Question

The denominator of a rational number is greater than its numerator by 3.if 3 is subtracted from the numerator and 2 is added to its denominator ,the new number becomes 1/5.find the original number.check your solution.

Answer 1

let n=numerator
n+7=denominator
..
n-3/(n+7)+2=1/5
n-3/n+9=1/5
5n-15=n+9
4n=24
n=6
n+7=13
number: 6/13 

Answer 2

Given:-

• The denominator of a rational number is greater than its numerator by 3. If 3 is subtracted from its numerator and 2 is added to its denominator, the new number becomes 1/5.

To Find:-

• Original rational number.

Solution:-

Let,

$\mapsto \sf{Numerator = x}$

$\mapsto \sf{Denominator =\: x + 3}$

Hence, the required original rational number is :

$\mapsto \sf \dfrac{Numerator}{Denominator}$

$\mapsto{\boxed{\sf{\dfrac{x}{x + 3}}}}\red\bigstar$

According to the question,

$\begin{gathered}:\implies \sf \dfrac{Numerator - 3}{Denominator + 2} =\: New \: Number\\\end{gathered}$

$:\implies \sf \dfrac{x - 3}{x + 3 + 2} =\: \dfrac{1}{5}$

$:\implies \sf \dfrac{x - 3}{x + 5} =\: \dfrac{1}{5}$

By doing cross multiplication we get,

$:\implies \sf 5(x - 3) =\: 1(x + 5)$

$:\implies \sf 5x - 15 =\: x + 5)$

$:\implies \sf 5x - x =\: 5 + 15$

$:\implies \sf 4x =\: 20$

$:\implies \sf x =\: \dfrac{\cancel{20}}{\cancel{4}}$

$:\implies \sf x =\: \dfrac{5}{1}$

$: \implies   \underline{\boxed{ \frak{x=5}}} \blue \bigstar$

Hence, the required original rational number is;

$:\implies\sf \dfrac{x}{x + 3}$

$:\implies \sf \dfrac{5}{5 + 3}$

$:\implies\underline{\boxed{\frak{\dfrac{5}{8}}}}\pink\bigstar$

• The original rational number is$\underline{\underline{\bf{\dfrac{5}{8}}}}.$

Thank you!!

@itzshivani