Question

A number consists of two digits whose sum is 9 if 27 subtracted from the number then the digits of numbers are reversed then find the numbers

Answer 1

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

Given

A number consists of two digits whose sum is 9.

$\implies x + y = 9--(1)$

If 27 is subtracted from the number, then the digits of numbers are reversed.

$\implies 10x+y-27=10y+x\\\implies 10x-x+y-10y=27\\\implies 9x-9y=27\\\implies x -y=3--(2)$

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

$x+y=9\\\underline{x-y=3}\\\underline{\underline{2x=12}}\\\implies x = 6\\\implies y = 3$

$\boxed{\boxed{\bold{Therefore, \ the \ required \ number \ is \ 63}}}}}}}$

                                                       

Answer 2

$\sf Let \ the \ digit \ in \ ten's \ place \ be \ x \ and \ digit \ in \ one's \ place \ be \ y \ .\\\\\sf Original \ number = 10x+y \\\\\sf Reversed \ number = 10y+x \\\\\sf According \ to \ the \ first \ condition\\\\ \sf\longrightarrow x+y=9 \ \ \ \ \ \ \ \ \ \ \ ........(1)\\\\\\ \sf According \ to \ the \ second \ condition \\\\\sf \Longrightarrow 10x + y - 27 = 10y + x\\\\\sf\Longrightarrow 9x - 9y = 27\\\\\sf\Longrightarrow x-y=3\ \ \ \ \ \ \ \ \ \ \ \ ...........(2)\\\\\sf Eq(1) + Eq(2)$

$\sf\longrightarrow 2x = 12\\\\\sf\longrightarrow \bf x=6$

$\sf y = 9-6=\bf 3$

$\bf\LARGE Two \ digit \ number = 63$