Question

the sum of the two digits of a two digit number is 9 if the difference between the number and the number formed by reversing the positions of its digits is 27, find the original number​

Answer 1

Answer:63

Step-by-step explanation:

Let x and y be the two digits of a two digit number .

Then the two digit number so formed will be (10x+y) where x is at 10's place and y is at unit place.

Now by question,

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

And, (10x+y)-(10y+x)=27......(2)

Here (10y+x) is the reversed two digit number.

From (2), 10x+y-10y-x=27

Or, 9x-9y=27

Or, x-y=3..............(3)

Now adding (1) & (3) we get

2x=12

Therefore x=6

Putting x=6 in (3) we get

6-y=3

Therefore y=3

Hence required two digit no. is (10x+y)=10×6+3=63

Answer 2

Let the units digit of the original number be x and

the tens digit of the original number be y.

Original number = 10y+x

On reversing the digits, you will get

New number = 10x+y.

Given that

(10x+y)-(10y+x) = 27

10x+y-10y-x=27

10x-x-10y+y=27

9x-9y=27

9 (x-y) = 27

x-y = 27/9=3

x-y = 3

x=y+3. -(1)

Substitute x value and you will get your answer

y = 6 and x = 3