In a hotel, 60% has 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?

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RRB ASSISTANT LOCO PILOT (Allahabad) 2010 click here to view all questions

In a hotel, 60% has 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?

A20

B24

C26

D28

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Correct answer is: 24 Explanation: $\begin{array}{r}n(A)=(\frac{60}{100}\ast 96)=\frac{288}{5}\\ n(B)=(\frac{30}{100}\ast 96)=\frac{144}{5}\\ n(A\cap B)=(\frac{15}{100}\ast 96)=\frac{72}{5}\\ \text{People who have either or both lunch}\\ n(A\cup B)=\frac{288}{5}+\frac{144}{5}-\frac{72}{5}\\ =\frac{360}{5}=72\end{array}$ So People who do no have either lunch were = 96 -72 = 24 biru ? answered Dec 23 '2019 at 17:53 All Comments

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