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The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number.

(A) 39 (B) 57 (C) 66 (D) 93

1 Answer

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Let two digit number be $ab$, it's actual value is $10a + b$

$a + b = 12$ (sum of digits is $12$)

Number formed by reversing the digits is greater than the original number by $54$

$\Rightarrow 10b +a - (10a + b) = 54$

$\Rightarrow b -a = 6$

Solving $a+b=12$ and $b - a = 6$, We get $a = 3,b = 9$

Thus $39$ is the required number.
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