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The problem presented involves percentages of candidates passing and failing in different subjects (English and Maths) and asks for the total number of candidates who appeared in the examination.

**Given dataThe problem presented involves percentages of candidates passing and failing in different subjects (English and Maths) and asks for the total number of candidates who appeared in the examination.

**Given **
1. 60% of the candidates pass in English.
2. 70% of the candidates pass in Maths.
3. 20% of the candidates fail in both subjects.
4. 2500 candidates pass in both subjects.

**To find:** The total number of candidates who appeared in the examination.

Let's denote:
- Total number of candidates = \( x \)
- \( E \cap M \) represents the candidates who passed in both English and Maths.

### Step-by-step solution:

1. **Relation for failure and pass rates:**
   - 20% failed in both subjects, so the remaining 80% passed in at least one subject.
   - This means 80% of \( x \) passed in at least one subject.

   Therefore:
   \[
   0.80x = \text{Candidates who passed in at least one subject}
   \]

2. **Formula using the number of students who passed in both subjects:**
   From the data, 2500 candidates passed in both English and Maths:
   \[
   \text{Candidates passing in both subjects} = 2500
   \]
   We know that \( E \cap M = 2500 \), i.e., \( |E \cap M| = 2500 \).

3. **Using Venn diagram logic:**
   According to the principle of inclusion and exclusion for sets:
   \[
   |E \cup M| = |E| + |M| - |E \cap M|
   \]
   where:
   - \( |E| = 0.60x \) (since 60% pass in English),
   - \( |M| = 0.70x \) (since 70% pass in Maths).

   Substituting the values:
   \[
   0.80x = 0.60x + 0.70x - 2500
   \]
   Simplifying:
   \[
   0.80x = 1.30x - 2500
   \]
   Rearranging to isolate \( x \):
   \[
   0.80x - 1.30x = -2500
   \]
   \[
   -0.50x = -2500
   \]
   \[
   x = \frac{2500}{0.50} = 5000
   \]

Thus, the total number of candidates who appeared in the examination is **5000**.

Would you like more details on the solution or any specific part of the process explained further?

### Related questions:
1. How would the problem change if 65% of the candidates passed in English?
2. What would be the total number of candidates if 25% failed in both subjects?
3. If 3000 candidates passed only one subject, how does this affect the total count?
4. What is the relationship between passing percentages in each subject and the total number of candidates?
5. How can this problem be solved using set theory principles?

**Tip:** When dealing with percentages in exams or group problems, using set theory concepts like inclusion-exclusion can simplify complex relationships.

In an examination, 60% of the candidates pass in English and 70% of the candidates pass in Maths, but 20% failed in both the subjects. If 2500 candidates pass in both the subjects, find the number of candidates who appeared in the examination.

Solve a problem involving pass rates in English and Maths using set theory and percentage calculations. Given 60% pass in English, 70% in Maths, and 2500 pass in both subjects, find the total number of candidates. Suitable for Grades 8-10.

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