Question:1

Solution:

Let the fraction $= \dfrac{x}{y}$

Then, according to the first condition, we have

$\dfrac{x+1}{y-1} = 1$

$\Rightarrow x + 1 = y - 1$

$\Rightarrow x - y = -2$ ……..(i)

According to the second condition

$\dfrac{x}{y + 1} = \dfrac{1}{2}$

$\Rightarrow 2x = y + 1$

$\Rightarrow 2x - y = 1$ …….(ii)

Subtracting eq (i) to eq (ii)

$(x - y)-(2x - y) = -2 - 1$

$\Rightarrow x - y - 2x + y = -3$

$\Rightarrow -x = -3$

$\Rightarrow x = 3$

Putting the value of x in (i)

$x - y = -2$

$\Rightarrow 3 - y = -2$

$\Rightarrow 3 + 2 = y$

$\Rightarrow y = 5$

Hence, the fraction $=\dfrac{3}{5}$