Question

In a class of $42$ students, $23$ are studying Mathematics, $24$ are studying Physics, $19$ are studying Chemistry. If $12$ are studying both Mathematics and Physics, $9$ are studying both Mathematics and Chemistry, $7$ are studying both Physics and Chemistry and $4$ are studying all the three subjects, then the number of students studying exactly one subject, is

• A. $15$
• B. $30$
• C. $22$
• D. $27$

Answer 1

The correct option is C $22$

Let $M,P,C$ denote the set of students studying Mathematics, Physics and Chemistry respectively.
$n\left(M\right)=23,n\left(P\right)=24,n\left(C\right)=19n(M\cap P)=12,n(M\cap C)=9n(P\cap C)=7,n(M\cap P\cap C)=4$

Now, $n(M\cap P^{\prime }\cap C^{\prime })$
$n(M\cap (P\cup C{)}^{\prime })$
$n\left(M\right)-n(M\cap (P\cup C\left)\right)$
$=n\left(M\right)-n\left[\right(M\cap P)\cup (M\cap C\left)\right]$
$=n\left(M\right)-\left[n\right(M\cap P)+n(M\cap C\left)\right]+n(M\cap P\cap C)=23-[12+9]+4=6$

Similarly, $n(P\cap M^{\prime }\cap C^{\prime })$
$=n\left(P\right)-\left[n\right(P\cap M)+n(P\cap C\left)\right]+n(M\cap P\cap C)=24-[12+7]+4=9$

and $n(C\cap P^{\prime }\cap M^{\prime })$
$=n\left(C\right)-\left[n\right(C\cap P)+n(C\cap M\left)\right]+n(M\cap P\cap C)=19-[9+7]+4=7$
Hence, number of students studying exactly one subject $=6+9+7=22$

Alternatively, using Venn diagram,


Given, $a+b+c+d=23\cdots \left(1\right)$
$b+c+e+f=24\cdots \left(2\right)$
$d+c+f+g=19\cdots \left(3\right)$
$b+c=12,d+c=9,c+f=7$
and $c=4$
$\therefore b=8,d=5,f=3$
From $\left(1\right),$ $a=6$
From $\left(2\right),$ $e=9$
From $\left(3\right),$ $g=7$
Required number $=a+e+g=22$