Let's Unite Them!

Probability Level 5

Let us say A 1 , A 2 , A 3 , … , A 30 A_1,A_2,A_3,\ldots ,A_{30} are thirty sets containing 6 elements each. While B 1 , B 2 , B 3 , … , B n B_1,B_2,B_3,\ldots ,B_n are n n sets containing 3 elements each. Now consider the following;

⋃ k = 1 30 A k = μ = ⋃ k = 1 n B k \large\displaystyle \mathop{\bigcup}_{k=1}^{30}A_k=\mu=\displaystyle \mathop{\bigcup}_{k=1}^{n}B_k

Such that, each element of μ \mu belongs to exactly 10 elements of A k A_k 's and exactly 9 elements of B k B_k 's, then find the value of n n .

Notation : The symbol ∪ \cup denotes set union , and ⋃ k = 1 m X k = X 1 ∪ X 2 ∪ X 3 ∪ ⋯ ∪ X m . \large\displaystyle\mathop{\bigcup}_{k=1}^{m}X_k=X_1\cup X_2\cup X_3\cup\dots\cup X_m.


The answer is 54.

Sravanth C. by Sravanth C. , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sravanth C.
May 27, 2016

Firstly, the total number of elements in A k A_k 's = 30 × 6 = 180 =30\times 6=180 , and the total number of elements in B k B_k 's = n × 3 = 3 n =n\times 3=3n . But each element of μ \mu is present in 10 10 elements of A k A_k 's and 9 9 elements of B k B_k 's, we can say; 180 10 = 3 n 9 n = 180 × 3 10 n = 54 \begin{aligned} \dfrac{180}{10}&=\dfrac{3n}{9}\\ n&=\dfrac{180\times 3}{10}\\ n&=54 \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...