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 number ​

Answer 1

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

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

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

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

$\sf\therefore \blue{let\:the\:numerator\:be\:x}$

$\sf\dashrightarrow \blue{denominator= x-5}$

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

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

$\sf\dashrightarrow \red{numerator= x+5}$

THE EQUATION FORM IS,

$\rm{\boxed{\sf{\green{ \circ\:\: \dfrac{x+5}{x-5} = \dfrac{11}{6}\:\: \circ}}}}$

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

$\sf\dashrightarrow \red{\:THE\:ORIGINAL\:RATIONAL\:NUMBER}$

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

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

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

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

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

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

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

$\sf\implies \orange{x = 17}$

$\rm{\boxed{\sf{ \circ\:\: x= 17 \:\: \circ}}}$

THE ORIGINAL NUMBERS ARE,

$\sf\implies \red{numerator=x =17}$

$\sf\implies \red{d enominator= x-5}$

$\sf\implies \red{denominator=17-5}$

$\sf\implies \pink{denominator=12}$

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

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

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

$\rm\underline\blue{NUMERATOR\:IS\:17\:AND\: DENOMINATOR\:IS\:12}$

$\rm{\boxed{\sf{ \circ\:\: \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12} \:\: \circ}}}$

Answer 2

Answer:

.

\large\underline\red{GIVEN,}

GIVEN,

\sf\dashrightarrow \blue{THE\:GIVEN\:FRACTION\:IS\:A\:RATIONAL\:NUMBER.}⇢THEGIVENFRACTIONISARATIONALNUMBER.

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

\sf\therefore \blue{let\:the\:numerator\:be\:x}∴letthenumeratorbex

\sf\dashrightarrow \blue{denominator= x-5}⇢denominator=x−5

\sf\dashrightarrow \blue{\dfrac{x}{x-5}}⇢

x−5

x

\sf\dashrightarrow \bold\pink{if \:5 \:is \:added\: to \:the \:numerator, \:numerator \:becomes\: ; \dfrac{11}{6}}⇢if5isaddedtothenumerator,numeratorbecomes;

6

11

\sf\dashrightarrow \red{numerator= x+5}⇢numerator=x+5

THE EQUATION FORM IS,

\rm{\boxed{\sf{\green{ \circ\:\: \dfrac{x+5}{x-5} = \dfrac{11}{6}\:\: \circ}}}}

∘

x−5

x+5

=

6

11

∘

\large\underline\purple{TO\:FIND,}

TOFIND,

\sf\dashrightarrow \red{\:THE\:ORIGINAL\:RATIONAL\:NUMBER}⇢THEORIGINALRATIONALNUMBER

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

x−5

x+5

=

6

11

\sf\implies \green{6 \times (x+5)= 11 \times (x-5)}⟹6×(x+5)=11×(x−5)

\sf\implies \green{6x+30= 11x-55}⟹6x+30=11x−55

\sf\implies \green{30+55=11x-6x}⟹30+55=11x−6x

\sf\implies \green{85= 5x}⟹85=5x

\sf\implies \green{x= \dfrac{85}{5}}⟹x=

5

85

\sf\implies \green{x= \cancel \dfrac{85}{5}}⟹x=

5

85

\sf\implies \orange{x = 17}⟹x=17

\rm{\boxed{\sf{ \circ\:\: x= 17 \:\: \circ}}}

∘x=17∘

THE ORIGINAL NUMBERS ARE,

\sf\implies \red{numerator=x =17}⟹numerator=x=17

\sf\implies \red{d enominator= x-5}⟹denominator=x−5

\sf\implies \red{denominator=17-5}⟹denominator=17−5

\sf\implies \pink{denominator=12}⟹denominator=12

\large\underline\orange{FRACTION,}

FRACTION,

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

x−5

x

=

12

17

\sf\dashrightarrow \purple{\dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12}}⇢

DENOMINATOR

NUMERATOR

=

12

17

\rm\underline\blue{NUMERATOR\:IS\:17\:AND\: DENOMINATOR\:IS\:12}

NUMERATORIS17ANDDENOMINATORIS12

\rm{\boxed{\sf{ \circ\:\: \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12} \:\: \circ}}}

∘

DENOMINATOR

NUMERATOR

=

12

17

∘