Question

Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey. 24 played all the three games. The number of boys who did not play any game is
A) 128
B) 216
C) 240
D) 160

Answer 1

We can consider the number of boys playing each game as different sets and equate the given numbers as the number of elements in each set. Then we can find the intersections of the sets, taking 2 at a time and equate the numbers. Then we can find the intersection of the 3 sets taken together. Then we can find the number of who played at least one of the game by taking the number of elements in the union of the 3 sets, which is obtained by substituting in the equation, $n\left( {C \cup H \cup B} \right) = n\left( C \right) + n\left( H \right) + n\left( B \right) - n\left( {B \cap H} \right) - n\left( {C \cap B} \right) - n\left( {C \cap H} \right) + n\left( {C \cap H \cap B} \right)$ . Then we can subtract this number from the total number of boys to get the number of boys who did not play any game.

Complete step by step solution:
Let C represent the boys who played cricket. Then the number of boys played cricket is given as 224.
$\Rightarrow n\left( C \right) = 224$
Let H represent the boys who played hockey. Then the number of boys played hockey is given as 240.
$\Rightarrow n\left( H \right) = 240$
Let B represent the boys who played basketball. Then the number of boys played basketball is given as 336.
$\Rightarrow n\left( B \right) = 336$
According to the intersections of sets,
Let $B \cap H$ represent the boys playing both basketball and hockey. The number of boys played both basketball and hockey are 64.
$\Rightarrow n\left( {B \cap H} \right) = 64$ .
Let $C \cap B$represents the boys playing cricket and basketball. The number of boys playing both cricket and basketball are 80.
$\Rightarrow n\left( {C \cap B} \right) = 80$ .
Then $C \cap H$ represents the boys playing both cricket and hockey. The number of boys playing both cricket and hockey are 40.
$\Rightarrow n\left( {C \cap H} \right) = 40$ .
Then $C \cap H \cap B$ represents the boys playing all the three games. The number of boys playing all the three games are 24.
$\Rightarrow n\left( {C \cap H \cap B} \right) = 24$ .
The union of the three sets will give the set of all the boys that play either of the 3 games. So, the number of boys who played at least any one of the games is given by $n\left( {C \cup H \cup B} \right)$ .
By properties of sets we know that, $n\left( {C \cup H \cup B} \right) = n\left( C \right) + n\left( H \right) + n\left( B \right) - n\left( {B \cap H} \right) - n\left( {C \cap B} \right) - n\left( {C \cap H} \right) + n\left( {C \cap H \cap B} \right)$
On substituting the values, we get,
$n\left( {C \cup H \cup B} \right) = 224 + 240 + 336 - 64 - 80 - 40 + 24$
On further calculations, we get,
$n\left( {C \cup H \cup B} \right) = 640$
So, the number of boys played at least one of the games is 640.
It is given that there are 800 boys in total. Then the number of boys who did not play any game is given by subtracting the number of boys played at least one of the games from the total number of boys.
$\Rightarrow 800 - 640 = 160$

Therefore, the number of boys who did not play any game is 160.
So, the correct answer is option D.

Note:
The concept used here is set theory. Union of two sets gives a set of all elements that are at least in one of the two sets. Intersection of two sets gives the set of all elements that are in both the sets.
While writing the equation to find the number of elements in the union of the 3 sets, first we add the number of elements in each set, then we subtract the number of elements in the intersection of the sets two taken at a time and again add the number of elements in the intersection of the three sets. We must take care of the signs while writing the equation.