Question

if 1 is subtracted from the numerator and from the denominator of a fraction, the value obtained is 1/2. if 1 is added to it's numerator and to it's denominator, the resulting value is 2/3. Find the fraction​

Answer 1

Let numerator be "N" and denominator be "D".

If 1 is subtracted from the numerator and from the denominator of a fraction, the value obtained is 1/2.

$\bold{Subtract \:1\:from\:both\:=>}\: \begin{cases} \text{Numerator = N - 1} \\ \text{Denominator = D - 1} \end{cases}$

According to question,

=> $\sf{\frac{N\:-\:1}{D\:-\:1}\:=\:\frac{1}{2}}$

=> $\sf{2(N\:-\:1)\:=\:1(D\:-\:1)}$

=> $\sf{2N\:-\:2\:=\:D\:-\:1}$

=> $\sf{2N\:-\:1\:=\:D}$ ___ (eq 1)

if 1 is added to it's numerator and to it's denominator, the resulting value is 2/3.

$\bold{Add \:1\:to\:both\:=>}\: \begin{cases} \text{Numerator = N + 1} \\ \text{Denominator = D + 1} \end{cases}$

According to question,

=> $\sf{\frac{N\:+\:1}{D\:+\:1}\:=\:\frac{2}{3}}$

=> $\sf{3(N\:-\:1)\:=\:2(D\:-\:1)}$

=> $\sf{3N\:-\:1\:=\:2D\:-\:2}$

=> $\sf{3N\:=\:2D\:-\:1}$

=> $\sf{3N\:=\:2(2N\:-\:1)\:-\:1}$ [From (eq 1)]

=> $\sf{3N\:=\:4N\:-\:2\:-\:1}$

=> $\sf{-N\:=\:-3}$

=> $\sf{N\:=\:3}$

Substitute value of N in (eq 1)

=> $\sf{2(3)\:-\:1\:=\:D}$

=> $\sf{6\:-\:1\:=\:D}$

=> $\sf{D\:=\:5}$

•°• Fraction = $\sf{\frac{Numerator}{Denominator}}$

=> $\sf{\dfrac{3}{5}}$

Answer 2

$\huge\underline\blue{\sf Answer:}$

$\large\red{\boxed{\sf Fraction={\frac{3}{5}}}}$

$\huge\underline\blue{\sf Solution:}$

Let ,

N = Numerator

D = Denominator

As per first condition :

1 is subtracted from numerator and denominator , the value obtain is 1/2

Therefore ,

$\large\implies{\sf {\frac{N-1}{D-1}}=1(D-1)}$

$\large\implies{\sf 2(N-1)=1(D-1)}$

$\large\implies{\sf 2N-2=D-1}$

$\large\implies{\sf 2N-1=D →(1) }$

As per Second condition :

1 is added to its Numerator and Denominator we get 2/3

Therefore ,

$\large\implies{\sf {\frac{N+1}{D+1}}={\frac{2}{3}}}$

$\large\implies{\sf 3(N+1)=2(D+1) }$

$\large\implies{\sf 3N+3=2D+2 }$

From Equation 1

$\large\implies{\sf 3N=2(2N-1)-1 }$

$\large\implies{\sf 3N=4N-2}$

$\large\implies{\sf -N=-3}$

$\large{\boxed{\sf N=3}}$

Substituting value of N ,

$\large\implies{\sf 2(3)-1=D}$

$\large\implies{\sf 6-1=D}$

$\large{\boxed{\sf D=5 }}$

_______________________________________

$\large{\sf Fraction={\frac{Numerator}{Denominator}}}$

$\huge\red{♡}$$\large\red{\boxed{\sf Fraction={\frac{3}{5}}}}$