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

number obtained by reversing the order of the digits. Find the number.

Let unit digit  = x

Tens digit  = y

Number will 10 times the tens digit + unit times the unit digit

Hence number will 10 y + x


Sum of digits are 9

So that

X + y  = 9        ………….(1)

 nine times this number is twice the number obtained by reversing the order of the digits


9 (10 y + x )  =  2 (10 x + y )

90 y + 9 x  = 20 x + 2y

88 y – 11 x  = 0

Divide by 11 we get

8 y  - x  = 0     …………..(2)

X + y  = 9        ………….(1)


Adding both equations we get

9 y = 9

Y = 9/9 = 1

Plug this value in equation first we get

X+ y = 9

X + 1 = 9

X = 8

So our original number is 10 y  + x    = 10*1 + 8 = 18

Answer 2

Here is your solution

Given :-

The sum of the digits of a two digit number is 9

Let
The digits be x and y, and the number be 10x + y.

I. e

x + y = 9 ------ (1)

Now
A/q

9×(10x + y) = 2×(10y + x)

90x + 9y = 20y + 2x

88x = 11y

y = 8x ---- (2)

putting the value of y in equation (1.) we get,

x + 8x = 9
9x = 9.
x = 1

y = 8×1 = 8.

Hence

The number is 10x +y = 10×1 + 8 = 18.

Hope it helps you