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

Total number of people=96
In hotel 60% people are vegetarian then =60% of 96
$=\frac{60}{100}\times 96=57.6$
Non-vegetarian people=30% of 96
$\frac{30}{100}\times 96=28.8$
15% people are both then =15 % of 96
$\frac{15}{100}\times 96=14.4$
$\therefore$Total number of people had lunch=
$(57.6-14.4)+(28.8-14.4)+14.4$
$43.2+14.4+14.4$
$72$
$\therefore$Required number of people who did not eat either type of lunch=$96-72=24$