Question

When the digits in a two-digit number are interchanged, the sum of the two numbers is $88$. If the tens digit is $2$ more than the ones digit in the original number, what is the original number?

• A. $35$
• B. $42$
• C. $64$
• D. $53$

Answer 1

The correct option is D $53$

Let the ones digit of the original number be $x$.
The tens digit is $2$ more than the ones digit, $x+2$.

So the original number is
$10(x+2)+x$

When the digits are interchanged, the number becomes
$10x+(x+2)$

The sum of these two numbers is $88$.
⇒ $10(x+2)+x+10x+(x+2)=88$
⇒ $22x+22=88$
⇒ $22x=88-22=66$
⇒ $x=\frac{66}{22}=3$

Therefore, the original number is $10(x+2)+x=10\left(5\right)+3=53$