A fraction becomes 7/8, If 5 is added to both the numerator and the denominator , If 3 is added to both the numerator and the denominator It becomes `6/7.` find the fraction.

Options

A. `8/11`
B. `9/11`
C. `10/11`
D. Connot of the determined

Answer

To solve the problem, we need to find the fraction based on the conditions given. Let's denote the numerator of the fraction as \( x \) and the denominator as \( y \).

### Step-by-Step Solution:

1. **Set up the equations based on the conditions:**
- According to the first condition, when 5 is added to both the numerator and the denominator, the fraction becomes \( \frac{7}{8} \):
\[ \frac{x + 5}{y + 5} = \frac{7}{8} \]
- According to the second condition, when 3 is added to both the numerator and the denominator, the fraction becomes \( \frac{6}{7} \):
\[ \frac{x + 3}{y + 3} = \frac{6}{7} \]

2. **Cross-multiply to eliminate the fractions:**
- For the first equation:
\[ 8(x + 5) = 7(y + 5) \]
Expanding this gives:
\[ 8x + 40 = 7y + 35 \]
Rearranging it leads to:
\[ 8x - 7y = -5 \quad \text{(Equation 1)} \]

- For the second equation:
\[ 7(x + 3) = 6(y + 3) \]
Expanding this gives:
\[ 7x + 21 = 6y + 18 \]
Rearranging it leads to:
\[ 7x - 6y = -3 \quad \text{(Equation 2)} \]

3. **Solve the system of equations:**
- We have the two equations:
\[ 8x - 7y = -5 \quad \text{(1)} \]
\[ 7x - 6y = -3 \quad \text{(2)} \]
- To eliminate \( y \), we can multiply Equation (1) by 6 and Equation (2) by 7:
\[ 6(8x - 7y) = 6(-5) \implies 48x - 42y = -30 \quad \text{(3)} \]
\[ 7(7x - 6y) = 7(-3) \implies 49x - 42y = -21 \quad \text{(4)} \]

4. **Subtract Equation (4) from Equation (3):**
\[ (48x - 42y) - (49x - 42y) = -30 + 21 \]
This simplifies to:
\[ -x = -9 \implies x = 9 \]

5. **Substitute \( x \) back to find \( y \):**
- Substitute \( x = 9 \) into Equation (1):
\[ 8(9) - 7y = -5 \]
\[ 72 - 7y = -5 \]
\[ -7y = -5 - 72 \implies -7y = -77 \implies y = 11 \]

6. **Find the fraction:**
- The fraction is:
\[ \frac{x}{y} = \frac{9}{11} \]

### Final Answer:
The fraction is \( \frac{9}{11} \).

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