Question

In a class of 35 students, 26 students like to eat apples and 16 students like to eat oranges. Also, each student likes to eat at least one of the two fruits. How many students like to eat both the fruits?

Answer 1

Hint: In this problem we consider the number of student who like apples as set A and number of students who like oranges as set B then total number of students will be their union and by using set theory formula we calculate intersection of set A and B to get the required number of students who like both fruits.
F $ n\left( {A \cup B} \right) = n(A) + n(B) - n(A \cap B) $

Complete step-by-step answer:
Total number of students in class = $ 35 $
Number of students who like apples = $ 26 $
Number of students who like oranges = 16
We will solve given problems by the help of set theory.
Here, in set theory we take as many students who like apples as set ‘A’.
And the number of students who like oranges as set ‘B’.
Therefore from above discussion we can say that
 $ n\left( A \right) = 26 $ and $ n\left( B \right) = 16 $
Since, it is given that each student like to eat at least one of the two fruits.
Therefore $ n\left( U \right) = n\left( {A \cup B} \right) $
Hence, $ n\left( {A \cup B} \right) = 35 $
Now, let number of students who like to eat both apple and oranges be = x
 $ \Rightarrow n(A \cap B) = x $
By set theory formula we have
 $ n\left( {A \cup B} \right) = n(A) + n(B) - n(A \cap B) $
Substituting given values in above mentioned formula we have
 $
  35 = 26 + 16 - x \\
   \Rightarrow 35 = 42 - x \\
   \Rightarrow x = 42 - 35 \\
   \Rightarrow x = 7 \\
  $
Therefore, $ n\left( {A \cap B} \right) = 7 $
Hence, from above we see that number of students who like to eat both apples and oranges $ 7 $ .

Note: Here, as there are no students who did not like any fruit. Therefore $ n(U) = n(A \cup B) $ and then using the union formula of set theory we can find the intersection as the value of students who like both fruits as individuals are given.