The sum of the digits of a two-digit number is $12$. When the digits are interchanged, the resulting number is $36$ more than the original number. What is the original two-digit number?

Let the digit in the tens place be $x$ and the digit in the units place be $y$.
The original number is $10x+y$.

From the first condition:

$$x+y=12\text{— (i)}$$

When the digits are interchanged, the new number is $10y+x$.
According to the second condition:

$$(10y+x)-(10x+y)=36$$

$$9y-9x=36$$

$$y-x=4\text{— (ii)}$$

Adding equations (i) and (ii):

$$(x+y)+(y-x)=12+4$$

$$2y=16⟹y=8$$

Substitute $y=8$ into equation (i):

$$x+8=12⟹x=4$$

Thus, the original two-digit number is $10(4)+8=48$.