Suppose A1-A2-A30 are thirty sets having 5 elements and B1-B2-Bn are n sets each with 3 elements- Let -i-130Ai-i-1nBj-S and each elements of S belongs to exactly 10 of the Ai-s and exactly 9 of the Bj-s- Then n is equal to

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Standard XII

Mathematics

Properties of Set Operation

Suppose A1,...

Question

Suppose $A_{1}, A_{2} . . . . . A_{30}$ are thirty sets having 5 elements and $B_{1}, B_{2} . . . . B_{n}$ are n sets each with 3 elements. Let $30 ⋃ i = 1 A i = n ⋃ i = 1 B j = S$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to

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Solution

The correct option is B 45
Since each element of S belongs to $10$ $A_{i} s$ , there are only $15$ elements in S $[\frac{30 \times 5}{10}]$

Since each of the $15$ elements belongs to 9 $B_{j} s$ there are 135 elements in the second union.

No. of sets $= \frac{135}{3} = 45$

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Q. Suppose

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of the

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and exactly

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is equal to

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Suppose A1,A2,.....................A30 are thirty sets each having 5 elements and B1,B2,................Bn are n sets each with 3 elements, let $30 ⋃ i = 1$ $A_{i}$ = $n ⋃ j = 1$ $B_{j}$ = S and each element of S belongs to exactly 10 of $A_{i}$ 'S and exactly 9 of the $B_{j}$ 'S. Then n is equal to

Q. Suppose

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(a) 15 (b) 3 (c) 45 (d) 35

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Suppose A1,A2.....A30 are thirty sets having 5 elements and B1,B2....Bn are n sets each with 3 elements. Let 30⋃i=1Ai=n⋃i=1Bj=S and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to

The correct option is B 45Since each element of S belongs to 10 Ais, there are only 15elements in S [30×510]Since each of the 15 elements belongs to 9 Bjs there are 135 elements in the second union.No. of sets =1353=45

Suppose $A_{1}, A_{2} . . . . . A_{30}$ are thirty sets having 5 elements and $B_{1}, B_{2} . . . . B_{n}$ are n sets each with 3 elements. Let $30 ⋃ i = 1 A i = n ⋃ i = 1 B j = S$ and each elements of S belongs to exactly 10 of the Ai's and exactly 9 of the Bj's. Then n is equal to

The correct option is B 45
Since each element of S belongs to $10$ $A_{i} s$ , there are only $15$ elements in S $[\frac{30 \times 5}{10}]$

Since each of the $15$ elements belongs to 9 $B_{j} s$ there are 135 elements in the second union.

No. of sets $= \frac{135}{3} = 45$