Question

The tens digit is three less than the units digit. If the digits are reversed, the sum of the reversed number and the original number is 121. Find the original number.

Answer 1

Let the ten's and one's digit be x and y respectively.

Given

The tens digit is three less than the units digit.

$\implies x = y-3\\\implies x-y=-3--(1)$

If the digits are reversed, the sum of the reversed number and the original number is 121.

$\implies 10x+y+10y+x=121\\\implies 11x+11y=121\\\implies x +y=11--(2)$

Solving (1) and (2), we get,

$x-y=-3\\\underline{x+y=11}\\\underline{\underline{2x=8}}\\\implies x = 4\\\implies y = 7$

$\boxed{\boxed{\bold{Therefore, \ the \ number \ is \ 47}}}}}}}$

                                                             

Answer 2

Answer:

Let the Ones Digit be a, and Tens Digit be (a - 3) of the Original Number.

⠀⠀✩ Original Number : 10(a - 3) + a

⠀⠀✩ Reverse Number : 10a + (a - 3)

⠀

☯ $\underline{\boldsymbol{According\: to \:the\: Question :}}$

$\dashrightarrow\sf\:\: Original\:No.+Reverse\:No.=121\\\\\\\dashrightarrow\sf\:\:[10(a-3)+a]+[10a+(a-3)]=121\\\\\\\dashrightarrow\sf\:\:10a - 30 + a + 10a + a - 3 = 121\\\\\\\dashrightarrow\sf\:\:22a - 33 = 121\\\\\\\dashrightarrow\sf\:\:22a = 121 + 33\\\\\\\dashrightarrow\sf\:\:22a = 154\\\\\\\dashrightarrow\sf\:\:a = \dfrac{154}{22}\\\\\\\dashrightarrow\sf\:\:a = 7$

⠀⠀⠀$\rule{160}{1}$

☯ $\underline{\textsf{Original Number Formed :}}$

$:\implies\sf Original\:Number=[10(a-3)+a]\\\\\\:\implies\sf Original\:Number=[10(7-3)+7]\\\\\\:\implies\sf Original\:Number=[10(4)+7]\\\\\\:\implies\sf Original\:Number=40+7\\\\\\:\implies\underline{\boxed{\sf Original\:Number=47}}$

⠀

$\therefore\:\underline{\textsf{Hence, Original Number formed will be \textbf{47}}}.$