Question

⇒ Solve this question​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:n(A_1) = 5$

$\rm :\longmapsto\:n(A_2) = 5$

$\rm :\longmapsto\:n(A_3) = 5$

.

.

.

.

$\rm :\longmapsto\:n(A_{30}) = 5$

So,

It implies,

$\sf \: A_1 \cup \:A_2\cup \:A_3 \cup \:- - - \cup \:A_{30} = 30 \times 5 = 150$

Since, each element is used 10 times,

So,

$\rm :\longmapsto\:10 \times S = 150$

$\bf\implies \:S = 15 - - - (1)$

Also,

$\rm :\longmapsto\:n(B_1) = 3$

$\rm :\longmapsto\:n(B_2) = 3$

$\rm :\longmapsto\:n(B_3) = 3$

.

.

.

.

$\rm :\longmapsto\:n(B_n) = 3$

So,

It implies,

$\sf \: B_1\cup \:B_2\cup \:B_3\cup \: - - - \cup \:B_n = 3 \times n = 3n$

Since, each element is used 9 times.

$\rm :\implies\:9 \times S = 3n$

$\bf\implies \:S = \dfrac{n}{3} - - - (2)$

From equation (1) and equation (2), we get

$\rm :\longmapsto\:15 = \dfrac{n}{3}$

$\bf\implies \:n \: = \: 45$

$\boxed{ \bf \: Hence, \: Option \: (c) \: is \: correct}$

Answer 2

ANSWER:

Given:

• A₁, A₂. . . . . A₃₀ are 30 sets with 5 elements each.
• B₁, B₂. . . . . B$_n$ are n sets with 3 elements each.
• $^{^{30}}U_{_{i=1}}\,A_i=\:^{^n}U_{_{j=1}}\,B_j = S}$
• Each element of S∈10 elements of A$_i$'s and 9 elements of B$_j$'s

To Find:

• Value of n

Solution:

We are given that, A₁, A₂. . . . . A₃₀ all contain 5 elements each.

So,

⇒ Total elements in A₁, A₂. . . . . A₃₀ = 30 * 5 = 150elements.

As each element of S belongs to 10 of A$_i$'s,

⇒ 10 × S = 150

⇒ S = 15 ------(1)

Now,

We are given that, B₁, B₂. . . . . B$_n$ all contain 3 elements each.

So,

⇒ Total elements in B₁, B₂. . . . . B$_n$ = n * 3 = 3n elements.

As each element of S belongs to 9 of B$_j$'s,

⇒ 9 × S = 3n

From (1),

⇒ 9 × 15 = 3n

⇒ n = 135/3

⇒ n = 45

Therefore, the correct option is c) 45.