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. $26$
• B. $20$
• C. $24$
• D. $28$

Answer 1

Answer: (C)

Let $n\left(A\right)=\frac{60}{100}\times 96=\frac{288}{5}$

$n\left(B\right)=\frac{30}{100}\times 96=\frac{144}{5}$

$n(A\cap B)=\frac{15}{100}\times 96=\frac{72}{5}$

People who have either or both lunch is

$n(A\cup B)=\frac{288}{5}+\frac{144}{5}-\frac{72}{5}$

$=\frac{360}{5}$

$=72$

So, people who do not have either lunch were $=96-72=24$.