Question

In a survey of 500 TV viewers, it was found that 285 watch cricket, 195 watch football and 115 watch tennis. Also, 45 watch both cricket and football, 70 watch both cricket and tennis and 50 watch football and tennis. If 50 do not watch any game on TV, how many watch all the three games?
A.10
B.20
C.30
D.15

Answer 1

Hint: Consider the number viewers of 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 games by taking the number of elements in the union of the 3 sets which is obtained by subtracting the number of people who do not watch any games in the TV from the total number of viewers. Then we can find the number of viewers watching all the three games by finding the number of elements in the intersection of the 3 sets. It is determined by using the equation \[n\left( {C \cup F \cup T} \right) = n\left( C \right) + n\left( F \right) + n\left( T \right) - n\left( {F \cap T} \right) - n\left( {C \cap F} \right) - n\left( {C \cap T} \right) + n\left( {C \cap F \cap T} \right)\] .

Complete step-by-step answer:
Let C represent the viewers watch that cricket. Then the number of viewers that watch cricket is given as 285.
 $ \Rightarrow n\left( C \right) = 285$
Let F represent the viewers that watch football. Then the number of viewers that watch football is given as 195.
 $ \Rightarrow n\left( F \right) = 195$
Let B represent the viewers that watch tennis. Then the number of viewers that watch tennis is given as 115.
 $ \Rightarrow n\left( T \right) = 115$
According to the intersections of sets,
Let $C \cap F$ represent the viewers that watch both cricket and football. Then the number of viewers that watch both cricket and football are 45.
 $ \Rightarrow n\left( {C \cap F} \right) = 45$ .
Let $C \cap T$ represent the viewers that watch both cricket and tennis. Then the number of viewers that watch both cricket and tennis are 70.
 $ \Rightarrow n\left( {C \cap T} \right) = 70$ .
Then $F \cap T$ represents the viewers that watch both football and tennis. Then the number of viewers that watch both football and tennis are 50.
 $ \Rightarrow n\left( {F \cap T} \right) = 50$ .
The union of the three sets will give the set of all the viewers that watch either of the 3 games. So, the number of viewers that watch at least any one of the games is given by $n\left( {C \cup F \cup T} \right)$ .
It is given that out of 500 people 50 does not watch any of the 3 games. So, we can obtain the number of viewers that watch at least of any one of the games by finding the difference. So, we can write,
 $ \Rightarrow n\left( {C \cup F \cup T} \right) = 500 - 50$
 $ \Rightarrow n\left( {C \cup F \cup T} \right) = 450$
Then $C \cap F \cap T$ represents the viewers that watch all the three games. Then the number of viewers that watch all the three games is given by $n\left( {C \cap F \cap T} \right)$ .
We know that,
 \[n\left( {C \cup F \cup T} \right) = n\left( C \right) + n\left( F \right) + n\left( T \right) - n\left( {F \cap T} \right) - n\left( {C \cap F} \right) - n\left( {C \cap T} \right) + n\left( {C \cap F \cap T} \right)\]
Now we can substitute the values,
 \[ \Rightarrow 450 = 285 + 195 + 115 - 50 - 45 - 70 + n\left( {C \cap F \cap T} \right)\]
On simplification, we get,
 \[ \Rightarrow 450 = 430 + n\left( {C \cap F \cap T} \right)\]
On rearranging, we get,
 \[ \Rightarrow n\left( {C \cap F \cap T} \right) = 450 - 430\]
 \[ \Rightarrow n\left( {C \cap F \cap T} \right) = 20\]
Therefore, the number of viewers that watch all the three games is 20.
So, the correct answer is option B which is 20.

Note: 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 sets, 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 and while simplification.
In 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.