Suppose A1,A2,…,A30are thirty sets each having 5 elements and B1,B2,…,Bn are n sets each with 3 elements let ∪i=130 Ai=∪j=1n Bi=S and each element of S belongs to exactly 10 of the Ai′s and exactly 9 of the Bi′s.Then n is equal to

# Suppose ${A}_{1},{A}_{2},\dots ,{A}_{30}$are thirty sets each having 5 elements and ${B}_{1},{B}_{2},\dots ,{B}_{n}$ are n sets each with 3 elements let $\underset{i=1}{\overset{30}{\cup }} {A}_{i}=\underset{j=1}{\overset{n}{\cup }} {B}_{i}=S$ and each element of $S$ belongs to exactly 10 of the ${A}_{i}^{\mathrm{\prime }}s$ and exactly 9 of the ${B}_{i}^{\mathrm{\prime }}s.$Then n is equal to

1. A
2. B
3. C
4. D

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.

### Solution:

$S={A}_{1}\cup {A}_{2}\cup \dots \cup {A}_{30}={B}_{1}\cup {B}_{2}\cup \dots \cup {B}_{n}$

Since each ${A}_{i}$ contains 5 elements, number of elements in (with repetitions) = 5 x 30 = 150

But each element of $S$ appears exactly 10 times.

$⇒$ Number of distinct elements in $S=\frac{150}{10}=15$

Similarly by other union of ${B}_{i}^{\mathrm{\prime }}{\mathrm{s}}_{\mathrm{\prime }}$ number of distinct elements in $S=\frac{3n}{9}$

Now, $\frac{3n}{9}=15⇒n=45$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.