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

  1. $10$
  2. $20$
  3. $30$
  4. $40$
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Answer is B

$n(A\cup B) = n(A)+n(B)-n(A\cap B)$

n(persons who do not prefer either chicken or fish) = n(persons who do not prefer chicken) + n(persons who do not prefer fish) – n(persons who do not prefer both chicken and fish)

n(persons who do not prefer either chicken or fish) = (100 – 60) + 50 – 10 = 80

ׯ The number of persons who prefer both fish and chicken = 100 – 80 = 20

 

Answer:

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