Question

The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.

​

Answer 1

Step-by-step explanation:

Let x be the digit at unit’s place and y be the digit at ten’s place.

Since y is at ten’s place, then the number formed is 10y+x.

By reversing the digits, it becomes 10x+y.

As the difference of the numbers is 18, so,

(10y+x)−(10x+y)=18

9(y−x)=18

y−x=2 .. (1)

As the sum of digits is 8, so,

x+y=8 (2)

On adding equations (1) and (2), we get

2y=10⇒y=5

Putting this in (2), we get x=8−5=3

x=3,y=5

Hence, number =10y+x=10×5+3=53.

Answer 2

AnswEr :

53.

$\bf{\green{\underline{\underline{\bf{Given\::}}}}}$

The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18.

$\bf{\green{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

$\bf{\green{\underline{\underline{\bf{Explanation\::}}}}}$

Let the digit on ten's place be r

Let the digit on one's place be m

$\leadsto{\underline{\sf{The\:Original\:number\:=\:\red{10r+m}}}}\\\leadsto{\underline{\sf{The\:Reversed\:number\:=\:\red{10m+r}}}}$

$\bf{\orange{\underline{\underline{\tt{A.T.Q\::}}}}}$

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

&

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

$\star$ Putting the value of m in equation (1), we get;

$\mapsto\sf{r=8-3}\\\\\mapsto\sf{\green{r=5}}$

Thus,

$\underbrace{\sf{The\:original\:number\:=\:10r+m=10(5)+3=50+3=\red{53.}}}$