Question

The units digit of a two digit number is twice its tens digit. If the number formed by reversing the digits is added to the original number the sum is 132. Find the original number​

Answer 1

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Answer 2

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

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{digit \: at \: tens \: place \: be \: x} \\ &\sf{digits \: at \: ones \: place \: be \: 2x} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: So-\begin{cases} &\sf{number \: formed = 10 \times x + 2x \times 1 = 12x} \\ &\sf{reverse \: number = 10 \times 2x + x \times 1 = 21x} \end{cases}\end{gathered}\end{gathered}$

According to statement,

The number formed by reversing the digits is added to the original number the sum is 132.

$\rm :\longmapsto\:21x + 12x = 132$

$\rm :\longmapsto\:33x = 132$

$\bf\implies \:x = 4$

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{digit \: at \: tens \: place \: be \: 4} \\ &\sf{digits \: at \: ones \: place \: be \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So- \begin{cases} &\sf{number \: formed = 12 \times 4 = 48}\end{cases}\end{gathered}\end{gathered}$

Aliter Method :-

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

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

According to statement,

The unit digit of a two digit number is twice its tens digit.

$\rm :\implies\:\boxed{ \rm{ y\: = \: 2x}} - - - (1)$

According to statement again,

The number formed by reversing the digits is added to the original number the sum is 132.

$\rm :\longmapsto\:10y + x + 10x + y = 132$

$\rm :\longmapsto\:11x + 11y = 132$

$\rm :\longmapsto\:11(x + y) = 132$

$\rm :\longmapsto\:x + y = 12$

$\rm :\longmapsto\:x + 2x = 12 \: \: \: \: \{ \: using \: (1) \: \}$

$\rm :\longmapsto\:3x = 12$

$\bf\implies \:x = 4$

On substituting x = 4, in equation (1), we get

$\bf\implies \:y = 8$

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{digit \: at \: tens \: place \: be \: 4} \\ &\sf{digits \: at \: ones \: place \: be \: 8} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: So- \begin{cases} &\sf{number \: formed = 10 \times 4 + 8 \times1 = 48}\end{cases}\end{gathered}\end{gathered}$