In a two-digit number, the units digit is twice the ten's digit. If $27$ is added to the number, the digits are reversed. Find the number.

The correct option is D: $36$

Let the digit in the units place be $x$ and digit in the tens place be $y.$

According to first condition,we have

$\Rightarrow x=2y\dots \dots \left(i\right)$ [$\because$ units digit $=2\times$tens digit]

$\therefore$ the two digit number is $10y+x.$

When the digits are reversed, the reversed number will be $10x+y$

When $27$ is added, the digits are reversed. [Given]

Original number $+27=$ Reversed number

$\therefore 10y+x+27=10x+y$

$\Rightarrow 9x-9y=27$

$\Rightarrow x-y=3\dots \dots \left(ii\right)$

Putting $x=2y$ in equation $\left(ii\right),$ we get

$2y-y=3$

$\Rightarrow y=3$

Putting $y=3$ in equation $\left(ii\right),$ we get

$x-3=3$

$\Rightarrow x=6$

Original number $=10y+x=10\times 3+6=36$

Hence, the number is $36.$