Question

The sum of the digit of 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 two digit number is 10x + y.

Sum of the digits of to digit number is

x + y = 9 ………….. (i)

Number obtained by reversing the order of the digits, we get 10y + x.

Twitching the 10y + x, it is nine times of first number.

∴ 2(10y + x) = 9(10x + 4)

20y + 2x = 90x + 9y

2x – 90x + 20y – 9y = 0

-88x + 11y = 0

88x – 11y = 0

8x – y = 0 ………….. (ii)

By adding eqn. (i) to eqn. (ii)

$x + y = 9 \\ \underline {+ 8x - y = 0} \\ 9x \: \: \: \: \: = 9$

Substituting the value of ‘x’ in eqn. (i),

x + y = 9

1 + y = 9

y = 9 – 1

y = 8

∴ x = 1, y = 8

∴ Two digit number = 10x + y

= 10 × 1 + 8

= 10 + 8

=18

$so \: no. \: are \boxed{18}$

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