Question

Suppose $A_1,A_2,....,A_{30}$ are thirty sets each having 5 elements and $B_1,B_2,....,B_n$ are n sets each with 3 elements. Let $\bigcup _{i=1}^{30}{A}_i=\bigcup _{j=1}^n{B}_j=S$ and each elements of S belongs to exactly 10 of the $A_i$ and exactly 9 of the $B_j$. Then n is equal to-

• A. 35
• B. 45
• C. 55
• D. 65

Answer 1

Answer: (B)

45

$A_0,A_1,.............,A_{30}⟹$ each of 5 elements

$B_1,B_2,B_3...........n⟹$ each of 3 elements

The number of elements in the union of the A sets is $5\left(30\right)-r$ where 'r' is the number repeats likewise the number of elements in the B sets $3n-rB$

Each element in the union (in5) is repeated 10 times in A which means if x was the real number of elements in A (not counting repeats) then q out of those 10 should be thrown away or 9x .likewise on the B side 8x of those elements should be thrown away So, $⟹150-9x=3n-8x$

$n=50-3x$

Now in figure out what x is we need to use the fact that the union of a group of sets contains every member of each sets . If every element in 'S' is repeated 10 times that means every element in the union of the n's is repeated 10 times .

This means that $10/10⟹15$ is the number of in the A's without repeats counted (same for the B's aswell ) So now

$\frac{50-15}{3}=n$

$n=45$

Subset:- A proper subset is nothing but it contain atleast one more element of main set .

Ex:$\{3,4,5\}$ is a set then the possible subsets are

$\left\{3\right\},\left\{4\right\},\left\{5\right\},\{1,5\},\{3,4\}$