An organization awarded $48$ medals in event 'A', $25$ in event 'B' and $18$ in event 'C'. If these medals went to total $60$ men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
(A)
$10$
(B)
$15$
(C)
$21$
(D)
$9$
HARD
Correct Answer:C
Explanation
Let $|A|$ be the number of medals in event A, $|B|$ be the number of medals in event B, and $|C|$ be the number of medals in event C. We are given $|A| = 48$, $|B| = 25$, and $|C| = 18$. We are also given that the total number of people who received medals is $60$, so $|A \cup B \cup C| = 60$, and the number of people who received medals in all three events is $5$, so $|A \cap B \cap C| = 5$.
We want to find the number of people who received medals in exactly two of the three events. Let $x$ be the number of people who received medals in exactly two events. Using the Principle of Inclusion-Exclusion, we have:
$|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$
$60 = 48 + 25 + 18 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$
$60 = 91 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$
$|A \cap B| + |A \cap C| + |B \cap C| = 91 + 5 - 60 = 36$
Now, let $N_{2}$ be the number of people who received medals in exactly two events. Then:
$|A \cap B| + |A \cap C| + |B \cap C| = N_{2} + 3|A \cap B \cap C|$
$36 = N_{2} + 3(5)$
$N_{2} = 36 - 15 = 21$
Therefore, the number of people who received medals in exactly two of the three events is $21$.