Question Detail
In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?
- 27
- 26
- 25
- 24
Answer: Option D
Explanation:
\begin{aligned}
n(A) = \left(\frac{60}{100}*96\right) = \frac{288}{5} \\
n(B) = \left(\frac{30}{100}*96\right) = \frac{144}{5} \\
n(A\cap B) = \left(\frac{15}{100}*96\right) = \frac{72}{5} \\
\text{People who have either or both lunch} \\
n(A\cup B) = \frac{288}{5}+\frac{144}{5}-\frac{72}{5} \\
= \frac{360}{5} = 72
\end{aligned}
So People who do no have either lunch were = 96 -72
= 24