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 ?

Answer: Option B

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