In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both types of lunch. If… · ABC26GN2490

Subject: General Aptitude · Chapter: Percentage

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

Answer (as printed): B

Explanation

$n(\mathrm{A})=\left(\frac{60}{100} \times 96\right)=\frac{288}{5}, n(\mathrm{B})$ $$=\left(\frac{30}{100} \times 96\right)=\frac{144}{5}, n(\mathrm{A} \cap \mathrm{B})=\left(\frac{15}{100} \times 96\right)=\frac{72}{5} .$$ $\therefore \quad n(\mathrm{A} \cup \mathrm{B})=n(\mathrm{A})+n(\mathrm{B})-n(\mathrm{A} \cap \mathrm{B})$

Study loop for Percentage

1. Practise the PYQsPrevious-year questions, with answers

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