Question

The sum of the digits of a two -digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digit. Find the number.

Answer 1

Let the no. in units place be x.
and the no. in tens place be (9-x).

The original no. = 10 (9-x)+x
= 90-9x
According to the question,
=>9 (90-9x)=2{10x+(9-x)}
=>810-81x=2 (9x+9)
=>-81x-18x= 18-810
=> -99x= -792
=>x= -792/-99
=>x=8

Therefore ,the No. = 10 (9-8)+8
=18

Hope this will help.

Answer 2

Answer:

the answer is 18

Step-by-step explanation:

Let the no. in units place be x.

and the no. in tens place be (9-x).

The original no. = 10 (9-x)+x

= 90-9x

NOW,

9 (90-9x)=2{10x+(9-x)}

810-81x=2 (9x+9)

-81x-18x= 18-810

-99x= -792

x= -792/-99

x=8

Therefore ,the No. = 10 (9-8)+8

=18