Question

What is the original number, if the sum of the digits of a two-digit number is $7$. By interchanging the digits is $27$ more than the original number?

• A. $43$
• B. $25$
• C. $61$
• D. $80$

Answer 1

The correct option is B $25$

Let the original number be $10a+b$.
Then, $‘a^{\prime }$ is the tens digit and $‘b^{\prime }$ is the unit digit.
Since the sum of the digits is $7$.
Therefore, $a+b=7$
$b=7–a$
So, the original number is $10a+(7–a)$.
Therefore, the number obtained by interchanging the digits is
$10(7–a)+a$,
and so we have $\left\{10\right(7-a)+a\}-\{10a+(7-a\left)\right\}=27$
Solving this equation, we get
$a=2$
And so, $b=7–2=5$
Therefore, the original number is $10a+b=20+5=25$.