The digit of tens place of a 2 digit is 3 times the digit at a unit place of the digit are reversed, the new number will be 36 less the original number. Find the originalnumber.

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Solution

The digit in unts digit be $x$
Unit in tens digit is $3 x$
The numbers is $3 x \times 10 + x 31 x$
The digits are reversed
So the number is $x \times 10 + 3 x = 13 x$
The difference is $36$
$⟹ 31 x - 13 x = 36 18 x = 36 x = 2$
The original number is $31 (2) = 62$

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The digit of tens place of a 2 digit is 3 times the digit at a unit place of the digit are reversed, the new number will be 36 less the original number. Find the originalnumber.

The digit in unts digit be xUnit in tens digit is 3xThe numbers is 3x×10+x31xThe digits are reversed So the number is x×10+3x=13xThe difference is 36⟹31x−13x=3618x=36x=2The original number is 31(2)=62

The digit in unts digit be $x$
Unit in tens digit is $3 x$
The numbers is $3 x \times 10 + x 31 x$
The digits are reversed
So the number is $x \times 10 + 3 x = 13 x$
The difference is $36$
$⟹ 31 x - 13 x = 36 18 x = 36 x = 2$
The original number is $31 (2) = 62$