Question

A number consists of two digits whose sum is 9. IF 27 is subtracted from the number its
digits are reversed. Find the number ? ( please give ans fast ) ​

Answer 1

Answer:

Step-by-step explanation :-

Let the ones digit be x and the tens digit be y.

Now, A/Q,

°•° x + y = 9................(i)

Original number = 10y + x .

And, the number obtained on reversing the digits = 10x + y .

And,

°•°10y + x - 27 = 10x + y

==> 10y - y + x - 10x = 27

==> 9y - 9x = 27

==> 9 ( y - x ) = 27

==> y - x = 3...............(ii)

Now, add in eq. (i) and (ii), we get

x + y = 9

- x + y = 3

-....+......+

----------------

==> 2y = 12 .

•°• y = 6 .

Now, put the value of y = 6 in eq. (i) , we get

==> x + y = 9 .

==> x + 6 = 9 .

==> x = 9 - 6 .

\therefore∴ x = 3 .

Therefore, original Number = 10y + x .

= 10 ( 6 ) + 3 .

= 60 + 3 .

= 63.

Hence, The required number is 63

Answer 2

Let the ones digit be x.

So tens digit = (9-x)

According to condition,

{10(9-x)+x}-{10x+(9-x)} = 27

or, (90-10x+9)-(10x+9-x) = 27

or, (90-9x)-(9x+9) = 27

or, 90-9x-9x-9 = 27

or, -18x+81 = 27

or, -18x = 27-81

or, -18x = -54

or, 18x = 54

or, x = 54/18

or, x = 3

So, ones digit = x = 3.

And tens digit = (9-x) = (9-3) = 6.

Hence, the number is 63.