In a committee $50$ people speak French, $20$ speak Spanish and $10$ speak both Spanish and French. The number of persons speaking at least one of these two languages is

60

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50

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58

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30

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Solution

The correct option is B $60$
Let $F$ and $S$ denote people speaking French and Spanish respectively
Given, $n (F) = 50, n (S) = 20, n (S \cap F) = 10$
We know that,
$n (S \cup F) = n (S) + n (F) - n (S \cap F)$
$\Rightarrow n (S \cup F) = 20 + 50 - 10 = 60$

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In a committee 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of persons speaking at least one of these two languages is

The correct option is B 60Let F and S denote people speaking French and Spanish respectivelyGiven, n(F)=50,n(S)=20,n(S∩F)=10We know that,n(S∪F)=n(S)+n(F)−n(S∩F)⇒n(S∪F)=20+50−10=60

In a committee $50$ people speak French, $20$ speak Spanish and $10$ speak both Spanish and French. The number of persons speaking at least one of these two languages is

The correct option is B $60$
Let $F$ and $S$ denote people speaking French and Spanish respectively
Given, $n (F) = 50, n (S) = 20, n (S \cap F) = 10$
We know that,
$n (S \cup F) = n (S) + n (F) - n (S \cap F)$
$\Rightarrow n (S \cup F) = 20 + 50 - 10 = 60$