Question

the denominator of a rational number is gearter than its numerator by 7.if the numerator is increased by 17 and the denominator decreased by 6.the new number becomes 2.find the original number​

Answer 1

☯GIVEN :

• The denominator of a rational number is 7 more than its numerator.

• The new number becomes 2, if the numerator is increased by 17 and the denominator is decreased by 6.

☯TO FIND :

• Original Number.

➲SOLUTION :

Let the numerator be x.

Then , denominator = x+7

• After increasing the numerator by 17 and decreasing the denominator by 6, the new number becomes 2.

$: \implies{\sf{ \dfrac{Numerator + 17}{Denominator - 6} = 2 }}$

$: \implies{\sf{ \dfrac{x + 17}{(x + 7) - 6} = 2 }}$

$: \implies{\sf{ \dfrac{x + 17}{x + 7 - 6} = 2 }}$

$: \implies{\sf{ \dfrac{x + 17}{x + 1} = 2 }}$

• By cross Multiplication:

$: \implies{\sf{x + 17 = 2(x + 1) }}$

$: \implies{\sf{x + 17 = 2x + 2 }}$

• Transposing 2 to LHS and x to RHS:

$: \implies{\sf{17 - 2= 2x - x }}$

$\implies{\sf{15= x }}$

$\implies{ \underline{ \huge{ \boxed{\red{ \bf{x= 15 }}}}}}$

★Numerator :

• $\bf{x = 15}$

★ Denominator :

• Substituting the value of x :

$:\leadsto {\sf{x+7}}$

$:\leadsto{\sf{15+7}}$

$:\leadsto\ {\bf{22}}$

$\huge {\pink{\therefore}}$ The Original Number is $\bf{\dfrac{15}{22}}$.

Answer 2

Given : The denominator of a rational number is greater than its numerator by 7 & 17 and the denominator decreased by 6, the new becomes 2.

To Find : The Original Rational number ?

________________________

❍ Let's consider the numerator of fraction be p & denominator of the fraction is q.

• $\boxed{\sf\pink{\therefore~Original_{(Rational ~Number ~or ~Fraction)}~=~\dfrac{p}{q}}}$

$~$

Given that,

• The denominator of a rational number is greater than its numerator by7.

$\therefore{\sf{Denominator~=~Numerator~+~7}}$

$~~~~~~~~~~{\sf:\implies{Denominator~ =~ Numerator~+~7}}$

$~~~~~~~~~~{\sf:\implies{q~=~p~+~7}}$

$~~~~~~~~~~{\sf:\implies{q~=~p~+~7~~~~~~~~~~~~~~~~~~~~\bigg\lgroup{Eqⁿ~1}\bigg\rgroup}}$

$~~~~~~~~~~{\sf\purple\therefore{New_{(Rational ~Number ~or ~Fraction)}~=~\dfrac{p}{p~+~7}}}$

$~$

$\underline{\frak{According~ to ~the~ Given ~Question~:}}$

$~$

• The numerator is increased by 17 & the denominator decreased by 6 then, the new number becomes 2.

$~$

$~~~~~~~~~~{\sf:\implies{\dfrac{Numerator~+~17}{Denominator~-~6}~=~2}}$

$~~~~~~~~~~{\sf:\implies{\dfrac{p~+~17}{p~+~7~-~6}~=~2}}$

$~~~~~~~~~~{\sf:\implies{\dfrac{p~+~17}{p~+~1}~=~2}}$

$~~~~~~~~~~{\sf:\implies{p~+~17~=~2(p~+~1)}}$

$~~~~~~~~~~{\sf:\implies{p~+~17~=~2p~+~2}}$

$~~~~~~~~~~{\sf:\implies{p~=~2p~+~2~-~17}}$

$~~~~~~~~~~{\sf:\implies{p~=~2p~-~15}}$

$~~~~~~~~~~{\sf:\implies{p~-~2p~=~- 15}}$

$~~~~~~~~~~{\sf:\implies{- p~=~- 15}}$

$~~~~~~~~~~{\sf:\implies{\cancel{-}p~=~\cancel{-}15}}$

$~~~~~~~~~~{\sf:\implies{p~=~15}}$

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{\therefore~p~=~15}}}}}$★

$~$

Here,

• p denotes the numerator which is 15.

$~$

$\underline{\bf{Now~By~Substituting ~the ~values ~of~pin~Eqⁿ~ 1 :}}$

$~~~~~~~~~~{\sf:\implies{q~=~p~+~7~~~~~~~~~~~~~~~~~~~~\bigg\lgroup{Eqⁿ~1}\bigg\rgroup}}$

$~~~~~~~~~~{\sf:\implies{q~=~p~+~7}}$

$~~~~~~~~~~{\sf:\implies{q~=~15~+~7}}$

$~~~~~~~~~~{\sf:\implies{q~=~22~+~7}}$

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\purple{\therefore~q~=~22}}}}}$★

$~$

Here,

• q denotes the denominator which is 22.

$~$

Now,

• Finding Original Fraction :

$~~~~~~~~~~{\sf:\implies{Original_{(Rational ~Number ~or ~Fraction)}~=~\dfrac{p}{q}}}$

$~~~~~~~~~~{\sf:\implies{Original_{(Rational ~Number~ or~ Fraction)}~=~\dfrac{15}{22}}}$

• $\underset{\blue{\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{Original_{(Rational ~Number ~or~ Fraction)}~=~\dfrac{15}{22}}}}}}$

$~$

$\therefore\underline{\sf{The ~Original ~Number ~is~\bf{\dfrac{15}{22}}}}$