QJEEM19Ma08a7: JEE Main 2019 Mathematics - Class 11 JEEM19Ma08a7 Solution: In a class of 140 students numbered 1 to 140, all even numbered …

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