Modern Math - Sets - Previous Year CAT/MBA Questions
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1. CAT 2018 QA Slot 2 | Modern Math - Sets
If A = {62n - 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, A1 = 0
For n = 2, A2 = 1225 = 35 × 35
Also, 62n- 35n - 1 = (36n - 1) - 35n.
The term in the bracket 36n - 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, B1 = 35(1 - 1) = 0
For n = 2, B2 = 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).
2. 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
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).
3. 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).
4. 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?
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).
5. 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?
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).
6. 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
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).
7. CAT 2003 QA - Retake | Modern Math - Sets
Consider the sets Tn = {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, ...)?
Answer: Option
A
Explanation :
n = 1, 2, 3, … , 96 and Tn = {n, n + 1, n + 2, n + 3, n + 4}
n could be either 6k or 6k + 1 or 6k + 2 or 6k + 3 or 6k + 4 or 6k + 5.
When n = 6k, then set Tn will definitely contain a multiple of 6 as it contains n
When n = 6k + 5, then set Tn will contain a multiple of 6 as it contains n + 1 = 6k + 6
When n = 6k + 4, then set Tn will contain a multiple of 6 as it contains n + 2 = 6k + 6
When n = 6k + 3, then set Tn will contain a multiple of 6 as it contains n + 3 = 6k + 6
When n = 6k + 2, then set Tn will contain a multiple of 6 as it contains n + 4 = 6k + 6
However, for every n = 6k + 1, the set Tn will not contain any multiple of 6There will be 16 such sets for k = 0 to 15, for which Tn will not contain a multiple of 6
∴ (96 – 16) = 80 sets contain multiples of 6
Hence, option (a).
8. 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).
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