A fraction becomes 4 when 1 is added to both | vedclass

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Question

(D) Let the required fraction be $\frac{x}{y}$.According to the first condition,adding $1$ to both the numerator and denominator gives $4$:$\frac{x+1}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) Let the required fraction be $\frac{x}{y}$.According to the first condition,adding $1$ to both the numerator and denominator gives $4$:$\frac{x+1}{y+1} = 4 \Rightarrow x+1 = 4y+4 \Rightarrow x-4y = 3$ (Equation $1$).According to the second condition,subtracting $1$ from both the numerator and denominator gives $7$:$\frac{x-1}{y-1} = 7 \Rightarrow x-1 = 7y-7 \Rightarrow x-7y = -6$ (Equation $2$).Subtracting Equation $2$ from Equation $1$:$(x-4y) - (x-7y) = 3 - (-6)$$3y = 9 \Rightarrow y = 3$.Substituting $y=3$ into Equation $1$:$x - 4(3) = 3 \Rightarrow x - 12 = 3 \Rightarrow x = 15$.The fraction is $\frac{15}{3}$. The numerator is $15$.

$A$ fraction becomes $4$ when $1$ is added to both the numerator and denominator; and it becomes $7$ when $1$ is subtracted from both the numerator and denominator. The numerator of the given fraction is

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