In a class of 140 students numbered 1 to 140, all even-numbered students opted for Mathema…

Question

In a class of 140 students numbered 1 to 140, all even-numbered students opted for Mathematics course, those whose number is divisible by 3 opted for Physics course, and those whose number is divisible by 5 opted for Chemistry course. Then the number of students who did not opt for any of the three courses is:

Accepted Answer

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

Solutions · 1