Question

In a class of 55 students, the number of students studying different subjects is 23 in Mathematics and 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is
A. 20
B. 24
C. 23
D. 22

Answer 1

Hint: Let the no. of students studying mathematics be $ n\left( M \right) $ , studying physics be $ n\left( P \right) $ and studying chemistry be $ n\left( C \right) $ . To find the number of students who have taken exactly one subject, we have to subtract the number of students who have taken more than one subject and add the no. of students who have taken all the three subjects from the no. of students studying respective subjects. We have to take a separate case each for mathematics, physics and chemistry and add the three together to find the required number.

Complete step-by-step answer:
We are given that in a class of 55 students, $ n\left( M \right) $ is 23, $ n\left( P \right) $ is 24 and $ n\left( C \right) $ is 19.
And no. of students who study both mathematics and physics is 12, $ n\left( {M \cap P} \right) = 12 $ ; both mathematics and chemistry is 9, $ n\left( {M \cap C} \right) = 9 $ ; both physics and chemistry is 7, $ n\left( {P \cap C} \right) = 7 $ ; and the no. of students who study all the subjects is 4, $ n\left( {M \cap P \cap C} \right) = 4 $
So to find the number of students who have taken exactly one subject, we have to find the no. of students who have taken only mathematics, only physics and only chemistry and add them all.
No. of students who have taken only mathematics is $ n\left( M \right) - \left[ {n\left( {M \cap P} \right) + n\left( {M \cap C} \right)} \right] + n\left( {M \cap P \cap C} \right) $
 $ \Rightarrow 23 - \left( {12 + 9} \right) + 4 = 27 - 21 = 6 $
No. of students who have taken only physics is $ n\left( P \right) - \left[ {n\left( {M \cap P} \right) + n\left( {P \cap C} \right)} \right] + n\left( {M \cap P \cap C} \right) $
 $ \Rightarrow 24 - \left( {12 + 7} \right) + 4 = 28 - 19 = 9 $
No. of students who have taken only chemistry is $ n\left( C \right) - \left[ {n\left( {M \cap C} \right) + n\left( {P \cap C} \right)} \right] + n\left( {M \cap P \cap C} \right) $
 $ \Rightarrow 19 - \left( {9 + 7} \right) + 4 = 23 - 16 = 7 $
Therefore, the number of students who have taken exactly one subject is $ 6 + 9 + 7 = 22 $
Hence, the correct option is Option D, 22.
So, the correct answer is “Option D”.

Note: So the students who have taken exactly one subject means taking either mathematics or physics or chemistry, not more than one. So we had to subtract the no. of students who took more than one subject. And we added $ n\left( {M \cap P \cap C} \right) $ because it is subtracted twice in each case indirectly. But it should only be subtracted once.