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A school awarded 38 medals in football, 15 in basketball and 20 in cricket. Suppose these medals went to a total of 58 students and only three students got medals in all three sports. If only 5 students got medals in football and basketball, then the number of medals received in exactly two of three sports is :

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Detailed Solution

$\begin{array}{l} 58 = n (F \cup B \cup C) = n (f) + n (B) + n (C) - n (F \cap B) - n (B \cap C) - n (F \cap C) + n (F \cap B \cap C) \\ = 38 + 15 + 20 - n (F \cap B) - n (B \cap C) - n (F \cap C) + 3 \\ \Rightarrow n (F \cap B) + n (B \cap C) + n (F \cap C) = 18 \end{array}$

Number of medals received in exactly two of three sports is $= n (F \cap B \cap C^{'}) + n (B \cap C \cap F^{'}) + n (F \cap C \cap B^{'})$

$= n (F \cap B) + n (B \cap C) + n (F \cap C) - 3 n (F \cap B \cap C)$

$= 18 - 9 = 9$

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