The average mathematics marks of two Sections A and B of Class IX in the annual examination is 74. The average marks of Section A is 77.5 and that of Section B is 70. The ratio of the number of students of Section A and B is

Options

A. `7:8`
B. `7:5`
C. `8:7`
D. `8:5`

Answer

To solve the problem, we need to find the ratio of the number of students in Section A and Section B based on the given average marks.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The average marks of both sections combined (A and B) = 74
- The average marks of Section A = 77.5
- The average marks of Section B = 70

2. **Let the Number of Students in Section A be \( x \)** and in Section B be \( y \)**:
- We need to find the ratio \( \frac{x}{y} \).

3. **Set Up the Equation for the Combined Average**:
- The total marks for Section A = \( 77.5x \)
- The total marks for Section B = \( 70y \)
- The combined average can be expressed as:
\[ \frac{77.5x + 70y}{x + y} = 74 \]

4. **Cross-Multiply to Eliminate the Fraction**:
- Multiply both sides by \( (x + y) \):
\[ 77.5x + 70y = 74(x + y) \]

5. **Expand the Right Side**:
- This gives us:
\[ 77.5x + 70y = 74x + 74y \]

6. **Rearrange the Equation**:
- Move all terms involving \( x \) to one side and \( y \) to the other:
\[ 77.5x - 74x = 74y - 70y \]
- Simplifying this gives:
\[ 3.5x = 4y \]

7. **Find the Ratio \( \frac{x}{y} \)**:
- Rearranging gives:
\[ \frac{x}{y} = \frac{4}{3.5} \]
- Simplifying \( \frac{4}{3.5} \) gives:
\[ \frac{x}{y} = \frac{4 \times 2}{3.5 \times 2} = \frac{8}{7} \]

8. **Conclusion**:
- The ratio of the number of students in Section A to Section B is \( \frac{8}{7} \).