Question

In a survey of $200$ students of a school it was found that $120$ study Mathematics, $90$ study Physics and $70$ study Chemistry, $40$ study mathematics and physics, $30$ study Physics and Chemistry, $50$ study chemistry and mathematics and $20$ of the subjects. The number of students who study all the three subjects.
(A) $30$
(B) $20$
(C) $22$
(D) $25$

Answer 1

Hint: To solve this type question we need to know the concept of sets. Set is defined as the collection of distinct elements. In this question we basically use our knowledge of intersection and union. In this question the number of students in a school is considered to be an Universal set and rest other conditions are the small sets under the universal set.

Complete step by step solution:
In this question we are supposed to find a number of students who are interested in studying all the three subjects, which are chemistry, mathematics and physics. There are many conditions given in the question as per the choice of the students to study the subjects. There are in total $200$ students in which $120$ student study Mathematics, $90$ is interested in Physics and $70$in Chemistry, $40$ students are such that they like both Mathematics and Physics, $30$ like both Physics and Chemistry and $50$ study book chemistry and mathematics and $20$ are those who do not study any of the subjects. Considering Mathematics, Physics and Chemistry to be $\text{M, P}$ and $\text{C}$. On writing these fact in mathematical form we get:
$n\left( U \right)=200$, $n\left( M \right)=120$, $n\left( P \right)=90$ , $n\left( C \right)=70$, $n\left( M\cap P \right)=40$, $n\left( P\cap C \right)=30$,$n\left( C\cap M \right)=50$, $n\left( M\cup P\cup C \right)=20$
Also, for the students who study any of the subject or all the subjects:
 $\Rightarrow n\left( U \right)-n\left( M\cup P\cup C \right)$
$\Rightarrow 200-20$
$\Rightarrow 180$
Now to find the number of students who are interested in all the three subjects are:
\[n\left( M\cup P\cup C \right)=n\left( M \right)+n\left( P \right)+n\left( C \right)-n\left( M\cap P \right)-n\left( P\cap C \right)-n\left( C\cap M \right)+n\left( M\cap P\cap C \right)\]
\[\Rightarrow 180=120+90+70-40-30-50+n\left( M\cap P\cap C \right)\]
\[\Rightarrow 180=160+n\left( M\cap P\cap C \right)\]
\[\Rightarrow 180-160=n\left( M\cap P\cap C \right)\]
\[\Rightarrow 20=n\left( M\cap P\cap C \right)\]
$\therefore $ Number of students interested in all the three subjects are $\left( B \right)20$.

So, the correct answer is “Option B”.

Note: Do remember the difference between Union and intersection of the set. The union of two sets contains all the elements contained in either set( or both set). The union is noted by $A\cup B$. The intersection of two sets contains only the elements that are in both sets. Intersection is denoted by $A\cap B$. Here $A$ and $B$ are the two sets.