In a hotel,60 % of people had vegetarian lunc | vedclass

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(B) Let $V$ be the set of people who had vegetarian lunch and $N$ be the set of people who had non-vegetarian lunch.Given:$P(V) = 60 \%$$P(N) = 30 \%$

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Let $V$ be the set of people who had vegetarian lunch and $N$ be the set of people who had non-vegetarian lunch.Given:$P(V) = 60 \%$$P(N) = 30 \%$$P(V \cap N) = 15 \%$Using the principle of inclusion-exclusion,the percentage of people who had at least one type of lunch is:$P(V \cup N) = P(V) + P(N) - P(V \cap N)$$P(V \cup N) = 60 \% + 30 \% - 15 \% = 75 \%$Therefore,the percentage of people who did not have either type of lunch is:$100 \% - 75 \% = 25 \%$Given the total number of people is $96$,the number of people who did not have either type of lunch is:$25 \% 	ext{ of } 96 = \frac{25}{100} 	imes 96 = \frac{1}{4} 	imes 96 = 24$

In a hotel,$60 \%$ of people had vegetarian lunch,$30 \%$ had non-vegetarian lunch,and $15 \%$ had both types of lunch. If $96$ people were present,how many did not have either type of lunch?

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