IPMAT Indore 2025 (MCQ) - Let A and B be two finite sets such that n(A - B), n(A B), n(B - A) are in an arithmetic progression. Here n(X) denotes the number of elements in a finite set X. If n(A B) = 18, then n(A) + n(B) is ____ | PYQs + Solutions

IPMAT Indore 2025

Modern Math

Set Theory

Medium

Let A and B be two finite sets such that $n(A - B)$ , $n(A\cap B)$ , $n(B - A)$ are in an arithmetic progression. Here $n(X)$ denotes the number of elements in a finite set $X$ . If $n(A\cup B) = 18$ , then $n(A) + n(B)$ is ____

✅ Correct Option: 1

Let

n(A-B) = x

n(A\cap B) = y

, and

n(B-A) = z

Since these three values form an arithmetic progression, we have:

\newline

y - x = z - y

This gives us:

2y = x+z

We know that

n(A\cup B) = x + y + z = 18

Substituting

2y = x+z

into

x + y + z = 18

x + \dfrac{x+z}{2} + z = 18

\dfrac{3x}{2} + \dfrac{3z}{2} = 18

3(x+z) = 36

x + z = 12

Since

2y = x+z = 12

, we have

y = 6

Using the set theory formulas:

\newline

n(A) = n(A-B) + n(A\cap B) = x + y = x + 6

\newline

n(B) = n(B-A) + n(A\cap B) = z + y = z + 6

Therefore:

\newline

n(A) + n(B) = x + 6 + z + 6 = x + z + 12 = 12 + 12 = 24

The answer is

24