Question

In a survey of $200$ students of a higher secondary school, it was found that $120$ studied mathematics; $90$ studied physics and $70$ studied chemistry; $40$ studied mathematics and physics; $30$ studied physics and chemistry; $50$ studied chemistry and mathematics, and $20$ studied none of these subjects. If $M$, $P$ and $C$ represent the set of students who studied mathematics, physics and chemistry respectively, then

• A. $n(M\cup P\cup C)=180$
• B. $n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=20$
• C. $n(M\cap P\cap C)=40$
• D. $n(M\cap P\cap C)=20$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $n(M\cup P\cup C)=180$
B $n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=20$
D $n(M\cap P\cap C)=20$

Let $U$ be the universal set.
$n\left(U\right)=200$,
$n\left(M\right)=120,n\left(P\right)=90,n\left(C\right)=70$,
$n(M\cap P)=40,n(P\cap C)=30,n(C\cap M)=50,$
$n(M^{\prime }\cap P^{\prime }\cap C^{\prime })=n(M\cup P\cup C{)}^{\prime }=20$

Now, $n(M\cup P\cup C{)}^{\prime }=n(U)-n(M\cup P\cup C)$
$\Rightarrow n(M\cup P\cup C)=180$
$n(M\cup P\cup C)=n\left(M\right)+n\left(P\right)+n\left(C\right)-n(M\cap P)-n(P\cap C)-n(C\cap M)+n(M\cap P\cap C)$
$\Rightarrow n(M\cup P\cup C)=120+90+70-40-30-50+n(M\cap P\cap C)$
$\Rightarrow n(M\cap P\cap C)=180-160=20$