Question

the denominator of a rational number is less than its ne numerator than 5 if the five is added to the numerator the new number become 11 by 6 find the original rational number​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

Let x be the numerator of the required rational number

Then ( x - 5) is it's denominator

Then the rational number is

$= \displaystyle \: \frac{x }{x - 5}$

By the given condition

$\displaystyle \: \frac{x + 5}{x - 5} = \frac{11}{6}$

$\implies \: 11x - 55 = 6x + 30$

$\implies \: 5x = 85$

$\therefore \: x = 17$

RESULT

SO THE REQUIRED RATIONAL NUMBER

$= \displaystyle \: \frac{17 }{17 - 5}$

$= \displaystyle \: \frac{17 }{12}$

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow THE\:GIVEN\:FRACTION\:IS\:A\:RATIONAL\:NUMBER.$

$\sf\dashrightarrow {\red{\mathbb{\text{ denominator is less than its numerator by 5.}}}}$

$\sf\therefore let\:the\:numerator\:be\:x$

$\sf\dashrightarrow denominator= x-5$

$\sf\dashrightarrow \dfrac{x}{x-5}$

$\sf\dashrightarrow \bold\red{if \:5 \:is \:added\: to \:the \:numerator, \:numerator \:becomes\: ; \dfrac{11}{6}}$

$\sf\dashrightarrow numerator= x+5$

THE EQUATION FORM IS,

$\large{\boxed{\bf{\green{ \star\:\: \dfrac{x+5}{x-5} = \dfrac{11}{6}\:\: \star}}}}$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow \:THE\:ORIGINAL\:RATIONAL\:NUMBER$

$\sf\implies \dfrac{x + 5}{x - 5} = \dfrac{11}{6}$

$\sf\implies 6 \times (x+5)= 11 \times (x-5)$

$\sf\implies 6x+30= 11x-55$

$\sf\implies 30+55=11x-6x$

$\sf\implies 85= 5x$

$\sf\implies x= \dfrac{85}{5}$

$\sf\implies x= \cancel \dfrac{85}{5}$

$\sf\implies x = 17$

$\large{\boxed{\bf{ \star\:\: x= 17 \:\: \star}}}$

THE ORIGINAL NUMBERS ARE,

$\sf\implies numerator=x =17$

$\sf\implies denominator= x-5$

$\sf\implies denominator=17-5$

$\sf\implies denominator=12$

$\large\underline\bold{FRACTION,}$

$\sf\dashrightarrow \dfrac{x}{x-5}= \dfrac{17}{12}$

$\sf\dashrightarrow \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12}$

$\large\underline\bold{NUMERATOR\:IS\:17\:AND\: DENOMINATOR\:IS\:12}$

$\large{\boxed{\bf{ \star\:\: \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12} \:\: \star}}}$

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