The sum of the digits of a two digit number is $y$ when the digit are reversed, then the number is decreased by $9$ . Find the number.

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Solution

Let required number at two digit number 10y+x and reverse number 10x+y
$∴$ $\begin{matrix} x + y = y x = 0 \end{matrix}$
$∴$ $\begin{matrix} 10 x + y = 10 y + x - 9 0 + y = 10 y + 0 - 9 y - 10 y = - 9 - 9 y = - 9 y = 1 \end{matrix}$
$∴$ Number = $10$

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The sum of the digits of a two digit number is y when the digit are reversed, then the number is decreased by 9. Find the number.

Let required number at two digit number 10y+x and reverse number 10x+y∴ x+y=yx=0∴ 10x+y=10y+x−90+y=10y+0−9y−10y=−9−9y=−9y=1∴ Number = 10

The sum of the digits of a two digit number is $y$ when the digit are reversed, then the number is decreased by $9$ . Find the number.

Let required number at two digit number 10y+x and reverse number 10x+y
$∴$ $\begin{matrix} x + y = y x = 0 \end{matrix}$
$∴$ $\begin{matrix} 10 x + y = 10 y + x - 9 0 + y = 10 y + 0 - 9 y - 10 y = - 9 - 9 y = - 9 y = 1 \end{matrix}$
$∴$ Number = $10$