Question

In a class of $55$ students, the number of students studying different subjects are $23$ in Mathematics, $24$ 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. $6$
• B. $9$
• C. $7$
• D. $22$

Answer 1

Answer: (D)

$n(M)=23, n(P)=24, n(C)=19$
$n(M \cap P)=12, n(M \cap C)=9, n(P \cap C)=7$
$n(M \cap P \cap C)=4$
We have to find $n\left(M \cap P^{\prime} \cap C^{\prime}\right), n\left(P \cap M^{\prime} \cap C^{\prime}\right)$,
$n\left(C \cap M^{\prime} \cap P^{\prime}\right)$
Now $n\left(M \cap P^{\prime} \cap C^{\prime}\right)=n\left[M \cap(P \cup C)^{\prime}\right]$
$=n M-n M \cap P \cup C$
$=n(M)-n[(M \cap P) \cup(M \cap C)]$
$=n(M)-n(M \cap P)-n(M \cap C)+n(M \cap P \cap C)$
$=23-12-9+4=27-21=6$
$n\left(P \cap M^{\prime} \cap C^{\prime}\right)=n\left[P \cap(M \cup C)^{\prime}\right]$
$=n(P)-n[P \cap(M \cup C)]=n(P)-n[(P \cap M) \cup(P \cap C)]$
$=n(P)-n(P \cap M)-n(P \cap C)+n(P \cap M \cap C)$
$=24-12-7+4=9$
$n\left(C \cap M^{\prime} \cap P^{\prime}\right)=n(C)-n(C \cap P)-n(C \cap M)+n(C \cap P \cap M)$
$=19-7-9+4=23-16=7$
Hence, the total no. of students studying exactly one subject are $6+9+7=22$