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 candi dates who appeared at the ex amination was

Options

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

Answer

To solve the problem step by step, we will use the information provided in the question along with some basic principles of percentages.

### Step 1: Understand the given data
- 60% of candidates passed in English.
- 70% of candidates passed in Mathematics.
- 20% of candidates failed in both subjects.
- 2500 candidates passed in both subjects.

### Step 2: Calculate the percentage of candidates who passed in at least one subject
Since 20% failed in both subjects, it means that 80% of the candidates passed in at least one subject (either English or Mathematics or both).

### Step 3: Set up the Venn diagram relationships
Let:
- \( P(E) \) = Percentage of candidates who passed in English = 60%
- \( P(M) \) = Percentage of candidates who passed in Mathematics = 70%
- \( P(F) \) = Percentage of candidates who failed in both = 20%

Using the principle of inclusion-exclusion for the union of two sets, we can express the percentage of candidates who passed in at least one subject as:
\[ P(E \cup M) = P(E) + P(M) - P(E \cap M) \]
Where \( P(E \cap M) \) is the percentage of candidates who passed both subjects.

### Step 4: Substitute the known values
From the information given:
- \( P(E \cup M) = 80\% \) (since 20% failed in both)
- \( P(E) = 60\% \)
- \( P(M) = 70\% \)

Substituting these values into the equation:
\[ 80\% = 60\% + 70\% - P(E \cap M) \]
Now, rearranging the equation to find \( P(E \cap M) \):
\[ P(E \cap M) = 60\% + 70\% - 80\% \]
\[ P(E \cap M) = 50\% \]

### Step 5: Relate the percentage of candidates who passed both subjects to the total number of candidates
We know that 2500 candidates passed both subjects, which corresponds to 50% of the total candidates. Let \( N \) be the total number of candidates who appeared for the examination.

From the relationship:
\[ 50\% \text{ of } N = 2500 \]
This can be expressed mathematically as:
\[ 0.5N = 2500 \]

### Step 6: Solve for \( N \)
To find \( N \), we can rearrange the equation:
\[ N = \frac{2500}{0.5} \]
\[ N = 2500 \times 2 \]
\[ N = 5000 \]

### Conclusion
The total number of candidates who appeared at the examination is **5000**.

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