. If 
is added in the denominator, the fraction becomes . Find the fraction so obtained.Let the numerator be A and the denominator be B. So, the required fraction is $A/B$. ATQ, The sum of the numerator and denominator of the fraction is $12$. $A+B=12$ ⇒ $A+B–12=0$ ATQ, If $3$ is added...
and numerator is increased by , a fraction becomes . It also becomes 
if we only increase the denominator by . So now find the fraction?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 $1$,...
is subtracted from both its numerator and denominator then fraction becomes 
. If numerator and denominator are added by , it becomes 
. Find the fraction.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, Thus, the equation so formed is, $(A–1)/(B−1)=1/3$ ⇒ $3(A–1)=(B–1)$ ⇒...
more than the numerator. If the denominator is eight times the numerator then the numerator is lessen by 
and denominator is increased by . Find the original fraction calculated.Let the numerator of the fraction to be A and the denominator of the fraction to be B. So, fraction is $A/B$. The numerator of the fraction is $4$ less the denominator. Thus, the equation so formed...
is added to both numerator and the denominator then the fraction becomes 
. If is added to both the numerator and the denominator it becomes 
. Find the fraction.Let’s assume 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 , the equation so formed is, $A+2B+2=9/11$ ⇒ $11\left( A+2...