When the numerator of a certain fraction is increased by 2 and denominator by 1 , its value changes to 1/2 , but when the numerator is increased by 3 and denominator by 5 ,its value then equals to 2/5 ,what is the original fraction ?

Options

A. 1/6
B. 1/4
C. 1/3
D. 1/5

Answer

To solve the problem, we will denote the original fraction as \( \frac{X}{Y} \), where \( X \) is the numerator and \( Y \) is the denominator.

### Step 1: Set up the first equation
According to the problem, when the numerator is increased by 2 and the denominator by 1, the new fraction equals \( \frac{1}{2} \). We can express this as:
\[ \frac{X + 2}{Y + 1} = \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(X + 2) = 1(Y + 1) \]
Expanding this, we have:
\[ 2X + 4 = Y + 1 \]
Rearranging gives us our first equation:
\[ 2X - Y + 3 = 0 \quad \text{(Equation 1)} \]

### Step 2: Set up the second equation
Next, the problem states that when the numerator is increased by 3 and the denominator by 5, the new fraction equals \( \frac{2}{5} \). This can be expressed as:
\[ \frac{X + 3}{Y + 5} = \frac{2}{5} \]
Cross-multiplying gives:
\[ 5(X + 3) = 2(Y + 5) \]
Expanding this, we have:
\[ 5X + 15 = 2Y + 10 \]
Rearranging gives us our second equation:
\[ 5X - 2Y + 5 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have the following system of equations:
1. \( 2X - Y + 3 = 0 \)
2. \( 5X - 2Y + 5 = 0 \)

We can solve these equations simultaneously. First, let's express \( Y \) from Equation 1:
\[ Y = 2X + 3 \]

Now, substitute \( Y \) in Equation 2:
\[ 5X - 2(2X + 3) + 5 = 0 \]
Expanding this gives:
\[ 5X - 4X - 6 + 5 = 0 \]
This simplifies to:
\[ X - 1 = 0 \]
Thus, we find:
\[ X = 1 \]

### Step 4: Substitute back to find \( Y \)
Now substitute \( X = 1 \) back into the expression for \( Y \):
\[ Y = 2(1) + 3 = 5 \]

### Step 5: State the original fraction
The original fraction is:
\[ \frac{X}{Y} = \frac{1}{5} \]

### Final Answer
The original fraction is \( \frac{1}{5} \).