Question

The sum of the digits of a two-digit number is $9$. When we interchange the digits it is found that the resulting new number is greater than the original number by $27$. What is the two-digit number?

Answer 1

Let the digits at tens place and ones place be $x$ and $(9-x)$ respectively.

Therefore, original number $=10x+(9-x)=9x+9$

On interchanging the digits, the digits at one's place and tens place will be $x$ and $(9-x)$ respectively.

Therefore, new number after interchanging the digits $=10(9-x)+x$

$=90-10x+x$

$=90-9x$

According to the given question,

New number $=$ Original number $+27$

$90-9x=9x+9+27$

$90-9x=9x+36$

Transposing $9x$ to R.H.S and 36 to L.H.S, we obtain

$90-36=18x$

$54=18x$

Dividing both sides by $18$, we obtain

$3=x$ and $9-x=6$

Hence, the digits at tens place and one's place of the number are $3$ and $6$ respectively.

Therefore, the two-digit number is $9x+9=9\times 3+9=36$