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Home » Suppose are thirty sets each having 5 elements and are sets each with 3 elements, let and each element of belongs to exactly 10 of the ‘s and exactly 9 of the B,’S. then is equal to A. 15 B. 3 C. 45 D. 35
Suppose A_{1}, A_{2}, \ldots, A_{30} are thirty sets each having 5 elements and B_{1}, B_{2}, \ldots, B_{n} are n sets each with 3 elements, let \bigcup_{i=1}^{30} A_{i}=\bigcup_{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,’S. then n is equal to A. 15 B. 3 C. 45 D. 35
CBSE | Class 11 | Maths | NCERT Exemplar | Sets
Suppose A_{1}, A_{2}, \ldots, A_{30} are thirty sets each having 5 elements and B_{1}, B_{2}, \ldots, B_{n} are n sets each with 3 elements, let \bigcup_{i=1}^{30} A_{i}=\bigcup_{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,’S. then n is equal to A. 15 B. 3 C. 45 D. 35

Solution:

As per the question,

\bigcup_{i=1}^{30} A_{i}=\bigcup_{j=1}^{n} B_{j}=S

As the elements are not repeating, no. of elements in \mathrm{A}_{1} \cup \mathrm{A}_{2} \cup \mathrm{A}_{3} \cup \ldots \ldots \ldots . \cup \mathrm{A}_{30}=30 \times 5

As each element is used 10 times, so now

We have,

10 \times S=3

\Rightarrow 10 \times \mathrm{S}=150

\Rightarrow \mathrm{S}=15

As the elements are not repeating, no. of elements in \mathrm{B}_{1} \cup \mathrm{B}_{2} \cup \mathrm{B}_{3} \cup \ldots \ldots \ldots . \cup \mathrm{B}_{\mathrm{n}}=3 \times \mathrm{n}

As each element is used 9 times, so now

We have,

9 \times S=3 \times n

\Rightarrow 9 \times S=3 n

\Rightarrow S=n / 3

\Rightarrow \mathrm{n} / 3=15

\Rightarrow \mathrm{n}=45

Hence, the value of n is 45

As a result, the correct answer is option (C).

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