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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?
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?
CBSE | Class 10 | Exercise 3.8 | Maths | Pair of Linear Equations In Two Variables | RD Sharma
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 $1$, 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 $3/ 5$.

i

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