Question

Let $A_1,A_2,A_3...A_{40}$ are 40 sets each with 7 elements and $B_1,B_2,......B_n$ are n sets each with 7 elements. If ${\bigcup }_{i=1}^{40}{A}_i={\bigcup }_{j=1}^n{B}_j=S$ and each element of S belongs to exactly ten of $A_i^{\prime }s$ and exactly 9 of $B_j^{\prime }s$, then n equals

• A. $42$
• B. $35$
• C. $28$
• D. $36$.

Answer 1

The correct option is D $36$.

$\sum n\left(A_i\right)=7\times 40=280$

But, as each element of S occurs in 10 of $A_i$'s,

$n\left(S\right)=\frac{280}{10}=28$

Also, $\sum n\left(B_j\right)=7n$

But, each element of S occurs in 9 of $B_j$'s

ie, $28=\frac{7n}{9}\Rightarrow n=\frac{28\times 9}{7}=9\times 4=36$

Option D is the answer.