Question

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

Answer 1

Answer:

18

Step-by-step explanation:

let the digit at tens place= x

let the digit at ones place= y

number= 10x + y

acc to the ques:

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

on reversing: 10y+x

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

90x+9y = 20y + 2x

88x = 11y

8x = y ----------2)

after substituting:

x + 8x = 9

9x = 9

x=1

y = 8x

y = 8 × 1

y = 8

Number= 10x+y

=10×1+8

=18

Answer 2

Answer:

The two digit number is 18.

Step-by-step explanation:

Let the two-digit number be, 10x + y.

Condition 1: x + y = 9...(i)

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

Solve the equation in condition 2:

$9(10x+y)=2(10y+x)\\90x+9y=20y+2x\\88x-11y=0...(ii)$

Solve equations (i) and (ii) simultaneously:

x + y = 9      } × 11

88x - 11y = 0

⇒

11x + 11y = 99

88x - 11y = 0

99x = 99

x = 1

Substitute x = 1 in equation (i) and solve for y as follows:

x + y = 9

y = 9 - x

  = 9 - 1

  = 8

y = 8.

Thus, the number is:  

10x + y = (10 × 1) + 8

            = 10 + 8

            = 18.

Check:

     9 (10x + y) = 2 (10y + x)

9 ((10 × 1) + 8) = 2 ((10 × 8) + 1)

            9 × 18 = 2 × 81

                162 = 162

Thus, the two digit number is 18.