Let the numerator of the fraction to be A and the denominator of the fraction to be B.
So, the required fraction is $A/B$.
ATQ,
If denominator is decreased by $1$ and numerator is increased by $1$, a fraction becomes $1$
Thus, the equation so formed is,
$(A+1)/(B-1)=1$
⇒ $(A+1)=(B–1)$
⇒ $A+1–B+1=0$
⇒ $A–B+2=0$ …….. (i)
And also it’s given in the question as,
if we only increase the denominator by $1$, the fraction becomes $1/2$.
Expressing the above condition in an equation, we will get,
$A/(B+1)=1/2$
⇒ $2A=(B+1)$
⇒ $2A–B–1=0$ …… (ii)
Solving (i) and (ii), to find the fraction
BB using cross-multiplication, we have
$A\left( -1 \right)\times \left( -1 \right)-1\times 2=-B-1\times \left( -2 \right)-2\times 2=11\times -1-2\times -1$
$A1+2=-B-1-4=1-1+2$
$A-3=-B-5=11$
$A-3=B-5=1$
$A=3,B=5$
Hence, the fraction is $3/ 5$.