If 1 is added to the denominator of a fraction, the fraction becomes $\frac{1}{2}$. If 1 is added to the numerator of the fraction, the fraction becomes 1. The fraction is

### Concept & Logic This problem requires setting up a system of two linear equations based on independent conditions applied to a single fraction. By assigning variables to the numerator and denominator, each sentence in the problem translates into a distinct algebraic equation that can be solved using substitution. ### Step-by-Step Solution * **Given:** * Condition 1: Adding 1 to the denominator results in $\frac{1}{2}$. * Condition 2: Adding 1 to the numerator results in 1. * **Calculation:** 1. Let the fraction be $\frac{x}{y}$. 2. Translate Condition 1 into an equation: $$ \frac{x}{y + 1} = \frac{1}{2} $$ Cross-multiply: $$ 2x = y + 1 \Rightarrow y = 2x - 1 $$ 3. Translate Condition 2 into an equation: $$ \frac{x + 1}{y} = 1 $$ Cross-multiply: $$ x + 1 = y $$ 4. Substitute the expression for $y$ from the second condition into the first: $$ x + 1 = 2x - 1 $$ 5. Solve for $x$: $$ 1 + 1 = 2x - x $$ $$ x = 2 $$ 6. Substitute the value of $x$ back into the second equation to find $y$: $$ y = 2 + 1 = 3 $$ 7. The original fraction is $\frac{x}{y} = \frac{2}{3}$. ### Exam Strategy & Shortcut **Option Elimination (Fastest Method):** Instead of solving a system of equations, plug the given options into the second condition (adding 1 to the numerator makes the fraction equal to 1). A fraction equals 1 when its numerator and denominator are exactly equal. * (a) $\frac{1+1}{3} = \frac{2}{3} \neq 1$ * (b) $\frac{2+1}{3} = \frac{3}{3} = 1$ (Valid) * (c) $\frac{3+1}{4} = \frac{4}{4} = 1$ (Valid) * (d) $\frac{3+1}{2} = \frac{4}{2} = 2 \neq 1$ Now test the remaining valid options (b and c) against the first condition (adding 1 to the denominator makes it $\frac{1}{2}$): * Try (b) $\frac{2}{3} \rightarrow \frac{2}{3+1} = \frac{2}{4} = \frac{1}{2}$. This is a perfect match! You have your answer instantly. ### Common Pitfall Students often get confused and try to combine both conditions into a single operation, such as adding 1 to both the numerator and denominator simultaneously. It is crucial to recognize that the problem presents two separate, independent scenarios to form two distinct equations. ### Final Answer **Therefore, the correct answer is $\frac{2}{3}$.**