Question

There are $100$ students in a class. In the examination, $50$ of them failed in Mathematics, $45$ failed in Physics, $40$ failed in Biology and $32$ failed in exactly two of the three subjects. Only one student passed in all the subjects. Then, the number of students failing in all three subjects

• A. $12$
• B. $4$
• C. $2$
• D. Cannot be determined

Answer 1

The correct option is C

$2$

Compute the number of students:

Let $\mathrm{M}$ denote the number of students who failed in Mathematics, $\mathrm{P}$ denote the number of students who failed in Physics and $\mathrm{B}$ denote the number of students who failed in Biology.

$\mathrm{n}\left(\mathrm{M}\right)=50,\mathrm{n}\left(\mathrm{P}\right)=45,\mathrm{n}\left(\mathrm{B}\right)=40$

Exactly $32$ failed in two of the three subjects

$\mathrm{n}(\mathrm{M}\cap \mathrm{P})+\mathrm{n}(\mathrm{M}\cap \mathrm{B})+\mathrm{n}(\mathrm{P}\cap \mathrm{B})-3\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=32$……………..$\left(1\right)$

Total number of students $=100$

Since only one student passed in all the subjects.

$\mathrm{n}(\mathrm{M}\cup \mathrm{P}\cup \mathrm{B})=99$

We can write,

$\mathrm{n}\left(\mathrm{M}\right)+\mathrm{n}\left(\mathrm{P}\right)+\mathrm{n}\left(\mathrm{B}\right)-\left\{\mathrm{n}\right(\mathrm{M}\cap \mathrm{P})+\mathrm{n}(\mathrm{M}\cap \mathrm{B})+\mathrm{n}(\mathrm{P}\cap \mathrm{B}\left)\right\}+\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=99$

$50+45+40-32-3\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})+\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=99$ [ derived from equation $\left(1\right)$ ]

$135-32-2\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=99$

$-2\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=-4$

$\mathrm{n}(\mathrm{M}\cap \mathrm{P}\cap \mathrm{B})=2$

Therefore, there are only $2$ students failing in all three subjects.

Hence, option $\left(\mathrm{C}\right)$ is the correct option.