# 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: Option B **

__Solution(By Examveda Team)__

$$\eqalign{
& n\left( A \right) = \left( {\frac{{60}}{{100}} \times 96} \right) = \frac{{288}}{5} \cr
& n\left( B \right) = \left( {\frac{{30}}{{100}} \times 96} \right) = \frac{{144}}{5} \cr
& n\left( {A \cap B} \right) = \left( {\frac{{15}}{{100}} \times 96} \right) = \frac{{72}}{5} \cr
& \therefore n\left( {A \cup B} \right): \cr
& = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr
& = \frac{{288}}{5} + \frac{{144}}{5} - \frac{{72}}{5} \cr
& = \frac{{360}}{5} \cr
& = 72 \cr} $$So, people who had either or both types of lunch = 72

Hence, people who had neither type of lunch = (96 - 72) = 24

## Join The Discussion

## Comments ( 1 )

Related Questions on Percentage

A. $$\frac{1}{4}$$

B. $$\frac{1}{3}$$

C. $$\frac{1}{2}$$

D. $$\frac{2}{3}$$

100-(60+30-15)=25

25% of 96=24