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...
If denominator is decreased by $1$ and numerator is increased by $1$, a fraction becomes $1$. It also becomes $\frac{1}{2}$ if we only increase the denominator by $1$. 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 $1$,...
If $1$ is subtracted from both its numerator and denominator then fraction becomes $1/3$ . If numerator and denominator are added by $1$, it becomes $1/2$. 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, $(A–1)/(B−1)=1/3$ ⇒ $3(A–1)=(B–1)$ ⇒...
The denominator of a fraction is $4$ more than the numerator. If the denominator is eight times the numerator then the numerator is lessen by $2$ and denominator is increased by $1$. 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...
If $2$ is added to both numerator and the denominator then the fraction becomes $9/11$. If $3$ is added to both the numerator and the denominator it becomes $5/6$. 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...