by q, then the maximum possible difference between the minimum

possible and maximum possible remainder can be? (a) p –q (b) p –1 (c) q – 1 (d) None of these

[Number System]


Solution: ;q

p minimum possible remainder = 0 (when q exactly divides P)

Maximum possible remainder = q – 1 So, required maximum possible difference = (q – 1) – 0 = (q – 1)

option (c ) 4. Find the remainder when 2256 is divided by 17.

(a) 0 (b) 1 (c) 3 (d) 5 (e) 16

Solution:. 1 R17

6416

17

64)42(

17

2562

1a

an

; R = 1

when n even

option (b) 5. Find the difference between the remainders when 784 is divided by 342 & 344.

(a) 0 (b) 1 (c) 3 (d) 5 (e) 16

Solution: 1 R342

28343

342

28)37(

342

847

also, 1 R344

28343

344

28)37(

344

847

The required difference between the remainders = 1 – 1 = 0

Option (a)

6. What will be the value of x for 0 R ;9

)3410()117100(

x

(a) 3 (b) 6 (c) 9 (d) 4 (e) 8

Solution: 9

)3410()117100( x

10017 – 1= nines 16

99............9999

zeroes 17

100...........1000

divisible by 9 R = 0

Since the first part of the expression is giving a remainder of 0, the second part should also give 0 as a remainder if the entire remainder of the expression has to be 0. Hence, we now evaluate the second part of the numerator.

1034 + x = xxzeroes 33

00............1000

zeroes 34

00...........1000

[Number System]


with x at the right most place. In order for this number to be divisible by 9, the sum of digits should be divisible by 9.

1 + 0 + 0 - - - - - + 0 + x should be divisible by 9.

1 + x should be divisible by 9 x = 8

option (e)

1.7 Highest power dividing a factorial Factorial: - It is defined for any +ve integer (or natural number). For any natural number N,

N! = 1 2 3 - - - - - (n – 1) n. Where! or L in the symbol for factorial. Always remember 0! = 1 & 1! = 1 Problems related to factorials appear regularly in all aptitude exams. The highest power of a prime number p, which divides n! exactly is given by

32 p

n

p

n

p

nwhere [ x ] denotes the greatest integer less than or equal to

x.

ILLUSTRATIONS SET 1.7:- 1) Find the largest power of 5 contained in 200! Solution: Divide 200 successively by 5 (since 5 is a prime number). The required answer will be got through: [200/5] 40 + (40/5 =8 ) + (8/5 =1) + (1/5= 0) = 40+8+1+0=49. Since 1 cannot be divided by 5 anymore, we stop here and add all the quotients 40 + 8 + 1 to give 49. Hence 49 is the largest power of 5 that can divide 200! without leaving any remainder.

2. How many zeroes are there in 200!

Solution: Zero is got through the multiplication of 5 and 2. (Since 10 = 51 21. ) Hence, in order to find the number of zeroes in any expression we need to find out the number of 5‘s and the number of 2‘s in that expression. Then the number of zeroes will be equal to the smaller value amongst the number of 5‘s and 2‘s. Also, we should know that in any factorial the number of 5‘s will always be less than the number of 2‘s. Hence, in order to find the number of zeroes in 200! we will just need to find out the number of 5‘s in the value of 200!. Hence, find the maximum power of 5 in 200! to get the required answer. So, No of zeroes in 200! is 49. (from the solution to the previous question)

1. The number of zeroes in the expansion of 229! are (a) 45 (b) 55 (c) 54 (d) 56 (e) 65

[Number System]


Solution: To find out number of zeroes, we divide 229 first by 5 and then by increasing power to 5 till 229 is not less than the denominator, which is 5 with power on it. When finding results of various divisions we consider greater integer of the result ie. only integral part of results. After that we add all the results and hence get the required result i.e number of zeroes, look at it step wise.

Step (i)

45

5

229 where [ ] denotes greatest integer.

Maximum 925

229

25

229

again 1125

229

35

229

again

625

229

45

229 (No need to do it because D > N is 625 > 229)

Step (ii) Required number of zeroes = 45 + 9 + 1 = 55 Option (b) Alternative method for calculation: -

455

229

now, 95

45

now 15

9

The process to solve this question will be exactly the same as that of the above and

hence the same answer. 2. The maximum value of n for which 229! is exactly divisible by 5n is.

(a) 65 (b) 45 (c) 54 (d) 55 (e) 56 Solution: To find out the required maximum value of n we divide 229 successively by 5 with its increasing power and lastly add all the values found. Use the same procedure as in the previous question.

option (d) 3. The maximum value of n for which 982! is exactly divisible by 12n is

(a) 491 (b) 325 (c) 975 (d) 485 (e) 729

[Number System]


Solution: 12 is not a prime number. So, first we‘ve to write down 12 into standard

form i.e. 12 = 22 3the required answer will be equal to the minimum of the number of 22 and the number of 3‘s that are contained in 982! Now, the number of 2‘s contained in 982! are:

=

42

982

32

982

22

982

2

982 and so on

=

512

982

256

982

128

982

64

982

32

982

16

982

8

982

4

982

2

982

= 491 + 245 + 122 + 61 + 30 + 15 + 7 + 3 + 1 = 975 This means that there are [975//2] = 487 22s contained in 982! Similarly the maximum power of 3 present in 982! Are:

729

982

243

982

81

982

27

982

9

982

3

982

= 324 + 108 + 36 + 12 + 4 + 1 = 485 So, Required maximum value of n

= minimum

485,

2

975where [ ] denotes greater integer.

i.e. Minimum (487, 485) = 485

option (d) 4. The maximum value of n for which 982! Is exactly divisible by 36n, is

(a) 491 (b) 161 (c) 324 (d) 815 (e) 242

Solution: 36 = 22 32 So, required maximum power (value of n) will be equal to

MinimumNumber of 2's contained in 982!

2,number of 3's contained in 982!

3

ìíî

üýþ

Now, number of 2‘s contained in 982! can be got by:

4912

982

+ 245

2

491

+ 122

2

245

+ 61

2

122

+ 30

2

61

+ 15

2

30

+

72

15

+ 3

2

7

+ 1

2

3

= 975

Hence, number of 22s =487 Number of 3s in 982!

3243

982

+ 108

3

324

+ 36

3

108

+ 12

3

36

+ 4

3

12

+ 1

3

4

[Number System]


= 485 Hence, number of 32s are: [485/2]= 242.

So, the required answer = minimum

2

485,

2

975

= minimum (487, 242)

option (e) 5. The maximum value of n for which 1982! is exactly divisible by 75n is

a) 660 b) 396 c) 246 d) 493 e) 327

Solution:

75 Not a prime number

So, the standard form; 75 = 3 52 So, the required maximum value of power ‗n‘ will be equal to minimum

2

1982!in contained 5s ofnumber ,1982!in contained 3s ofnumber

Now, maximum value of power ‗n‘ in case of 3 equals and maximum value of power ‗n‘ in case of 5 equals

6603

1982

396

5

1982

now, 2203

660

now, 79

5

396

Now, 733

220

Now, 15

5

79

now, 243

73

now, 3

5

15

now, 83

24

sum = 493

now, 23

8

------------------- Sum = 327 ------------------- So, the required value of n equals

Minimum

2

493,327 where [ ] denotes greatest integer value.

i.e. minimum (327, 246) = 246 option (c)

[Number System]


6. How many consecutive digits from right side in the expansion of 179! will have to be removed so that the remaining part of the expanded number is not divisible by 10? (a) 35 (b) 42 (c) 36 (d) 34 (e) 43

Solution: If we remove all the consecutive zeroes from the right end, the unit digit will not be zero and hence the remaining part of the expanded number will not be divisible by 10. So, we need to find the number of consecutive zeroes from the right and for the number of consecutive zeroes from the right end in the expansion of 179! we need to find out the number of 5s in 179!.

355

179

now, 7

5

35

now, 1

5

7

= 35 + 7 + 1 = 43

option (e) Successive Division: A successive division process can continue upto any number of steps i.e. the quotient in the first division is taken and divided in the second division, the quotient in the second division is taken as the dividend in the third division and so on. Illustration: A number when divided successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number? Solution: To solve this problem we start from the last step. Instead of taking the general form of the number that will satisfy the above conditions. We write down the divisors one after the other and their respective remainders below them. 3 4 7 2 1 4

Then starting from the last remainder we go diagonally left upwards to first row multiplying and then directly below adding the figure already obtained. We continue this process till we reach the figure in the third row on the extreme left. Arrows above indicate the sequence of numbers taken. Solution: (4x4+1)x3 +2 = 53. Hence, the number will be = 53

[Number System]


Exercise on LCM and HCF 1. How many 3 digit numbers are there which when divided by 7, 5, 3 give remainder

5 in each case? (a)7 (b) 8 (c) 9 (d) 10

2. 2. The HCF of two numbers is 10 and their LCM is 45. Find the product of the

numbers. (a) 4500 (b) 2250 (c) 3600 (d) Data Inconsistent

3. If the HCF of two numbers is 12 and their sum is 144, find the maximum value of

their product. (a)5184 (b) 5040 (c) 4608 (d) 3888

4. The LCM of two numbers is 46 and the HCF of the numbers is 4. If one of the

numbers is 23, what is the other number? (a)2 (b) 4 (c) 8 (d) 16

5. Find the largest 4-digit number which when divided by 5 and 7 leave remainders of

3 and 5 respectively. (a)9977 (b) 9973 (c) 9971 (d) 9969

6. Find the number of 3 digit-numbers which when divided by 11 leave a remainder

as 8 and when divided by 7 leave a remainder as 4. (a)11 (b) 12 (c) 13 (d) 14

7. The LCM and HCF of two numbers are 880 and 16 respectively. The sum of the

numbers is 256. The difference between the numbers is____ (a) 80 (b) 96 (c) 112 (d) None of these

8. The LCM of A, B, C, D = A×B×C×D. Thse HCF of B and D =____

(a)1 (b) 2 (c) B (d) Cannot be determined 9. The LCM of the numbers 5/9, 2/3, 7/27 is___

(a)1/27 (b) 4/5 (c) 3/70 (d) 70/3 10. The HCF of the numbers 5/8, 3/16, 9/24 is__?

(a)1/48 (b) 1/5 (c) 1/6 (d) 1/7 11. Find the largest number which when divides 708 and 482, the respective

remainders left are 8 and 2? (a)20 (b) 30 (c) 40 (d) 50

12. A bell rings every 12 minutes. A second bell rings every 16 minutes. A third bell

rings every 28 minutes. If all the three bells ring at the same time at 5PM in the morning, at what other time will they all ring together? (a)10 :56 PM (b) 10 :36 PM (c) 10 :46 PM (d) 11: 00 PM

[Number System]


13. Which of the following is the smallest number which when decreased by 8 is divisible by 25, 45 and 60? (a)928 (b) 918 (c) 908 (d) There is no such number

14. In an attempt to find the HCF of two numbers by division method, the quotients

obtained were 19, 1, 2 and 5 in that order. The final divisor was 3. Find the two numbers. (a)48 and 678 (b) 68 and 495 (c)48 and 945 (d) 68 and 945

15. Find the largest number which when divides 576, 876 and 1206, the

remainders left are same. (a)20 (b) 30 (c) 40 (d) 50

16. What is the LCM of the numbers 12x3y2z, 15x2y3z2, 18x3y2z4

(a)360x6y6z4 (b) 360x6y2z4 (c)180x6y6z4 (d)180x3y3z4 17. What would be the HCF in the previous question?

(a)3xyz (b) 12xyz (c) 3x2y2z (d) 12x2y2z 18. What is the smallest number that must be subtracted from 9999 so that the

result is divisible by 20, 25, 40 and 70 is? (a)99 (b) 199 (c) 299 (d) 399

19. Three numbers are in the ratio 3: 4: 5. Their LCM is 3600. Find the HCF of the

numbers. (a)20 (b) 40 (c) 60 (d) 80

20. The product of two numbers is 4050 and their HCF is 15. Find the number of

such pairs. (a)0 (b) 1 (c) 2 (d) 3

21. The least number which when divided by 3, 4, 5, 6 and 7 leaves a remainder 2,

but when divided by 9 leaves remainder as 8 is (a)422 (b) 602 (c) 503 (d) 701

22. What is the LCM of 1.08, 0.96 and 0.72.

(a)8.64 (b) 4.32 (c) 12.96 (d) 17.28 23. What is the least number which when doubled is exactly divisible by 12, 13, 15

and 16? (a)1560 (b) 3120 (c) 780 (d) 2340

24. Three numbers which are co-prime are taken. The product of first two numbers is

143 and the product of last two numbers is 273. What is the sum of the numbers? (a)45 (b) 49 (c) 53 (d) 59

[Number System]


25. Find the greatest number that will divide 131, 158 and 183 so as to leave the same remainder in each case. (a)13 (b) 26 (c) 36 (d) 28

26. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13

and 14. The larger of the two numbers is: (a)127 (b) 256 (c) 128 (d) 322

27. Find the smallest number that leaves a remainder of 3 on division by 4, 4 on

division by 5, 5 on division by 6, 6 on division by 7 and 7 on division by 8. (a)839 (b)1679 (c) 2519 (d) 3079

28. A sum of two numbers x, y =1050. What is the maximum value of the HCF

between x and y if x and y are distinct? (a)150 (b) 250 (c) 350 (d)450

29. The sum of two numbers is 528 and their HCF is 33. The number of pairs of such

numbers satisfying the above conclusion is? (a)2 (b) 3 (c) 4 (d) 5

30. The traffic lights at three different road crossings change every 48 seconds, 72

seconds and 108 seconds respectively. If they all change simultaneously at 8:20:00 hours, then they will again change simultaneously at (a)8:27:12 hours (b) 8:27:24 hours (c) 8:27:36 hours (d) 8:27:48 hours

Answer Key

1 c 2 d 3 b 4 c 5 b 6 b 7 b 8 a 9 d 10 a

11 a 12 b 13 c 14 c 15 b 16 d 17 c 18 b 19 c 20 c

21 a 22 a 23 b 24 a 25 b 26 d 27 a 28 d 29 c 30 a

Solutions 1. LCM of 7, 5 and 3 = 105.

All the numbers of the form (105n + 5) will satisfy the conditions. There will be 9 such three digit numbers.

2. LCM is always a multiple of HCF. So, the data is inconsistent. 3. The numbers possible are (84, 60) and (12, 132). The product is maximum in

the first case and is equal to 5040. 4. Let the unknown number be a.

We know that HCF × LCM = Number 1 × Number. 46 × 4 = 23 × a

[Number System]


So, a = 8 5. LCM of 5 and 7 = 35. All the numbers of the form 35k - 2. The largest such

number of four digits = 9973. 6. LCM of 7 and 11 is 77.

The number leaves remainder 8 when divided by 11. So, it is of the form 11a + 8

The number leaves remainder 2 when divided by 10. So, it is of the form 7b + 4 11a + 8 = 7b + 4 a = (7b-4)/11 The smallest integral values of a and b satisfying the equations are b = 10, a = 6

Putting a = 6 in 11a+8, we get 74 The numbers of the form 77 n + 74 will satisfy the conditions. There will be 12 such 3 digit numbers, from n = 1 to n = 12.

7. Let the numbers be 16a and 16b, where a and b are relatively prime numbers.

So, 16(a × b) = 880 ab = 55 …………..(i)

Also, 16a + 16b = 256 a + b = 16………. (ii)

Solve (i) and (ii) to get a=11 b=5 or a=5, b=11 The numbers are 80 and 176 Difference = 96

8. As the LCM of A, B, C, D = A×B×C×D, the numbers do not have any common

factor but 1. So, the HCF of B and D = 1 9. LCM of fractions = LCM of numerators/ HCF of denominators

LCM of 5, 2 and 7 is 70. HCF of denominators =3 So, the answer is 70/3.

10. HCF of fractions = HCF of numerators/ LCM of denominators

HCF of 5, 3 and 9 = 1

LCM of 8, 16 and 24 = 48 So, the answer = 1/48

11. HCF (708-8, 482-2) = 20

So, the required answer is 20. 12. The LCM of 12, 16 and 28 = 336 minutes.

So, the bell rings after 336 minutes = 5 hours 36 minutes. 5 + 5: 36 = 10: 36 PM

13. LCM of 25, 45 and 60 = 900

We need to add 8 to this. So, the smallest such number is 900 + 8 = 908 14. Method 1:

The final remainder will be zero as the numbers have a HCF.

[Number System]


The final dividend will be the final divisor × final quotient + final remainder = 3×5 + 0 =15 Similarly, the last but one previous dividend will be = 15 × 2 + 3 = 33 The last but two previous dividend will be = 33 × 1 + 15 = 48 The first dividend will be = 48 × 19 + 33 = 945 The numbers are 48 and 945. Method 2: Solve using options. Only 48 and 945 give the respective values.

15. Take the differences between any two pairs out of the given numbers.

1206 – 876 = 330

1206 – 576 = 630 The required number is the HCF of the two differences. HCF of 330 and 630 = 30

16. LCM of (12,15,18) = 180

LCM of 12x3y2z, 15x2y3z2, 18x3y2z4 = 180x3y3z4 17. HCF of (12,15,18) = 3

LCM of 12x3y2z, 15x2y3z2, 18x3y2z4 = 3x2y2z 18. Greatest number of 4 digits is 9999.

LCM of 20, 25, 40 and 70 is 1400. Remainder when 9999 is divided by 1400 is 199. So, 199 has to be subtracted to make it divisible by all the four numbers.

19. Let the numbers be 3a, 4a, 5a.

LCM is 60a 60a = 3600 So, a = 60 Numbers are 180, 240 and 300. Their HCF is 60.

20. Let the numbers be 15a and 15b.

15a × 15b = 4050 a × b = 18 Now, the co-prime with product 12 are (2, 9) and (1, 18) The numbers are (15×2, 15×9) and (15×1, 15×18) = (30, 135) and (15, 270). There are two such pairs.

21. The LCM of the numbers is 420.

Required number is of the form 420k + 2. The lowest such number is 422. 22. The LCM of 108, 96 and 72 is 864. So, the LCM of the required numbers is

8.64. 23. The LCM of 12, 13, 15 and 16 is 3120. So, half of the number is 1560. 24. The HCF of 143 and 273 is 13. This is the middle number.

Smallest number = 143/13 = 11 Largest Number = 273/13 = 21

[Number System]


Sum of numbers = 45 25. Required number = H.C.F. of (183 - 131), (183 - 157) and (157 - 131)

= H.C.F. of 52, 26 and 26 = 26 26. Clearly, the numbers are (23 × 13) and (23 × 14)

So, the larger number = 23 × 14 = 322 27. LCM (4, 5, 6, 7, 8) - 1 = 839 28. If we take x=350, y=700 then the HCF is 350 which is maximum

29. Required numbers are 33x and 33y. Then,

33x + 33y = 528 or x + y = 16 Now, two prime numbers with sum 16 are: (1, 15) (3, 13) (5, 11) (7, 9) Numbers with sum 528 and HCF 33 are (33 × 1, 33 × 15) (33 × 3, 33 × 13) (33 × 5, 33 × 11) (33 × 7, 33 × 9) There are 4 such pairs

30. Interval of change = LCM of (48, 72, 108) = 432 seconds. The lights will change

simultaneously after every 432 seconds = 7 minutes and 12 seconds So, they will change at 8: 27: 12 hrs.

[Number System]


Exercise on Properties of Numbers 1. How many 4-digit perfect square numbers are there whose last two digits are

identical? (a)7 (b)8 (c)9 (d)10

2. Rahim started adding all the natural numbers from 1 to n. The sum that he

calculated equals 200. However, while adding the numbers he missed one number. What is the missing number? (a)9 (b)10 (c)11 (d)12

3. In the previous question, had Rahim added a number twice to get the sum 200,

what could be the number? (a)7 (b)8 (c)9 (d)10

4. Which of the following is definitely not a perfect square?

(a)ab56 (b)aa44 (c)aa86 (d)ab25 5. If the three digit number A9B is a perfect square, then (A+B) =_____

(a)6 (b)7 (c)8 (d)Can‘t Say 6. The difference between a two digit number and sum of its digits is not necessarily a

multiple of (a)3 (b)6 (c)9 (d)Cannot say

7. If a and b are both prime numbers greater than 10, which of the following CANNOT

be true? a. ab is an even number. b. The difference between a and b equals 117. c. The sum of a and b is even.

(a) I only (b)I and II only (c)I and III only (d)II and III only 8. Rahim writes a number of the form abcabcabc. Which of the following is definitely

true about the number? (a)It is divisible by 9 (b)It is divisible by 11 (c)It is divisible by 7 (d)It is divisible by 5

9. The product of three consecutive natural numbers, the first of which is an even

number, is always divisible by? (a)6 (b)12 (c)24 (d)All of these

10. There are two numbers a and b such that their sum is 40. What is the maximum

value of the product of a and b? (a)394 (b)399 (c)400 (d)410

11. The number 6666666…….12 times is divisible by?

(a)13 (b)14 (c)15 (d)16

[Number System]


12. A three digit number abc is such that abc = a! + b! + c! How many such numbers are possible? (a)0 (b)1 (c)2 (d)3

13. A perfect square number has n factors. What can be definitely true about the

value of n? (a)n is a prime number (b)n is an odd number (c)n is an even number (d)We cannot say anything about n

14. Divya wrote all the numbers from 700 to 800. Then she started counting the

number of seven‘s that has been used while writing all these numbers. What is the

number that she got? (a)119 (b)120 (c)121 (d)122

15. If each of the three nonzero numbers a, b and c is divisible by 4, then abc must

not be divisible by which of the following numbers? (a)4 (b)16 (c)64 (d)128

16. If x and y are any natural numbers, then which of the following is an odd

number? (a)xy+ yx+ (x-y) (xy + x) (b)xy(x+y) (xy + x) (c)xy+ yx+ (x-y) (xy + x) (d)None of these

17. Which whole numbers, greater than one, can divide all the nine three digit

numbers 111, 222, 333, 444, 555, 666, 777, 888and 999? (a)3 and 37 (b)3 and 111 (c)3, 37 and 111 (d)3, 9 and 111

18. If A and B are twin primes with B>17, then which of the following numbers would

always divide A + B? Twin prime numbers are the prime numbers which differ by 2, example 11 and 13. (a) 6 (b)12 (c)18 (d)24

19. A number 15B is divisible by 6. Which of these will be true about the positive

number B? (a)B will be even (b)B will be odd (c)B will be divisible by 6 (d)Both (a) and (c).

20. Anjali had to multiply two positive integers. Instead of taking 27 as one of the

multipliers, she incorrectly took 72. As a result, the product went up by 360. What is the new product?? (a)216 (b)360 (c)576 (d)720

21. Let x, y and z be distinct integers. x and y are odd and positive and z is even and

positive. Which one of the following cannot be true? (a)(x-z)2y is even (b)(x-z)y2 is odd (c)(x-z)y is odd (d)(x-y)2z is even

[Number System]


22. If two fractions, each of which has a value between 0 and 1, are multiplied together, the product will be ___? (a)always greater than either of the original fractions (b)always less than either of the original fractions (c)sometimes greater and sometimes less than either of the original fractions (d)remains the same.

23. If x, y, z are chosen from the three numbers, -3, 1/2 and 2, what is the largest

possible value of the expression (x/y) z2? (a)16 (b)24 (c)36 (d)44

24. There is a perfect square number abcd. Which of the following cannot be the value of d? (a)2 (b)3 (c)7 (d)All of these

25. There is a number xyz. We get a new number p such that p = (xyz)5. What will be

the units digit of p – xyz. (a)0 (b)1 (c)2 (d)Depends on z

26. The product of two numbers a and b is 1600. What is the minimum value of a+

b? (a)76 (b)80 (c)84 (d)88

27. If the three digit number A6B is a perfect square, then A is_____

(a)Even (b)Odd (c)Prime number (d)Can‘t say 28. Consider the sum of two numbers x and y whose sum ends in 9. Which of the

following will necessarily end in 9? (a)x6 + y6 (b)x7 + y7 (c)x13 + y13 (d)x18 + y18

29. A frog is located at vertex A of an octagon ABCDEFGH. It can jump to any of the

adjacent vertices. If it makes (2n+1) jumps to reach the vertex E, then how many ways are possible (a)0 (b)1 (c)2 (d)3

30. There are 11 coins placed on a table. 6 are showing Heads and 5 are showing

Tails. Rajiv flips the coins 11 times in no particular order. Then he places his palm over one coins so as to hide it. If 5 coins are showing Heads and 5 are showing Tails, then which of the following is true? (a)The hidden coin is definitely showing Heads (b)The hidden coin is definitely showing Tails (c)The hidden coin can show either Heads or Tails (d)Cannot be determined

31. There are natural numbers from 1 to 20 written on a board. Ramesh erases any

two numbers a and b and replaces the numbers by (a+b-1). The operation is repeated until only one number remains. What is the number that remains in the board? (a)161 (b)191 (c)201 (d)210

[Number System]


Answer Key

1 c 2 b 3 d 4 c 5 b

6 b 7 c 8 c 9 d 10 c

11 a 12 b 13 b 14 b 15 d

16 a 17 c 18 b 19 a 20 c

21 a 22 b 23 c 24 d 25 a

26 b 27 b 28 a 29 a 30 b

31 b

[Number System]


Exercise On Divisibility, Sum Of Divisors, Number Of Divisors 1. Find the number of divisors of 1728.

(a)22 (b) 28 (c) 30 (d) None of these 2. Find the sum of divisors of 544.

(a)1134 (b) 1124 (c) 589 (d) Data Inconsistent 3. What digit should be in place of c in 38c to make it divisible by 4?

(a)2 (b) 4 (c) 6 (d) 9 4. Find the number of even divisors of 7200?

(a)40 (b) 42 (c) 43 (d) 45 5. Find the number of numbers from 1 to 100(both included) which are not divisible

by any one of 2, 3 and 5. (a)26 (b) 27 (c) 28 (d) 29

6. Out of 99 consecutive whole numbers a, a+1, a+2,…….,a+98, how many numbers

are divisible by 99 : (a)0 (b) 1 (c) 2 (d) 9

7. A number m1n082 is divisible by 11 where m, n are single digit positive integers

then the value of m+ n =? (a) 4 (b) 5 (c) 6 (d) None of these

8. Find the sum of all even divisors of 4096.

(a)8192 (b) 6144 (c) 8190 (d) 6142

9. What number should be subtracted from if it is to be perfectly divisible by x + 3?. (a)41 (b) 42 (c) 43 (d) 44

10. Find the sum of all odd factors of 1065. (a)1228 (b) 1428 (c) 1628 (d) 1728

11. If a1, a2, a3 ,………… a99 are 99 consecutive integers then

a1×a2×a3×a4×………×a99 is divisible by? (a)98 (b) 98! (c) 99 (d) 99!

12. Which of the following number has least number of divisors?

(a)99 (b) 101 (c) 176 (d) 182

13. If each of the four nonzero numbers a, b, c and d is divisible by 3, then must be divisible by which one of the following the numbers? (a) 27 (b) 8 (c) 81 (d) 121

14. What is the value of M×N if M39048458N is divisible by 8 and 11, where M and N

are single digit integers?

[Number System]


(a)24 (b) 18 (c) 54 (d) 36 15. Which digits should come in place of @ and # if the number 51574@# is divisible

byboth 8 and 5? (a)4, 5 (b) 5, 5 (c) 6, 5 (d) 4, 0

16. Find the number of numbers from 1 to 110 which are divisible by 3.

(a)54 (b) 44 (c) 34 (d) 36 17. Find the number of numbers from 1 to 200 which are not divisible by 5.

(a)160 (b) 24 (c) 40 (d) None of these

18. Find the number of numbers from 1 to 100 which are divisible by 2, 3?

(a)66 (b) 17 (c) 67 (d) 33 19. In a 5 digit number sum of 1st four digits are equal to sum of all five digits then

the number not necessarily divisible by : (a)2 (b) 4 (c) 5 (d) 10

20. 987654321x- x123456789 is always divisible by

(a)8 (b) 9 (c) None of above (d) Cannot be determined 21. 47n312 is divisible by 7 where n is a single digit natural number and n>2 then

possible value of n =? (a)2 (b) 4 (c) 3 (d) 5

22. 43a5b is divisible by 36 then how many numbers such numbers are possible?

(a)3 (b) 6 (c) 0 (d) None of these.

23. It is given that is exactly divisible by a certain number. Which one of the following is also divisible by the same number?

(a) (b) (c) (d) 24. If x and y are the two digits of the number 543xy such that this number is

divisible by 80, then x + y=? (a) 2 (b) 3 (c) 4 (d) 5

25. The product of 3 consecutive even numbers is always divisible by:

(a) 41 (b) 32 (c) 23 (d) 48

Answer Key

1 b 2 a 3 b 4 d 5 a

6 b 7 c 8 c 9 b 10 d

11 d 12 b 13 c 14 a 15 d

16 d 17 a 18 c 19 b 20 b

21 c 22 a 23 c 24 a 25 d

[Number System]


Exercise on Number of zeroes 1. Find the number of zeroes at the end of 58!

(a)10 (b) 11 (c) 12 (d) 13 2. What power of 7 will divide 80! exactly?

(a)11 (b) 12 (c) 13 (d) 14 3. What power of 35 will divide 62! exactly?

(a)9 (b) 10 (c) 11 (d) 12 4. Which of the following cannot be the number of zeroes at the end of any factorial?

(a)26 (c) 27 (c) 28 (d) 29 5. What will be the number of zeroes at the end of (36!)4!

(a)96 (b) 168 (c) 192 (d) None of these 6. Find the number of zeroes at the end in the expression 125 × 64 + 25 × 8 + 35 ×

12 (a)0 (b) 1 (c) 2 (d) 3

7. Find the maximum value of n for which the expression

is an integer.

(a)8 (b) 12 (c) 23 (d) 32 8. Kashish took a number N such that N! has 31 zeroes at the end. Also, the

number N is a prime number. What is the sum of digits of the number N? (a)9 (b) 10 (c) 11 (d) Data Inadequate

9. If 490! = 7p × q, where q is not a multiple of 7, then find the value of p.

(a)79 (b) 80 (c) 81 (d) 82 10. If N = (4!)4!× (6!)6!× (8!)8!× (10!)10!, then find the number of zeroes at the end of N

(a)6! + 8! + 10! (b) 2 × 6! + 8! + 10! (c) 6! + 2 × 8! + 10! (d) 6! + 8! + 2 × 10!

11. If in the previous question, all × are replaced by +, then the number of zeroes at

the end of N will be….. (a)6! + 8! + 2 × 10! (b) 0 (c) 4! (d) None of these

A = Product of first 50 multiples of 2. a. B = Product of first 8 multiples of 5 12. Find the number of zeroes at the end of A/B.

(a)3 (b) 4 (c) 5 (d) 6

[Number System]


Answer Key

1 c 2 b 3 a 4 d 5 c 6 c 7 a 8 b 9 c 10 d 11 b 12 c

Solutions

1. Number of zeroes = [58/5] + [58/25] = 10 + 2 = 12. [x] indicates the integer just

lower than the fraction x.

2. [80/7] + [80/49] = 11 + 1 = 12 3. 35 = 5 × 7, hence every time we can form a pair of one 5 and one 7, we count 1.

62! contains [62/7] + [62/49] = 8 + 1 = 9 sevens Also 62! Contains [62/5] + [62/25] = 12 + 2 = 12 fives. Hence 35 will divide 62! 9 times as the restrictions is on the power is because of the number of 7s and not on the number of 5s.

4. To solve this question we have to remember some points that the number of

zeroes depends upon the number of pairs of 2 × 5.We know that number of zero in 5! = 1 5!to 9! = 1 10! to 14 ! = 2 15! to 19 ! = 3 20! to 24 ! = 4 25! to 29 ! = 6 Here the order of zero shifted to 4 to 6 because in 25! 25 itself is square of 5 i.e. 25 = 5 × 5 therefore there are six 5‘s in 25! so the number of zeroes are 6 . Come to the question So we know that the number of zeroes in 100! are 24 101! to 104 ! = 24 105! to 109 ! = 25

110! to 114 ! = 26 115! to 119 ! = 27 120! to 124 ! = 28 125! to 129 ! = 31. From this 125 is a multiple of 25 and can be written as 5 × 5 × 5. So there will be the addition of 3 more zeroes in it so numbers of zeroes will shift from to 28 to 31. Now see the options, option number (d) is not possible in any case.

5. Number of zeroes at the end of 36! = [36/5] + [36/25] = 7 + 1 = 8.

Number of zeroes at the end of (36!)4! = 8 × 4! = 8 × 24 = 192 6. 25 × 32 + 225 × 8 + 105 × 12 = 52× 25 + 32× 52× 23 + 3 × 52× 7 × 22.

[Number System]


The number of zeroes is decided by the minimum number of pairs of 2 and 5 in the expression. Among 52× 25, 32× 52× 23 and 3 × 52× 7 × 22, minimum pairs are in the third part. So, the number of zeroes = 2

7. 105 = 3 × 5 × 7. The maximum power of 105 will be governed by 7 as it is the

largest prime number among the four. So, n = [52/7] + [52/49] = 7 + 1 = 8

8. The number N has 31 zeroes at the end. So, N can take any value from 125 to

129. Further, N is a prime number, so N = 127 as rest of the numbers in the range are not prime.

So, sum of digits of N = 1 + 2 + 7 = 10 9. p = [490/7] + [490/49] + [490/343] = 70 + 10 + 1 = 81 10. Number of zeroes in 4! = 0

Number of zeroes in 6! = 1 Number of zeroes in 8! = 1 Number of zeroes in 10! = 2 Number of zeroes in (4!)4!× (6!)6!× (8!)8!× (10!)10! = 0 × 4! + 1 × 6! + 1 × 8! + 2 × 10!

= 6! + 8! + 2 × 10! 11. In this case, the number of zeroes will be governed by the smallest value.

So, the number of zeroes = 0 as (4!)4! will have no zeroes.

12. A = 2 × 4 × 6 × 8 × 10 × ……………… 100 = 250 (1 × 2 × 3 × 4 × ……….50) = 250× 50! = 250× 512× p = 12 zeroes B = 5 × 10 × 15 × ……. 40 = 58 (1 × 2 × 3 × ….. × 8) = 58× 8! = 58× 5 × 27× q = 59× 27× q = 7 zeroes A/B = (250× 512× p) / (59× 27× q) = 5 zeroes

[Number System]


Exercise on Remainder Theorem 1. 1421×1423×1427 is divided by 6 then leaves remainder:

(a)1 (b) 2 (c) 3 (d) None of these. 2. Find the remainder when 51×53 is divided by 54:

(a)1 (b) 2 (c) 3 (d) 6 3. Find the remainder of (80×81×84)/85?

(a)20 (b) 45 (c) 65 (d) 39

4. Find the remainder of

(a)1 (b) 2 (c) 3 (d) 6


.

(a)1 (b) 2 (c) 3 (d) 6 6. (20×23×24×26)/ 100 will give a remainder =?

(a)15 (b) 10 (c) 20 (d) 40 7. Find the remainder when (23+ 24+ 25+…….+ 32)/14 ?

(a) 1 (b) 4 (c) 6 (d) 9


.


9. Find the remainder when is divided by 17. (a)5 (b) 7 (c) 6 (d) 8

10. What is the remainder when -23 is divided by 7?


11. Find the difference between the remainders when is divided by 511 and 513? (a)0 (b) 1 (c) 2 (d) 3

12. What will be the value of a for which ( ) ( )

leaves remainder 0?

(a)0 (b) 1 (c) 7 (d) 8 13. What exact power of 5 divides 90!?

(a) 19 (b) 20 (c)21 (d)None of these. 14. What is the power of 15 that divides 90! exactly?


15. What is the value of n for which the expression

( ) be an integer.

(a)33 (b) 40 (c) 122 (d)123

[Number System]


16. Find the last two digits of ( ) (a)01 (b) 11 (c) 21 (d) 31

17. Find the last two digits of ( ) . (a)11 (b) 31 (c) 41 (d) 01

18. Find the last two digits of ( ) : (a)25 (b) 35 (c) 45 (d) 55

19. Find the last two digits of ( ) : (a)76 (b) 36 (c) 16 (d) 44

20. Find the last two digits of ( ) : (a)24 (b) 48 (c) 76 (d) None of these.

21. 21. Find the remainder for

:

(a) 1 (b) 2 (c) 4 (d) 5

22. Find remainder of is divided by 41: (a) 1 (b) 7 (c) 40 (d) None of these.

23. Find the remainder when 30! is divided by 31.

(a) 30 (b) 29 (c) -1 (d) 1 24. What is the remainder left after dividing 1!+2!+3!.....+1000! by 7?

(a) 3 (b) 4 (c) 5 (d) 6

Answer Key

1 a 2 c 3 c 4 a 5 d

6 d 7 d 8 a 9 a 10 a

11 a 12 d 13 c 14 c 15 b

16 c 17 c 18 a 19 a 20 a

21 c 22 b 23 a 24 c

[Number System]


Exercise on Unit’s Digit 1. What is the unit‘s digit of the number 1978 + 3539 + 4367?

(a) 1 (b) 2 (c) 3 (d) 4 2. What is the unit‘s digit of the number 2334× 3457× 5761?

(a) 9 (b) 8 (c) 7 (d) 6 3. Aman wrote product of all natural numbers from 1 to 997 on a board. Naman

came and erased all the multiples of 5. Find the unit‘s digit of the remaining number. (a) 6 (b) 7 (c) 8 (d) None of these

4. What is the unit‘s digit of the product of first 40 odd natural numbers?

(a)5 (b) 7 (c) 9 (d) Cannot be determined

5. Let us define function f (x) such that f(x) = Unit digit of x.

Also, A = f (4278), B = f (4526), C = f (1839), then which of the following is true? (a)A > B > C (b) A > C > B (c) B > C > A (d) B > A > C

6. Find the unit‘s digit of N, where N = 1616! + 1717! + 1818!

(a)2 (b) 3 (c) 4 (d) 5 7. Rahim wrote a number A such that A = (xyz)4k + 2, where x, y, z and k are positive

integers. For how many values of z, will the number A have unit‘s digit as 1? (a)0 (b) 1 (c) 2 (d) 3

8. Devraj took a six digit number ending in 4 and raised it to an even power greater

than 1000. He then took the number 12 and raised it to a power which leaves the remainder 2 when divided by 8. If he now multiplies both the numbers, what will be the unit‘s digit of the number he so obtained? (a)3 (b) 4 (c) 5 (d) 6

9. The last digits in the expansion of the numbers (584a)149 and (584a)151 are 8 and

2 respectively. What can be said about the value of a? (a) a = 8 (b) a = 6 (c) a = 4 (d) a = 2

10. What is the unit‘s digit of the number 1! + 2! + 3! + 4! + 5! + …….. 20!?

(a) 1 (b) 3 (c) 5 (d) 7 11. Ravi takes two numbers M and N such that, M = (ab)ab and N = (ba)ba where a and

b are distinct integers with a + b = 7 and a > b. The unit‘s digit of (M-N) is 1.Find a-b. (a) 1 (b) 2 (c) 3 (d) Cannot be determined.

[Number System]


12. Find the unit‘s digit of the number (a)2 (b) 3 (c) 4 (d) 5

13. If the unit‘s digit of a number N = (1258a)b remains a for all the values of b, then

which of the following is true? (a)The number of values of a can be 2 (b)The number of values of a can be 3 (c)The number of values of a can be 4 (d)The number of values of a can be 5

14. Let us define ak = Unit‘s digit of [(15b)k]. If a1 + a2 + a3 + a4 = 24, what is the value

of b? (a)4 (b) 6 (c) 8 (d) None of these

Answer Key

1 c 2 b 3 C 4 a 5 d

6 b 7 c 8 b 9 a 10 b

11 c 12 b 13 C 14 b

[Number System]


Word Based Problems 1. 5 years ago Kate was 5 times as old as her son. 5 years hence her age will be 8

less than three times the corresponding age of her son. Find the present age of Kate. (a)30 years (b) 35 years (c) 40 years (d) 45 years

2. Girish's youth lasted one sixth of his life. He grew a beard after one twelfth more.

After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father and Girish died four years after his son. How many years did Girish live? (a)76 years (b) 80 years (c) 84 years (d) 88 years

3. The ages of a father and son add up to 66. The father's age is the son's age

reversed. How many possible combinations of their ages are possible? (a)1 (b) 2 (c) 3 (d) 4

4. Raju takes a two digit number. He then reverses the digits to get a new number.

If he subtracts the new number from the original number, he gets 54. The original number is prime. What is the sum of digits of the number? (a)14 (b) 12 (c) 10 (d) 8

5. There is a 6 digit number starting with 1. If we multiply it by 3, then 1 gets

shifted from the left most place to the right most place and all the digits are shifted one place to left. The hundred‘s digit of the original number is……. (a)8 (b) 7 (c) 5 (d) 4

Questions 6-8: There are three positive integers a, b and c such that a < b < c. b is a two digit prime number. The maximum sum of the integers taken two at a time is 39 and the minimum sum is 20. Now, answer the following questions. 6. What is the maximum possible value of a + c?

(a)35 (b) 36 (c) 37 (d) 38 7. What is the minimum value of b – a?

(a)1 (b) 2 (c) 3 (d) 4

8. For how many values of a, b and c,

an integer?

(a)2 (b) 3 (c) 4 (d) 5 9. Mark, Frank and Kristen decided to go to a fair. The cost of transport from their

home to the fair is $150. $100 was given by Mark and $50 was by Frank. The entry ticket was $10 per head which was given by Mark. In the fair, the trio ate 3 plates of roasted chicken costing $20 per plate. The entire amount was given by Frank. They returned to their home with $100 given by Frank and $50 by Mark as the return fare. How much money should Kristen give to Mark and Frank respectively so that they all spend equal amount of money in the excursion? (a)$80, $50 (b) $180, $210 (c) $210, $180 (d) $50, $80

[Number System]


10. The sum of ages of 5 children born at the intervals of 4 years each is 60 years. What is the age of the eldest child? (a)12 years (b) 16 years (c) 20 years (d) 24 years

11. Ajay is two years older than Binay who is twice as old as Charan. If the total of

the ages of Ajay, Binay and Charan be 17, the how old is Ajay? (a)3 years (b) 6 years (c) 8 years (d) 12 years

Directions for questions 12-15 Amit, Bharat, Chirag and Daler are four friends living in a flat. One day Amit‘s father can and brought a box of sweets. Since no one was around, Amit divided the sweets in

four equal parts and ate his share after which he put the rest of the sweets in the box. As he was closing the box, Bharat walked in and took the box. He again divided the sweets in four parts. Amit and Bharat ate one part each and kept the remaining sweets in in the box. Suddenly, Chirag appeared and snatched the box. He again divided the sweets in four equal parts, the three of them ate one part each and kept the remaining sweets in the box. Later when Daler came, he again divided the sweets in four equal parts and all four ate their respective share. In total, Daler ate 3 sweets. 12. How many sweets in total did Chirag eat?

(a) 12 (b) 15 (c) 39 (d) None of these 13. How many sweets in total did Bharat eat?

(a) 24 (b) 15 (c) 39 (d) None of these 14. How many sweets did Amit eat the first time?

(a) 32 (b) 24 (c) 15 (d) None of these 15. How many sweets were given to Amit by his father?

(a) 128 (b) 64 (c) 32 (d) None of these

Answer Key

1 b 2 c 3 c 4 d 5 a

6 c 7 b 8 a 9 d 10 c

11 c 12 b 13 c 14 a 15 a

[Number System]


Past Exam Questions 1. If a natural number is divided by 7, the remainder is 5. What is the remainder

when the cube of that natural number is divided 7? (a) 2 (b) 3 (c) 5 (d) 6

2. What is the smallest natural number by which 1792 must be divided to make it a

perfect square? (a) 2 (b) 4 (c) 7 (d) 8

3. Suppose x is a 10digit number with leftmost digit being 5 and is divisible by 3

and 11. Suppose y is the number obtained by reversing the digits of x then which of the following statements is not necessarily true? (a) y is divisible by 33 (b) y is divisible by 45 (c) y is divisible by 55 (d) y is divisible by 165

4. HCF of two expressions is (a + 1) and LCM is (a3+ a 2 – a – 1). If one expression is

(a2 - 1), then what is the second expression? (a) a + 1 (b) (a - 1)2 (c) (a + 1)2 (d) (a - 1)( a + 1)

5. What is the square root of

(a) 4 (b) 3 (c) 2 (d) 1 6. What is the HCF of (x4 – x2 - 6) and (x4 – 4x2 + 3)?

(a) x2 – 3 (b) x + 2 (c) x + 3 (d) x2 + 3 7. If the HCF of three numbers 144, x and 192 is 12, then the number x cannot

be……….. (a) 180 (b) 84 (c) 60 (d) 48

8. What is the HCF of 3.0, 1.2 and 0.06?

(a) 0.6 (b) 0.06 (c) 6.0 (c) 6.06

9. If x5 – 9x2 + 12x –14 is divided by (x – 3), what is the remainder? (a) 0 (b) 1 (c) 56 (d) 184

10. If 3y× 27y = 9(y+4), find the value of y?

(a) 4 (b) 5 (c) 6 (d) 7 11. What is the number of prime factors of 30030?


If p is a prime number such that p + 2 is also a prime, then (a) p (p + 2) + 1 is a perfect square (b) 12 is a divisor of p + (p + 2), if p > 3.

12. Which of the above statements is/are correct?

(a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

[Number System]


13. When a positive integer n is divided by 5, the remainder is 2. What is the remainder when the number 3n is divided by 5? (a) 1 (b) 2 (c) 3 (d) 4

14. Which one of the following 3 digit numbers divides 9238 and 7091 with the same

remainder in each case? (a) 113 (b) 209 (c) 317 (d) 191

15. If a and b are positive integers, x and y are non-negative integers and a = bx + y,

then which of the following is correct? (a) 0 ≤ y < a (b) 0 < y ≤ b (c) 0 < y < a (d) 0 ≤ y < b

16. The sum of two numbers is 80. If the larger number exceeds four times the

smaller by 5, what is the smaller number? (a) 5 (b) 15 (c) 20 (d) 25

Consider the numbers 1) 247 and 2 ) 203.

17. Which of the numbers is/are prime? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2

18. The number 2784936 is divisible by which of the following numbers?

(a) 86 (b) 87 (c) 88 (d) 89 19. X is twice as old Y 3 years ago, when X was as old Y today. If the difference

between their ages is 3 years, how old is X at present? (a) 18 years (b) 12 years (c) 9 years (d) 8 year

20. For a positive integer n, define d(n) = the number of positive divisors of n. What is

the value of d(d(d(12)))? (a) 1 (b) 2 (c) 4 (d) None of these

21. If 1 is added to the denominator of a fraction, it becomes

and if 1 is added to the

numerator, the fraction becomes 1. What is the fraction?

(a)

(b)

(c)

(d)

22. The product of a rational number and an irrational number is…

(a) a natural number (b) an irrational number (c) a composite number (d) a rational number

23. If x varies as the mth power of y, y varies as the nth power of z and x varies as the

pth power of z, then which of the following is correct? (a) p = m+n (b) p = m – n (c) p = mn (d) None of these

24. If x = (b – c)(a – d), y = (c – a)(b – d), z = (a – b)(b – c), then what is x3 + y3 + z3

equal to? (a) xyz (b) 2xyz (c) 3xyz (d) –3xyz

[Number System]


25. If n is a positive integer, then what is the digit in the unit‘s place of 32n+1 + 22n+1? (CDS 2010) (a) 0 (b) 3 (c) 5 (d) 7

26. If k is any positive integer, then (k2 + 2k) is……….

(a) divisible by 24 (b) divisible by 8 but may not be divisible by 24 (c) divisible by 4 but may not be divisible by 8 (d) divisible by 2 but may not be divisible by 4

27. Which of the following is a non-terminating and repeating decimal?

(a)

(b)

(c)

(d)

28. What are the factors of x2 + 4y2 + 4y – 4xy – 2x – 8?

(a) (x – 2y – 4) and (x – 2y + 2) (b) (x – y + 2) and (x – 4y + 4) (c) (x – y + 2) and (x – 4y – 4) (d) (x + 2y – 4) and (x + 2y + 2)

29. What is the sum of digits of the least number which, when divided by 52 leaves

remainder 33, when divided by 87 leaves remainder 59 andwhen divided by 117 leaves remainder 98? (a) 17 (b) 18 (c) 19 (d) 21

30. If f(x) and g(x) are two polynomials with integral coefficients which vanish at

x=1/2, then what is the factor of HCF of f(x) and g(x)? (a) x – 1 (b) x – 2 (c) 2x – 1 (d) 2x + 1

31. If (px3 + x2– 2x– q) is divisible by (x – 1) and (x + 1), what are the values of p and q

respectively? (a)2, –1 (b) –2, 1 (c) –2, –1 (d) 2, 1

32. What is the last digit in the expansion of 34798?

(a) 1 (b) 3 (c) 7 (d) 9

33. What is the smallest positive number positive integer which when divided by 4, 5, 8, 9 leaves remainders 3, 4, 7, 8 respectively? (a) 119 (b) 319 (c) 359 (d) 719

34. What is the value of x for which x, x + 1, x + 3 are all prime numbers?

(a) 0 (b) 1 (c) 2 (d) 101 35. If px = ry = m and rw = pz = n, then which of the following is correct?

(a) xw = yz (b)xz = yw (c) x + y = w + z (d) x – y = w – z 36. Five persons fire bullets at a target at an interval of 6, 7, 8, 9 and 12 seconds

respectively. The number of times they would fire the bullets together at the target in an hour is (a)6 (b)7 (c) 8 (d) 9

[Number System]


37. A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 8 o'clock in the morning, at what other time will they all ring together? (a)12 :40 PM (b) 12 :48 PM (c)12 :56 PM (d)13 : 04 PM

38. A group of 630 children is seated in rows for a group photo session. Each row

contains three less childrenthan the row infrontofit.Which one of the following number of rows is not possible? (a) 3 (b) 4 (c) 5 (d) 6

39. A gardener has 1000 plant. He wants to plant them in such a way that the

number of rows is equal to the number of columns. What is the minimum number of plants that he needs more for this purpose? (a) 14 (b) 24 (c) 32 (d) 34

40. A person has only Rs. 1 and Rs. 2 coins with her. If the total number of coins

that she has is 50 and the amount that she has is Rs. 75, then the number Rs. 1 and Rs. 2 coins are respectively……. (a) 15, 35 (b) 35, 15 (c) 30, 20 (d) 25, 25

41. A number of cats got together and decided to kill between them 999919 mice.

Every cat killed an equal number of mice. Each cat killed more mice than there were cats. How many cats do you think there were? (a) 1 (b) 991 (c) 1099 (d) 999919

42. A wizard named Nepo says ―I am only three times my son‘s age. My father is 40

years more than twice my age. Together the three of us are a mere 1240 years old.‖ How old is Nepo? (a) 120 years (b) 160 years (c) 360 years (d) 720 years

43. The difference between 6 times and 8 times of a figure is 14. What is the figure?

(a) 12 (b) 9 (c) 7 (d) 6

44. The greatest common divisor of 123456789 and 987654321 is

(a) 1 (b) 3 (c) 9 (d) greater than 9 45. A number G236G0 can be divided by 36 if G is:

(a) 8 (b) 6 (c) 1 (d) More than one values are possible 46. The symbol (25)b represents a two digit number in base b. I the number (52)b is

double the number (25)b, then the value of b is……….. (a)7 (b) 8 (c) 9 (d) 11

47. If the digit 1 is placed after a two digit number whose tens‘ digit is t, and units‘

digit is u, the new number is: (a) 10t + u + 1 (b) 100t + 10u +1 (c) 1000t + 10u + 1 (d) t + u + 1

[Number System]


48. In our number system the base is ten. If the base were changed to four, you would count as follows: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30………. The twentieth number would be: (a) 110 (b)104 (c) 44 (d) 38

49. If the square of a number of two digits is decreased by the square of the number

formed by reversing the digits, then the result is not always divisible by: (a) 9 (b) the product of the digits (c) the sum of the digits (d) the difference of the digits

50. A number which when divided by 10 leaves a remainder of 9, when divided by 9

leaves a remainder of 8, by 8 leaves a remainder of 7, etc., down to where, when divided by 2, it leaves a remainder of 1, is: (a) 59 (b) 419 (c) 1259 (d) 2519

51. Four different integers form an increasing AP. If one of these numbers is equal to

the sum of the squares of the other three numbers, then the numbers are: (a) -2, -1, 0, 1 (b) 0, 1, 2, 3 (c) -1, 0, 1, 2 (d) None of these

52. A man arranges to pay off a debt of Rs 3,600 by 40 annual installments which

are in AP. When 30 of the installments are paid he dies leaving one-third of the debt unpaid. The value of the 8th installment is (a) Rs35 (b) Rs50 (c) Rs65 (d) None of these

53. A father with 8 children takes 3 children at a time to the zoological garden, as

often as he can without taking the same 3 children together more than once. Then: (a)Number of times he will go to the zoological garden is 56. (b)Number of times each child will go to the zoological garden is 21. (c)Number of times a particular child will not go to the zoological garden is 35. (d)All of the above.

54. Consider the following statements:

A. Mode can be computed from histogram. B. Median is not independent of change of scale. C. Variance is independent of both the change of origin and scale. D. Which of these is/are correct?

(a) Only A and B (b) Only A (c) Only A (d) A, B and C 55. A uniform one metre long rod AB of weight 17kg is suspended horizontally from

fixed supports by two vertical strings attached to points C and D on the rod at distances if 12cm and 16cm from A and B respectively. The stings at C and D can support weights of 10kg and 9kg respectively without breaking. Then the position where a weight of 2kg can be attached to the rod without breaking either of the strings is: (a) 10cm from A (b) 12cm from A (c) 13cm from A (d) None of these

56. Which of the following is true?

A.99/101<97/99<95/97 B.95/97<97/99<99/101

[Number System]


C.{95/97}2>{97/99}2>{99/101}2 D.{95/101}2>{97/99}2>{95/97}2

(a) Only A (b) Only B (c)Only B and C (d)Only B and D 57. In figure below, how many paths are there from A to X if the only ways to move

are up and to the right?

(a) 4 (b) 5 (c)6 (d)9 58. Three times of a number is 122 less than the five times of another number. If the

sum of both the numbers be 74, find the two numbers (a) 31, 43 (b) 71, 133 (c) 19, 55 (d) 28, 46

59. A juggler keeps three balls going with one hand, so that at any instant, two are in

the air and one in hand. If each ball rises to a height of x metres, then each ball stays in the hand for:(MAT 2006)

(a) 0.45√ s (b) 0.33√ s (c) 0.24√ s (d) None of these 60. After being set up, a company manufactured 6000 scooters in the third year and

7000 scooters in the seventh year. Assuming that the production increases uniformly by a fixed number every year, what is the production in the tenth year? (a) 7850 (b) 7650 (c) 7750 (d) 7950

61. If the number 34x68 is divisible by 9, find the value of x?

(a) 3 (b) 4 (c) 5 (d) 6 62. The HCF and LCM of two numbers are 21 and 4641 respectively. If one of the

numbers lies between 200 and 300, then the two numbers are: (a) 273, 357 (b)273, 361 (c)273, 359 (d)273, 363

63. A railways half ticket costs half of the full fare and the reservation charge is the

same on the half ticket as on full ticket. A person received one full ticket from Hyderabad to Bangaluru and paid 262. Another received one full ticket and one half ticket and paid‘ 405. What is the reservation charge? (a) 24 (b) 48 (c) 12 (d)34

64. In a certain laboratory, chemicals are identified by a color coding system. These

are 20 different chemicals. Each one is coded with either a single color or a unique two color pair. If the order of colors in the pairs doesn‘t matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors? (a)5 (b)7 (c)6 (d)20

[Number System]


65. If the numbers from 1 to 45 which are exactly divisible by 3 are arranged in ascending order, minimum number being on the top, which would come at the ninth place from the top? (a)18 (b)24 (c)21 (d)27

66. Rohit, Harsha and Sanjeev are three typists who, working simultaneously can

type 216 pages in four hours. In one hour Sanjeev can type as many pages more than Harsha as Harsha can type more than Rohit. During a period of five hours, Snajeev can type as many pages as Rohit can during seven hours. How many pages does each of them type per hour respectively? (a)14, 17, 20 (b)16, 18, 22 (c)15, 17, 22 (d)15, 18, 21

67. HCF of 3240, 3600 and a third number is 36 and their LCM is 24 35 52 72. The

third number is:

(a)22 35 72 (b) 22 53 72 (c) 23 35 72 (d)25 52 72

68. Running at the same constant rate, 6 identical machines can produce a total of

270 bottles per minute. At this rate, how many bottles could 10 such machines produce in 4, minutes? (a)1,800 (b)648 (c)2,700 (d)10,800

69. There are two examination rooms A and B. If 10 candidates are sent from A to B,

the number of students in each room is the same. If 20 candidates are sent from B to A, the number of students in A is double the number of students in A is double the number of students in B. find the number of students in each room. (a)100 in A and 80 in B (b)80 in A and 100 in B (c)120 in A and 100 in B (d)100 in A and 120 in B

70. The charges for a five-day trip by a tourist bus for one full ticket and half-ticket

are Rs.1,440 inclusive of boarding charges which are same for a full ticket and a half-ticket. The charges for the same trip for 2 full tickets and one half-ticket inclusive of boarding charges are Rs.2,220. The fare for a half-ticket is 75% of the full ticket. Find the fare and the boarding charges separately for one full ticket. (a)Rs. 580, Rs. 400 (b)Rs. 280, Rs.200

(c)Rs.480, Rs.300 (d)Rs. 380, Rs.400 71. The LCM of two numbers is 4800 and their HCF is 160. If one of the numbers is

480, then the order number is: (a)16 (b)16000 (c)160 (d)1600

72. The numbers 1 to 29 are written side by side as follows 1234567891011…..28 29

if the number is divided by 9, then what is the remainder? (a)3 (b)1 (c)0 (d)None of these

73. The number 6n2+6n for natural number n is always divisible by:

(a)6 only (b)18 only (c)12 only (d)6 and 12

[Number System]


74. Rs.6500 were divided equally among a certain number of persons. Had there been 15 more persons each would have got Rs.30 less. Find the original number of persons: (a)45 (b)50 (c)55 (d)48

75. One-fourth of herd of camels was seen in the forest. Twice the square root of the

herd had gone to mountains and the remaining 15 camels were seen on the bank of the river. Find the total number of camels: (a)32 (b)34 (c)35 (d)36

76. Determine the value of…

√ √

√ √

√ √

√ √

(a)√ (b)10 (c)12√ (d)8 77. How can the relationship between x and y be best defined, if values of x and y are

as follows?

X 2 3 4 5 6

Y 0 2 6 12 20

(a)y = 2x-4 (b)x2 – 3x + 2 (c)y = x2 – 4x (d) y = x2 – 4

78. A man received a cheque. The rupees has been transposed for paise and vice

versa. After spending 5 rupees 42 paise, he discovered that he now had exactly six times the value of the correct cheque amount. What amount should he have received? (a)Rs.3.22 (b)Rs.6.44 (c)Rs.18.25 (d)Rs.8.36

79. When Rajeev was born, his father was 32 years older than his brother and his

mother was 25 years older than his sister. If Rajeev‘s brother is 6 years older than Rajeev and his mother is 3 years younger than his father, how old was Rajeev‘s sister when he was born? (a) 15 years (b) l 4 years (c) 7 years (d) 10 years?

80. The sum of two numbers is 462 and their highest common factor is 22. What is

the maximum number of pairs that satisfy these conditions? (a)1 (b)3 (c)2 (d)6

81. What is the value of M and N respectively if M39048458N is divisible by 8 and 11,

where M and N are single digit integers? (a)7, 8 (b)8, 6 (c)6, 4 (d)5, 4

82. The LCM of two numbers is 280 and their ratio is 7:8. The two numbers are:

(a)70, 80 (b)42, 48 (c)35, 40 (d)28, 32 83. In a class of 50 students, 23 speak English, 15 speak Hindi and 18 speak

Punjabi. 3 speak only English and Hindi, 6 speak only Hindi and Punjabi and 6 speak only English and Punjabi. If 9 can speak only English, then how many students speak all the three languages?

[Number System]


(a)1 (b)2 (c)3 (d)5 84. If Dennis is 1/3rd the age of his father Keith now, and was 1/4th the age of his

father 5 years ago, then how old will his father Keith be 5 years from now? (a)20 years (b)45 years (c)40 years (d)50 years

85. Two spinning machines A and B can together produce 3,00,000 metres of cloth in

10 hours, if machine B alone can produce the same amount of cloth in 15 hours, then how much cloth can machine A produce alone in 10 hours? (a)2,00,000m (b)1,00,000m (c)1,50,000m (d)50,000m

86. The ages of two persons differ by 20 years. If 5 years ago, the older one be 5 times as old as the younger one, then their present ages, in years are: (a)25, 5 (b)30, 10 (c)35, 15 (d)50, 30

87. A man has 1044 candles. After burning, he can make a new candle from 9 stubs

left behind. Find the maximum number of candles that can be made: (a)226 (b)120 (c)130 (d)140

88. What is the third term in a sequence of numbers that leave remainder of 1, 2 and

3 when divided by 2, 3 and 4 respectively? (a)11 (b)17 (c)19 (d)35

89. A sum of Rs. 36.90 is made up of 180 coins which are either 10paise coins or

25paise coins. Determine the number of each type of coins: (a)126 of 10p coins and 54 of 25p coins (b)54 of 10p coins and 126 of 25p coins (c)90 of 10p coins and 90 of 25p coins (d)None of these

90. If x+y>5 and x-y>3, then which of the following gives all possible values of x?

(MAT 2004) a)x>3 (b)x>4 (c)x>5 (d)x<5

91. Which of the following integers is the square of an integer of every integers n?

(a)n2+1 (b)n2+n (c)n2+2n (d)n2+2n+1 92. If x and y are negative, then which of the following statements is/are always

true? I.x+ y is positive II.xy is positive III.x – y is positive (a)I only (b)II only (c)III only (d)I and III only

93. The LCM of two numbers is 4800 and their HCF is 160. If one of the numbers is

480, then the second number is: (a)16 (b)16000 (c)160 (d)1600

94. A student was asked to divide a number by 6 and add 12 to the quotient. He,

however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been:

[Number System]


(a)112 (b)118 (c)114 (d)124 95. The sum of the digits of a 3 digit number is subtracted from the number. The

resulting number is always: (a)divisible by 6 (b) not divisible by 6 (c) divisible by 9 (d)not divisible by 9

96. In a group of 15 women, 7 have nose studs, 8 have ear rings and 3 have neither. How many of these have both nose studs and ear rings? (a)0 (b)2 (c)3 (d)7

97. The number of rectangles that you can find on a chessboard is:

(a)1764 (b) 1600 (c)1825 (d)1296 98. Pinto dealt some cards to Mintoo and himself from a full pack of playing cards

and laid the rest aside. Pintoo then said to Mintoo, ―If you give me a certain number of your cards, I will have 4 times as many cards as you have. If I give you the same number of cards, I will have thrice as many cards as you have.‖ How many cards did Pintoo have? (a)31 (b)32 (c)29 (d)30

99. If x is a positive number, then which of the following fractions has the greatest

value? (a)x/x (b)(x+1)/x (c)x/(x+1) (d)(x+2)/(x+3)

100. HCF of 3240, 3600 and a third number is 36 and their LCM is 24 35 52 72. The

third number is (a) 24 53 72 (b) 22 35 72 (c) 23 35 72 (d) 25 52 72

101.

(a)2477 (b) 2478 (c) 2467 (d) 2476 (e) None of these.

102. ( √ √ ) ( √ √ ) ( )

(a)143 (b) 134 (c) 145 (d) 514 (e) None of these

103. ( ) ( ) ( ) (a)280.4 (b) 290.4 (c) 295.4 (d) 285.4 (e) None of these

104. 23% of 6783 + 57% of 8431 = ? +5.76

(a)6460 (b) 6420 (c) 6320 (d) 6630 (e) 6360 105. When X subtracted from number 9, 15, 27 the remainders are in continued

proportion. What is the value of X? (a)8 (b) 6 (c) 4 (d) 5 (e) None of these.

106. Rachita enters a shop to buy ice-creams, cookies, and pastries. She has to buy at

least 9 units of each. She buys more cookies than ice creams and more pastries than cookies. She picks up a total 32 items. How many cookies does she buy? (a) Either 12 or 13 (b) Either 11 or 12 (c) Either 10 or 11 (d) Either 9 or 13 (e) Either 9 or 10

[Number System]


107. The product of three consecutive even numbers is 4032. The product of 1st and

3rd number is 252. What is the 5 times of the second number? (a) 80 (b) 100 (c) 60 (d) 70 (e) 90

108. The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-in-law, the sum of their ages is 92. Assuming that there has been no other change in the family structure and all the people are alive, what is the difference in the age of the daughter and daughter in law? (a) 22 (b) 11 (c) 25 (d) 19 (e) 15

109. Sum of three consecutive numbers is 2262. What is 41% of the highest number?

(a) 301.51 (b) 303.14 (c) 308.73 (d) 306.35 (e) 309.55

110. Rubina could get equal number of Rs. 55, Rs. 85, Rs.105 tickets for a movie. She

spends Rs.2, 940 for all the tickets. How many of each did she buy? (a) 12 (b) 14 (c) 16 (d) Cannot be determined. (e) None of these.

111. The ratio of present age of Manisha and Deepali is 5:x. Manisha is 9 years

younger than Parineeta. Parineeta‘s age after 9 years will be 33 years. The difference between Deepali‘s and Manisha‘s age is the same as the present age of Parineeta. What should come in place of x? (a) 23 (b) 39 (c) 15 (d) Cannot be determined. (e) None of these.

112. If the cost of 8 erasers and 5 sharpeners is Rs.31. What will be the cost of 21

erasers and 10 sharpeners? (a) Rs.62 (b) Rs. 84 (c) Rs. 78 (d) Cannot be determined. (e) None of these.

113. √

(a) 17 (b) 19 (c) √ (d) 289 (e) None of these. 114. Radha‘s present age is 3 years less than twice her age 12 years ago. Also the

respective ratio between Raj‘s present age and Radha‘s present age is 4:9. What will be the Raj‘s age after 5 years? (a) 12 yr (b) 7 yr (c) 21yr (d) Cannot be determined (e) None of these

115. The present age of Romila is one fourth of that of her father. After 6 years the

father‘s age will be twice the age of Kapil. If Kapil celebrated fifth birthday 8 years ago, what is Romila‘s present age? (a) 7 years (b) 7.5 years (c) 8 years (d) 8.5 years (e)None of these

116. .find approximate value of (a) 6 (b) 9 (c) 7 (d) 10 (e) 12

[Number System]


117. A father tells his son "I was your present age when you were born". If the father is

64 years old now, the age of son after 10 years will be (a) 42 years (b) 44 years (c) 38 years (d) 50 years

118. (a) 840 (b) 990 (c) 570 (d) 680 (e) 720

119. 31.999 (a) 6600 (b) 6720 (c) 6480 (d) 6070 (e) 6270

120. (a) 381 (b) 347 (c) 372 (d) 311 (e) None of these.

121. ( ) ( ) ( ) (a) 28 (b) 120 (c) 10 (d) 65 (e) 140

122.

of

of

of a number is 15. What is 30 percent of that number?

(a) 45 (b) 60 (c) 75 (d) 30 (e) None of these 123. The present ages of Vishal and Shekhar are in the ratio of 14:17. 6 years from

now, their ages will be in the ratio of 17:20. What is Shekhar‘s present age? (a) 17 years (b) 51 years (c) 34 years (d) 28 years (e) None of these

124. At present Meena is eight times her daughter‘s age. Eight years from now, the

ratio of ages of Meena and her daughter will be 10:3. What is the Meena‘s present age? (a) 32 years (b) 40 years (c) 36 years (d) Cannot be determined (e)None of these

125. The present age of Amit and his father are in ratio of 2:5 respectively. Four year

from hence the ratio of their ages will become 5:11 respectively. What was the

father‘s age five years ago? (a) 40 years (b) 45 years (c) 30 years (d) 35 years (e) None of these

126. The smallest number that must be added to 803642 in order to obtain multiple of

11 is: (a) 1 (b) 4 (c) 7 (d) 9

127. In a division sum, the divisor is 3 times the quotient and 6 times the remainder. If the remainder is 2, then the dividend is: (a) 50 (b) 48 (c) 36 (d) 28

128. A rational number between 3/4and 3/8 is:

(a) 12/7 (b) 7/3 (c) 16/9 (d) 9/16 129. In a class there are z students out of them x are boys. What part of class

composed by girls?

[Number System]


(a) x/z (b) z/x (c)

(d)

130. The unit digit in is (a) 4 (b) 2 (c) 6 (d) 8

131. If the sum of 5 consecutive integers is S, then the largest of those integers in terms of S.

(a)

(b)

(c)

(d)

132. ‗a‘ divides 228 leaving a remainder 18. The biggest two digit value of ‗a‘ is:

(a) 70 (b) 21 (c) 35 (d) 30 133. A number x when divide by 289 leaves 18 as a remainder. The same number

when divided by 17 leaves y as a remainder: (a)5 (b)2 (c) 3 (d) 1

134. A positive integer when divided by 425 gives a remainder 45. When the same

number is divided by 17, the remainder will be: (a) 11 (b) 8 (c) 9 (d) 10

135. How many numbers between 400 and 800 are divisible by 4, 5, 6?

(a) 7 (b) 8 (c) 9 (d) 10

136. Greatest common divisor of

is:

(a) 2 (b) 1 (c) (d) 20

137. Divide 37 in two parts so that 5 times the one part and 11 part the other are

together 227. (a) 15, 22 (b) 20, 17 (c) 25, 12 (d) 30, 7

138. When n is divided 6, the remainder is 4. When 2n is divided by 6, the remainder

is: (a) 2 (b) 0 (c) 4 (d) 1

139. A fraction becomes 9/11, if 2 is added to both the numerator and denominator. If

3 is added to both it becomes 5/6. What is the fraction? (a) 7/9 (b) 3/7 (c) 5/9 (d) 7/10

140. In a two digit number the digit at unit place is 1 less than twice the digit at the

ten‘s place. If the digit at the unit‘s and ten‘s place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is: (SSC 2011) (a) 59 (b) 23 (c) 35 (d) 47

141. The sum of squares of 3 consecutive positive numbers is 365. The sum of the

numbers: (a) 30 (b) 33 (c) 36 (d) 50

[Number System]


142. If a and b are odd number then which of the following is even? (a) a+ b+ ab (b) a+b-1 (c) a+ b+ 1 (d) a+ b+ 2ab

143. I multiplied a natural number by 18 and another by 21 and added the product.

Which one of the following could be the sum? (a) 2007 (b) 2008 (c) 2006 (d) 2002

144. The last digit of is: (a) 1 (b) 3 (c) 7 (d) 9

145. The sum of a two digit number and the number obtained by reversing its digits is a square number. How many such numbers are there? (a) 5 (b) 6 (c) 7 (d) 8

146. If [n] denotes the greatest integer < n and (n) denotes the smallest integer > n

where n is any real number then (

) *

+ (

) *

+ ( ) :

(a) 1.5 (b) 2 (c) 2.5 (d) 3.5

147. When is divided by 6, the remainder is: (a) 5 (b) 3 (c) 2 (d) 1

148. When a number is divided by 357 the remainder is 39. If that number is divided

by 17, the remainder will be: (a) 0 (b) 3 (c) 5 (d) 11

149. The remainder of

is:

(a) 0 (b) 1 (c) 2 (d) 3

150. The expression , where n is the natural number always divisible by: (a) 15 (b) 18 (c) 36 (d) 48

[Number System]


Answer Key

1 d 2 c 3 b 4 c 5 d

6 a 7 d 8 b 9 d 10 a

11 c 12 c 13 a 14 a 15 c

16 b 17 d 18 c 19 c 20 d

21 b 22 b 23 c 24 c 25 c

26 b 27 c 28 a 29 a 30 c

31 d 32 d 33 c 34 c 35 a

36 b 37 b 38 d 39 b 40 d

41 b 42 c 43 c 44 c 45 a

46 b 47 b 48 a 49 b 50 d

51 c 52 c 53 d 54 a 55 c

56 d 57 c 58 a 59 a 60 c

61 d 62 a 63 a 64 c 65 d

66 d 67 a 68 a 69 a 70 c

71 d 72 a 73 d 74 b 75 d

76 b 77 b 78 b 79 d 80 d

81 c 82 c 83 d 84 d 85 b

86 b 87 c 88 d 89 b 90 b

91 d 92 b 93 d 94 a 95 c

96 c 97 d 98 a 99 b 100 b

101 d 102 a 103 b 104 e 105 e

106 c 107 a 108 a 109 e 110 a

111 e 112 d 113 a 114 e 115 b

116 b 117 a 118 e 119 b 120 e

121 a 122 a 123 c 124 a 125 d

126 c 127 a 128 d 129 c 130 d

131 d 132 a 133 d 134 a 135 a

136 a 137 d 138 a 139 a 140 d

141 b 142 d 143 a 144 a 145 d

146 b 147 b 148 c 149 a 150 d


Chapters 2 - Progressions

INTRODUCTION

A succession of numbers a1, a2 . . . . . . an formed according to some definite rule is called a sequence or a progression. In this chapter we deal with the following types of progression - Arithmetic Progression - Geometric Progression

2.1 ARITHMETIC PROGRESSION Definition: A succession of numbers is said to be in arithmetic progression (A. P.), if the difference between any two consecutive terms is always constant. Assumption: the given arithmetic sequence is a1, a2, a3 . . . . an-1, an

Term number 1st 2nd 3rd . . . xth . . . pth. . .qth . . . (n-1)th nth (last)

(d common difference) Arithmetic mean can also be obtained by considering any two terms which are equidistant from both ends of A.P & taking their average. Thus, The average of the second term from the beginning and second term from the end will be equal to A.M. The average of the third term from the beginning and the third term from the end will also be equal to A.M and so on. In general the average of Kth term from the beginning and Kth term from the end will be equal to A.M of the AP. nth TERM OF A.P.

nth term = an = any term value + (n – term number) d Corresponding e.g. if first term value is know, then an = a1 + (n – 1)d since term number = 1 (refer, assumption) COMMON DIFFERENCE (d)

Common difference = numbers termingcorrespond twoof difference

values term twoof difference

e.g. Above assumption indicates that


d = 1

1...)1(

1......3

3

12

12

n

an

a

nn

na

na

pq

pa

qa

x

ax

aaa

SUM UP TO nth TERM (Sn) Sum up to nth term = Sn = 1st term + 2nd term + ….. + nth term

Sn = ] d . number) term. 2-1(n valueany term .2 [2

n

(corresponding)

e.g. If only the value of the first term is known, then term number = 1, and term value = a1

[Refer, assumption] Sn = n/2 [2a1 +(n-1) d]

Till now we have studied APs in their mathematical context. This was important for you to understand the basic mathematical construct of A.P‘s. However, you need to understand that questions on A.P. are seldom solved on a mathematical basis, (Especially under the time pressure that you are likely to face under the CAT and other aptitude exams). In such situations the mathematical processes for solving progressions based questions are likely to fail. Hence, understanding the following logical aspects about Arithmetic Progressions is likely to help you solve questions based on AP‘s in the context of an aptitude exam. Let us look at these issues one by one:

1. Process for finding the nth term of an A.P: Suppose you have to find the 17th term of the A.P. 3,7,11…….

The conventional mathematical process for this question would involve using the formula. Tn = a + (n – 1) d Thus, for the 17th term we would do

T17 = 3 + (17 –1) × 4 = 3 + 16 × 4 = 67 Most students would mechanically insert the values for a, n & d & get this answer. However, if you replace the above process with a thought algorithm, you will get the answer much faster. The algorithm goes like this: In order to find the 17th term of the above sequence add the common difference to the first term, sixteen times. (Note: Sixteen, since it is one less than 17). Similarly, in order to find the 37th term of the A.P. 3, 11 …, All you need to do is add the common difference (8 in this case), 36 times. Thus, the answer is 288 + 3 = 291. (Note: You ultimately end up doing the same thing, but you are at an advantage since the entire solution process is reactionary.)


2. Average of an A.P. & Corresponding terms of the A.P: Consider the A.P, 2, 6, 10, 14, 18, 22. If you try to find the average of these six numbers you will get: Average = 2+6+10+14+18+22/6 = 12 Notice that 12 is also the average of the first and the last terms of the A.P. In fact, it is also the average of 6 & 18 (which correspond to the second and 5th terms of the A.P.). Further, 12 is also the average of the 3rd and 4th terms of the A.P. (Note: In this A.P. of six terms, the average was the same as the average of the 1st and 6th term. It was also given by the average of the 2nd and the 5th term, as well as that of the 3rd and 4th term. ) We can call each of these pairs as ―CORRESPONDING TERMS‖ in an A.P.

What you need to understand is that every AP has an average. And for any A.P., the average of any pair of corresponding terms will also be the

average of the A.P.

If you try to notice the sum of the term numbers of the pair of corresponding terms given above:

1st& 6th (so that 1 + 6 = 7) 2nd& 5th(hence, 2 + 5 =7) 3rd& 4th (hence, 3 + 4 = 7)

Note that: In each of these cases, the sum of the term numbers for the terms in a corresponding pair is one greater than the number of terms of the A.P.

This rule will hold true for all APs.

For example, if an A.P. has 23 terms then for instance, you can predict that the 7th term will have the 17th term as its corresponding term, or for that matter the 9th term will have the 15th term as its corresponding term. (Since 24 is one more than 23 and 7+17=9+15=24.)

3. PROCESS FOR FINDING THE SUM OF AN A.P. Once you can find a pair of corresponding terms for any A.P., you can easily find the sum of the A.P. by using the property of averages: i.e. Sum = number of terms × Average. In fact, this is the best process for finding the sum of an A.P. It is much more superior than the process of finding the sum of an A.P. using the expression [n (2a+(n–1)d)]/2

4. Finding the common difference of an A.P., given 2 terms of an A.P. Suppose you were given that an A.P. had its 3rd term as 8 & its 8th term as 28. You should visualize this A.P. as –,–,8,–,–, –, –, 28. From the above figure, you can easily visualize that to move from the third term to the eighth term, (8 to 28) you need to add the common difference five times. The net addition being 20, the common difference should be 4. Illustration: Find the sum of an A.P. of 17 terms, whose 3rd term is 8 and 8th term is 28. Solution: Since we know the third term and the eight terms, we can find the common difference as 4 by the process illustrated above. The total = 17 × Average of the A.P.


Our objective now shifts to finding the average of the A.P. In order to do so, we need to identify either the 10th term (which will be the corresponding term for the 8th term) or the 15th term (which will be the corresponding term for the 3rd term.) Again: Since the 8th term is 28 and d = 4, the 10th term becomes 28 + 4 + 4=36. Thus, the average of the A.P. = Average of 8th and 10th term = (28 + 36)/2 = 32. Hence, the required answer is sum of the A.P. = 17 × 32 = 544. The logic that has applied here is that the difference in the term numbers will give you the number of times the common difference is used to get from the first to the second term. For instance, if you know that the difference between the 7th term and 12th term of an AP is -30, you should realize that 5 times the common difference will be

equal to -30. (Since 12– 7= 5). Hence, d = -6. Note Replace this algorithmic thinking in lieu of the mathematical thinking of: 12th term = a + 11d 7th term = a + 6d Hence, difference = –30 = (a + 11d) – (a + 6d) – 30 = 5d

d = -6.

Important Facts : When three numbers are in A.P we can take the these terms to be (a – d) a & (a + d) If four numbers are in A.P then these numbers can be assumed as: (a - 3d ), ( a – d ), ( a + d ) and (a + 3d) or if five numbers are in A.P then

(a - 2d) (a – d) a (a + d) (a + 2d)

ILLUSTRATIONS SET 2.1 1. According to a bet between two players playing chess, the loser will have to pay

Rs(ninj) for each nth block (where i and j represents the row and column number

of nth block) to the winner. How much money will winner get? (a) 64 (b) 2080 (c) 1296 (d) 512 (e) None of these

There are 8 rows and 8 columns present on a chessboard starting from 1st row and 1st column.

So, Required amount =

8,8

1,1

jn

jnj

ni

n

= (1 + 2 + 3 + …+ 8) + (2 + 4 + 6 - - - + 16) + (3 + 6 + 9 + - - - - + 24) + - - - - - + (8 + 16 + 24 + - - - 64)

= (1 + 2 + 3 + …+ 8) x (1 + 2 + 3 + …+ 8) = 36 36 = 1296

Option (c) 2. What will be the 12th term and sum upto 12th term of the following series?


1,6,11,………..

(a) 56, 342 (b) 611, 5

1612 (c) 56, 684

(d) 512, 4

1512

(e) None of these

The given series 1, 6, 11. . . . ie, series is in A.P. with 1st term = 1 and common difference = 5

so, 12th term i.e. t12 = a + (n – 1) d

= 1 + (12 -) 5 = 56 and sum upto 12th term i.e., S12 = n x average = 12x28.5 = 342

Option (a) 3. Find the sum of the series upto 90 terms 1+2+3+3+4+6+5+6+9+………

(a) 1613 (b) 9675 (c) 1612 (d) 6450 (e) 3225 1 + 2 + 3 + 3 + 4 + 6 + 5 + 6 + 9 + - - - - upto 90 terms. ie (1 + 3 + 5 + - - - - - - - upto 30 terms) (c.d. = 2) + (2 + 4 + 6 + - - - - - - - upto 30 terms) (c.d. = 2) + (3 + 6 + 9 + - - - - - - - upto 30 terms) (c.d. = 3) Calculate the sum of each of three APs independently and add to get the value as = 3225

Option (e) 4. Find the sum of the series to 30 terms 1-50+4-45+7-40+…….

(a) 605 (b)945 (c)1605 (d)105 (e)-945 1 – 50 + 4 – 45 + 7 – 40 + - - - - - - up to 30 terms ie. (1 + 4 + 7 + - - - - - up to 15 terms) (c.d. = 3) – (50 + 45 + 40 - - - - up to 15 terms) (c.d. =-5)

= )}]5)(115(502{2

15}3)115(12[{

2

15

= )701002

15)422(

2

15

= 15 22 – 15 15 = 105

option (e) 5. The salary of an employee of a reputed Company increases by Rs. 500 per week.

How much salary (weekly) will he receive just after one year if he joins the company with initial salary of Rs. 10,000 per week? (The salary of any week is given on the 1st day of next week). (a) 35,000 (b) 35,500 (c)36,000 (d)36,500 (e)37,000

The series of weekly salary of the employee will be as follows: - 10,000, 10,500, 11,000, - - - - - - - - - upto 1 year (ie. Up to 52 weeks) So, Required salary = Salary of 52nd week (which he will receive on 1st day of 53rd week). In other words,you need the 52nd term of the AP.

= 10,000 + (52 – 1) 500 = 10,000 + 25,500 = 35,500


option (b) 6. Let the positive numbers a, b, c, d be in A.P. then abc, abd, acd, bcd are :

(a) not in A.P./ G.P./ H.P. (b) in A.P. (c) in G.P. (d) in H.P.

a,b,c,d, are in A.P.abcd

a,

abcd

b,

abcd

c,abcd

d, are in A.P.

bcd

1,cda

1,abd

1,abc

1 are in A.P.

bcd, cda, abd, abc are in H.P.

Option (d) (Alternately, take any 4 values of a,b,c and d so that they are in AP and solve through values to see that the given series is an HP.) 7. Suppose a, b, c are in A.P. and a2, b2, c2 are in G.P. If a < b < c and a + b + c =

,2

3then the value of a is ;

(a) 22

1 (b)

32

1 (c)

3

1

2

1

(d) 2

1

2

1 (e) none of these

a, b, c are in A.P. So let them be, a = A – D, b = A, c = A + D

a + b + c = 2

3

(A – D) + A + (A + D) = 2

3

3A = 2

3

A = 2

1

the numbers are DD 2

1,

2

1,

2

1

also,

2

4

1,

4

1,

2

2

1

DD are in GP. Hence,

2

2

12

2

12

4

1

DD

4

12

4

1 D

2

12 D


D = 2

1

a = 2

1

2

1

So, out of the given values, a = 2

1

2

1 is the right choice.

option (d) 8. The sum of integers from 1 to 100 that are divisible by 2 or 5 is.

a) 3050 b) 2550 c) 1050 d) 3600 e) 1400

Integers divisible by 2 are {2, 4, 6, 8, 10, . . . . . , 100} Integers divisible by 5 are {5, 10, 15, . . . . . , 100}

Thus sum of integers divisible by 2 = 25505150)1002(2

50

Sum of integers divisible by 5 = 105010510)1005(2

50

However, in order to get the correct sum we need to remove any double counting between the two series. For this we also need to find the sum of the terms that are simultaneously divisible by 2 and 5. Thus sum of integers divisible by 10 =

5501105)10010(2

10

Sum of integers from 1 to 100 divisible by 2 or 5 = 2550 + 1050 – 550 = 2550 + 500 = 3050 9. The sum of first n terms of the series 12 + 2.22 +32 + 2.42 + 52 + 2.62 +- - - - - is

2

2)1( nn, when n is even. When n is odd the sum is.

(a) 2

2)1( n (b) n

nn

2

)1(2 (c)

2

)1(2 nn

(d) 2

2)1(2 nn (e)

2

)1( nnn

Here 12 + 2.22 +32 + 2.42 + 52 + . . . . . up to n terms.

= 2

2)1( nn (when n is even)

When n is odd 12 + 2.22 +32 + 2.42 + 52 + . . . . . . . + n2 = {12 + 2.22 +32 + 2.42 + . . . . . + 2 (n – 1)2} + n2

= 2

2

2))(1(n

nn

Since we have n as an odd number, the formula given for the

sum of the series when n is even will need to be used with the value of n as (n-1).


= 2

)1(212

12

n

nn

n

The sum of the series 12 + 2.22 +32 + 2.42 . . . . . . up to n terms when n is odd =

2

)1(2 nn

option (C) 10. The third term of a geometric progression is 4. The product of the first five terms

is: a) 42 b) 43 c) 44 d) 45 e) none of these

Here, t3 = 4 {third term of G.P.}

Thus product of first five terms = a .ar . ar2 . ar3. ar4

= a5r10 = (ar2)5 = 45

Product of first five terms = 45

option (d)

11. Sum of the first ‗n‘ terms of the series 16

15

8

7

4

3

2

1is equal to :

a) 2n – n – 1 b) 1- 2– n c) n + 2-n – 1 d) 2n + 1 e) None of these Sum of the first ‗n‘ terms of the series can be written as

16

15

8

7

4

3

2

1up to n terms

= ....16

11

2

11

8

11

4

11

2

11

up to n terms

= n -

G.P. of n terms ........

8

1

4

1

2

1

= 12

2

11

2

11

2

1

nnn

n

option (c) [Alternately solve using options] 12. The internal angles of a plane polygon are in A.P. The smallest angle is 1200 and

the common difference is 50. Find the number of sides of the polygon and the sum of the internal angles.

Sum of interior angles of a polygon of n sides = (n – 2) 1800. Here a = 1200, d = 50.

000 180)2(5)1(2402

nnn

i.e. n2 – 25n + 144 = 0 or n = 9, 16.


Since the interior angle of a polygon < 1800 and T16 = 1950, T9 = 1600, so n 16. Hence n = 9. 13. After striking a floor, a certain ball rebounds (4/5)th of the height from which it

has fallen. Find the total distance that it travels before coming to rest, if it is gently dropped from a height of 120 metres.

After striking the floor, the ball, falling from a height of 120 mt. rebounds to a height

of (4/5)120mt.

Again it falls from this height and rebounds to a height of (4/5)2120mt. and so on. Hence the total distance traveled

= 120+2[120(4/5)+120(4/5)2+…to ]

= 120+2{120(4/5)}/{1–(4/5)} = 120+960=1080 meters. 14. A person was appointed in the pay scale of Rs.700 – 40 – 1500. In how many

years will he reach the maximum of the scale? Let the required number of years be n. Given tn = 1500, a = 700, d = 40, to find n, tn = a + (n – 1)d

1500 = 700 + (n – 1)40 or, (n – 1)40 = 800 or, n – 1 = 20 or, n = 21. 15. A sum of money kept in a bank amounts to Rs.600/- in 4 years and Rs.800/- in

12 years. Find the sum and the interest carried over every year. Let the required sum be a and the interest carried every year be d, then t1 = a. From question t5 = 600, t13 = 800. To find a and d. tn = a + (n – 1)d …(i) or, 600 = a + (5 – 1)d = a + 4d …(ii) And 800 = a + (13 – 1)d = a + 12d …(iii) Subtracting (ii) from (iii) we get

200 = 8d or d = 25 Putting d = 25 in (ii), we get a = 600 – 4.25 = 500 Sum = Rs.500/- and the interest carried every year = Rs.25/- 16. Is 55 a term of the sequence 1, 3, 5, 7, …? If yes, find which term isit? If possible let the nth term of the sequence be 55. Now tn = a + (n – 1)d, Here tn = 55, a = 1, d = 2 55 = 1 + (n – 1)2 or 2n = 56, or, n = 28 Hence 55 is 28th term of the given sequence. Note: If n does not come out to be an integer then 55 cannot be a term of the given

sequence. Alternatively, you can project the sequence and get the value.


17. If m times the mth term of an A.P. is equal to n times the nth term, find its (m + n)th term.

Let ‗a‘ be the first term and d be the common difference. Given ntn = mtm n[a + (n – 1)d] = m[a + (m – 1)d] or, (m – n)a = d[n(n – 1) – m(m – 1) or, (m – n)a = d[(m – n) – (m2 – n2)] or, (m – n)a = d(m – n)[1 – (m + n)]

or, a = d[(1 – (m + n)] [ m n] or, –a = d(m + n – 1) tm+n = a + (m+n–1)d = a – a = 0 [from (i)]


Exercise on Arithmetic Progressions

1. Find the 12th term of the series 2, 8, 14, 20, ……

(a)64 (b)66 (c)68 (d)70

2. Find the sum of first 10 terms of 2, 6, 10, 14, 18,………..

(a)100 (b)200 (c)300 (d)400

3. Find the number of terms of an A.P. 5, 8, 11, 14… 155.

(a)49 (b)50 (c)51 (d)52

4. Find the 10th term from end of an A.P. 1, 5, 9, 13…121.

(a)77 (b)81 (c)85 (d)None of these

5. Find the sum of all terms of an A.P. having 1st term =5 last term = 50 & number

of terms=10.

(a)250 (b)275 (c)300 (d)325

6. Find the sum of all terms of the series having 1st term 5 total numbers of terms

10 & common difference 5.

(a)270 (b)275 (c)280 (d)None of these

7. Find the sum of last 5 terms of the series 3, 8, 13, ……..103.

(a)465 (b)560 (c)2040 (d)1560

8. Find the sum of 1st 100 natural numbers.

(a)5000 (b)5050 (c)5100 (d)5150

9. Find the sum of series

(a)2400 (b)2415 (c)2485 (d)2565

10. Find the sum of cubes of 1st 20 natural numbers.

(a)44000 (b)44100 (c)44300 (d)44600

11. Find the value of the expression 1– 6 + 3 –10+ 5 -14 +7– 18+…….. to 100 terms.

(a)2700 (b)-2700 (c)-3200 (d)-3100

12. What will be the value of terms?

(a) ( ) ( ) ( )

13. Que13, 14: If the term of an A.P. is n and nth term is m. Then answer the

following questions:

term of the series is:


(a)m+ n+ p (b)m+ n-p (c)m+ n+2p (d)m+ n-2p

14. ( ) term of the series is:

(a)0 (b)m+ n (c)m-2n (d)

15. How many terms are identical in the two APs 2, 5, 8, 11, .... 60 terms and 3, 5, 7,

9 ,..... , 50 terms..

(a)10 (b)14 (c)17 (d)None of these.

16. What is the sum of all 2 digit numbers that leave a remainder of '2' when divided

by 3?

(a)1615 (b)1635 (c)1645 (d)1655

17. How many keystrokes are needed to type numbers from 1 to 500 on a standard

keyboard?

(a)1392 (b)1292 (c)1497 (d)None of thes

Answer Key

Arithmetic progressions

1 C 2 b 3 c 4 c 5 b

6 B 7 a 8 b 9 c 10 b

11 B 12 b 13 a 14 b 15 c

16 B 17 a


Chapters 3 – Permutations and Combinations

Permutation is the arrangement of things whereas combination is selection of things. Again, in Permutation, ‗order‘ becomes important while ‗order‘ has no significance in combination. e.g. Event of hand-shaking in a party – Combination (order is not important). Event of exchanging cards in a party – Permutation (order is important). Definitions: 1. An arrangement without replacement is called a permutation. Arrangement of

cards, number problems without repetitions are examples of permutation. 2. A selection of n objects without replacement is called a combination. Fundamental rules: 1.The Sum Rule: Suppose a work A can be done in a ways and B can be done in b

ways and bothcannot occur simultaneously. Then A or B (at least one of them) can occur in (a + b) ways. Thisrule is also applicable for two or more exclusive events.

2. The Product Rule: Suppose there are two works A and B. Let the work A be done in a ways andthe work B in b ways. And the ways of happening of A and B are

independent to each other.Then, both A and B can be done in ab ways. For example, let there be two works A and B which can be done in 4 ways and 3 ways respectively. Then either of the work (A or B) can be done in 4 + 3 = 7

ways and both the works (A and B) can be done in 4 3 = 12 ways.

Important results 1. n! = 1·2·3·…n; 0! = 1

2. Number of permutation of n distinct things taken r at a time, 0 r n = n(n – 1) (n – 2) …(n – r + 1). = n!/(n – r)! = nPr. 3. The number of permutations of n distinct objects taken all at a time = n!.

4. The numbers of combinations of n objects taken r at a time, 0rn.

= ( 1)...( 1) !

1.2.3... ( )!. !

n

r

n n n r nC

r n r r

5. Number of permutations of n things, out of which p are alike and are of one type, q are alike and are of second type, r are alike and are of third type and rest are all different = n!/p!q!r!

6. Number of selections of r things (r n) out of n identical things is 1. 7. Number of permutations (arrangements) of different things taking r at a time when

things can be repeated any number of times = n n …r times = nr 8. Total number of selections of zero or more things from p identical things = p + 1. 9. Total number of selections of zero or more things from n different things

= nC0 + nC1 + nC2 + … + nCn = 2n. 10. Number of ways of distributing n identical things among r persons when each

person may get any number of things = n + r – 1Cr – 1

11. Distribution into Groups: i. The number of was in which n distinct objects can be split into three groups

containing respectively r, s and t objects, r, s and t are distinct and r + s + t = n, is given by


nCr·n – r Cs·n – r – s Ct = .!!!

!

tsr

n

ii. If 3n things are to be divided equally between 3 persons (i.e. division of 3n things into 3 equal groups with permutation of groups) then the number of ways

= 3!

!3

n

n

iii. If 3n things are divided into three equal groups, then the number of ways =

.)!(!3

)!3(

!3!!!

)!3(3n

n

nnn

n

Since for any one way the three groups can be placed in 3! ways without obtaining

the new division. So it is divided by 3!. 12. The greatest value of nCr i. When n is even, nCr is greatest when r = n/2. ii. When n is odd, nCr is greatest when r = (n + 1) /2 or (n – 1)/2. 13. If nCx = nCy, then either x = y or x + y = n 14. nCr – 1 + nCr = n + 1Cr 15. Restricted Permutations and Combinations:

a. Number of combinations of n distinct things taken r at a time when p-particular things always occur = (n – p)C(r – p). The reason is that since p particular things are to be taken in each selection, so to make the selection of r things, we are to select only (r – p) things from the remaining ( n – p) things which can be done in (n – p)C(r – p) ways.

b. The number of permutations of n distinct things taken r at a time when p

particulars things always occur = (n – p)C(r – p) r ! Since number of combinations = n – pCr – p and each combination consists of r

objects which can be permuted in r! ways. The result follows by the product rule.

c. Number of combinations of n distinct things taken r at a time when p particular things never occur = (n – p)Cr.

d. Number of permutations (arrangements) of n distinct things taken r at a time

when p particular things never occur = (n – p)Cr r! Example. Let there be 10 boys and 6 girls in a class and we have to select a group

of 6 in which three particular boys never come and two particular girls are

always selected. Here we have to select a group of 4 from remaining 11 students (7 boys and 4 girls), as 2 particular girls are already included in the group, which can be done in

11C4 ways. 16. Circular Permutations:

a. Arrangements round a circular table: A circular table has no fixed starting point or ending point and arrangements like those given below in the figure are considered identical.

If n persons are to be arranged in a straight line, there are n! different ways in which this can be done. When n persons are to sit round a circular table, each arrangement will be repeated n times, so there are (n – 1)! different arrangements. b. Arrangement of beads (all different) around a circular wire: A circular wire differs from a circular table because when we turn it

over we see that the other side


presents an arrangement of diferent beads different from that on the first side.

c. If the wire on the left is turned over we obtain the arrangements on the right. We can see that the only two different arrangements of three coloured beads on a circular wire can be found on opposite sides of same wire, i.e. there is actually one arrangement. More generally, n beads on a circular wire can be arranged in

)!1(2

1n different ways.


Practice Exercise on Permutations and Combinations

1. If nC1 = 8 and nCr = 28, find the value of n+ r (a) 9 (b) 10 (c) 11 (d) 12

2. From a group of 5 men and 3 women, three persons are to be selected to form a committee so that at least 2 men are there on the committee. In how many ways can it be done? (a) 30 (b) 35 (c) 40 (d) 45

3. In how many different ways can the letters of the word 'MINDWORKZZ' be

arranged in such a way that the vowels always come together? (a) 10! (b) 9! (c) 2 × 9! (d) 8!

4. There are 5 routes to go from Kanpur to Lucknow and 6 ways to go from Lucknow to Faizabad. How many ways are possible for going from Kanpur to Faizabad. (a) 30 (b) 11 (c) 56 (d) 65

5. In how many ways can a person send invitation cards to 5 of his friends if he has three servants to deliver the cards? (a) 53 (b) 35 (c) 120 (d) 6

6. If nC3 = nC7. What is the value of n? (a) 8 (b) 9 (c) 10 (d) 11

7. Karan draws a polygon of n sides. The polygon has 44 diagonals. Find the value of n (a) 11 (b) 10 (c) 9 (d) 8

8. In how many ways can the letters of the word HIPPOPOTAMUS be arranged?

(a) 12! (b)

(c)

(d)

9. How many three digit integers can be formed using the digits 0, 1, 2, 5, 8 and 9 (repetition of digits not allowed). (a) 60 (b) 100 (c) 160 (d) 720

10. What will be the number of integers formed in the previous question if repetition of digits is allowed? (a) 305 (b) 180 (c) 125 (d) None of these

11. There are 10 boys and 8 girls in a group. Two boys are to be selected and two girls are to be selected for the play. In how many ways can they be selected? (a) 18 (b) 73 (c) 80 (d) 1260

12. Karim throws two dice. In how many ways can he get the sum of numbers on the dice as 8? (a) 3 (b) 4 (c) 5 (d) 6


13. Hemant draws three cards from a standard deck of cards (without replacement). How many possibilities are there of selecting the cards of alternating colours? (a) 16900 (b) 33800 (c) 50700 (d) None of these

14. There are 7 red balls and 6 white balls in a bag. Two balls are taken out. Find the number of ways of selecting the balls such that the balls are of different colours? (a) 42 (b) 84 (c) 126 (d) 168

15. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? (a) 5 (b) 10 (c) 15 (d) 20

16. Badri has 9 pairs of dark Blue socks and 9 pairs of Black socks. He keeps them all in a same bag. If he picks out three socks at random what is the probability he will get a matching pair? (a) 1 (b) (2 × 9C2 × 9C1)/18C3 (c) (9C2 × 9C1)/18C3 (d) None of these

17. How many ways can 10 letters be posted in 5 post boxes, if each of the post

boxes can take more than 10 letters? (a) 510 (b) 105 (c) 10P5 (d) 10C5

18. A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel? (a) 60 (b) 98 (c) 126 (d) 158

19. How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes? (a) 12 (b) 64 (c) 81 (d) None of these

20. 36 identical chairs must be arranged in rows with the same number of chairs in each row. Each row must contain at least three chairs and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible?

(a) 4 (b) 5 (c) 6 (d) 7

21. A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made? (a) 210 (b) 630 (c) 840 (d) 1260

22. 10 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points? (a) 140 (b) 210 (c) 280 (d) 350

23. There are 6 non-collinear points. How many triangles can be drawn by joining these points? (a) 720 (b) 120 (c) 35 (d) 20


24. A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 7 from part P and 5 from part Q, in how many ways can he do that? (a) 30240 (b) 32400 (c) 40230 (d) 42300

25. Find out the number of ways in which 4 rings of different types can be worn in 3 fingers? (a) 12 (b) 24 (c) 64 (d) 81

26. In a birthday party, every person shakes hand with every other person. If there was a total of 55 handshakes in the party, how many persons were present in the party?

(a) 9 (b) 10 (c) 11 (d) 12

27. If the letters of the word ‗JITEN‘ are permuted in all possible ways and the words thus obtained are arranged alphabetically, then what is the rank of ‗JITEN‘? (a) 54 (b) 58 (c) 61 (d) 59

28. In how many ways can 4 man draw water from 4 taps if no tap can be used more than once? (a) 6 (b) 16 (c) 24 (d) 32

29. Consider the equation:- a + b + c = 10 Let the number of positive integral solutions be p and the number of non-negative integral solutions be q. Then the value of q-p (a) 30 (b) 36 (c) 66 (d) None of these

30. In how many ways can 6 beads be arranged to form a necklace? (a) 720 (b) 360 (c) 120 (d) 60

31. In how many ways can 9 people be seated at a round table? (a) 720 (b) 5040 (c) 40320 (d) None of these

32. 10 persons we invited to a party. In how many ways can they be seated in a round table such that two particular persons sit on either side of the host?

(a) 9! × 2 (b) 9! (c) 8! × 2 (d) 8!

33. If 7 people are going to sit around a circular table, but Kriti will not sit next to Heena, how many different ways can the group of 7 sit? (a) 240 (b) 480 (c) 720 (d) None of these

34. 5 directors, the vice chairman and the chairman are to be seated around a circular table. If the chairman should sit between a director and the vice chairman, in how many ways can they be seated? (a) 720 (b) 240 (c) 120 (d) None of these

35. In how many ways 12 boys and 4 girls can sit around a circular table, so that no two girls sit together.

(a)

(b)

(c)

(d)


Answer Key

1 b 2 c 3 c 4 a 5 b

6 c 7 a 8 d 9 c 10 a

11 d 12 c 13 b 14 b 15 d

16 a 17 a 18 c 19 c 20 b

21 d 22 b 23 d 24 a 25 d

26 c 27 d 28 c 29 a 30 d

31 c 32 c 33 a 34 b 35 d


Chapters 4 – Probability

1. Probability is defined as the chance that an event would occur.

2. Events are defined with the help of ANDs and ORs and the key to solving probability is the conversion of the Questionstatement into the Answerstatement. The following example would illustrate what we mean by this best: There are two boxes, one with 3 red, 4 blue and 5 white balls the other with 4 red, 5 blue and 6 white balls. A ball is drawn at random from one of the two boxes. Find the probability that the ball drawn is white. The description of the event above is what we can call as the ―Question Statement.‖ The ―Answer statement or the Event Definition‖ of the above question can be defined as below: Choose the First Box AND Choose a white ball OR Choose the Second Box AND Choose a white ball Hence, the required Probability would be: P (First Box)× P( White Ball) + P(Second Box) × P (White Ball) = (½)× (5/12) + ½ × (6/15) = 49/120 3. The Non Event and Its Use in Probability: For every event, we also can define the Non-event which is exactly the opposite of the event. The Non Event is also denoted by E‘. Thus, n(E) + n(E‘) = Total Sample Space i.e. the sum of the number of occurrences of the Event + the number of occurrences of the Non-Event = The total number of possible occurrences. Also, P(E) + P(E‘) = 1 Some examples of events and their non events: i) In a throw of a die, the chances of getting a number greater than 4. Non-Event: A number Less than or equal to 4. ii) In a throw of 2 dice, the chances of getting a number less than 10. Non-Event: A number Equal to or greater than 10. In other words 10 OR 11 OR 12. iii) The probability that Rahul passes at least 1 out of 4 exams. Non-Event: He fails all. iv) When a coin is tossed 8 times, the probability that heads turns up at least 2 times. Non-Event: Heads turns up once OR Heads does not turn up at all.

4. Key Definitions of Probability: i) Random Experiment An experiment whose outcome has to be among a set of events that are completely known but whose exact outcome is unknown is a random experiment (e.g. Throwing of a dice, tossing of a coin). ii) Sample Space This is defined in the context of a random experiment and denotes the set representing all the possible outcomes of the random experiment.[e.g. Sample space when a coin is tossed is(Head, Tail). iii) Impossible Event An event that can never occur is an impossible event. The probability of an impossible event is 0. e.g.(Probability of the occurrence of 7 when a dice with 6 faces numbered 1-6 is thrown). iv) Mutually Exclusive Events Two or more events are mutually exclusive if they cannot occur together. (E.g. If an even number appears on a die, an odd number would not appear.)


v) Equally Likely Events If two events have the same probability or chance of occurrence they are called equally likely events. (in a throw of a dice, the chance of 1 showing on the dice is equal to 2 is equal to 3 is equal to 4 is equal to 5 is equal to 6 appearing on the dice.) vi) Exhaustive Set of Events A set of events that includes all the possibilities of the sample space is said to be an exhaustive set of events. (e.g. In a coin toss, Head or tail is an exhaustive list of possibilities.) vii) Independent Events An event is described as such if the occurrence of an event has no effect on the probability of the occurrence of another event. (If the first 100 coin tosses are heads, there is no change to the fact that the probability of a heads in the 101st throw remains 0.5) viiii) Conditional Probability It is the probability of the occurrence of an A given that

the event B has already occurred. This is denoted by P(A/B). (Example: The probability that in two throws of a die we get a total of 7 or more, given that in the first throw of the die the number 5 had occurred) 5. The concept of Odds For and Odds Against Sometimes, probability is also viewed in terms of odds for and odds against an event. Odds in favour of an event E is defined as: P(E)/P(E)‘ Odds against an event is defined as: P(E)‘/P(E)

Worked Out Examples 1. If two cards are simultaneously drawn from a pack of well shuffled cards, then find the probability of both being diamonds? Solution. 13c2/52c2

2. If four dice are thrown simultaneously, what is the probability that the sum of the numbers is exactly 20? Solution .We have to check for the different combinations to check that the sum formed is 20. A sum of 20 can be formed in the following ways: (5,5,5,5)---------4!/4!=1 (6,5,5,4)---------4!/2!=12 (6,6,6,2)--------- 4!/3!=04 (6,6,5,3)---------4!/2! =12 (6,6,4,4)---------4!/2!.2!=06 Number of favourable cases are 35 Total number of cases=64=1296. Hence, the required probability =35/1296 3. How many numbers can be formed with digits1,3,5,7 without repetition? Solution. We have to form 1digit,2 digit,3digit,4 digit numbers. Case1: There are 4 numbers of 1 digit. Case 2: There are 4*3=12 numbers of 2 digits Case 3: There are 4*3*2= 24 numbers of 3 digits Case 4: There are 4*3*2*1= 24 numbers of 4 digits 4. A test paper consists of 10 questions and each question has 5 choices. If each question is necessarily attempted, then find the number of ways of answering the test paper?


Solution. Since each question can be answerd in 5 ways, so the total number of ways of answering the question is. 5*5*5*5*5……..10 times=510. 5. In how many ways can the letters of the word REPEAT be arranged? Solution.6!/2!=720/2=360 6. In how many ways can the letters of the word SATNA can be re-arranged? Solution.5!/2!-1= 59 7. When four cards are drawn in succession from a pack of cards with replacement, what is the probability that all the cards are from different suits? Solution.13/52*13/52*13/52*13/52 =1/4*1/4*1/4*1/4=1/256 8. A bag contains 2 red ,3 blue and 6 white balls. If a ball is drawn at random, a) What is the probability that it is not a blue ball

Solution. One ball can be drawn out of the 11 balls from the bag in 11C1 ways. Similarly, a ball other than a blue ball(2+6=8) can be drawn in 8C1 ways.

Hence, the probability that the ball drawn is not a blue ball is 8C1/11C1=8/11 9. For the above question, what is the probability that it is a white ball? One white ball out of 6 white balls in the bag can be drawn in 6C1 ways . Hence, the probability that the ball drawn is white = 6/11. 10. A person is moving from Patna to Agra and then to from Agra to Kolkatta. There are 3 ways to move from Patna to Agra and 5 ways to move from Agra to Kolkatta. In how many ways can he move from Patna to Kolkatta. Solution. He can move from Patna to Agra in 3 ways and in 5 ways he can move from Agra to Kolkatta. So in total 3*5=15 ways 11. Twenty eight games were played in a football tournament with each team playing once against each other. How many teams were there?

Solution: Let the number of teams is n, then number of matches to be played = nC2 = 28

056nor 282

1 2 n)n(n-

or (n – 8)(n + 7) = 0

n = 8 as n –7.

12. Find the number of ways in which 6 boys and 6 girls can be seated in a row so that

i. no two girls sit together.

ii. boys and girls sit alternately.

iii. all the girls sit together and all the boys sit together

iv. all the girls are never together.

Solution:

i. gbgbgbgbgbgbg


Let b denotes boys, then g denotes the possible places for girls which are 7. So selection of 6 places for girls = 7C6

Permutation of 6 boys and 6 girls among themselves = 6!6!

ii. There are two possible arrangements:

bg bg bg bg bg bg = 6!6!

And gb gb gb gb gb gb = 6!6!

Since only one of these may happen, so by the sum rule

Number of arrangements = 2·(6!)2

iii. Considering boys and girls as seperate units, their permutations

= 2!·6!·6! = 2·(6!)2

iv. Total arrangements that all the girls are not together = (12)! – 7!·6!.

13. How many different words can be formed from the letters of the word LUCKNOW when

a. all the letters are taken,

b. words begin with L,

c. the letters L and W respectively occupy the first and last places,

d. the vowels are always together,

e. the letters U, C, K are never together,

f. the vowels always occupy even places?

Solution:

In Lucknow,

Number of letters – 7 (all distinct)

Vowels (u, o) = 2

Consonants = 5

a. If all the letters are given, number of words = number of arrangements of 7 letters

= 7! = 5040

b.Having fixed up L at first place, remaining 6 letters can be arranged in 6! = 720 ways.

c.Fixing L at first and W at last place, remaining 5 letters can be arranged in 5! = 120.

d.Considering the two vowels u and o as one, we are left with 6 letters and permutation of

6 letter = 6!

u and o can be interchanged in 2!


number of words in this case

= 6! 2! = 1440.

e.Numbers of words in which U, C, K are never together (all three not together)

= total number of words – number of words in which U, C, K are together.

Consider U, C, K as one letter than permutation of (UCK), LNOW = 5!

U, C, K can be permuted in 3!

So number of words = 5! 3! In which U, C K are together.

number of words in which U, C, K are never together 5040 – 5!3!= 5040 – 720 = 4320

f. There are 7 letters so 7 places are required. Number of even places is 3 and vowels are 2. Now 2 vowels can be placed in 3C2·2! = 6 and 5 consonants can be placed in remaining places in 5! ways.

Number of words = 6.5! = 720.


Practice Exercise on Probability

1. A die is rolled, find the probability that a prime number is obtained (a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3

2. Three coins are tossed, find the probability that two heads are obtained. (a) 1/4 (b) 3/8 (c) ½ (d) 5/8

3. Two dice are rolled, find the probability that the sum is 4 (a) 1/6 (b) 1/9 (c) 1/12 (d) 1/18

4. A die is rolled and a coin is tossed, find the probability that the die shows an even number and the coin shows a tails (a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3

5. A card is drawn at random from a deck of cards. Find the probability of getting a queen or a spade. (a) 1/13 (b) 17/52 (c) 3/13 (d) 4/13

6. A jar contains 2 red marbles, 4 green marbles and 7 white marbles. If a marble is drawn from the jar at random, what is the probability that this marble is white? (a) 7/13 (b) 2/13 (c) 4/13 (d) None of these

7. The blood groups of 200 people is distributed as follows: 45 have type A blood, 60 have B blood type, 65 have O blood type and rest have type AB blood. If a person from this group is selected at random, what is the probability that this person has AB blood type? (a) 0.10 (b) 0.15 (c) 0.20 (d) 0.25

8. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 4? (a) 1/5 (b) 1/4 (c) 1/3 (d) ½

9. In a box, there are 4 red, 5 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? (a) 1/3 (b) 1/4 (c) 1/5 (d) 1/6

10. Three unbiased coins are tossed. What is the probability of getting at most two tails? (a) 5/8 (b) 3/4 (c) 7/8 (d) None of these

11. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even? (a) 5/8 (b) 3/4 (c) 7/8 (d) None of these

12. In a class, there are 10 boys and 5 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is: (a) 43/91 (b) 44/91 (c) 45/91 (d) 46/91


13. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings? (a) 1/221 (b) 2/221 (c) 3/221 (d) 4/221

14. A bag contains 3 white, 4 red and 5 blue balls. Two balls are drawn at random from the bag. The probability that both of them are blue is_______ (a) 6/21 (b) 4/21 (c) 1/7 (d) 2/21

15. A bag contains 5 black and 6 white balls. James takes a ball out and places it in the box. He again takes out a ball. What is the probability that both the balls are white? (a) 25/121 (b) 36/121 (c) 3/11 (d) 49/121

16. What is the probability of getting at least one four in a single throw of three

unbiased dice? (a) 1/6 (b) 125/216 (c) 1/36 (d) 91/216

17. A four digit number is formed using the digits 0, 1, 2, 5, 8, 9. Find the probability that the number formed is a multiple of four (repitiition of digit not allowed). (a) 3/10 (b) 2/5 (c) 1/2 (d) 3/5

18. In the previous question, what will be the probability if repition of digits is allowed (a) 1/3 (b) 2/5 (c) 11/30 (d) None of these

19. In a game there are 50 people in which 30 are boys and 20 are girls, out of which 10 people are selected at random. One from the total group, thus selected is selected as a leader at random. What is the probability that the person, chosen as the leader is a boy? (a) 1/3 (b) 2/5 (c) 3/5 (d) 2/3

20. A bag contains 21 toys numbered 1 to 21. A toy is drawn and then another toy is drawn without replacement. Find the probability that both toys will show even numbers

(a) 3/14 (b) 3/7 (c) 4/7 (d) 5/14

21. A speaks truth in 60% of cases and B in 55% of cases. In what percent of cases are they likely to contradict each other in narrating the same event? (a) 1/2 (b) 49/100 (c) 147/400 (d) None of these

22. The odds against an event are 3:4 and the odds in favour of another independent event are 3:5. Find the probability that at least one of the two events will occur. (a) 5/7 (b) 11/14 (c) 6/7 (d) None of these

23. In a charity show tickets numbered consecutively from 101 through 350 are placed in a box. What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2? (a) 0.30 (b) 0.35 (c) 0.40 (d) 0.45


24. The probability of success of three students A, B and C in the one examination are 1/3, 1/2 and 1/4 respectively. Find the probability of all three students failing in the examination. (a) 0.25 (b) 0.35 (c) 0.45 (d) 0.55

25. A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack? (a) 4/13 (b) 9/13 (c) 5/13 (d) 10/13

26. There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel? (a) 3/8 (b) 1/2 (c) 5/8 (d) 3/8

27. In a race, the odd favour of horses P,Q,R,S are 1:3, 1:4, 1:5 and 1:6 respectively.

Find the probability that one of them wins the race. (a) 3/7 (b) 311/420 (c) 16/21 (d) 319/420

28. Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice. (a) 1/3 (b) 13/36 (c) 11/36 (d) None of these

29. Urn A contains 4 white and 5 black balls and Urn B contains 6 white and 7 black balls. One ball is transferred from Urn A to Urn B and then one ball is drawn from the Urn B. Find the probability that the ball drawn is white. (a) 3/7 (b) 29/63 (c) 4/7 (d) 31/63

30. Three cards are drawn from a deck of cards. Find the probability such that one of

them is a spade, one is a diamond and one is a heart. (a) 3/13 (b) 1/133 (c) 6/133 (d) 10/133

Answer Key

1 a 2 b 3 c 4 c 5 d 6 a 7 b 8 d 9 a 10 c

11 b 12 c 13 a 14 d 15 b 16 d 17 a 18 c 19 c 20 a

21 b 22 b 23 c 24 a 25 b 26 a 27 d 28 c 29 b 30 c


Chapters 5 – Set Theory

Set theory is an important concept of mathematics which is open asked in aptitude exams. There are two types of questions in this chapter:

(i) Numerical questions on set theory based on venn diagrams

(ii) Logical questions based on set theory

Let us first take a look at some standard theoretical inputs related to set theory. Set Theory Look at the following diagrams:

Figure 1: Refers to the situation where there are two attributes A and B. (Let‘s say A refers to people who passed in Hindi and B refers to people who passed in English.) Then the shaded area shows the people who passed both in Hindi and English.

In mathematical terms, the situation is represented as: Total number of people who

passed at least 1 subject = A + B – A B

Figure 2:Refers to situation where are three attributes being measured. In the figure below, we are talking about people who passes English, Hindi and/or Sanskriti.

In the above figure, the following explain the respective areas: Area 1:People who passed in English only. Area 2:People who passed in English and Hindi only. (in other words-People who passes English and Hindi but not Sanskriti) Area 3: People who passed Hindi only Area 4: People who passed in Hindi and Sanskriti only. (also, can be describe as people who passes Hindi and Sanskriti but not English) Area 5: People who passed in Sanskriti and English only. (also, can be describe as people who passes Sanskriti and English but not Hindi) Area 6: People who passed English, Hindi and Sanskriti Area 7: People who passed Sanskriti


Area 8:People who passed in no subjects Also take note of the following language which can be confusing to interpret: People passing English and Hindi – Represented by the sum of area 2 and 6 People passing English and Sanskriti – Represented by the sum area 5 and 6 People passing Hindi and Sanskriti – Represented by the sum area 4 and 6 People passing English – Represented by the sum of the area 1, 2, 5 and 6 In Mathematical terms , this means:

Total number of people who passed at least 1 subject = Let us consider the following questions and see how these figures work in terms of real tie problem solving: Illustration 1: In Mindworkzz club all the members like either in Superman or Batman. 320 like in the Batman, 350 like in Superman and 220 like in both. How many members does the club have? (a)410 (b)550 (c)440 (d)None of these

The total number of people = Options (d) is correct. Illustration 2:Directions for question 1 to 3: Refer to the data below and answer the questions that follow. Last year, there were 3 sections in ASCC, a mock CAT paper. Out of them 33 students cleared the cutoff in Section A, 34 students cleared the cutoff in Section B and 32 cleared the cutoff in Section C. 10 students cleared the cutoff in Section A and Section B, 9 cleared the cutoff in Section B and Section C, 8 cleared the cutoff in Section A

and Section C. The number of people who cleared each section alone was equal and was 21 for each section. 1. How many cleared all the three sections?

(a)7 (b)6 (c)5 (d)7

2. How many cleared only one of the three sections? (a)21 (b)63 (c)42 (d)52

3. The ratio of the member of students clearing the cutoff in one or more of the sections to the number of students clearing the cutoff in Section A alone is?

(a)78/21 (b)3 (c)73/21 (d)None of these


From the Venn Diagram, 21 + a + 8 – a + 9 – a = 32 a = 6

The answers would be: 1. 6. Option (b) is correct. 2. 21+21+21=63. Option (b) is correct.

3. (21+21+21+6+4+3+2)/21 = 78/21. Option (a) is correct.

Illustration 3: In a party, four different drinks – Pepsi, Limca, Sprite, Mirinda- are available. 20% of the guests do not have any drink. The four drinks given in the above order are taken by 230, 180, 180 and 220 guests respectively. The number of guests taking exactly 2 drinks for any two drinks is 20. There are 30 guests who take all the four drinks but there is nobody who takes exactly three out of four drinks. 1. How many guests have exactly two drinks? 2. How many guests have exactly one drink? 3. How many guests do not have any drink at all? 4. What percentage of the guests taking Limca also have at least one other drink? Solution:


1. Guests exactly have two drinks = 20+ 20+ 20+ 20+ 20+ 20 = 120.

2. Guests exactly have one drink= 140+ 130+ 90+ 90 =450.

3. Total guests who have one or more drinks=90+ 20+ 140+ 90+ 20+ 130+ 20+

20+ 20+ 20+ 30= 600.

80% of the total students = 600 So 20% of total students = 600/4 =150 So total number of students who do not have any drink at all = 150.

4. Total guests having Limca = 180

Total number of the guests taking Limca also have at least one other drink = 30+ 20+ 20+ 20 =90.

So the required percentage =

Illustration 4: In a college, where every student follows at least one of the three teams- CSK, RCB, or KKR- 60% follow CSK, 82% follow RCB, and 66% follow KKR. What can be the maximum and minimum percentage of students who follow

i) all three teams

ii) exactly two teams

Solution: Percentage of students who follow CSK + Percentage of students who follow RCB + Percentage of students who follow KKR = 60% + 82% + 66% = 208% i.e. surplus = 108%. This surplus can be accommodated throughadding elements either to intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of 8% will still be left.

This surplus of 8% can be accommodated by adding elements to intersection of three sets. For that we have to take 8% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three = 8%. In this case the percentage of students who follow exactly two teams will be maximum = 92%. The surplus of 108% can also be accommodated through adding elements to only intersection of three sets. As adding 1 element to intersection of three sets give a surplus of 2 sets, adding 54% to intersection of three sets will give a surplus of 108%. Therefore, the maximum value of students who follow all three teams is 54%. In this case the percentage of students who follow exactly two teams will be minimum = 0%.


Practice Exercise on Set Theory

Directions for questions 1-2: In a school of 350 students, 100 students are in the band, 200 students are on sports teams, and 50 students participate in both activities.

1. How many students are involved neither in band nor sports?


2. What is the ratio of the number of students who participate only in band to the number of students who participate only in sports?

(a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 1

Directions for questions 3-4: There are 60 students in a class. 60% of the students fail in English. 30% of the students pass in Maths. 20% of the students pass in both.

3. How many students failed in either of the subjects?

(a) 30 (b) 12 (c) 6 (d) 24

4. How many students who passed only in Maths? (a) 30 (b) 12 (c) 6 (d) 24

Directions for questions 5-7: A veterinarian surveys 52 of his patrons. He discovers that 28 have dogs, 20 have cats, and 10 have parrots. Further, 8 have a dog and a cat, 6 have a dog and a parrot, and 2 have a cat and a parrot. If no one has all three kinds of pets? 5. How many patrons own only a dog?

(a)2 (b) 10 (c) 12 (d) 14

6. How many patrons have at least 2 pets? (a) 16 (b) 14 (c) 10 (d) 8

7. How many patrons have neither of the three pets?

(a) 16 (b) 14 (c) 10 (d) 8

Directions for questions 8-10: A teacher is planning language classes for 105 students. 48 students say they want to take French, 48 want to take Spanish, and 33 want to take Latin. 15 say they want to take both French and Latin, and of these, 9 wanted to take Spanish as well. 15 want only Latin, and 24 want only Spanish.15 students want none of the subjects.

8. How many students want French only?


9. How many students want Spanish but not Latin? (a) 24 (b) 33 (c) 36 (d) None of these

10. How many students want at most two subjects?


(a) 96 (b) 81 (c) 66 (d) 60

Directions for questions 11-14:A survey of faculty and graduate students at the Mumbai Film Academy revealed the following information: 51 admire Aamir Khan, 49 admire Shahrukh Khan, 60 admire Salman Khan, 34 admire Aamir Khan and Shahrukh Khan, 32 admire Shahrukh Khan and Salman Khan, 36 admire Aamir Khan and Salman Khan, 24 admire all three of the Khans and 3admire none of the three Khans

11. How many people were surveyed?

(a) 82 (b) 83 (c) 84 (d) 85

12. How many admire Shahrukh, but not Salman or Aamir? (a) 5 (b) 7 (c) 16 (d) 24

13. How many admire either Salman or Aamir but not Shahrukh? (a) 33 (b) 38 (c) 43 (d) 48

14. How many admire not more than one of the Khans? (a) 28 (b) 31 (c) 84 (d) 39

Directions for Questions 15-17: There are 79 Grade 10 learners at school. All of these take some combination of Maths, Geography and History. The number who take Geography is 41; those who take History is 36; and 30 take Maths. The number who take Maths and History is 16; the number who take Geography and History is 6, and there are 8 who take Maths only and 16 who take History only.

15. How many learners take Maths and Geography but not History?

(a) 6 (b) 10 (c) 16 (d) 22

16. How many learners take Geography only? (a) 24 (b) 29 (c) 34 (d) 39

17. How many learners take all three subjects?

(a) 8 (b) 6 (c) 4 (d) 2

Directions for Questions 18-20: In a group of 200 children, 30% play hockey, 40% play football and 40% play cricket. 18% play Hockey and Football, 10% play football and cricket, 15% play hockey and cricket. Only 5% play all the three games.

18. How many children play only one game?

(a) 39 (b) 65 (c) 78 (d) 91

19. How many children play atleast two games? (a) 66 (b) 70 (c) 76 (d) 82

20. How many children play hockey or football but not cricket? (a) 20 (b) 26 (c) 34 (d) 64


Answer Key

1 b 2 c 3 a 4 c 5 d

6 a 7 c 8 b 9 c 10 a

11 d 12 b 13 a 14 b 15 a

16 b 17 d 18 c 19 a 20 d


Chapters 6 – Logical Reasoning

QUESTIONS FOR PRACTICE Directions for questions 1 and 2: Answer the questions on the basis of the information given below. Five executives of a company, namely Arun, Barun, Charu, Dharam and Ehsan are to be seated around a circular table for a meeting. Two of them are from Jaipur and the remaining three are from Udaipur. They must be seated under the following constraints: (i) Both the persons from Jaipur cannot be seated together.

(ii) The persons adjacent to Ehsan must be either both Arun and Charu or neither of Arun and Charu.

(iii) Ehsan must have a person from Jaipur to his Immediate right. Any additional information provided in a particular question pertains to that

individual question only.

1. If both the persons adjacent to Arun are from Jaipur, then which of the following

statements is/are definitely TRUE?

(i) Barun and Dharam are adjacent to Ehsan.

(ii) Charu is to the immediate right of Arun.

(iii) Arun is two places away to the right of Ehsan.

(a) Only (i) (b) Only (i) and (ii) (c) Only (ii) and (iii)

(d) (i), (ii) and (iii)

2. If Arun is from Udaipur and is sitting to the immediate left of Ehsan, who among

the following is definitely from Jaipur?

(a) Barun (b) Charu (c) Dharam (d) None of the above

Directions for questions 3 and 4: The questions given below are followed by two

statements, 1 and 11. Study the information given in the two statements and assess

whether the statements are sufficient to answer the questions and choose the

appropriate option from among the choices given below.

Mark (A) If the question cannot be answered on the basis of the two statements.

Mark (B) if the question can be answered by using both the statements together but

not by either of the statements alone.

Mark (C) if the question can be answered by using either of the statements alone.

Mark (D) if the question can be answered by using statement II alone but not by using

statement I alone.

Mark (E) if the question can be answered by using statement I alone but not by using statement II alone. 3. Six persons – A, B, C, D, E and F – participated in a race in which every

participant finished in a different time. At least two persons finished before B. The number of persons who finished before D is same as the number of persons who finished after F. A finished before E. Who finished the in second position?


I. E finished the race in third position. II. Only D finished the race after C 4. Mohan and Naman are the father and mother of Pulkit respectively. Pulkit has

four uncles and three aunts. None of the siblings of Mohan and Naman are married. Naman has two siblings. How many sisters does Mohan have?

I. Mohan has a total of 5 siblings. II. Naman has two brothers.

Directions: Each question is followed by two statements, I and II. Answer each question using the following instructions: Mark (A) if the question can be answered by using statement I alone but not by using

statement II alone.

Mark (B) if the question can be answered by using statement II alone but not by using

statement I alone.


Mark (D) if the question can be answered by using both the statements together but


Mark (E) if the question cannot be answered on the basis of the two statements.

5. Eight persons — Atul, Binny, Chatur, Dany, Ehsaan, Faroq, Gauti and Himesh —

are travelling in an aircraft which has four rows of seats. Each row has a capacity of two, one on the left side and one on the right side. All the seats on the left hand side are in one column while all the seats on the right hand side are in another column. Atul and Faroq sit on the left side in the first row and the third row respectively. Ehsaan sits immediately behind Faroq and Binny sits on the right side In the second row. Find Dany's position.

I. Dany does not sit in the same row or the same column which has a person whose name begins with a vowel.

II. Dany does not sit on the right side in any row.

Directions: The question is followed by two statements, I and II. Answer the question

using the following instructions.

Mark (A) if the question can be answered by using statement I alone but not by using

statement II alone.

Mark (B) If the question can be answered by using statement II alone but not by using

statement I alone.


Mark (D) if the question can be answered by using both the statements together but


Mark (E) if the question cannot be answered on the basis of the two statements.

6. Top six rankers, of a class, Ajay, Bina, Celina, Dona, Evan and Fredrick, are

sitting in a row, facing the same direction, not necessarily in the same order,

such that no two persons whose ranks are consecutive are sitting together. Ajay

is sitting to the immediate left of Celina. Among the six, the number of persons

who got a rank better than Ajay's rank is the same as the number of persons who


got a rank worse than Celina's rank. What Is the rank of the person who is sitting

at the extreme right of the row?

I. Dona, who is the first ranker, is not sitting at any of the extremes but is three

places away to the right of the third ranker.

II. Fredrick, who is the fourth ranker, Is sitting third from the left end of the row.

Directions: Select the correct alternative from the given choices.

7. Mr. Gupta has three children, but does not remember their ages or the months of

their birth. The following information is available about the ages of his children:

(a) The girl who was born in July is 8 years old.

(b) One of the children is 5 years old, but it is not Guddi.

(c) One of the children was born in October, but it is not Sreyashi.

(d) Samar's birthday is in May.

(e) The youngest child is only one year old.

Based on the above information, which of the following is true?

(A) Shreyashi is the oldest, followed by Guddi, who was born in October, and the

youngest is Samar, who was born in May.

(B) Guddi is the oldest, born in July, followed by Samar, who is 5 years old and the

youngest is Shreyashi, who is one year old.

(C) Shreyashi is the oldest, being 8 years old, followed by Samar, who was born in May

and the youngest is Guddi. who was born in October.

(D) Samar is the oldest, born in May, followed by Shreyashi, who was born in July and

Guddi, who was born in October.

8. Directions for questions: Select the correct alternative from the given choices.

One evening, two friends – Amar and Bimal – were sitting in a lawn, facing each

other. If Amar‘s shadow was falling exactly on her left, then which directions was

Bimal facing?

(a)North (b) South (c)West (d)East

Directions for questions 9 and 10: Answer these questions on the basis of the

information given below.

In Gum Shum Island, every inhabitant gives two statements as an answer to any

question. One of these two statements is always true and the other is always false, but

in no particular order. I came across three inhabitants of the Island, Alka, Binni and

Chaya, while searching for the local priest. When I asked them about how I could find

the priest, their replies were as follows:

A: Only the priest wears white pants. The priest always wears a blue shirt.

B: Only the priest wears white pants. Chaya is my dad.

C: The priest always wears a blue shirt. Binni is my son.

9. From the above statements, which of the following can be confirmed?

I. There is at least one male and at least one female among Alka, Binni and Chaya


II. Chaya is a parent of Binni

III. The priest wears either a blue shirt or white pants, but not both

(a) Only II (b) Only Ill (c) Only II and III (D) Only I and II

(e) I, II and III

10. If it is known that the priest is male, at most how many of the following

statements can be TRUE simultaneously?

I. The priest is wearing a blue shirt today.

II. The priest is wearing white pants today.

III. The priest was not wearing white pants yesterday.

IV. The priest was not wearing a white shirt yesterday.

V. Alka is wearing a blue shirt today.

VI. Chaya was wearing white pants yesterday.

(a) 6 (b) 5 (c) 4 (d) 3 (e) 2

11. Five players — Ajay, Aryan, Asit, Alok and Abhi — played five overs of cricket

among themselves. Each of the five players bowled exactly one over and also

batted for exactly one over. The runs conceded by players in the respective overs

bowled by them are 1, 2, 3, 4 and 5, not necessarily in the same order. Ajay

bowled to Alok and conceded 1 run and he scored 2 runs in Aryan's over. Asit

neither scored 3 runs nor conceded 3 runs. Alok did not bowl to Asit. If Aryan

batted when Asit bowled, then he scored 3 runs.

Which of the following statements CANNOT be true?

(a) Abhi conceded 4 runs to Asit.

(b) Alok neither bowled to Abhi nor batted in Abhi's over.

(c) Abhi did not score 3 runs.

(d) Alok scored 2 runs less than what he conceded.

(e) both (b) and (c)

Directions for questions 12 and 13: Answer the questions on the basis of the

information given below.

Mr. Gupta wants to buy a book and is confused between four books of different

subjects - Physics, Chemistry, biology and computer. The books are written by L, M, N

and O and published by P, Q, R and S, not necessarily in the same order. The

Chemistry book is published by Q and the computer book is written by N. Each book

is written by a different author and published by a different publisher. It is also known

that L and M get their books published by P or Q only.

12. If the Physics book is written by O then who can be the publisher of the biology

book?

(a) P or Q (b) Only P (c) P or R (d) P or R or S

13. How many combinations of publisher and author are possible for the Physics

book?


(a) 6 (b) 3 (c) 4 (d) 5

14. Each one of the three friends Bimal, Lalit and Shahir is either from Kanpur or

from Jaipur. Bimal and Shahir make statements which are given below.

Bimal: Both Lalit and Shahir are from Jaipur.

Shahir: Lalit says that he is not from Jaipur.

If it is known that the residents of Kanpur never speak a lie and the residents of

Jaipur never speak the truth, then who is definitely from Jaipur?

(a) Bimal (b) Shahir (c) Both Bimal and Shahir (d) Lalit

15. A, B, C, D and E are statements such that, if A is true then both B and D are

true, and if both C and D are true then E is false. So we can conclude that

(a) if E is true then both A and C must be true.

(b) if E is true then both A and C must be false.

(c) if E is true then atleast one of A and C must be true.

(d) if E is true then atleast one of A and C must be false.

16. A team of three, comprising two boys and one girl, has to be selected from a

group of seven people A, B, C, D, E, F and G in which there are three girls and

four boys. Some of the possible teams are shown below:

Team 1 Aman Chris Ehsan

Team 2 Baman Dona Floura

Team 3 Ehsan Floura Gaytri

Team 4 Aman Baman Dona

Who among the following is/are girl(s)?

(a) Aman (b) Chris (c) Gaytri (d) Both Chris and Gaytri

Directions for questions 17 to 19: Answer the following questions on the basis of

information given below:

Four friends MOHIT, MANAN, MAYA and MOHAN own four different brands of bikes

SPLENDER, PASSION, PULSUR and DISCOVER and four different brands of cars

BENZ, BMW, AUDI and TOYOTA respectively. Each person likes exactly one of the four

bikes and one of the four cars mentioned. Each bike and each car is liked by exactly

one of the four persons. Further it is known that:

i. MOHIT likes the bike of the person who likes BENZ and DISCOVER.

ii. The person who likes PULSUR is the only person who likes his own car and he is

not MAYA.

17. Which bike is liked by MANAN?

(a) SPLENDER (b) PASSION (c) PULSUR (d) DISCOVER

18. Which car is liked by MAYA?

(a) BENZ (b) BMW (c) AUDI (d) TOYOTA


19. Which of the following statements is correct?

(a) MOHIT likes MANAN‘s bike and MOHAN‘s car.

(b) DISCOVER‗s owner likes BENZ.

(c) The person who likes AUDI also likes PASSION.

(d) TOYOTA is liked by the owner of PULSUR.

Directions for questions 20 and 21: Answer the following questions on the basis of

information given below:

Six friends A, B, C, D, E and F who are to be seated on six chairs numbered 1 to 6,

each facing North, with 1 being the chair lying on the extreme left and 6 on the

extreme right. Some other information is also given regarding their seating plan.

• C is sitting to the right of A and E.

• B is sitting to the left of D.

• E is sitting immediately to the left of F.

• D is not sitting in any of the two rightmost positions.

20. How many positions are possible for A?

(a) 2 (b) 3 (c) 4 (d) 5

21. If there are exactly two people between B and F, who is sitting second from the

left?

(a) B (b) D (c) Either (a) or (b) (d) None of these

Directions for questions 22 and 24:

Each of these questions are based on the information given below :

1. 8 persons E, F, G, H, I, J, K and L are seated around a square table - two on each side.

2. There are 3 ladies who are not seated next to each other.

3. J is between L and F.

4. G is between I and F.

5. H, a lady member is second to the left of J.

6. F, a male member is seated opposite to E, a lady member.

7. There is a lady member between F and I.

22. Who among the following is to the immediate left of F ?

(a) G (b) I (c) J (d) H

23. What is true about J and K?

(a) J is male, K is female (b) J is female, K is male

(c) Both are female. (d) Both are male

24. Who among the following is seated between E and H?

(a) F (b) I (c) K (d) Cannot be determined


Directions for questions 25 to 27: Answer the questions on the basis of the

information given below. Salma, Sujit, Sushma and Sanjit visited Fariq one at a time

in the following manner.

I. At least one person visited Fariq between Salma and Sujit.

II. At least one of Sushma and Sanjit visited Fariq before Salma.

III. Sushma did not visit Fariq between Sujit and Sanjit.

25. If Sujit was the last person to visit Fariq, then who visited Fariq first?

(a) Salma (b) Sanjit (c) Sushma (d) Cannot be determined

26. Who could not be the last person to visit Fariq?

(a) Salma (b) Sujit (c) Sushma (d) Sanjit

27. In how many ways could the four people visit Fariq?

(a) 2 (b) 3 (c) 4 (d) 5

28. Four friends, one of whom has stolen a gold ring, are interrogated by the police.

Each of them makes two statements, one of which is true and the other is false.

Ramesh said: It was me. It wasn‘t Rahman.

Rahman said: It was Sheela. It wasn‘t Ramesh.

Shyama said: It was Rahman. It wasn‘t Sheela.

Sheela said: It was Shyama. It was Ramesh.

Who has stolen the gold ring?

(a) Ramesh (b) Shyama (c) Rahman (d) Sheela

29. Five books A, B, C, D and E are to be distributed among five people P, Q, R, S

and T such that each person receives exactly one book. Further information is

given below:

P cannot receive A, B or C.

Q can only receive B or C.

D can only be received by R or S.

A can only be received by S or T.

In how many of the following cases can the distribution of books be uniquely

determined? Case I. R receives C.

Case II. S receives D.

Case III. T receives B.

(a) 0 (b) 1 (c) 2 (d) 3

30. M, N, O and P are playing a game of ‗passing the coins‘. Each of them has an

equal number of coins ‗n‘ at the beginning of the game. In the first round, M

starts by passing one coin to N; N passes two coins to O; O passes three coins to

P; and P passes four coins to M. In the second round, M passes five coins to N; N


passes six coins to O; O passes seven coins to P; and P passes eight coins to M.

The process continues till one person gets all the coins. Who is that person?

(a) M (b) N (c) P (d) Depends on the value of n

Answer Key

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

6. a 7. c 8. a 9. d 10. b

11. e 12. b 13. c 14. a 15. b

16. d 17. d 18. b 19. c 20. c

21. c 22. c 23. d 24. c 25. c

26. d 27. d 28. b 29. c 30. c


Chapters 7 – Data Interpretation

DI Based on Percentages Directions for Questions for 1-4: The following pie chart shows the hourly distribution (in degrees) of all the major activities of a student.

1. The percentage of time which he spends in school is:

(a) 38% (b) 30% (c) 40% (d) 25%

2. How much time (in per cent) does he spend in games in comparison to sleeping? (a) 30% (b) 40% (c) 25% (d) None of these

3. If he spends equal time in games and in homework and remains constant in other activities, then the percentage decrease in time of sleeping is: (a) 15% (b) 12.5% (c) 20% (d) None of these

4. If he spends 1 hour of homework on Mathematics, then what percent of his time on homework is spent on rest of the subjects in home work: (a) 33.33% (b) 40% (c) 50% (d) 66.67%

Directions for Questions for 5-8: The bar graph below shows the marks obtained by 10 students of a class in three subjects, English, Maths and Science. All the subjects are marked out of 100. Answer the questions based on the graph below.

85 78 64 80 88 92 45 38

91 77

87 82 60

73 89 97

41 39

92 73

83 84

54

74

92 90

48 49

95

72

0

50

100

150

200

250

300

Aman Bikas Chitra David Eram Fahim Gaurav Harshit Ishita Jameel

English

Maths

Science


5. What is the percentage of marks scored by Fahim? (a) 92% (b) 93% (c) 94% (d) 95%

6. Marks obtained by Ishita in Maths as a percentage of total marks obtained by her is ________ (a) 30.67% (b) 32.33% (c) 32.83% (d) 33.1%

7. If the passing percentage is 60% of the total, then how many students failed in the examination? (a) 1 (b) 2 (c) 3 (d) 4

8. Marks obtained by David is what percent of marks obtained by Fahim?

(a) 84.4% (b) 82.2% (c) 81.4% (d) 80.2%

Directions for Questions for 9-12: The production of perfumes (in thousand) of five companies is represented by the graph below. The graph also shows the total number of perfumes exported. Assume that the perfumes that are not exported are sold in the domestic market.

9. What percent of perfumes of Company D are exported? (a) 54.1% (b) 55.6% (c) 57.7% (d) 58.9%

10. Which company sells the maximum percentage of its production in the domestic market? (a) Company A (b) Company B (c) Company C (d) Company E

11. The perfumes exported by Company C as a percentage of the total perfumes exported by all the companies is _____________ (a) 15.8% (b) 16.6% (c) 18.7% (d) 19.9%

12. The perfumes sold by Company A in the domestic market is what percentage of the perfumes sold by Company E in the domestic market? (a) 62% (b) 65% (c) 68% (d) 71%

124

148

180

104

128

64 80

45 60

36

0

20

40

60

80

100

120

140

160

180

200

Company A Company B Company C Company D Company E

Production

Exports


Directions for Questions 13-16: Study the following table and answer the questions based on it.

13. The percentage of candidates qualified from Punjab over those appeared from

Punjab is highest in the year? (a) 2011 (b) 2012 (c) 2013 (d) 2014

14. In the year 2011, which state had the lowest percentage of candidates selected over the candidates appeared? (a) Delhi (b) H.P. (c) U.P. (d) Punjab

15. The number of candidates selected from Haryana during the period under review is approximately what percent of the number selected from Delhi during this period? (a) 86.5% (b) 87.5% (c) 88.5% (d) 89.5%

16. The percentage of candidates selected from U.P over those qualified from U.P is highest in the year? (a) 2011 (b) 2012 (c) 2013 (d) 2014

Answer Key

1 b 2 c 3 b 4 d 5 b

6 d 7 c 8 c 9 c 10 c

11 a 12 b 13 d 14 d 15 c


Chapters 8 – DI Based on Averages

Directions for Questions for 1-4: The graph below shows the runs scored in 20 overs by Chennai Daredevils in the Indian Super Twenty League. The dots represent the number of wickets falling in the over. The first six overs are called Powerplay. Openers are the first two players that start the batting. Run rate = (Total runs scored) / Overs needed to score these runs

1. What was the run rate in the power play? (a) 8.8 (b) 9.0 (c) 9.5 (d) 10.8

2. What is the minimum run rate required by the opposition to win the match? (a) 9.8 (b) 9.85 (c) 9.9 (d) 10

3. MS Dhani was dismissed for 84 runs in the match. His average was 45.75 in 8 matches before the match. What is his average after the match? (a) 48 (b) 49 (c) 50 (d) 51

4. What is the average runs scored for every wicket fallen? (a) 32.33 (b) 32.50 (c) 32.67 (d) 32.83

Directions for Questions 5-8: The line graph below shows the production of rice in different states and the respective area covered in the states. The following terms are defined

Productivity =


5. The average production of all the states is (in million tons)? (a) 19 (b) 19.2 (c) 19.4 (d) 19.6

6. If State F having rice production of 32 million tons is also added to the data, the new average production of all the states is (in million tons)? (a) 21 (b) 21.5 (c) 22 (d) 22.5

7. The productivity of State A is how much more/less than that State D? (a) 0.354 million tons per acre less (b) 0.354 million tons per acre more (c) 0.454 million tons per acre less (d) 0.454 million tons per acre more

8. The average productivity of all the states is closest to the productivity of ________. (a) State A (b) State B (c) State C (d) State D

Directions for Questions 9-12: Study the following table and answer the questions based on it. Expenditures of a Company (in Lakh Rupees) per Annum Over the given Years.

9. What is the average of salary distributed by the company from 2011 to 2015 (in

lakh Rupees)? (a) 312 (b) 340 (c) 332 (d) 342

10. What is the difference of averages of taxes paid and bonus distributed by the company (in lakh Rupees)? (a) 80.6 (b) 86.8 (c) 96.8 (d) None of these

12 16

19 21

29

20

25 26 22

19

0

5

10

15

20

25

30

35

State A State B State C State D State E

Production (in milliontons)

Area (in acres)


11. If the average of salary distributed by the company from 2011 to 2014 is x, and average of salary distributed by the company from 2012 to 2015 is y, then x:y=? (a) 0.91 (b) 0.77 (c) 1.09 (d) 1.5

12. What is the average of interest of loans paid by the company from 2011 to 2015? (a) 35.66 (b) 36.66 (c) 37.66 (d) None of these

Directions for Questions 13-16: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie-chart and answers the question that follow.

13. What is the average of all the sources of funds collected by NHAI?


14. Which of the following source of fund is closest to the average of all sources of fund? (a) Market Borrowing (b) SPVS (c) External Assistance (d) Toll

15. If we remove the market borrowing from the source of fund then what will be the new average of all the remaining sources of fund? (a) 5912 (b) 6912 (c) 6978 (d) None of these

16. What is the difference of average of Market borrowing and External Assistance and average of remaining? (a) 15331.66 (b) 14789.66 (c) 13333.33 (d) None of these

Answer Key

1 b 2 c 3 b 4 d 5 C 6 b 7 a 8 C 9 D 10 B

11 b 12 b 13 B 14 c 15 b 16 a 7 a 8 C 9 D 10 b


Chapters 9 – DI Based on Ratio and Proportion Directions for questions 1 to 4: The following bar chart gives information about marks secured (out of 100) by three students in three subjects.

1. What is the ratio of marks scored in Mathematics by Rashmi and Reshma?

(a) 3: 9 (b) 13: 19 (c) 13: 15 (d) 17: 15

2. What is the ratio of marks scored by Rashmi and Reshma in all subjects?

(a) 40: 39 (b) 20: 23 (c) 46: 39 (d) None of these.

3. If the average score of Reshma in all the three subjects is x and the average score

of Rajini in all the three subjects is y, then x: y = ?

(a) 40: 39 (b) 20: 23 (c) 46: 39 (d) None of these.

4. What is the ratio of marks scored by all the three students (Rashmi, Reshma,

Rajini) in social?

(a) 1: 1.5: 2 (b) 1: 1.1:1.5 (c) 1: 1.1:1.2 (d) None of these.

Directions for questions 5 and 8: These questions are based on the following table which shows the sales figures of pharmaceutical companies in India from 2011-12 through 2014-15.

2011-12

2012-13

2013-14

2014-15 Tanbaxy

6000

7000

8700

8000

Pipla

4000

4000

6500

7000

Dr.Meddy‘s

4000

6000

6300

6000

Mupin

2000

3000

3000

4000

Rockhardt

3000

5000

5800

6000


5. What is the ratio of sales of Tanbaxy and Pipla in 2011- 12? (a) 2: 1 (b) 7: 4 (c) 87: 65 (d) 3: 2

6. Ratio of sales of Tanbaxy and Pipla is maximum in: (a) 2011-12 (b) 2012- 13 (c) 2013-14 (d) 2014-15

7. Find the ratio of total sales of Dr. Meddy‘s, Mupin, Rockhardt in 2011-12 &

2014-15. (a) 3: 5 (b) 9: 16 (c) 30: 58 (d) None of these.

8. Ratio of sales of Dr. Meddy and Mupin is minimum in:

(a) 2011-12 (b) 2012- 13 (c) 2013-14 (d) 2014-15

Directions for questions 9 to 12: Answer the questions based on the information given below. The pie charts give the distribution of monthly expenditure of Mr. Shyamlal in two years.

A: Food B: Rent C: Provisions D: Entertainment E: Medical Expenses F: Travelling G Clothes

9. What is the ratio of Mr. Shyamlal‘s expenditure on provision in 2013 to 2014?

(a) 23: 42 (b) 25: 42 (c) 15: 42 (d) None of these

10. What is the ratio of Mr. Shyamlal‘s expenditure on food in 2014 to 2013?


11. What is the ratio of Mr. Shyamlal‘s combined expenditure on food & rent in 2014

to that in 2013? (a) 49: 30 (b) 30: 49 (c) 30: 29 (d) None of these

Directions for questions 12 to 13: The given line chart represents the discharge rate of Red river in Fargo station in different months of 2015:


12. What is the ratio of discharge rate in May to June?

(a) 5: 4 (b) 4: 5 (c) 3: 5 (d) 5: 3

13. What is the ratio of discharge rate in May to August?


Directions for questions 14 to 16: Ratio of Exports to Imports (in terms of money in Rs. crores) of Two Companies over the Years:

14. In how many of the given years were the exports more than the imports for

Company A? (a) 2 (b) 3 (c) 4 (d) 5

15. If the imports of Company A in 2012 were increased by 40 percent, what would be the ratio of exports to the increased imports? (a) 1.20 (b) 1.25 (c) 1.30 (d) 1.4

16. What is the ratio of export to import of company A and ratio of import to export of company B in 2015? (a) 1.25: 1 (b) 1: 1.25 (c) 1: 0.75 (d) 0.75:1

Answer key

1 b 2 b 3 c 4 d 5 d 6 b 7 b 8 d

9 b 10 b 11 a 12 d 13 b 14 b 15 b 16 a