In a two digit number, the ten's digit is three times the unit's digit. When the number is decreased by 54, the digits are reversed. Find the number.

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Solution

Let the digit in the unit;s place be $x$ and the digit in ten's place be $y$ . Then,
Number = $10 y + x$
According to question,
$y = 3 x$
Also,
$10 y + x - 54 = 10 x + y$
$x - y = - 6$
Solving these equations, we get,
$x = 3$ and $y = 9$
Hence, number is $10 y + x = 93$

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In a two digit number, the ten's digit is three times the unit's digit. When the number is decreased by 54, the digits are reversed. Find the number.

Let the digit in the unit;s place be x and the digit in ten's place be y. Then,Number = 10y+xAccording to question,y=3xAlso,10y+x−54=10x+yx−y=−6Solving these equations, we get,x=3 and y=9Hence, number is 10y+x=93

Let the digit in the unit;s place be $x$ and the digit in ten's place be $y$ . Then,
Number = $10 y + x$
According to question,
$y = 3 x$
Also,
$10 y + x - 54 = 10 x + y$
$x - y = - 6$
Solving these equations, we get,
$x = 3$ and $y = 9$
Hence, number is $10 y + x = 93$