Question

In a class, 52% of the students failed in English, 40% failed in Mathematics, and 20% failed in both the subjects. What percent of the students passed in both the subjects?

• A. 18%
• B. 28%
• C. 36%
• D. None of these

Hint:

Passed both = passed English + passed Maths minus passed at least one, and passed at least one equals 100% minus the percent who failed both.

Answer 1

The Correct Option is B

Step 1: Understand what the question asks.
We know the percent of students who failed in each subject and the percent who failed in both. We need the percent who passed in both English and Mathematics.

Step 2: Find the percent who passed each subject on its own.
If 52% failed in English, the rest passed English, so percent who passed English \( = 100 - 52 = 48\% \).
If 40% failed in Mathematics, percent who passed Mathematics \( = 100 - 40 = 60\% \).

Step 3: Find the percent who passed at least one subject.
A student fails to pass at least one subject only when they fail both subjects together. So the 20% who failed both is exactly the group that passed neither subject.
So the percent who passed English or Mathematics or both \( = 100 - 20 = 80\% \).

Step 4: Apply the union rule for the two passed groups.
For any two groups, \( P(E \cup M) = P(E) + P(M) - P(E \cap M) \), where E is passed English and M is passed Mathematics.
Putting in the numbers: \( 80 = 48 + 60 - P(E \cap M) \).
So \( P(E \cap M) = 48 + 60 - 80 = 28\% \).

Final Answer:
28% of the students passed in both subjects. \[ \boxed{28\%} \]