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Question

# If 1 is added to both of the numerator and denominator of a fraction, it becomes $\frac{4}{5}$. If however, 5 is subtracted from both numerator and denominator, the fraction $\frac{1}{2}$. Find the fraction.

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Solution

## Let the required fraction be $\frac{x}{y}$. Then, we have: $\frac{x+1}{y+1}=\frac{4}{5}$ ⇒ 5(x + 1) = 4(y + 1) ⇒ 5x + 5 = 4y + 4 ⇒ 5x − 4y = −1 ....(i) Again, we have: $\frac{x-5}{y-5}=\frac{1}{2}$ ⇒ 2(x − 5) = 1(y − 5) ⇒ 2x − 10 = y − 5 ⇒ 2x − y = 5 ....(ii) On multiplying (ii) by 4, we get: 8x − 4y = 20 ....(iii) On subtracting (i) from (iii), we get: 3x = (20 − (−1)) = 20 + 1 = 21 ⇒ 3x = 21 ⇒ x = 7 On substituting x = 7 in (i), we get: 5 × 7 − 4y = −1 ⇒ 35 − 4y = −1 ⇒ 4y = 36 ⇒ y = 9 ∴ ​x = 7 and y = 9 Hence, the required fraction is $\frac{7}{9}$.

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