JEE Main 2019 — Sets And Relations Question with Solution

We're given that there are 140 students numbered from 1 to 140.

1. Define the set $A$ to be the set of even numbered students. The cardinality of $A$ (the number of elements in $A$), denoted as $n(A)$, can be computed as the greatest integer less than or equal to $140/2$. Hence, $n(A)=\left[\frac{140}{2}\right]=70$. ([.] denotes greatest integer function)

2. Similarly, let $B$ be the set of students whose number is divisible by 3. Hence, $n(B)=\left[\frac{140}{3}\right]=46$. ([.] denotes greatest integer function)

3. Let $C$ be the set of students whose number is divisible by 5. Hence, $n(C)=\left[\frac{140}{5}\right]=28$.

So far, we've found the number of students who opted for Mathematics ($n(A)$), Physics ($n(B)$), and Chemistry ($n(C)$).

We also need to consider the students who have opted for multiple subjects :

1. $n(A\cap B)$ represents the count of numbers that are divisible by both 2 and 3 (i.e., divisible by 6).

So, $n(A\cap B)=\left[\frac{140}{6}\right]=23$.

2. $n(B\cap C)$ represents the count of numbers that are divisible by both 3 and 5 (i.e., divisible by 15).

So, $n(B\cap C)=\left[\frac{140}{15}\right]=9$.

3. $n(C\cap A)$ represents the count of numbers that are divisible by both 2 and 5 (i.e., divisible by 10).

So, $n(C\cap A)=\left[\frac{140}{10}\right]=14$.

Finally, $n(A\cap B\cap C)$ represents the count of numbers that are divisible by 2, 3, and 5 (i.e., divisible by 30).

So, $n(A\cap B\cap C)=\left[\frac{140}{30}\right]=4$.

Now we use the principle of inclusion and exclusion to compute the number of students who have opted for at least one subject. The principle states :

$$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$$

Substituting the values we calculated above :

$$n(A\cup B\cup C)=(70+46+28)-(23+9+14)+4=102$$

Hence, the number of students who opted for at least one subject is 102. Therefore, the number of students who did not opt for any of the subjects is

Total $$-$$ n(A $$\cup$$ B $$\cup$$ C)

= 140 $$-$$ 102 = 38