Question

The sum of the digits of a two-digit number is 9. If the new number formed by reversing the digits is greater than the original number by 63, then find the original number.​

Answer 1

$\large\underline{\sf{Solution-}}$

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

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

Now, According to first condition

The sum of the digits of a two-digit number is 9.

$\rm :\longmapsto\:x + y = 9$

$\rm :\longmapsto\: \boxed{ \bf{ \: x = 9 - y }}- - - - (1)$

According to second condition

The new number formed by reversing the digits is greater than the original number by 63.

$\rm :\longmapsto\:10x + y = 63 + 10y + x$

$\rm :\longmapsto\:10x + y - 10y - x= 63$

$\rm :\longmapsto\:9x - 9y = 63$

$\rm :\longmapsto\:9(x - y) = 63$

$\rm :\longmapsto\:x - y= 7$

$\rm :\longmapsto\:9 - y - y= 7$

$\rm :\longmapsto\:9 - 2y= 7$

$\rm :\longmapsto\: - 2y= 7 - 9$

$\rm :\longmapsto\: - 2y= - 2$

$\bf\implies \:y = 1$

On substituting y = 1 in equation (1), we get

$\rm :\longmapsto\:x = 9 - 1$

$\bf\implies \:x = 8$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So-\begin{cases} &\sf{digit \: at \: tens \: place \: = \: 1} \\ &\sf{digits \: at \: ones \: place \: = \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 10y + x = 10 \times 1 + 8 = 18} \\ &\sf{reverse \: number = 10 x + y = 10 \times 8 + 1 = 81} \end{cases}\end{gathered}\end{gathered}$

• Hence, Number formed = 18

Alternative Method :-

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{digit \: at \: tens \: place \: be \: x} \\ &\sf{digits \: at \: ones \: place \: be \: 9 - x} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 10 \times x + 9 - x = 9x + 9} \\ &\sf{reverse \: number = 10 (9 - x) + x = 90 - 9x} \end{cases}\end{gathered}\end{gathered}$

According to statement,

The new number formed by reversing the digits is greater than the original number by 63.

$\rm :\longmapsto\:90 - 9x = 63 + 9x + 9$

$\rm :\longmapsto\:90 - 9x = 72 + 9x$

$\rm :\longmapsto\: - 9x- 9x = 72 - 90$

$\rm :\longmapsto\: - 18x = - 18$

$\bf\implies \:x = 1$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So-\begin{cases} &\sf{digit \: at \: tens \: place \: = \: 1} \\ &\sf{digits \: at \: ones \: place \: = \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 18} \\ &\sf{reverse \: number = 81} \end{cases}\end{gathered}\end{gathered}$

• Hence, Number formed = 18