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Question

A dinner party is to be fixed for a group of 100 persons. In this party, 50 persons do not prefer fish, 60 prefer chicken, and 10do not prefer either chicken or fish. The number of persons who prefer both fish and chicken is.


A

20

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B

22

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C

25

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D

noneofthese

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Solution

The correct option is A

20


Explanation for the correct option:

Step1. Define events.

The number of people who prefer fish =n(F)

The number of people who prefer chicken =n(C)=60

People who do not prefer fish =n(F')

The number of people who prefer both fish and chicken =n(FC)

Total number of people =n(U)

The number of people who prefer either fish or chicken =n(FC)

The number of people who do not prefer either fish or chicken =n(FUC)'=10

Number of people who prefer neither fish nor chicken therefore=10

Therefore, The number of people who prefer either fish or chicken =n(FC)=100-10=90

Step2. Calculate the number of people who prefer both fish and chicken:

Given,

n(U)=100n(C)=60n(FC)'=10n(F')=50n(F)=100-50;[n(F)=n(U)-n(F')]=50n(FC)=100-10;n(FC)=n(U)-n(FC)'=90n(FC)=n(F)+n(C)-n(FC)

Substitute the value now,

90=50+60-n(FC)n(FC)=110-90n(FC)=20

Hence, Option(A) is the correct answer.


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