## Chapter 1: SET THEORY

### Introduction

Set: A well defined collection of objects is called a set.
Finite set: a set of finite elements.
Null set:  set with no elements.
Singleton set: set with one element only.
Equal set: each element of one set is in the other and vice versa.
Subset: Every element of A is element of B. Denote it as A⊆B A={1,2} and B={1,2,3} or B={1,2}. Proper subset is if there is at least one element in B but not in A but all elements in A are in B. A⊂B. E.g: A = {1,2,3} and B={1,2,3,4}.
Union of sets:
A∪B = {all elements of A and B but no duplicates};
Intersection of sets: A∩B = {common elements of A and B but no duplicates}
Difference of sets: A-B = {set of all elements of A not in B}
Complement of set: If A is the subset of X then the a set of all elements in X but not in A is Ac

Note: A set of 'n' elements can have total 2n

### Laws of Indices:

(i) am * an = am + n

(ii) am/an = am - n

(iii) (am)n = am * n

(v) a-n = 1/an

(vi) n$\sqrt{a}$m = am/n

(vii) (ab)m= am ∙ bm.

(viii) (a/b)m = am/bm

### Equations

• A∪B = n(A) + n(B) - n(A ∩B) and
if A ∩B = not empty then n(A-B) + n(B-A) + n(A ∩B)
• n(A∪B∪C) = n(A) + n(B) + n(C) - n(A ∩B) - n(B ∩C) -n(A ∩C) + n(A ∩B ∩C)

For the below Venn diagram if you want to find the formula for any region you can use simple addition and subtraction.

Sample questions that can be asked are:
A = People who love football, B = people who love basketball, C = people who love carrom, U = all people. Then find out
1. people who love only football
2. people who love only carrom
3. people who love only basketball
4. people who love carrom and basketball
5. people who love carrom and football
6. people who love football and basketball.
7. people who love all three.
8. people who love none.
9. Region that represents any of the above. In such cases the regions shall have numbers and not equations.

### Number System

Divisibility rules:

1. Divisible by 2: If its digit in units place is divisible by 2. E.g: 578 is divisible by 2 as units digit = 8 is divisible.
2. Divisible by 3: If sum of digits is divisible by 3. E.g: 372 = 3+7+2 = 12 which is divisible by 3 so 372 is also divisible.
3. Divisible by 4: If the number formed by the last two digits is divisible by 4. E.g: 340 is divisible as 40 is divisible.
4. Divisible by 5: If units digits is 5 or 0. E.g. 55 / 70.
5. Divisible by 6: If the number is divisible by both 2 and 3.
6. Divisible by 8: Number formed by last three digits is divisible by 8.
7. Divisible by 9: Sum of the digits is divisible by 9.
8. Divisible by 10: Last digit should be 0.
9. Divisible by 11: Difference between sum of digits at odd place and even place is 0 or divisible by 11. E.g: 121 i.e. 1+1 - 2 = 0 so divisible.
10. Divisible by 12: A number is divisible by 3 and 4.
11. Divisible by 14: A number is divisible by 2 and 7.
12. Divisible by 15: A number is divisible by both 3 and 5.
13. Divisible by 16: Last four digits form a number divisible by 16.

Co prime or relatively prime: Two numbers are said to be relatively prime if they have only 1 as the common factor. E.g: (4,5); (8,9). If a number is divisible by p,q where both are co primes then it is also divisible by p*q;

Test for Prime numbers: To find out if a number 'N' greater than 100 is prime; we find out the number 'x' whose square is as close but greater than that number. If the prime numbers below 'x' are not dividing 'N' then it is prime.
E.g: 191 is prime since 14^2 = 196 and prime numbers below 14 are 2,3,5,7,11,13 don't divide 191.

Multiplication short cuts:
A) a * ( b + c ) = a*b + a*c
B) a * ( b - c) = a*b - a*c;

For shortcut to solving tough looking problems use above tricks:
345345 * 9999 = 345345 * (1000 - 1) which is simpler.

Multiplication of a number by 5^n: Add 'n' zeros to the right of the number and divide it by 2^n.
E.g: 975436 * 625 = 975436 * 5^4 = 9754360000/2^4

### Progressions

Arithmetic Progressions

The sequence of numbers like 1,2,3,4... are said to be in arithmetic progression with common difference d = 1; Generalizing this the arithmetic series is of the form: a, a+d, a+2d, a+3d... a+(n-1)d.

Common difference d is Tn - T(n-1) i.e. next term minus previous term. This is uniform throughout.

Properties:

1. Corresponding terms of A.P = first and last, second and second last etc. So if an AP is 1,2,3,4,5,6 then corresponding terms are (1,6) , (2,5) , (3,4).
2. Average of the A.P = mean of the corresponding terms e.g: 1+6/2 or 2+5/2 etc
3. Sum of A.P = ( Average of A.P. ) * ( Number of terms )
4. Finding the Common difference given two terms in an A.P =If Tx and Ty are the terms in an AP at position 'x' and 'y'. The common difference = > ( Ty - Tx ) = Common difference * (y-x). E.g: So if 3rd term is 8 and 8th term is 28 the common difference is (28 - 8) = d * (8-3) = > 20 = 5*d and d = 4.

Geometric Progression

The series is in geometric progression if the numbers increase or decrease by a common ratio. So the series is a, ar, ar2, ar3, ar4.... arn-1

If r > 1 then Sum = a ( rn - 1) / ( r - 1)

If r < 1 then Sum = a ( 1 - rn) / ( 1 - r )

If an infinite geometric progression series is Sum = a / ( 1 - r )

### Theorem of Divisibility

Calculate the Sum of factors of a number:

Step 1:  Get prime factors of a number say 240

240 = 24 * 31 * 51

Step 2: Sum of factors formula is

240 = (20 + 21 + 22 + 23 + 24) * (30 + 31) * (50 + 51)

Step 3: 31*4*6 = 744

Calculate the Number of factors of a number:

Step 1: Get the prime factors of a number

240 = 24 * 31 * 51

Step 2: Number of factors of a number.

Number of factors = ( 4 + 1 ) * ( 1 + 1) * ( 1 + 1) = 5 * 2 * 2 = 20

Thus the powers of the numbers are increased by one and multiplied.

Calculate the sum and number of even factors of a number:

Step 1: Get the prime factors of a number

240 = 24 * 31 * 51

Step 2: Sum of even factors
Sum = (21 + 22 + 23 + 24) * (30 + 31) * (50 + 51)
Number of even factors = 4 * 2 * 2 = 16
Thus the powers of the numbers are increased by one and multiplied except 2.

Calculate the sum and number of odd factors of a number:

Step 1: Get the prime factors of a number

240 = 24 * 31 * 51

Step 2: Sum of odd factors
Sum = (20) * (30 + 31) * (50 + 51)
Number of odd factors = 1 * 2 * 2 = 4
Thus the powers of the numbers are increased by one and multiplied except 2.

Calculate the sum and number of factors of a number satisfying a condition:

Step 1:  Get prime factors of a number say 240

30 = 21 * 31 * 51

Step 2: Sum of factors formula is

30 = (20 + 21 ) * (30 + 31) * (50 + 51)

which is expanded as
= 20*30*50  + 20*30*50 + 20*31*51 + 20*31*51 + 21*30*50 + 21*30*50 + 21*31*51 + 21*31*51

These are all the factors of the number, We can apply any condition we want and remove unnecessary of them viz. remove factors that are perfect squares, not perfect squares, between higher and lower limit etc

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved.

Q. What is the number of projects in which Gyani alone is involved?

1. Uniquely equal to zero.

2. Uniquely equal to 1.

3. Uniquely equal to 4

4. Cannot be determined uniquely.

Ans . D

1. Putting the value of M in either equation, we get G + B = 17

2. Hence neither of two can be uniquely determined.

Q. What is the number of projects in which Medha alone is involved?

1. Uniquely equal to zero.

2. Uniquely equal to 1.

3. Uniquely equal to 4

4. Cannot be determined uniquely.

Ans . B

1. G + B = M + 16 Also, M + B + G + 19 = (2 × 19) – 1

2. i.e. (G + B) = 18 – M Thus, M + 16 = 18 – M

3. i.e. M = 1

### CAT Problems

There were a hundred schools in a town. Of these, the number of schools having a play – ground was 30, and these schools had neither a library nor a laboratory. The number of schools having a laboratory alone was twice the number of those having a library only. The number of schools having a laboratory as well as a library was one fourth the number of those having a laboratory alone. The number of schools having either a laboratory or a library or both was 35.

Q.How many schools had none of the three viz., laboratory, library or play – ground?

1. 20
2. 5
3. 30
4. 35

Ans.d

Q.What was the ratio of schools having laboratory to those having library?

1. 1/2
2. 5/3
3. 2/1
4. 2/3

Ans.b

Q.There are 3 clubs A, B & C in a town with 40, 50 & 60 members respectively. While 10 people are members of all 3 clubs, 70 are members in only one club. How many belong to exactly two clubs?

1. 20
2. 25
3. 50
4. 70

Ans.b

### CAT Problems for practice

Q.Amar, Akbar, and Anthony came from the same public school in the Himalayas. Every boy in that school either fishes for trout or plays frisbee. All fishermen like snow while no frisbee player likes rain. Amar dislikes whatever Akbar likes and likes whatever Akbar dislikes. Akbar likes rain and snow. Anthony likes whatever the other two like. Who is a fisherman but not a frisbee player?

1. Amar
2. Akbar
3. Anthony
4. None

Ans.b

Q.Let x, y and z be distinct positive integers satisfying x < y < z and x + y + z = k. What is the smallest value of k that does not determine x, y, z uniquely?

1. 9
2. 6
3. 7
4. 8

Ans.d

Q.The product of all integers from 1 to 100 will have the following numbers of zeros at the end.

1. 20
2. 24
3. 19
4. 22

Ans.b

Q.A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

1. 6666600
2. 6666660
3. 6666666
4. None of these

Ans.a

Q.Out of 100 families in the neighbourhood, 45 own radios, 75 have TVs, 25 have VCRs. Only 10 families have all three and each VCR owner also has a TV. If 25 families have radio only, how many have only TV?

1. 30
2. 35
3. 40
4. 45

Ans.c

Ghoshbabu is staying at Ghosh Housing Society, Aghosh Colony, Dighospur , Calcutta. In Ghosh Housing Society 6 persons read daily Ganashakti and 4 read Anand Bazar Patrika; in his colony there is no person who reads both. Total number of persons who read these two newspapers in Aghosh Colony and Dighospur is 52 and 200 respectively. Number of persons who read Ganashakti in Aghosh Colony and Dighospur is 33 and 121 respectively; while the persons who read Anand Bazar Patrika in Aghosh Colony and Dighospur are 32 and 117 respectively.

Q.Number of persons in Dighospur who read only Ganashakti is

1. 121
2. 83
3. 79
4. 127

Ans.b

Q.Number of persons in Aghosh Colony who read both of these newspapers is

1. 13
2. 20
3. 19
4. 14

Ans.a

Q.Number of persons in Aghosh Colony who read only one paper

1. 29
2. 19
3. 39
4. 20

Ans.c

Q.If the harmonic mean between two positive numbers is to their geometric mean as 12 : 13; then the numbers could be in the ratio

1. 12 : 13
2. 1/12 : 1/13
3. 4 : 9
4. 2 : 3

Ans.c

Q.Fourth term of an arithmetic progression is 8. What is the sum of the first 7 terms of the arithmetic progression?

1. 7
2. 64
3. 56
4. cant say

Ans.c

Q. The table below shows the agewise distribution of the population of Reposia. The number of people aged below 35 years is 400 million.If the ratio of females to males in the ‘below 15 years’ age group is 0.96, then what is the number of females in that age group?

1. 82.8 million

2. 90.8 million

3. 80 million

4. 90 million

Ans . B

### Quiz

Score more than 80% marks and move ahead else stay back and read again!