Question

The Sum Of Two Digits Is 9. Also , 9 times this number is twice the number obtained by reversing the order of the digits. Find the number.

Answer It Briefly On The Copy Or Sheet...​

Answer 1

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

Given that,

↝  Sum of two digits of a 2 digit number is 9.

Let assume that,

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

So,

Number formed = 1 × x + 10 × (9 - x) = x + 90 - 10x = 90 - 9x

Reverse number = 10 × x + (9 - x) × 1 = 10x + 9 - x = 9x + 9

Thus,

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

According to statement,

↝  9 times this number is twice the number obtained by reversing the order of the digits.

$\rm :\longmapsto\:9(90 - 9x) = 2(9 + 9x)$

$\rm :\longmapsto\:9(90 - 9x) = 2 \times 9(1 + x)$

$\rm :\longmapsto\:90 - 9x = 2 (1 + x)$

$\rm :\longmapsto\:90 - 9x = 2 + 2x$

$\rm :\longmapsto\: 2x+ 9x =90 - 2$

$\rm :\longmapsto\:11x =88$

$\bf\implies \:x = 8$

Hence,

• Number formed = 90 - 9x = 90 - 72 = 18