Question

the sum of a 2 digit number and the number formed by reversing the digits is 99 . the difference between the digit is 3. find the number....

Answer 1

Let the tens place digit be a

And Unit place digit be b

According to first condition,

10a + b + 10b + a = 99

=> 11a + 11b = 99

=> a + b = 9 --------(1)

Now,

According to second condition,

a - b = 3 -------(2)

On adding equation 1 and 2, we get

2a = 12

=> a = 6

On Substituting the value of a in equation 1, we get

6 +b = 9

=> b = 3

Required number = 63

Note :-

The number can be 63 and 36 both because both numbers satisfy the given conditions.

Answer 2

Let the digit in the tens place be x
Digit in ones place be y
x - y = 3
-y = 3 - x
y = x - 3
10x + y  +  10y + x = 99
Substituting,
10x + x - 3 +10(X - 3) + x = 99
10 x + x - 3 + 10x - 30 + x = 99
22x = 132
x = 132/22
x = 6
Hence y = 3 - 6
              = - 3 not posible
 Hence y = 3 
So no is 33