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Question: In an examination, \(35\% \) of the total students failed in Hindi and \(45\% \) failed in English a...

In an examination, 35%35\% of the total students failed in Hindi and 45%45\% failed in English and 20%20\% failed in both. The percentage of those who passed in both the subjects is:
A) 2020
B) 3030
C) 4040
D) 8080

Explanation

Solution

Let the total number of students be 100100
Let A be the set of students who failed in Hindi
So 35% of 100=3535\% {\text{ of 100}} = 35
So n(A)=35n(A) = 35
Similarly let B be the set of students who failed in English
n(B)=45% of 100=45n(B) = 45\% {\text{ of 100}} = 45
And also we know that 20%20\% failed in both
So n(A∩B)=20% of 100=20n(A \cap B) = 20\% {\text{ of 100}} = 20
We know that n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)

Complete step by step solution:
Here in this question, we are given that in an examination, 35%35\% of the total students failed in Hindi and 45%45\% failed in English and 20%20\% failed in both. So firstly we need to assume the total number of students in the examination. Let us assume that the total number of students be 100100
Total students who failed in Hindi=35% of 100=35100×100=35 = 35\% {\text{ of 100}} = \dfrac{{35}}{{100}} \times 100 = 35
Now we also know that 45%45\% failed in English=45% of 100=45100×100=45 = 45\% {\text{ of 100}} = \dfrac{{45}}{{100}} \times 100 = 45
And we know that 20%20\% of the students failed in both the subjects so number of students who failed in both the subjects=20% of 100=20 = 20\% {\text{ of 100}} = 20
Let A be the set of students who failed in Hindi
So 35% of 100=35100×100=3535\% {\text{ of 100}} = \dfrac{{35}}{{100}} \times 100 = 35
So n(A)=35n(A) = 35
Similarly let B be the set of students who failed in English
n(B)=45% of 100=45100×100=45n(B) = 45\% {\text{ of 100}} = \dfrac{{45}}{{100}} \times 100 = 45
Also let that A∩BA \cap B be the total students who failed in both the subjects.
And also we know that 20%20\% failed in both
So n(A∩B)=20% of 100=20n(A \cap B) = 20\% {\text{ of 100}} = 20
Now we need to find the percentage of the students who passed in both the subjects which is given by =total students − n(A∪B){\text{total students }} - {\text{ n(A}} \cup {\text{B)}}
Now as we know that
n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)
Putting the values of n(A),n(B),n(A∩B)n(A),n(B),n(A \cap B) we get
n(A∪B)=35+45−20=60n(A \cup B) = 35 + 45 - 20 = 60
Here the n(A∪B)n(A \cup B) represents the total students who failed in Hindi and English.
And n(A∩B)n(A \cap B) represents the students who failed in both the subjects.
So the total students passing in Hindi and English=100−n(A∪B) = 100 - n(A \cup B)
=100−60=40= 100 - 60 = 40
Now for percentage, number of studentstotal students×100\dfrac{{{\text{number of students}}}}{{{\text{total students}}}} \times 100
=40100×100=40%= \dfrac{{40}}{{100}} \times 100 = 40\%

Note: We can do it this way as it is given that 35%35\% of the total students failed in Hindi and 45%45\% failed in English and 20%20\% failed in both. It means that (35−20)%=15%(35 - 20)\% = 15\% failed exclusively in Hindi and (45−20)%=25%(45 - 20)\% = 25\% exclusively in English. So total %\% failed in Hindi, English and Both are
15%+25%+20%=60%15\% + 25\% + 20\% = 60\%
So the students passed in both subjects=(100−60)%=40%= (100 - 60)\% = 40\%