Question

The sum of a two digit number and the number obtained by reversing its digit is 143 . Find the number , if its tens place digit is greater than the units place digit by 3​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Let \: - \begin{cases} &\sf{digits \: at \: ones \: place \: be \: x} \\ &\sf{digits \: at \: tens \: place \: be \: y} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So \: - \begin{cases} &\sf{number \: is \: 10y + x} \\ &\sf{reverse \: number \: is \: 10x + y} \end{cases}\end{gathered}\end{gathered}$

$\bullet \: \red{ \bf \: According \: to \: question \: }$

• The sum of a two digit number and the number obtained by reversing its digit is 143.

$\rm : \implies \:10y + x + 10x + y = 143$

$\rm : \implies \:11x + 11y = 143$

$\rm : \implies \:x + y = 13$

$\bigstar \: \: \: \boxed{ \pink{ \: \rm : \implies \:x \: = \: 13 \: - \: y}} - - (i)$

Again,

$\bullet \: \red{ \bf \: According \: to \: question \: }$

• Tens place digit is greater than the units place digit by 3.

$\rm : \implies \:y \: = \: x \: + \: 3$

On substituting the value of x, from equation (i), we get

$\rm : \implies \:y = 13 - y + 3$

$\rm : \implies \:2y = 16$

$\bigstar \: \: \: \boxed{ \pink{ \: \rm : \implies \:y \: = \: 8}}$

Put value of y in equation (i), we get

$\bigstar \: \: \: \boxed{ \pink{ \: \rm : \implies \:x \: = \: 5}}$

Hence,

• Two digit number is

$\rm \: \rightarrow \: 10y + x = 10 \times 8 + 5 = 85$