Question

The sum of the digits of a two digit number is $5$. If the digits are reversed, the new number is $27$ less than the original. Find the number.

• A. $11$
• B. $12$
• C. $13$
• D. $41$

Answer 1

The correct option is D $41$

Let the unit's place number be $x$ and the tenth place number be $y$ of the two digit number

So the number will be $10y+x$

And the number formed by reversing the digits will be $10x+y$

Now given, sum of digits $=x+y=5$......(1)

And $10x+y=10y+x-27$

$\Rightarrow 9y-9x=27$

$\Rightarrow y-x=3$, divide both sides by $9$

$\Rightarrow y-(5-y)=3$, use (1)

$\Rightarrow 2y=3+5=8$

$\Rightarrow y=4$, so $x=1$

Hence the number is $=10y+x=10\times 4+1=41$