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 $\bigcup _{i=1}^{30}{A}_i=\bigcup _{i=1}^n{B}_i=S$ and each element of $S$ belongs to exactly $10$ of the $A_i$ and exactly $9$ of the $B_i$. then

• A. $n$ is an odd number
• B. $n$ is an odd prime number
• C. $n$ has two prime factors
• D. $n$ has only one prime factor

Answer 1

Answer: (A), (C)

$n$ has two prime factors

Number of elements in $A_1\cup A_2\cup ----\cup A_{30}$ is $30\times 5=150$
But each element is used 10 times.
So $S=\frac{30\times 5}{10}=15$
Total number of elemetns in $B_1\cup B_2\cup ----\cup B_n=3n$
But each element is repeated 9 times.
So, $S=\frac{3n}{9}=15$
$\Rightarrow n=45$