A fraction becomes 4 when 1 is added to both the numerator and denominator and it becomes 7 when 1 is subtracted from both the numerator and denominator. The numerator of the given fraction is :

Options

A. 2
B. 3
C. 7
D. 15

Answer

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

### Step 1: Set up the equations based on the conditions
1. According to the first condition, when 1 is added to both the numerator and denominator, the fraction becomes 4:
\[ \frac{x + 1}{y + 1} = 4 \]
Multiplying both sides by \( y + 1 \):
\[ x + 1 = 4(y + 1) \]
Simplifying this gives:
\[ x + 1 = 4y + 4 \]
Rearranging it, we get:
\[ x - 4y = 3 \quad \text{(Equation 1)} \]

2. According to the second condition, when 1 is subtracted from both the numerator and denominator, the fraction becomes 7:
\[ \frac{x - 1}{y - 1} = 7 \]
Multiplying both sides by \( y - 1 \):
\[ x - 1 = 7(y - 1) \]
Simplifying this gives:
\[ x - 1 = 7y - 7 \]
Rearranging it, we get:
\[ x - 7y = -6 \quad \text{(Equation 2)} \]

### Step 2: Solve the equations simultaneously
We have the two equations:
1. \( x - 4y = 3 \) (Equation 1)
2. \( x - 7y = -6 \) (Equation 2)

We can solve these equations by elimination. Subtract Equation 2 from Equation 1:
\[ (x - 4y) - (x - 7y) = 3 - (-6) \]
This simplifies to:
\[ -4y + 7y = 3 + 6 \]
\[ 3y = 9 \]
Dividing both sides by 3 gives:
\[ y = 3 \]

### Step 3: Substitute \( y \) back to find \( x \)
Now that we have \( y = 3 \), we can substitute this value back into either Equation 1 or Equation 2 to find \( x \). We'll use Equation 1:
\[ x - 4(3) = 3 \]
\[ x - 12 = 3 \]
Adding 12 to both sides gives:
\[ x = 15 \]

### Conclusion
The numerator of the given fraction is \( x = 15 \).