JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 16)
Let $$\mathop \cup \limits_{i = 1}^{50} {X_i} = \mathop \cup \limits_{i = 1}^n {Y_i} = T$$ where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi’s and exactly 6 of sets Yi’s, then n is equal to :
30
50
15
45
Explanation
$$\mathop \cup \limits_{i = 1}^{50} {X_i} = $$ X1, X2,....., X50 = 50 sets. Given each sets having 10 elements.
So total elements = 50 $$ \times $$ 10
$$\mathop \cup \limits_{i = 1}^n {Y_i} =$$ $$ Y1, Y2,....., Yn = n sets. Given each sets having 5 elements.
So total elements = 5 $$ \times $$ n
Now each element of set T contains exactly 20 of sets Xi.
So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
Also each element of set T contains exactly 6 of sets Yi.
So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
$$ \therefore $$ $${{50 \times 10} \over {20}}$$ = $${{5 \times n} \over {20}}$$
$$ \Rightarrow $$ n = 30
So total elements = 50 $$ \times $$ 10
$$\mathop \cup \limits_{i = 1}^n {Y_i} =$$ $$ Y1, Y2,....., Yn = n sets. Given each sets having 5 elements.
So total elements = 5 $$ \times $$ n
Now each element of set T contains exactly 20 of sets Xi.
So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
Also each element of set T contains exactly 6 of sets Yi.
So number of effective elements in set T = $${{50 \times 10} \over {20}}$$
$$ \therefore $$ $${{50 \times 10} \over {20}}$$ = $${{5 \times n} \over {20}}$$
$$ \Rightarrow $$ n = 30
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