Question

Sum of the digits of a two-digit number is 9. The number obtained by interchanging
the digits exceeds the given number by 27. Find the original number.​

Answer 1

Solution :

We suppose the ten's place digit be r & one's place digit be m respectively;

$\boxed{\bf{Original\:number=10r+m}}$

$\boxed{\bf{Reversed\:number=10m+r}}$

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

$\underbrace{\bf{1^{st}\:Condition\::}}$

$\mapsto\tt{r+m=9}$

$\mapsto\tt{r=9-m......................(1)}$

$\underbrace{\bf{2^{nd}\:Condition\::}}$

$\mapsto\tt{(Interchange\:number) =(Original\:number) +27}$

$\mapsto\tt{10m+r = 10r + m + 27}$

$\mapsto\tt{10m-m + r-10r = 27}$

$\mapsto\tt{9m -9r = 27}$

$\mapsto\tt{9(m-r) = 27}$

$\mapsto\tt{m-r = \cancel{27/9}}$

$\mapsto\tt{m-r =3}$

$\mapsto\tt{m-(9-m) = 3\:\:\:\:[from(1)]}$

$\mapsto\tt{m-9+m=3}$

$\mapsto\tt{2m=3+9}$

$\mapsto\tt{2m=12}$

$\mapsto\tt{m=\cancel{12/2}}$

$\mapsto\bf{m=6}$

∴ Putting the value of m in equation (1),we get;

$\mapsto\tt{r=9-6}$

$\mapsto\bf{r=3}$

Thus;

The original number = 10r + m

The original number = 10(3) + 6

The original number = 30 + 6

The original number = 36 .