Question

In a $\mathrm{B}$ School, there are $15$ teachers who teach marketing or finance. Of these, $8$ teach finance and $4$ teach both marketing and finance. How many teach marketing but not finance?

• A. $15$
• B. $20$
• C. $11$
• D. None of these.

Answer 1

The correct option is C

$11$

Explanation of the correct option:

Finding the number of teachers:

Let us consider $\mathrm{M}$ as group of teachers who are teaching marketing, and $\mathrm{F}$ as group of teachers who are teaching finance.

No. of teachers who are teaching marketing or finance $=\mathrm{n}\left(\mathrm{M}\cup \mathrm{F}\right)$

$=15$.

No. of teachers who are teaching marketing and finance both $=\mathrm{n}\left(\mathrm{M}\cap \mathrm{F}\right)$

$=4$.

No. of teachers who are teaching finance $=\mathrm{n}\left(\mathrm{F}\right)$

$=8.$

Now,

$\mathrm{n}(\mathrm{A}\cup \mathrm{B})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})$

Applying the above formula, we get,

$\mathrm{n}(\mathrm{M}\cup \mathrm{F})=\mathrm{n}\left(\mathrm{M}\right)+\mathrm{n}\left(\mathrm{F}\right)-\mathrm{n}(\mathrm{M}\cap \mathrm{F})$

$\Rightarrow 15=\mathrm{n}\left(\mathrm{M}\right)+8-4$

$\Rightarrow \mathrm{n}\left(\mathrm{M}\right)=11$.

Hence the correct option is C.