Question

The old sums of digits of the two digit number is 9. Also, nine times this number is twice the number the number obtained by reversing the order of the digits. Find the number.​

Answer 1

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Step-by-step explanation:

$\rightarrow \bold{Let \: the \: digit \: at \: tens \: place \: be \: x. \:} \\ \rightarrow \bold{ The \: digit \:at \: unit \: place \: be \: y.}$

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$\: \large\bold{Original \: number = 10x+y} \:$

$\large\mathtt{x + y = 9 \: \: \: eq(1)}$

$\: \large\bold{If \: the \: order \: of \: the \: digits \: are \: reversed..} \:$

$\rightarrow \: \bold{The \: digit \: at \: tens \: place \: be \: y.} \\ \rightarrow \: \bold{And \: the \: digit \: at \: unit \: place \: be \: x.}$

$\: \large\bold{New \: no. 10y+x} \:$

$\rightarrow \: \mathtt{9(10x + y)} = \mathtt{2(10y + x)} \\ \rightarrow \mathtt{90x + 9y = 20y + 2x} \\ \rightarrow \: \mathtt{90x - 2x = 20y - 9y} \\ \rightarrow \mathtt{ \cancel{88x} = \cancel{11y}} \\ \rightarrow \: \mathtt{8x = y} \\ \rightarrow \: \fbox\mathtt{8x - y = 0 \: \: eq(2)}$

$\: \: \: \mathtt{x + \cancel{y}= 9} \\ \mathtt{\frac{8x - \cancel{y} = 0} {9x = 9} } \\ \: \: \: \fbox\mathtt{x = \frac{ \cancel{9}}{ \cancel{9}} }$

$\bold{Put \: x = 1 \: \: eq(1)}$

$\rightarrow \: \mathtt{x + y = 9} \\ \: \: \: \: \: \: \mathtt{1 + y = 9} \\ \: \: \: \: \: \: \mathtt{y = 9 - 1} \\ \: \: \: \: \: \: \fbox\mathtt{y = 8}$

$\rightarrow \: \mathtt{Original \: no. \: = 10x + y} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt{10(1) + 8} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt{10 + 8} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \fbox \mathtt{18}$

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