In a fraction, when numerator increases by 1 and denominator increases by 2, then the fraction becomes `2/3` But when numerator increases by 5 and denominater increases by 1 then fraction becomes `5/4`. The original fraction is?

Options

A. a)`3/7`
B. b)`5/8`
C. c)`5/7`
D. d)`6/7`

Answer

To solve the problem, we need to set up two equations based on the information given in the question.

Let the original fraction be represented as \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.

### Step 1: Set up the first equation
According to the first condition:
- When the numerator increases by 1: \( x + 1 \)
- When the denominator increases by 2: \( y + 2 \)
- The fraction becomes \( \frac{2}{3} \)

This gives us the equation:
\[ \frac{x + 1}{y + 2} = \frac{2}{3} \]

Cross-multiplying gives:
\[ 3(x + 1) = 2(y + 2) \]
Expanding this, we have:
\[ 3x + 3 = 2y + 4 \]
Rearranging it results in:
\[ 3x - 2y = 1 \quad \text{(Equation 1)} \]

### Step 2: Set up the second equation
According to the second condition:
- When the numerator increases by 5: \( x + 5 \)
- When the denominator increases by 1: \( y + 1 \)
- The fraction becomes \( \frac{5}{4} \)

This gives us the equation:
\[ \frac{x + 5}{y + 1} = \frac{5}{4} \]

Cross-multiplying gives:
\[ 4(x + 5) = 5(y + 1) \]
Expanding this, we have:
\[ 4x + 20 = 5y + 5 \]
Rearranging it results in:
\[ 4x - 5y = -15 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have the two equations:
1. \( 3x - 2y = 1 \)
2. \( 4x - 5y = -15 \)

We can solve these equations using the elimination method. First, we can multiply Equation 1 by 5 and Equation 2 by 2 to align the coefficients of \( y \):

From Equation 1:
\[ 5(3x - 2y) = 5(1) \implies 15x - 10y = 5 \quad \text{(Equation 3)} \]

From Equation 2:
\[ 2(4x - 5y) = 2(-15) \implies 8x - 10y = -30 \quad \text{(Equation 4)} \]

### Step 4: Subtract the equations
Now, we subtract Equation 4 from Equation 3:
\[ (15x - 10y) - (8x - 10y) = 5 - (-30) \]
This simplifies to:
\[ 7x = 35 \]
Thus, we find:
\[ x = 5 \]

### Step 5: Substitute back to find \( y \)
Now, substitute \( x = 5 \) back into Equation 1 to find \( y \):
\[ 3(5) - 2y = 1 \]
This simplifies to:
\[ 15 - 2y = 1 \]
Rearranging gives:
\[ -2y = 1 - 15 \implies -2y = -14 \implies y = 7 \]

### Final Answer
The original fraction is:
\[ \frac{x}{y} = \frac{5}{7} \]