Question

In a class, 60% of the students pass in Hindi, and 45% pass in Sanskrit. If 25% of them pass in both subjects, what percentage of the students fails in both subjects?
(a) 80%
(b) 20%
(c) 25%
(d) 75%

Answer 1

Hint: We will take the students who passed in Hindi as $n\left( H \right)$ and students who passed in Sanskrit as $n\left( S \right)$ . Then, we will find $n\left( H \text{ only} \right)$ and $n\left( S \text{ only} \right)$ by subtracting $n\left( H \right)$ and $n\left( S \right)$ with $n\left( H\cap S \right)$ respectively. Then add $n\left( H \text{ only} \right)$, $n\left( S \text{ only} \right)$ and $n\left( H\cap S \right)$ and subtract from 100%.

Complete step-by-step solution:
In the question, we are given a situation of students who passed in Hindi and Sanskrit. It is said that 60% of the students pass Hindi and 45% pass in Sanskrit. Now if 25% passes in both the subjects then we have to find those students who failed in both the subjects.
Now, let’s represent number of students who passed in Sanskrit as $n\left( S \right)$ and number of students who passed in Hindi as $n\left( H \right)$ and number of students who passed in both $n\left( H\cap S \right)$.
So, according to the question we can say that,
$n\left( S \right)=45\%$
$n\left( H \right)=60\%$
$n\left( S\cap H \right)=25\%$
Now we will find percent of students who passed only in Hindi which is $60\%-25\%$ which is equal to 35% and percent of students who passed only in Sanskrit which is $45\%-25\%$ which is equal to 20%.
So, we say that $n\left( H \text{ only} \right)=35\%$ and $\left( S \text{ only} \right)=20\%$.
Now, we can find percent of student who passed by adding
$n\left( H \text{ only} \right)+n\left( S \right)+n\left( S\cap H \right)$
$=35\%+20\%+25\%$
$=80\%$
Here we can say that a total 80% passed in either of two subjects so the percent of students who failed in both the subjects are $100\%-80\%=20\%$.
Hence, 20% of students fail in both subjects. So, the correct option is ‘B’.

Note: Instead of finding terms related to only, one can directly find using formula $n\left( H\cup S \right)$ which is equal to $n\left( H \right)+\left( S \right)-n\left( H\cap S \right)$ and then subtract it from total.