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
Answers
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
∘