Question

In a class of $30$ pupils, $12$ take Chemistry, $16$ 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 the correct option:

Finding the number of pupils taking $2$ subject:

Total number of people in class$=30$

Number of people who take Chemistry, $\mathrm{n}\left(\mathrm{C}\right)$ $=12$

Number of people who take Physic, $\mathrm{n}\left(\mathrm{P}\right)=16$

Number of people who take History,$\mathrm{n}\left(\mathrm{H}\right)=18$

Also given,

$$\begin{gathered} \mathrm{n}(\mathrm{C}\cup \mathrm{P}\cup \mathrm{H})=30, \\ \mathrm{n}(\mathrm{C}\cap \mathrm{P}\cap \mathrm{H})=0 \end{gathered}$$

Now, as we know,

$(\mathrm{C}\cup \mathrm{P}\cup \mathrm{H})=\mathrm{n}\left(\mathrm{C}\right)+\mathrm{n}\left(\mathrm{P}\right)+\mathrm{n}\left(\mathrm{H}\right)-\mathrm{n}(\mathrm{C}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{H}\cap \mathrm{C})+\mathrm{n}(\mathrm{C}\cap \mathrm{P}\cap \mathrm{H})+\mathrm{n}(\mathrm{C}\cap \mathrm{P})+\mathrm{n}(\mathrm{P}\cap \mathrm{H})+\mathrm{n}(\mathrm{H}\cap \mathrm{C})$

$(\mathrm{C}\cup \mathrm{P}\cup \mathrm{H})=12+16+18-\mathrm{n}(\mathrm{C}\cap \mathrm{P})-\mathrm{n}(\mathrm{P}\cap \mathrm{H})-\mathrm{n}(\mathrm{H}\cap \mathrm{C})+0$

$$\begin{gathered} \mathrm{n}(\mathrm{C}\cap \mathrm{P})+\mathrm{n}(\mathrm{P}\cap \mathrm{H})+\mathrm{n}(\mathrm{H}\cap \mathrm{C})=(12+16+18)-30 \\ =46-30 \\ =16 \end{gathered}$$

Hence, the correct option is A.