Math, asked by saryka, 5 months ago

⇒ Solve this question​

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Answered by mathdude500
72

\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}

Answered by MrImpeccable
64

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.

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