Question

their ages
7. The denominator of a rational number is greater than its numerator by 8. If the
numerator is increased by 17 and the denominator is decreased by 1, the number
3.
obtained is Find the rational number.
NI​

Answer 1

GIVEN:-

• The denominator of a rational number is greater than Numerator by 8.

• if the Numerator is increased by 17 and denominator decreased by 1.

• The Number becomes 3.

TO FIND:-

• The Ration Number.

${\huge{\underline{\rm{\blue{Let}}}}}$

$\implies\rm{Numerator=x}$

$\implies\rm{Denominator=(x+8)}$

$\implies\rm{Original\:Rational\:Number=\dfrac{x}{x+8}}$

According to Question..

$\implies\rm{Numerator=(x+17)}$

$\implies\rm{Denominator=(x+8-1)=(x+7)}$

$\implies\rm{New\:Rational\:Number=\dfrac{x+17}{x+7}=3}$

$\implies\rm{3(x+7)=x+17}$

$\implies\rm{3x+21=x+17}$

$\implies\rm{2x=4}$

$\implies\rm{x=\dfrac{4}{2}}$

$\implies\rm{x=2}$

Hence,

$\implies\rm{Numerator=x=2}$

$\implies\rm{Denominator=(x+8)=10}$

Hence, The Rational Number is $\rm{\dfrac{2}{10}}$.

Answer 2

$\sf\large{\underline{\underline{Let:}}}$

$\sf{The\:numerator\:of\:the\:rational\:number\:be\:"x"}$

$\sf{Therefore,\: its\: denominator\:will\:be\:x + 8}$

$\sf{\underline{\underline{So:}}}$

$\sf{The\:rational\:number\:will\:be\: \dfrac{x}{x+8}}$

$\sf{\underline{\underline{According\:to\:the\:Question:}}}$

$\sf\red{ \implies x + \dfrac{17}{x} + 8 - 1 = \dfrac{3}{2}}$

$\sf\purple{\implies x + \dfrac{17}{x} + 7 = \dfrac{3}{2}}$

$\sf\orange{\implies 2(x + 17) = 3(x + 7)}$

$\sf\green{ \implies 2x + 34 = 3x + 21}$

$\sf\red{ \implies 34 - 21 = 3x - 2x}$

$\sf\purple{\implies 13 = x}$

$\sf{\underline{\underline{Therefore:}}}$

$\sf{Numerator\:of\:the\:rational\:number = x = 13}$

$\sf{Denominator\:of\:the\:rational\:number = x + 8}$$\sf{= 13 + 8 = 21}$

$\sf\underline{\underline{Hence:}}$

$\sf\red{Rational \:Number = \dfrac{13}{21}}$