# Modern Math - Sets - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic Modern Math - Sets. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**CAT 2018 QA Slot 2 | Modern Math - Sets**

If A = {6^{2n} - 35n - 1: n = 1, 2, 3, ...} and B = {35(n - 1) : n = 1,2,3,...} then which of the following is true?

- A.
Every member of A is in B and at least one member of B is not in A

- B.
At least one member of A is not in B

- C.
Every member of B is in A.

- D.
Neither every member of A is in B nor every member of B is in A

Answer: Option A

**Explanation** :

For n = 1, A_{1} = 0

For n = 2, A_{2} = 1225 = 35 × 35

Also, 6^{2n}- 35n - 1 = (36^{n }- 1) - 35n.

The term in the bracket 36^{n} - 1 is always divisible by (36 - 1) or by 35. Similarly 35n is also always divisible by 35. Therefore each term in the set A is divisible by 35. However, not all positive multiples of 35 are present in set A.

For n = 1, B_{1} = 35(1 - 1) = 0

For n = 2, B_{2} = 35(2 - 1) = 35 and so on.

Thus all whole number multiples of 35 are present in set B.

Thus every member of set A is present in every member of set B but at least one member of set B is not present in set A.

Hence, option (a).

Workspace:

**CAT 2018 QA Slot 2 | Modern Math - Sets**

For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is

- A.
6

- B.
8

- C.
9

- D.
7

Answer: Option D

**Explanation** :

The set (PΔQ) = {1,4,5,6}.

The set (RΔS) = {1,2,3,4,7,8,10}

Therefore the set (PΔQ) Δ (RΔS) = {2,3,5,6,7,8,10}.

Thus there are 7 elements in the set.

Hence, option (d).

Workspace:

**CAT 2007 QA | Modern Math - Sets**

Consider the set S = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y?

- A.
0

- B.
1

- C.
n/2

- D.
n + 1/2n

- E.
2008

Answer: Option B

**Explanation** :

Y = (2 + 4 + 6 + 8 + … + 2n)/n

Average of numbers in AP is same as average of first and the last terms.

⇒ Y = (2 + 2n)/2 = 1 + n

X = (3 + 5 + 7 + 9 + … + (2n + 1))/n

Average of numbers in AP is same as average of first and the last terms.

⇒ X = (3 + 2n+1)/2 = 2 + n

∴ X – Y = 2 + n - (1 + n) = 1

**Note**: The information that 'n is a positive integer larger than 2007' does not affect the answer in any way.

Hence, option (b).

Workspace:

**CAT 2006 QA | Modern Math - Sets**

Consider the set S = {1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?

- A.
3

- B.
4

- C.
6

- D.
7

- E.
8

Answer: Option D

**Explanation** :

Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first term and 1000 as the last. Let d be the common difference. Then,

1000 = 1 + (n – 1) × d

∴ 999 = (n – 1) × d ... (i)

∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999

Substituting in equation (i)

If d = 1, n = 1000

If d = 3, n = 334

If d = 9, n = 112

If d = 27, n = 38

If d = 37, n = 28

If d = 111, n = 10

If d = 333, n = 4

If d = 999, n = 2, which is not possible as n > 2

∴ 7 arithmetic progressions can be formed.

Hence, option (d).

Workspace:

**CAT 2005 QA | Modern Math - Sets**

Let S be a set of positive integers such that every element n of S satisfies the conditions

a. 1000 ≤ n ≤ 1200

b. every digit in n is odd

Then how many elements of S are divisible by 3?

- A.
9

- B.
10

- C.
11

- D.
12

Answer: Option A

**Explanation** :

n will be of the form 11ab, where a and b are odd numbers.

We are looking for all n’s divisible by 3.

∴ 1 + 1 + a + b = 3 or 6 or 9 or 12 or 15 or 18

∴ a + b = 1 or 4 or 7 or 10 or 13 or 16

∴ a + b = 1 or 7 or 13 is not possible as the sum of two odd numbers cannot be odd.

∴ (a, b) = (1, 3), (3, 1), (1, 9), (3, 7), (5, 5), (7, 3), (9, 1), (7, 9), (9, 7)

∴ 9 elements of S are divisible by 3.

Hence, option (a).

Workspace:

**CAT 2003 QA - Leaked | Modern Math - Sets**

Let T be the set of integers {3, 11, 19, 27, ..., 451, 459, 467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

- A.
32

- B.
28

- C.
29

- D.
30

Answer: Option D

**Explanation** :

The sum of the first and last terms in T is 470.

Likewise, the sum of the second and second-last terms is also 470.

In general the sum of the nth term from the beginning and the nth term from the end is 470.

∴ Only one of each of these pairs of terms will be in S. (For instance only one of 3 and 467 can be in S)

∴ The set S can have a maximum of half of the terms in T.

The terms in T are in A.P. with a common difference of 8.

Last Term = 467 = 3 + (n − 1) × 8

∴ n = 59

∴ Total number of terms in the set T = 59

∴ there are 29 pairs of numbers in T that add up to 470 and the 59th number is 235, which occurs in the middle of the series.

∴ S will be a set with 30 terms, with 29 terms which are from the pairs adding up to 470, and 235.

Hence, option (d).

Workspace:

**CAT 2003 QA - Retake | Modern Math - Sets**

Consider the sets *T _{n}* = {

*n*,

*n*+ 1,

*n*+ 2,

*n*+ 3,

*n*+ 4}, where

*n*= 1, 2, 3, … , 96.

How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?

- A.
80

- B.
81

- C.
82

- D.
83

Answer: Option A

**Explanation** :

*n *= 1, 2, 3, … , 96 and *T _{n}* = {

*n*,

*n*+ 1,

*n*+ 2,

*n*+ 3,

*n*+ 4}

*n* could be either 6*k* or 6*k + *1 or 6*k +* 2 or 6*k +* 3 or 6*k +* 4 or 6*k +* 5.

When *n* = 6*k*, then set *T _{n}* will definitely contain a multiple of 6 as it contains

*n*

When *n* = 6*k* + 5, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 1 = 6

*k*+ 6

When *n* = 6*k* + 4, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 2 = 6

*k*+ 6

When *n* = 6*k* + 3, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 3 = 6

*k*+ 6

When *n* = 6*k* + 2, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 4 = 6

*k*+ 6

However, for every *n* = 6*k* + 1, the set *T _{n}* will not contain any multiple of 6There will be 16 such sets for

*k*= 0 to 15, for which

*T*will not contain a multiple of 6

_{n}∴ (96 – 16) = 80 sets contain multiples of 6

Hence, option (a).

Workspace:

**CAT 2000 QA | Modern Math - Sets**

The set of all positive integers is the union of two disjoint subsets

{f(1), f(2) ....f(n),......} and {g(1), g(2),......,g(n),......}, where

f (1) < f(2) <...< f(n) ....., and g(1) < g(2) <...< g(n) ......., and

g(n) = f(f(n)) + 1 for all n ≥ 1.

What is the value of g(1)?

- A.
Zero

- B.
Two

- C.
One

- D.
Cannot be determined

Answer: Option B

**Explanation** :

The functions f(n) and g(n) are disjoint sets and union of these two sets is the set of all positive integers.

∵ g(n) = f(f(n)) + 1 for all n ≥ 1

and f (1) < f(2) <...< f(n) ....., and g(1) < g(2) <...< g(n) .......,

∴ f(1) = 1 or 2

If f(1) = 1

g(1) = f(f(1)) + 1

∴ g(1) = f (1) + 1 = 1 + 1 = 2

If f(1) = 2

g(1) = f(f(1)) + 1

∴ g(1) = f (2) + 1

∴ g(1) is greater than f(1), i.e. it is greater than 2.

But the set of all positive integers is the union of these two disjoint sets.

∴ This set has to include 1 which is not possible in this case as f(1) is 2 and g(1) will be greater than f(1).

∴ f(1) = 1 and g(1) = 2

Hence, option (b).

Workspace:

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