`45%` of the students of a class participated in Physics Olympaid and `65%` of the students of the class participated in Maths Olympiad. 4 students participated in neither of these two and 8 students participated in both. Find how many students
are there in the class?

Answer

To find the total number of students in the class, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
Let \( X \) be the total number of students in the class.

2. **Calculate Students Participating in Physics and Maths**:
- Students participating in Physics Olympiad = \( 45\% \) of \( X \) = \( \frac{45X}{100} \)
- Students participating in Maths Olympiad = \( 65\% \) of \( X \) = \( \frac{65X}{100} \)

3. **Account for Students Participating in Both Olympiads**:
- Students participating in both = 8
- Therefore, the total number of students participating in either Physics or Maths (or both) can be calculated using the principle of inclusion-exclusion:
\[ \text{Total participating} = \left(\frac{45X}{100} + \frac{65X}{100} - 8\right) \]

4. **Combine the Participation**:
- Simplifying the expression:
\[ \frac{45X}{100} + \frac{65X}{100} = \frac{110X}{100} \]
- Thus,
\[ \text{Total participating} = \frac{110X}{100} - 8 \]

5. **Account for Students Participating in Neither**:
- We know that 4 students participated in neither of the Olympiads. Therefore, we can set up the equation:
\[ X - \left(\frac{110X}{100} - 8\right) = 4 \]

6. **Solve the Equation**:
- Rearranging the equation gives:
\[ X - \frac{110X}{100} + 8 = 4 \]
- This simplifies to:
\[ X - 1.1X + 8 = 4 \]
- Combining like terms:
\[ -0.1X + 8 = 4 \]
- Subtracting 8 from both sides:
\[ -0.1X = 4 - 8 \]
\[ -0.1X = -4 \]
- Dividing by -0.1:
\[ X = \frac{-4}{-0.1} = 40 \]

### Final Answer:
The total number of students in the class is \( \boxed{40} \).

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