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The sum of numerator and denominator of a fraction is $4$ less than twice the denominator. If each of the numerator and denominator is decreased by $1$, the fraction becomes $\frac{1}{3}$. Find the fraction.
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Let the fraction be $\frac{x}{y}$ ($\frac{1}{2}$ Mark)
$x + y = 2y - 4$
$\Rightarrow x - y = -4$ (i) (1 Mark)
$\frac{x - 1}{y - 1} = \frac{1}{3}$ ($\frac{1}{2}$ Mark)
$\Rightarrow 3x - y = 2$ (ii) (1 Mark)
Solving equations (i) and (ii) to get
$x = 3$ and $y = 7$ (1 Mark)
Required fraction = $\frac{3}{7}$ ($\frac{1}{2}$ Mark)
$x + y = 2y - 4$
$\Rightarrow x - y = -4$ (i) (1 Mark)
$\frac{x - 1}{y - 1} = \frac{1}{3}$ ($\frac{1}{2}$ Mark)
$\Rightarrow 3x - y = 2$ (ii) (1 Mark)
Solving equations (i) and (ii) to get
$x = 3$ and $y = 7$ (1 Mark)
Required fraction = $\frac{3}{7}$ ($\frac{1}{2}$ Mark)