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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.

1. $39$
2. $57$
3. $66$
4. $93$

Let the two digits be $x$ and $y$.

Original number$=xy=10x+y$

$x + y = 12 \quad \to(1)$

Given that new number formed by reversing the digits is greater than the original number by $54,$

$y * 10 + x = 10 * x + y + 54$
$10 y + x = 10 x + y +54$
$9y-9x = 54$
$x - y = -6 \quad \to(2)$

From $(1)$ and $(2)$

$2 x = 6$
$x = 3$ and $y = 9$

Hence, Original number $=xy=10x+y= 39$

Correct Answer: $A$
by
Original no is 54 less than Reversed no.
66 Not possible as reversal of  66 is 66 itself.
93 and 57 also not possible because by adding 54  it will become 3 digit no.

Only option left is 39, which is correct answer. :-P
i.e. 39+54=93
by
x+y=12

10y+x=54+10x+y

solving the above equation we will get 9x-9y=-54

solving both equations we will get x=3,y=9