In an examination, 60% of the candidates passed in English and 70% of the candidates passed in Mathematics, but 20% failed in both of these subjects. If 2500 candidates passed in both the subjects, the number of candidates that appeared at the examination was

Options

A. 3000
B. 3500
C. 4000
D. 5000

Answer

To solve the problem step by step, we will use the information provided about the candidates who passed and failed in the examination.

### Step 1: Define the total number of candidates
Let the total number of candidates who appeared for the examination be \( N \).

### Step 2: Calculate the percentage of candidates who passed and failed
- 60% of the candidates passed in English, which means 40% failed in English.
- 70% of the candidates passed in Mathematics, which means 30% failed in Mathematics.
- 20% of the candidates failed in both subjects.

### Step 3: Set up the equation for candidates who passed in both subjects
Let’s denote:
- \( P(E) \) as the percentage of candidates who passed in English = 60%
- \( P(M) \) as the percentage of candidates who passed in Mathematics = 70%
- \( P(FB) \) as the percentage of candidates who failed in both subjects = 20%

Using the principle of inclusion-exclusion for the candidates who passed in at least one subject, we have:

\[ P(E \cup M) = P(E) + P(M) - P(E \cap M) \]

Where \( P(E \cap M) \) is the percentage of candidates who passed in both subjects.

### Step 4: Calculate the percentage of candidates who passed in at least one subject
From the information provided:
- The percentage of candidates who passed in at least one subject is:
\[ P(E \cup M) = 100\% - P(FB) = 100\% - 20\% = 80\% \]

### Step 5: Substitute the known values into the equation
Now substituting the known values into the inclusion-exclusion formula:
\[ 80\% = 60\% + 70\% - P(E \cap M) \]

### Step 6: Solve for \( P(E \cap M) \)
Rearranging the equation gives:
\[ P(E \cap M) = 60\% + 70\% - 80\% = 50\% \]

### Step 7: Relate the percentage of candidates who passed in both subjects to the total number of candidates
We know that 2500 candidates passed in both subjects, which corresponds to 50% of the total candidates \( N \):
\[ 0.5N = 2500 \]

### Step 8: Solve for \( N \)
To find \( N \), we can rearrange the equation:
\[ N = \frac{2500}{0.5} = 5000 \]

### Conclusion
Thus, the total number of candidates that appeared for the examination is \( \boxed{5000} \).

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