Question

The sum of 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

Answer:

Let tge unit digit and tens digits of tge number be x and y.

Number = 10y + x

The number after reversing the digits= 10x + y

So,

x+ y = 9. (i)

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

88y - 11x = 0

-x+8y = 0 (ii)

Adding equations (i) and (ii) we get

9y=9

y= 1

putting this value in the first equation we get,

x= 8

Hence the number is eqaul to

10y + x = 10 × 1 + 8

= 18

Hope it helps

Answer 2

Answer:

Let the unit digit and tens digit of a number be x and y respectively

According to the question

Then the number = 10y+x

Number after reversing the digits = 10x+y

Given the sum of two-digit number = 9

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

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

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

=> 90y+9x = 20x+2y

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

=> 88y - 11x = 0

Dividing the equation by 11

=> 8y - x = 0.......(ii)

Adding equation (i) and (ii)

x + y = 9

-x + 8y = 0

___________

9y = 9

y = 9/9

y = 1

Substituting , y = 1 in equation (i)

x+y = 9

x+ 1 = 9

x = 9-1

x = 8

Now, we got x = 8 and y=1

x is the unit digit and y is the tens digit

Therefore the number is 18

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