1, A2, A3, ..., A30 are thirty sets each..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> 1, A2, A3, ..., A30 are thirty sets each..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> 1, A2, A3, ..., A30 are thirty sets each..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 24

Mathematics Sets Easy

Suppose A<sub>1</sub>, A<sub>2</sub>, A<sub>3</sub>, ..., A<sub>30</sub> are thirty sets each having 5 elements with no common elements across the sets and B<sub>1</sub>, B<sub>2</sub>, ..., B<sub>n</sub> are n sets each with 3 elements with no common elements across the sets. Let&nbsp;<span class="math-tex">\(\rm \displaystyle\bigcup^{30}_{i = 1} A_i = \displaystyle\bigcup^n_{j = 1} B_j = S\)</span>&nbsp;and each elements of S belongs to exactly 10 of the A<sub>i</sub>&#39;s and exactly 9 of the B<sub>j</sub>&#39;s. Then n is equal to

(A) 15
(B) 30
(C) 40
(D) 45
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Count total memberships two ways.

  • A-sets: 30 sets × 5 elements = 150 memberships. Each element of S is in 10 A-sets, so \(|S| = \frac{150}{10} = 15\).
  • B-sets: \(3n\) memberships = 9 × 15 = 135, so \(n = 45\).

Answer: (D) 45

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