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

biru ? Dec 23 '2019 at 17:53

Correct answer is: 24

Explanation:

\(\begin{aligned} n(A) = \left(\frac{60}{100}*96\right) = \frac{288}{5} \\ n(B) = \left(\frac{30}{100}*96\right) = \frac{144}{5} \\ n(A\cap B) = \left(\frac{15}{100}*96\right) = \frac{72}{5} \\ \text{People who have either or both lunch} \\ n(A\cup B) = \frac{288}{5}+\frac{144}{5}-\frac{72}{5} \\ = \frac{360}{5} = 72 \end{aligned}\)

So People who do no have either lunch were = 96 -72
= 24

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Correct answer is: 24 Explanation: \(\begin{aligned} n(A) = \left(\frac{60}{100}*96\right) = \frac{288}{5} \\ n(B) = \left(\frac{30}{100}*96\right) = \frac{144}{5} \\ n(A\cap B) = \left(\frac{15}{100}*96\right) = \frac{72}{5} \\ \text{People who have either or both lunch} \\ n(A\cup B) = \frac{288}{5}+\frac{144}{5}-\frac{72}{5} \\ = \frac{360}{5} = 72 \end{aligned}\) 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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