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 digits. Find the number ​

Answer 1

let the two digit number be 10×+y

from the given condition

10x+y=9

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

90x+9y=20y+2x

90x-2x=20y-9y

88x=11y

88x=11y

88/11x=y

8x=y

sub y=8x in 10x+y=9

10x+8x=9

18x=9

x=1/2

sub x=1/2 in y=8x

y=8×1/2

y=4

the number is 10x+ y,

10×1/2+4=5+4=9

Answer 2

Answer:

Let the unit digit be x and tens digit be y.

So, the number will be = 10y + x

After interchanging the digits the number becomes = 10x + y

It is given that, The sum of the digits of two digit number is 9. Therefore we get :]

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

It is also given that, Nine times this number is twice the number obtained by reversing the order of the digits :]

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

90y + 9x = 20x + 2y

90y - 2y = 20x - 9x

188y = 11x

x = 8y.......(Equation ii)

____________________

Now,Putting the value of x = 8y in equation (i) we get :]

x + y = 9

8y + y = 9

9y = 9

y = 1

_____________________

Now, substitute the value of y in equation (ii) we get :]

x = 8y

x = 8(1)

x = 8

Therefore

The original number will be 10y + x = 10(1) + 8 = 18