Question

the sum of digit of the sum of digit of a two digit number is 9 also nine times this number is twice the number obtained by reversing the order of digit ​

Answer 1

Answer:

Let tens digit no.be X and unit digit no.be Y 

A/q

x+y=9

and 9 (10×x+y)=2 (10×y+x)

90x+9y=20y+2x

88x-11y=0

from 1st 

11x+11y=99

88x-11y=0

......

99x=99

x=1

and y=8 hence the no.is 18

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

Given :-

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

To Find :-

• The Number

Let the unit digit and tens digits of the number be x and y                  

Number = 10y + x                  

Number after reversing the digits = 10x + y    

A.T.Q                       

⇒  x + y = 9 ... (i)                  

⇒  9(10y + x) = 2(10x + y)                  

⇒  88y - 11x = 0                  

⇒ -x + 8y =0 ... (ii)                  

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

⇒  9y = 9                  

⇒  y = 1 ... (iii)                  

Putting the value in equation (i), we get                  

⇒  x = 8                

Hence

the number is 10y + x = 10 × 1 + 8 = 18.