Question

In a hotel, 60% had vegetarian lunch while 30% had non - vegetarian lunch and 15% had both types of lunch. If 96 people were present, how many did not eat either type of lunch?

• A. 20
• B. 24
• C. 26
• D. 28

Answer 1

The correct option is B 24


As can be seen in the above diagram, 25% of 96 people don't have either veg or non-veg which is 24.

Alternatively,
$n\left(A\right)-(\frac{60}{100}\times 96)=\frac{288}{5},n\left(B\right)=(\frac{30}{100}\times 96)=\frac{144}{5},n(A\cap B)=(\frac{15}{100}\times 96)=\frac{72}{5}\therefore n(A\cup B)=n\left(A\right)+n\left(B\right)-n(A\cap B)$
$\frac{288}{5}+\frac{144}{5}-\frac{72}{5}=\frac{360}{5}=72$
So, people who had either or both types of lunch = 72.
Hence, people who had neither type lunch
= (96 - 72 = 24)