Question

12) The sum of the digits of a 2 digit number is 8. The number obtained by
interchanging the digits exceeds the original number by 18. Find the number.​

Answer 1

Let the digits at ones place be x. Then,

the digits at tens place = (8-x)

Original number = 10(8-x) + x

= 80 - 10x + x

= 80 - 9x

On interchanging the digits

new number obtained = 10x + 8-x

= 9x + 8

According to question,

New number - Original number = 18

9x + 8 - (80-9x) = 18

=> 9x + 8 - 80 + 9x = 18

=> 18x - 72 = 18

=> 18x = 18 + 72

=> 18x = 90

=> x = 90/18

=> x = 5

Hence, the digits at ones place is 5.

The digits at tens place = (8-5) = 3.

So, the original number is 35 and the new number is 53.

Thanks

Answer 2

S O L U T I O N :

Let the ten's place digit be r

Let the one's place digit be m

$\boxed{\bf{The\:original\:number=10r+m}}}}}\\\boxed{\bf{The\:reversed\:number=10m+r}}}}}$

A/q

$\longrightarrow\sf{r+m=8}\\\\\longrightarrow\sf{m=8-r.......................(1)}$

&

$\longrightarrow\sf{10m+r-(10r+m)=18}\\\\\longrightarrow\sf{10m+r-10r-m=18}\\\\\longrightarrow\sf{10m-m+r-10r=18}\\\\\longrightarrow\sf{9m-9r=18}\\\\\longrightarrow\sf{9(m-r)=18}\\\\\longrightarrow\sf{m-r=\cancel{18/9}}\\\\\longrightarrow\sf{m-r=2}\\\\\longrightarrow\sf{8-r-r=2\:\:[from(1)]}\\\\\longrightarrow\sf{8-2r=2}\\\\\longrightarrow\sf{-2r=2-8}\\\\\longrightarrow\sf{-2r=-6}\\\\\longrightarrow\sf{r=\cancel{-6/-2}}\\\\\longrightarrow\bf{r=3}$

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

$\longrightarrow\sf{m=8-3}\\\\\longrightarrow\bf{m=5}$

Thus;

$\underbrace{\sf{The\:number\:(10r+m)=[10(3)+5]=[30+5]=\boxed{\bf{35}}}}}}$