Q: CAT 2000: 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)?

Answer: B · Two. The correct option is B f(x) is a set of all odd numbers and g(x) is a set of all even numbers. Thus, g(1) = f(f(1)) + 1 = 2.

Question 1

CAT 2025 Slot 3: In a class of 150 students, 75 students chose physics, 111 students chose mathematics and 40 students chose chemistry. All students chose at least one of the three subjects and at least one student chose all three subjects. The number of students who chose both physics and chemis

Accepted Answer

Answer: B · 35. Given Information: Physics students = 75, Mathematics students = 111 , Chemistry students = 40 Assume: x - number of students who chose both Physics and Mathematics , y - students who chose both Mathematics and Chemistry , z - student who chose both Physics and Chemistry , t - number of students who chose all three subjects And z = y , y = x/2 (Given) Using Set formula (the total number of students who chose at least one subject) |P ∪ M ∪ C| = |P| + |M| + |C| − |P ∩ M| − |M ∩ C| − |P ∩ C| + |P ∩ M ∩ C| Substitute the values: 150 = 75 + 111 + 40 - x - y - z + t Now, substitute z = y and y = x/2

Question 2

CAT 2018 Slot 2: 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

Accepted Answer

Answer: D · 7. To find the number of elements in (PΔQ)Δ(RΔS), let's break it down step by step. Find PΔQ (the set of elements belonging to P or Q but not both): PΔQ = {1, 2, 3, 4} Δ {2, 3, 5, 6} PΔQ includes elements that belong to either P or Q but not both. So, it contains {1, 4, 5, 6}. Find RΔS (the set of elements belonging to R or S but not both): RΔS = {1, 3, 7, 8, 9} Δ {2, 4, 9, 10} RΔS includes elements that belong to either R or S but not both. So, it contains {1, 3, 7, 8, 2, 4, 10}. Now, find (PΔQ)Δ(RΔS) (the set of elements belonging to PΔQ or RΔS but not both): (PΔQ)Δ(RΔS) = {1, 4, 5, 6} Δ {1, 3,

Question 3

CAT 2007: 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 th average of the even integers in S. What is the value of X – Y?

Accepted Answer

Answer: A · 0. Let's analyze the set S and calculate the values of X and Y: Set S = {2, 3, 4, …, 2n + 1} X is defined as the average of the odd integers in S, and Y is defined as the average of the even integers in S. The number of odd integers in the set S: There are n + 1 odd integers in the set S because it starts with 3 (an odd integer) and ends with 2n + 1 (also an odd integer). The number of even integers in the set S: There are n even integers in the set S because it starts with 2 (an even integer) and ends with 2n (also an even integer). Now, let's calculate the values of X and Y: X = (3 + 5 + 7 + …

Question 4

CAT 2006: 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?

Accepted Answer

Answer: C · 7. The nth term is a + (n-1)d 1000 = 1 + (n-1)d So, (n-1)d = 999 999 = 3^3 * 37 So, the number of factors is 4*2 = 8 Since there should be at least 3 terms in the series, d cannot be 999. So, the number of possibilities is 7...

Question 5

CAT 2005: 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?

Accepted Answer

Answer: D · 12. To find out how many elements of S are divisible by 3, we need to identify the positive integers within the range [1000, 1200] where every digit is odd. Let's break it down step by step. The conditions are: a. 1000 ≤ n ≤ 1200 b. Every digit in n is odd For a number to be divisible by 3, the sum of its digits must be divisible by 3. First, let's find the possible odd digits that can be used to create such numbers. The odd digits are 1, 3, 5, 7, and 9. Now, we will consider the number of ways to form a 4-digit number where every digit is an odd digit (1, 3, 5, 7, or 9). There are 5 choices for e

Question 6

CAT 2003: 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, …)?

Accepted Answer

Answer: A · 80. The correct option is A 80 Option (a) Calculating “n” which does not satisfy the requisite condition (i.e. the value of “n” which will not yield subsets with an integral multiple of 6).The terms will fall in an AP with n =1 as the first term. The general form of the AP is 6k+1.The last term will be 91. Thus, total number of sets which do not satisfy the condition= 90/6 = 15+1 = 16.Total number of terms = 96.Thus, number of sub sets possible = 96 - 16 = 80.

Question 7

CAT 2000: 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)?

Accepted Answer

Answer: B · Two. The correct option is B f(x) is a set of all odd numbers and g(x) is a set of all even numbers. Thus, g(1) = f(f(1)) + 1 = 2.