Question

Out of $64$ students the number of students taking mathematics is about $45$ and the number of students taking both mathematics and biology is $10$ . Then the number of students taking only biology is
A) 18
B) 19
C) 20
D) None of these

Answer 1

Hint: To find how many students are taking only biology, for that you have to write a given problem into small-small equations in terms of sets. Then find how many students from $64$ are taking biology. After finding how many students are in biology , subtract the number of students taking both mathematics and biology from total biology students and you will find the answer.

Complete step by step answer:
First of all, let’s consider $M$ refer to the number of students who take mathematics and $B$ refer to the number of students who take biology.
So we can say that,
$ \Rightarrow n(M) = 45$
And we have to find $n(B)$ .
Also we have given students take both mathematics and biology is $10$ we can write this sentence in terms of sets,
$ \Rightarrow n(M \cap B) = 10$
Also we have given total number of students is $64$ we can write this sentence in terms of sets,
$ \Rightarrow n(M \cup B) = 64$
Now, apply formula to find $n(B)$
$ \Rightarrow n(M \cup B) = n(M) + n(B) - n(M \cap B)$
Now, put values in above equation to find $n(B)$
$ \Rightarrow 64 = 45 + n(B) - 10$
$ \Rightarrow n(B) = 29$
So, we find a total number of students are in biology. Now, subtract it from number of students take both mathematics and biology and we will get our answer,
$ \Rightarrow n(B) = n(B) - n(M \cap B)$
$ \Rightarrow n(B) = 29 - 10$
$ \Rightarrow n(B) = 19$
Therefore, we can see that the number of students taking only biology is $19$. So, the correct option is option (B).

Note:
In this problem we use the union of sets and intersection of sets. More on union and intersection sets:
Union of sets: If set $A$ and set $B$ are two sets, then $A$ union $B$ is the set that contains all the elements of set $A$ and set $B$. It is referred to as $A \cup B$.
Intersection of sets: If set $A$ and set $B$ are two sets, then $A$ intersection $B$ is the set that contains only the common elements between set $A$ and set $B$. It is denoted as $A \cap B$.