Question

The sum of the digits of a two digit number is 9 on receiving its digits the new number obtained is 45 more than the original number find the number.

Answer 1

❍ Let's say, ten's place digit be x and unit place digit be y respectively.

Hence,

• Original number = (10x + y).

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀

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

• Sum of the digits of a two digit number is 9.

⠀⠀

$:\implies\sf x + y = 9 \\\\\\:\implies\sf y = 9 - x \qquad\qquad\bigg\lgroup\sf eq^{n}\;(1)\bigg\rgroup$

Also,

• After reversing the digits the new number obtained is 45 more than the original number. The number obtained by reversing the digits is — (10y + x).

⠀

$:\implies\sf 10y + x + 10x + y = 45 \\\\\\:\implies\sf 9y- 9x = 45 \\\\\\:\implies\sf 9\Big\lgroup\sf 9 - x\Big\rgroup - 9x = 45 \\\\\\:\implies\sf 9 - x - x = 5\\\\\\:\implies\sf 9 - 2x = 5\\\\\\:\implies\sf -2x = 5 - 9\\\\\\:\implies\sf -2x = -4\\\\\\:\implies\sf x = \cancel\dfrac{-4}{-2} \\\\\\:\implies\underline{\boxed{\pmb{\frak{x = 2}}}}\;\bigstar$

⠀⠀⠀⠀

⠀⠀⠀⠀$\underline{\bf{\dag} \:\mathfrak{Putting\;value\; of\:x\;in\;eq^{n}\;(1)\: :}}$⠀⠀⠀⠀

⠀⠀⠀⠀

$\dashrightarrow\sf y = 9 - x \\\\\\\dashrightarrow\sf y = 9 - 2 \\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{y = 7}}}}\;\bigstar$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

⠀⠀

❝O R I G I N A L⠀N U M B E R ❞

⠀⠀⠀

$\dashrightarrow\sf Original\;no. = 10x + y \\\\\\\dashrightarrow\sf Original\; no. = 10(2) + 7\\\\\\\dashrightarrow\sf Original\;no. = 20 + 7\\\\\\\dashrightarrow\underline{\boxed{\pmb{\frak{\purple{Original\;no. = 27}}}}}\;\bigstar$

⠀

$\therefore{\underline{\textsf{Hence, the required original number is \textbf{27}.}}}$

Answer 2

$\qquad \underline {\bf{\purple {\:\maltese \:Proper \:Question \:: \:}}}$

• The sum of digits of a two digit number is 9 . On reversing its digits the new number obtained is 45 more than the original number . Find the number .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given that , The sum of the digits of a two -- digit number is 9 & On reversing it's digits the new number obtained is 45 more than the original number .

Exigency To Find : The Original number ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Consider the the digit at ten's place & one's place be a & b , respectively.

Therefore,

⠀⠀⠀▪︎⠀ORIGINAL NUMBER : ( 10a + b ) .

As, Per the Question ;

⠀⠀CASE I : The sum of digits of a two digit number is 9 .

$\qquad \therefore \: \sf \Big\{ Digit_{(\:Ten's \:Place \:)} \:\Big\} \: + \:\Big\{ Digit_{(\:One's \:Place \:)} \:\Big\} \:=\:9\:$

$\qquad \dashrightarrow \: \sf \Big\{ Digit_{(\:Ten's \:Place \:)} \:\Big\} \: + \:\Big\{ Digit_{(\:One's \:Place \:)} \:\Big\} \:=\:9\:$

$\qquad \dashrightarrow \: \sf \Big\{ a \:\Big\} \: + \:\Big\{ b \:\Big\} \:=\:9\:$

$\qquad \dashrightarrow \: \sf \Big\{ a \:\Big\} \: + \:\Big\{ b \:\Big\} \:=\:9\:$

$\qquad \dashrightarrow \: \sf a \: \: + \:\:b \: \:=\:9\:$

$\qquad \dashrightarrow \: \sf b \: \: \:=\:9\:-\:a\:$

$\qquad \dashrightarrow \: \bf b \: \: \:=\:9\:-\:a\:\qquad \:\bigg\lgroup \sf{ Eq^n \: i }\bigg\rgroup$

⠀⠀CASE II : On reversing it's digits the new number obtained is 45 more than the original number .

On , reversing digits we get ,

• The digit at ten's place will be b &
• The digit at one's place will be a .

Therefore,

• NEW NUMBER = ( 10b + a )

$\qquad \therefore \: \sf \Big\{ New\:Number \:\Big\} \: \:=\:45\:+ \:\:\Big\{ Original \:Number \:\Big\}$

$\qquad \dashrightarrow \: \sf \Big\{ New\:Number \:\Big\} \: \:=\:45\:+ \:\:\Big\{Original \:Number \:\Big\}$

$\qquad \dashrightarrow \: \sf \Big\{ New \:Number \:\Big\} \: - \:\Big\{ Original \:Number \:\Big\} \:=\:45\:$

$\qquad \dashrightarrow \: \sf \Big\{ 10b + a \:\Big\} \: - \:\Big\{ 10a + b \:\Big\} \:=\:45\:$

$\qquad \dashrightarrow \: \sf 10b + a \:\: - \: 10a - b \: \:=\:45\:$

$\qquad \dashrightarrow \: \sf 9b - 9a \: \:=\:45\:$

⠀⠀[ Canceling each by 9 ]

$\qquad \dashrightarrow \: \sf 9b - 9a \: \:=\:45\:$

$\qquad \dashrightarrow \: \sf b - a \: \:=\:5\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Eq^n \: i \: as\: Value\:of \:b \::}}$

$\qquad \dashrightarrow \: \bf b \: \: \:=\:9\:-\:a\:\qquad \:\bigg\lgroup \sf{ Eq^n \: i }\bigg\rgroup$

$\qquad \dashrightarrow \: \sf b - a \: \:=\:5\:$

$\qquad \dashrightarrow \: \sf \Big\{ 9 - a \Big\} - a \: \:=\:5\:$

$\qquad \dashrightarrow \: \sf 9 - a - a \: \:=\:5\:$

$\qquad \dashrightarrow \: \sf 9 - 2a \: \:=\:5\:$

$\qquad \dashrightarrow \: \sf - 2a \: \:=\:5\:- 9$

$\qquad \dashrightarrow \: \sf - 2a \: \:=\:\:- 4$

$\qquad \dashrightarrow \: \sf a \: \:=\:\:\dfrac{- 4}{-2}$

$\qquad \dashrightarrow \: \sf a \: \:=\:\:2$

$\qquad \dashrightarrow \: \underline {\boxed {\pmb{\frak{\pink{ a \: ( \:or \:Ten's \:Place \:Digit \:)\: \:=\:\:2 }}}}}$

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

$\qquad \dashrightarrow \: \bf b \: \: \:=\:9\:-\:a\:\qquad \:\bigg\lgroup \sf{ Eq^n \: i }\bigg\rgroup$

$\qquad \dashrightarrow \: \sf b \: \: \:=\:9\:-\:a\:$

$\qquad \dashrightarrow \: \sf b \: \: \:=\:9\:-\:2\:$

$\qquad \dashrightarrow \: \sf b \: \: \:=\:7\:$

$\qquad \dashrightarrow \: \underline {\boxed {\pmb{\frak{\pink{ b \: ( \:or \:One's \:Place \:Digit \:)\: \:=\:\:7 }}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \underline {\:\pmb{\mathbb{ \maltese \:\:ORIGINAL \:NUMBER \:\::\:}}}$

$\qquad\sf \dashrightarrow Original \:Number \: \:=\: \Big\{ 10 a + b \:\Big\}\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \sf \dashrightarrow Original \:Number \: \:=\: ( 10 a + b \:)\:$

$\qquad\sf \dashrightarrow Original \:Number \: \:=\: ( 10 (2) + 7 \:)\:$

$\qquad\sf \dashrightarrow Original \:Number \: \:=\: ( 20 + 7 \:)\:$

$\qquad\sf \dashrightarrow Original \:Number \: \:=\: 27\:$

$\qquad \dashrightarrow \: \underline {\boxed {\pmb{\frak{\pink{ Original \:Number \: \:=\: 27 }}}}}$

$\qquad \therefore \: \underline {\sf Hence, \:The \:Original \:number \:is \:\pmb{\bf 27 } \:.}$