Question

A fraction becomes \(\frac{1}{2}\) when 1 is added to its numerator and denominator, and becomes \(\frac{1}{4}\) when 1 is subtracted from its numerator and denominator. Find the fraction.

• A. \(\frac{4}{9}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{4}{13}\)
• D. \(\frac{2}{5}\)

Hint:

Write the fraction as \(\frac{x}{y}\) and turn each condition into a linear equation in \(x\) and \(y\).

Answer 1

The Correct Option is D

Step 1: Write both conditions as linear equations.
Let the fraction have numerator $x$ and denominator $y$. Adding 1 to both parts gives $\frac{1}{2}$:
\[ 2(x+1) = 1(y+1) \]
\[ 2x - y = -1 \]

Step 2: Write the second condition the same way.
Subtracting 1 from both parts gives $\frac{1}{4}$:
\[ 4(x-1) = 1(y-1) \]
\[ 4x - y = 3 \]

Step 3: Eliminate $y$ by subtracting the equations.
Both equations have the same $-y$ term, so subtract the first from the second:
\[ (4x-y) - (2x-y) = 3 - (-1) \]
\[ 2x = 4 \]
\[ x = 2 \]

Step 4: Find $y$ using either equation.
Using $2x - y = -1$:
\[ 2(2) - y = -1 \]
\[ 4 - y = -1 \]
\[ y = 5 \]
So the fraction is $\frac{2}{5}$.

Step 5: Verify quickly.
Adding 1: $\frac{3}{6} = \frac{1}{2}$, matches. Subtracting 1: $\frac{1}{4}$, matches. Both conditions hold, so no other option needs checking.
\[ \boxed{\dfrac{2}{5}} \]