Question

In a committee, 50 people speak French, 20 speak Spanish, and 10 speak both French and Spanish. How many speak at least one of these two languages?

Answer 1

Hint: Use the formula of set theory for the number of items $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$ for any two items A and B. Also, note that the number of items for each type is given in the question which is only required for the solution.

Complete step by step answer:
Let say F be the set of people who speak French in the committee and let say S be the set of people who speak Spanish in the committee.
Number of people who speak French, $n\left( F \right) = 50$
Number of people who speak Spanish, $n\left( S \right) = 20$
Number of people who speak both Hindi and English, $n\left( {F \cap S} \right) = 10$
People who speak at least one language, $n\left( {F \cup S} \right)$
The formula of set theory for the number of items is,
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Replace A with F and B with S,
$ \Rightarrow n\left( {F \cup S} \right) = n\left( F \right) + n\left( S \right) - n\left( {F \cap S} \right)$
Substitute the values,
$ \Rightarrow n\left( {F \cup S} \right) = 50 + 20 - 10$
Simplify the terms,
$ \Rightarrow n\left( {F \cup S} \right) = 60$

Hence, 60 people in the committee in at least one of the two languages.

Note: The concept used here is set theory. Union of two sets gives a set of all elements that are at least in one of the two sets. If A and B are 2 sets, then its union is written as $A \cup B$ and it contains all the elements in A along with all the elements in B. Intersection of two sets gives the set of all elements that are in both the sets. If A and B are 2 sets, then its intersection is written as $A \cap B$ and it contains all the elements that are both in A and B. For set A, $n\left( A \right)$ represents the number of elements in the set, and for set B, $n\left( B \right)$ represents the number of elements in the set.