Suppose A _ 1 , A _ 2 , . . . , A _ 30 are thirty sets each having...

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

\mathop{\cup }_{i = 1}^{30}{A}_{i} = \mathop{\cup }_{j = 1}^{n}{B}_{j} = S

and each element of

S

belongs to exactly

10

of the

{A}_{i} ' s

~and exactly

9

of the

Bj '

s. Then

n

is equal to

Options

A $35$
B $45$
C $55$
D $65$

Correct answer

B. $45$

Step-by-step solution

$\mathop{∑}_{i = 1}^{30}n\left(Ai\right) = 5 \times 30 = 150$
Let $S$ consist of $m$ ~distinct elements. Since each elements of $S$ belongs to exactly $10$ of the $Ai^{'}$ ,~so,~we~have& $\mathop{∑}_{i = 1}^{30}n\left({A}_{i}\right) = 10m = 150 ⇒ m = 15$
Now each ${B}_{i}$ has $3$ elements and each elements of $S$ belongs to exactly $9$ of the ${B}_{i}$ 's.~So, we have~ $\mathop{∑}_{j = 1}^{n}n\left({B}_{j}\right) = 3n = 9m ⇒ 3n = 9 \times 15 ⇒ n = 45$

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