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The sum of the digits of a $2$-digit number is $12$. Seven times the number is equal to four times the number obtained by reversing the order of the digits. Find the number.
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Let the unit's place digit be $x$ and ten's place digit be $y$
$\therefore$ Number $= 10y + x$
According to question,
$x + y = 12$ ...(i)
and $7(10y + x) = 4(10x + y)$
$x - 2y = 0$ ...(ii)
Solving (i) and (ii), we get
$x = 8$ and $y = 4$
Hence, the required number is $48$
$\therefore$ Number $= 10y + x$
According to question,
$x + y = 12$ ...(i)
and $7(10y + x) = 4(10x + y)$
$x - 2y = 0$ ...(ii)
Solving (i) and (ii), we get
$x = 8$ and $y = 4$
Hence, the required number is $48$