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 unit digit be x
and ten's digit be y


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

Original number :- 10y + x
Reversed number :- 10x + y


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

90y + 9x = 20x + 2y

90y - 2y + 9x - 20x = 0

88y - 11x = 0

( dividing by 11 )

8y - x = 0

- x = - 8y

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

Putting value of x from ( ii ) in ( i )

x + y = 9

8y + y = 9

9y = 9

y = 1

Putting value of y in ( ii )

x = 8y

x = 8


Original number :- 10y + x

10 × 1 + 8

10 + 8

18

So,

18 is the required number !!

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