JEE Main 2019 — Sets and Relations Question with Solution

From the given $140$ students, all even numbered students opted Mathematics $\left(M\right),$ hence, the number of students opted Mathematics is $M\equiv \left[\frac{140}{2}\right]=70,$ where $\left[x\right]$ denotes the greatest integer function of $x.$

Similarly, the students who opted Physics are $P\equiv \left[\frac{140}{3}\right]=46$

And, the students who opted Chemistry are $C\equiv \left[\frac{140}{5}\right]=28$

Similarly, the students who opted Mathematics and Physics are the numbers which are divisible by both $2 \& 3$ i.e. $6,$ and their number is $\left[\frac{140}{6}\right]=23$

The students who opted Mathematics and Chemistry are the numbers which are divisible by both $2 \& 5$ i.e. $10,$ and their number is $\left[\frac{140}{10}\right]=14$

The students who opted Physics and Chemistry are the numbers which are divisible by both $3 \& 5$ i.e. $15,$ and their number is $\left[\frac{140}{15}\right]=9$

And, the students who opted all the three subjects are the numbers which are divisible by $2, 3 \& 5$ i.e. $30,$ and their number is $\left[\frac{140}{30}\right]=4$

Now, using
$n\left(P\cup C\cup M\right)=n\left(P\right)+n\left(C\right)+n\left(M\right)-n\left(P\cap C\right)-n\left(C\cap M\right)-n\left(M\cap P\right)+n\left(P\cap M\cap C\right)$

$=46+28+70-9-14-23+4$

$=102$

Thus, $102$ students have opted for atleast one of the three subjects.

Hence, number of students who did not opt for any of the three courses $=140-102=38.$