Suppose that A1, A2, A3….A30 are thirty sets each containing 6 elements and B1, B2, B3….Bn are n sets each containing 3 elements. If ∪i=130Ai=S=∪i=1nBi and each element of S belongs to exactly 10AiS ‵ and to exactly 9BiS ‵ then n =

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Suppose that $A_{1}, A_{2}, A_{3} \dots . A_{30}$ are thirty sets each containing 6 elements and $B_{1}, B_{2}, B_{3} \dots . B_{n}$ are n sets each containing 3 elements. If $\cup_{i = 1}^{30} A_{i} = S = \cup_{i = 1}^{n} B_{i}$ and each element of S belongs to exactly $10 A_{i} {}^{‵}S$ and to exactly $9 B_{i} {}^{‵}S$ then n =

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answer is 54.

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$\frac{30 \times 6}{10} = \frac{n \times 3}{9} \Rightarrow \frac{n}{3} = 18 \Rightarrow n = 54$

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