Let’s assume the digit at unit’s place is x and ten’s place is y. 

Thus from the question, the number we need to find is 10y + x. 

From the question since the two digits of the number are differing by 3. Therefore, 

x – y = ±3 …………. (i) 

And, after reversing the digits, the number so obtained is 10x + y. 

Again it’s given from the question that, the sum of the numbers obtained by reversing the digit’s places and the original number is 99. Thus, this can be written as; 

(10x + y) + (I0y + x) = 99 

⇒ 10x + y + 10y + x = 99 

⇒ 11x + 11y = 99 

⇒ 11(x + y) = 99 

⇒ x + y = 99/11

⇒ x + y = 9 …………… (ii) 

So, finally we have two sets of systems of equations to solve. Those are, 

x – y = 3 and x + y = 9 

x – y = -3 and x + y = 9 

Now, let’s solve the first set of system of equations; 

x – y = 3 ……….. (iii) 

x + y = 9 ………. (iv) 

Adding the equations (iii) and (iv), we get; 

(x – y) + (x + y) = 3 + 9 

⇒ x – y + x + y =12 

⇒ 2x = 12 

⇒ x = 12/2

⇒ x = 6 

Putting the value of x in equation (iii), we find y 

6 – y = 3 

⇒ y = 6 – 3 

⇒ y = 3 

Hence, when considering this set the required number should be 10 x 3 + 6 = 36 

Now, when solving the second set of system of equations, 

x – y = –3 ……….(v) 

x + y = 9 …………(vi) 

Adding the equations (v) and (vi), we get; 

(x – y) + (x + y) = –3 + 9 

x – y + x + y = 6 

2x = 6 

x = 3 

Putting the value of x in equation 5, we get; 

3 – y = -3 

⇒ y = 3 + 3 

⇒ y = 6 

Hence, when considering this set the required number should be 10 x 6 + 3 = 63 

Therefore, there are two such numbers for the given question.