Question

In a class of $30$ pupils, $12$ take needlework, $16$ take physics and $18$ take history. If all the $30$ students take at least one subject and no one takes all three then the number of pupils taking $2$ subjects is

• A. $16$
• B. $6$
• C. $8$
• D. $20$

Answer 1

The correct option is A

$16$

Explanation for correct option:

Let $\mathrm{N}$ be the pupils who take needlework, $\mathrm{P}$ be the pupils who take physics and $\mathrm{H}$ be the pupils who take history.

Then,

$$\begin{gathered} \mathrm{n}(\mathrm{N}\cup \mathrm{P}\cup \mathrm{H})=\mathrm{n}\left(\mathrm{N}\right)+\mathrm{n}\left(\mathrm{P}\right)+\mathrm{n}\left(\mathrm{H}\right)-\mathrm{n}(\mathrm{N}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{N}\cap \mathrm{H})+\mathrm{n}(\mathrm{N}\cap \mathrm{P}\cap \mathrm{H})\mathrm{n}(\mathrm{N}\cap \mathrm{P})+\mathrm{n}(\mathrm{P}\cap \mathrm{H})+\mathrm{n}(\mathrm{N}\cap \mathrm{H}) \\ 30=12+16+18-\mathrm{n}(\mathrm{N}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{N}\cap \mathrm{H})+0 \\ 46-30=\mathrm{n}(\mathrm{N}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{N}\cap \mathrm{H}) \\ 16=\mathrm{n}(\mathrm{N}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{N}\cap \mathrm{H}) \end{gathered}$$

Hence, the correct option is A.