A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Given :

A number consists of two digits whose sum is five.

When the digits are reversed, the number becomes greater by nine. 

To find :

We have to find the number.

Solution :

Let the number be $10x+y$.

$x+y=5$.....(i)

It is given that when the digits are reversed, the new number increases by 9.

Therefore,

$10y+x= (10x+y)+9$

$10y+x-10x-y = 9$

$10y-y+x-10x= 9$

$9y-x(10-1)=9$

$9y-9x=9$

$9(y-x)=9$

$y-x=\frac{9}{9}$

$y-x=1$.....(ii)

Adding equations (i) and (ii), we get,

$y-x+x+y=1+5$

$2y=6$

$y=\frac{6}{2}$

$y=3$

Therefore,

$x+3=5$

$x=5-3$

$x=2$

The original number is $10(2)+3=20+3=23$.

Therefore, the required number is 23. 

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

1K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started