Question

A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction.

Hint:

To solve for unknowns in word problems involving fractions, set up equations based on the given conditions and solve the system.

Answer 1

Let the fraction be \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator. We are given two conditions: 1. When 1 is subtracted from the numerator, the fraction becomes \( \frac{1}{3} \): \[ \frac{x-1}{y} = \frac{1}{3}. \] 2. When 8 is added to the denominator, the fraction becomes \( \frac{1}{4} \): \[ \frac{x}{y+8} = \frac{1}{4}. \] Now, solve these two equations: From the first equation: \[ \frac{x-1}{y} = \frac{1}{3} \implies x - 1 = \frac{y}{3} \implies x = \frac{y}{3} + 1. \] From the second equation: \[ \frac{x}{y+8} = \frac{1}{4} \implies x = \frac{y+8}{4}. \] Now, equate the two expressions for \( x \): \[ \frac{y}{3} + 1 = \frac{y+8}{4}. \] Multiply both sides by 12 to eliminate the denominators: \[ 4y + 12 = 3(y + 8). \] Simplifying: \[ 4y + 12 = 3y + 24 \implies 4y - 3y = 24 - 12 \implies y = 12. \] Now substitute \( y = 12 \) into the equation \( x = \frac{y}{3} + 1 \): \[ x = \frac{12}{3} + 1 = 4 + 1 = 5. \] Thus, the fraction is: \[ \frac{x}{y} = \frac{5}{12}. \]
Conclusion: The fraction is \( \frac{5}{12} \).