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

❍ Let's say, the numerator of the fraction be m and denominator of the fraction be n respectively.

Hence,

• The fraction is: m/n.

⠀

ATQ, the denominator 'n' of a rational number is greater than it's numerator 'm' by 7.

Therefore,

⇥ (Denominator = Numerator + 7)

⇥ n = (m + 7)⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ —eq.( I )

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀

⠀

$\underline{\bigstar\:{\pmb{ \sf{According \;to\; the\; given\; Question :}}}}$

⠀

• If the numerator 'm' is increased by 17 and the denominator 'n' is decreased by 6. The new number becomes 2.

⠀

$\dashrightarrow\sf \bigg\{\dfrac{m + 17}{m + 7 - 6}\bigg\} = 2 \\\\\\\dashrightarrow\sf \bigg\{\dfrac{m + 17}{m + 1}\bigg\} = 2 \\\\\\\dashrightarrow\sf m + 17 = 2(m + 1)\\\\\\\dashrightarrow\sf m + 17 = 2m + 2\\\\\\\dashrightarrow\sf m - 2m = 2 - 17\\\\\\\dashrightarrow\sf -m = -15\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{m = 15}}}}}\: \bigstar$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

¤ The value of 'm' is 15. Now we'll substitute this value in the eq. ( I ) to find out the denominator of the given fraction 'n' —

$\dashrightarrow\sf Denominator = \Big\{Numerator+ 7\Big\} \\\\\\\dashrightarrow\sf n = \Big\{m + 7 \Big\} \\\\\\\dashrightarrow\sf n = 15 + 7\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\purple{n = 22}}}}}\;\bigstar$

⠀

$\therefore{\underline{\sf{Hence,~ the~ required~ original~ number~ is~ \bf{\dfrac{15}{22}}.}}}$

Answer 2

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

Exigency To Find : The Original Rational number .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

$\qquad \therefore \sf \:Original_{(Rational \;Number \:or \: Fraction \:)}\: is \:: \bf \:\dfrac{\:\: p \:}{\:\:q\:}$

⠀⠀⠀⠀⠀Given that ,

⠀⠀》 The denominator of a rational number is greater than its numerator by 7 .

$\qquad \therefore \sf Denominator \:\:=\: Numerator \: + \: 7 \:$

$\qquad :\implies \sf Denominator \:\:=\: Numerator \: + \: 7 \:$

$\qquad :\implies \sf q \: \:\:=\: p \: + \: 7 \:$

$\qquad :\implies \pmb{ q \: \:\:=\: p \: + \: 7 \:} \qquad \:\bigg\lgroup \sf{ Eq^n \: 1 }\bigg\rgroup$

$\qquad \therefore \sf \:New_{(Rational \;Number \:or \: Fraction \:)}\: is \:: \bf \:\dfrac{\:\: p \:}{\:\:p + 7\:}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:According \:to \:the \:Question \::}}$

⠀⠀━━━━ The numerator is increased by 17 and the denominator decreased by 6 then , the new number becomes 2.

$\qquad \dashrightarrow \sf \dfrac{ Numerator \: + \: 17 \:}{Denominator - 6 \;}=\:\: 2 \:\:$

$\qquad \dashrightarrow \sf \dfrac{ p \: + \: 17 \:}{ p + 7 - 6 \;}=\:\: 2 \:\:$

$\qquad \dashrightarrow \sf \dfrac{ p \: + \: 17 \:}{ p + 1 \;}=\:\: 2 \:\:$

$\qquad \dashrightarrow \sf p \: + \: 17 \:=\:\: 2\:( p + 1 ) \:\:$

$\qquad \dashrightarrow \sf p \: + \: 17 \:=\:\: 2p + 2 \:\:$

$\qquad \dashrightarrow \sf p \: \:=\:\: 2p + 2 - 17\:\:$

$\qquad \dashrightarrow \sf p \: \:=\:\: 2p - 15\:\:$

$\qquad \dashrightarrow \sf p - 2p \: \:=\:\: - 15\:\:$

$\qquad \dashrightarrow \sf -p \: \:=\:\: - 15\:\:$

$\qquad \dashrightarrow \sf p \: \:=\:\: 15\:\:$

$\qquad \therefore \pmb{\underline{\purple{\frak{\:p = 15 }} }}\:\:\bigstar$

⠀⠀⠀⠀⠀▪︎⠀⠀⠀Here , p denotes the numerator which is 15 .

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of \: p \:in \:Eq^n \: 1 \::}}$

$\qquad \dashrightarrow \pmb{ q \: \:\:=\: p \: + \: 7 \:} \qquad \:\bigg\lgroup \sf{ Eq^n \: 1 }\bigg\rgroup$

$\qquad \dashrightarrow \sf q \: \:\:=\: p \: + \: 7 \:$

$\qquad \dashrightarrow \sf q \: \:\:=\: 15 \: + \: 7 \:$

$\qquad \dashrightarrow \sf q \: \:\:=\: 22 \: + \: 7 \:$

$\qquad \therefore \pmb{\underline{\purple{\frak{\:q = 22 }} }}\:\:\bigstar$

• Here q denotes the denominator which is 22 .

⠀⠀⠀⠀⠀Now , Finding Original Fraction :

$\qquad \dashrightarrow \sf \:Original_{(Rational \;Number \:or \: Fraction \:)}\: \:= \bf \:\dfrac{\:\: p \:}{\:\:q\:}$

$\qquad \dashrightarrow \sf \:Original_{(Rational \;Number \:or \: Fraction \:)}\: \:= \bf \:\dfrac{\:\: 15 \:}{\:\:22\:}$

$\qquad \therefore \pmb{\underline{\purple{\frak{\: \:Original_{(Rational \;Number \:or \: Fraction \:)}\: \:= \:\dfrac{\:\: 15 \:}{\:\:22\:}}} }}\:\:\bigstar$

$\qquad \therefore \underline { \sf Hence, \: The \: Original \:number \: is \: \bf 15/22 \:.}$