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