Question

A survey shows that 76% of the Indians like oranges, whereas 62% like bananas. What percentage of the Indians like both oranges and bananas?

Answer 1

Hint: Here, we will first consider the event of liking oranges as A and the event of liking bananas as B. Thus, we will use the theorem that $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$ to find the value of $n\left( A\cap B \right)$ which will give the number of people who like both oranges and bananas and using that we can find the percentage.

Complete step by step solution:

Here, region 1 represents the Indians liking oranges, region 2 represents Indians liking bananas and region 3 represents Indians liking both oranges and bananas.
A set is a well defined collection of distinct objects, considered as an object in its own right. The union of a collection of a set is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other. The union of two sets A and B is the set of elements which are in A, in B or in both A and B. It is denoted as $A\cup B$. The intersection of two sets A and B denoted by $A\cap B$ is the set containing all elements of A that also belong to B or equivalently, all elements of B that also belong to A. So, x is said to be an element of this intersection $A\cap B$if and only if x is an element of both A and B.
 Let A represent the set of people who like oranges and B is the set of people who like bananas.
Let the total number of Indians be x = $n\left( A\cup B \right).........\left( 1 \right)$
Then number of Indians liking oranges = $\dfrac{76}{100}\times x=n\left( A \right)............\left( 2 \right)$
And, number of Indians liking bananas = $\dfrac{62}{100}\times x=n\left( B \right).............\left( 3 \right)$
Now, we have to find $n\left( A\cap B \right)$, that is the number of Indians liking both oranges and bananas.
Since, $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)..........\left( 4 \right)$
On putting the values from equation (1), (2) and (3) in equation (4), we get:
$\begin{align}
  & x=\dfrac{76}{100}\times x+\dfrac{62}{100}\times x-n\left( A\cap B \right) \\
 & \Rightarrow x=\left( \dfrac{138}{100}\times x \right)-n\left( A\cap B \right) \\
 & \Rightarrow n\left( A\cap B \right)=\dfrac{138}{100}\times x-x \\
 & \Rightarrow n\left( A\cap B \right)=\dfrac{138x-100x}{100} \\
 & \Rightarrow n\left( A\cap B \right)=\dfrac{38}{100}\times x \\
\end{align}$
Hence, 38% of Indians like both oranges and bananas.

Note: Students should note here that since the value of $n\left( A\cap B \right)$ comes out to be $\dfrac{38}{100}\times x$, then by the definition of percentage, we can say that it is equal to 38%. Students should remember the formula that $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$. The calculations must be done properly to avoid mistakes.