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# A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all the three sports. Then the number of students who received medals in exactly two of the three sports is,(a) 18(b) 15(c) 9(d) 6

Last updated date: 21st Mar 2023
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Hint: Represent the football, basketball, cricket as the sets F, B, C. Then write down the values of

n(F), n(B), n(C)…….. $n\left( F\cup B\cup C \right)$and $n\left( F\cap B\cap C \right)$ .
Then further use the formula

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$ and get the desired results.

In the question it is said that a college awarded 38 medals in football, 15 medals in basketball and 20 in cricket. Then it is said that there medals went to a total of 58 men and only 3 men got the medals in all three sports and hence we have to find the number of men who received exactly in two of the three sports.

Let the number of men who got medals in football be represented as n(F), similarly for basketball let it be represented as n(B) and in cricket as n(C).

For number of people who won in all the three sports will be $n\left( F\cap B\cap C \right)$, and the number of people who got medals in two of three is $\left\{ n\left( F\cap C \right)+n\left( F\cap B \right)+n\left( B\cap C \right) \right\}$and for total number men be represented as $n\left( F\cup B\cup C \right).$

So now we will represent the data as,

n(F) = 38

n(B) = 15

n(C) = 20

$n\left( F\cap B\cap C \right)=3$
$n\left( F\cup B\cup C \right)=58$

So now we will use the formula which is

$n\left( A\cup B\cup C \right)=n\left( A \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( A\cap B \right)+n\left( B\cap C \right)+n\left( A\cap C \right) \right\}+n\left( A\cap B\cap C \right)$

Now instead of sets A, B, C we will put sets of football (F), Basketball (B) and cricket (C).
We will write as,

$n\left( F\cup B\cup C \right)=n\left( F \right)+n\left( B \right)+n\left( C \right)-\left\{ n\left( F\cap B \right)+n\left( B\cap C \right)+n\left( F\cap C \right) \right\}+n\left( F\cap B\cap C \right)$

So now putting the values of n(F) = 38, n(B) = 15, n(C) = 20, $n\left( F\cap B\cap C \right)=3$and$n\left( F\cup B\cup C \right)$ we get,

$58=38+15+20-\left\{ n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right) \right\}+3$

So,

$n\left( F\cap B \right)+n\left( F\cap C \right)+n\left( B\cap C \right)=\left( 38+15+20+3 \right)58=18$

Hence the number of men who got two out of three medals are 18.

Therefore the correct answer is option (a).

Note: Students are generally confused between the signs $\cup$ and $\bigcap$ . Actually $\left( A\cup B \right)$ means that we have to include both the elements of A and B. $\left( A\cap B \right)$ means that we have to consider only that element which is common to both A and B.