Question: In a school of 600 students, the intersection is equal to compliment. One quarter of set A is equal to set B and n(B) = n only (A) - 30. Find n only (A).

Answer: 150

Solution:
Let the set of the total number of students in the school be U.

Given,
$n(U) = 600$
$n(A \cap B) = \overline{ n (A \cup B)}$
$\dfrac{1}{4} of n(A) = n(B)$
$n(B) = n_o(A) - 30$

To find: $n_o(A) = ?$

Using formula,
 
$n(U) = n(A) + n(B) - n(A \cap B) + \overline{ n (A \cup B)} $

[$n(A \cap B) = \overline{ n (A \cup B)}$]

$or, 600 = n(A) + n(B) - n(A \cap B) + n(A \cap B)$

$or, 600 = n(A) + n(B)$

$or, 600 = n(A) + \dfrac{1}{4} of n(A) $

$or, 600 = \dfrac{4 n(A) + n(A)}{4}$

$or, 600 * 4 = 5 n(A)$

$or, \dfrac{2400}{5} = n(A)$

$\therefore n(A) = 480$


From given,
$\dfrac{1}{4} of n(A) = n(B)$ -- (i)
$n(B) = n_o(A) - 30$ -- (ii)

$\dfrac{1}{4} of n(A) = n_o(A) - 30$

$or, \dfrac{1}{4} * 480 = n_o(A) - 30$

$or, 120 = n_o(A) - 30$

$or, 120 + 30 = n_o(A)$

$\therefore n_o(A) = 150$

Hence, the required value of n only A is 150.


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