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Standard VI

Mathematics

Idea of a Set

Suppose A1,...

Question

Suppose $A_{1}, A_{2}, . . . . ., A_{30}$ are thirty sets each with five elements and $B_{1}, B_{2}, . . . . ., B_{n}$ are $n$ sets each with three elements such that $⋃_{i = 1}^{30} A_{i} = ⋃_{j = 1}^{n} B_{j} = S$ and each element of $S$ belongs to exactly $10$ of the $A_{i}$ 's and exactly $9$ of the $B_{j}$ 's. Then $n =$

15

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3

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45

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9

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Solution

The correct option is C $45$
Since, each $A_{i}$ has $5$ elements, we have
$\sum_{i = 1}^{30} n (A_{i}) = 5 \times 30 = 150$ ...(1)
Let set $S$ consists of $k$ elements.
Since, each elements in $S$ belongs to exactly $10$ of $A_{i}$ 's
Therefore,
$\sum_{i = 1}^{30} n (A_{i}) = 10 k$ ...(2)
Hence, from (1) and (2)
$10 k = 150 \Rightarrow k = 15$
Now, since each $B_{j}$ has $3$ elements and each $S$ belongs to exactly $9$ of $B_{j}$ 's
Therefore,
$\sum_{j = 1}^{n} n (B_{j}) = 3 n$ and $\sum_{j = 1}^{n} n (B_{j}) = 9 k$
$\Rightarrow 3 n = 9 k = 9 \times 15$
$\Rightarrow n = 45$

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Suppose A1,A2,.....,A30 are thirty sets each with five elements and B1,B2,.....,Bn are n sets each with three elements such that ⋃30i=1Ai=⋃nj=1Bj=S and each element of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n=