Difference between my age and my grandfather age 
In $1952$, I was as old as the number formed by last two digits of my
birth year. When I mentioned this to my grandfather he surprised me by
saying that the same applied to him also. The difference in our ages
is ?

Now the way it has been solved is that let us assume that I was born in the year $19xy$ and my grandfather was born in the year $18pq$. Now as per the question :-
$19xy+xy=1952$ where $xy$ is my current age.
$\Rightarrow 1900 + xy + xy = 1952$
$\Rightarrow xy = 26$
$18pq+pq=1952$ where $pq$ is my grandfather's current age.
$\Rightarrow 1800 + pq + pq = 1952$
$\Rightarrow pq = 76$
Difference between the ages = $76 - 26 = 50.$
Now you all can see that this solution applied little logic with the assumption of my age and my grandfather's age. But what if the year given would have been $1989$ or $1999$ then we can't just assume that I would have been born in some $19xy$ and my grandfather would have been born in some $18xy$. It can very well happen that both of us would have been born in the same century.
I tried solving this by assuming that I was born in the year $abcd$ and my grandfather was born in the year $pqrs$ and following the same approach but it didn't helped me and then I turned to this solution that I have put up above. 
So can anyone help me with this? Is it not possible to solve this problem by assuming years like I have or is it like we have to use some logic while assuming the birth years as the solution has done?
Thanks in advance !!!
 A: Expanding on @dxiv comment, the existence of a solution depends on the parity of the reference year, but if there is a solution, then it is necessarily $100/2 = 50$. Let us see this in more details.
Let $r$ be the reference year ($1952$ in your original question). Following your notation, let $x$ and $y$ be the last two digits of your birth year, and $p$ and $q$ the last two digits of your grandfather's birth year. As you already observed, you get the equations
$$
1900 + 2xy = r \text{ and } 1800 + 2pq = r. \qquad (*) 
$$
Subtracting the second one from the first, one obtains $(1900 - 1800) + 2(xy - pq) = 0$, whence $pq - xy = 100/2 = 50$, independently of the choice of $r$.
Note that the previous computation does not require to know your age.
However, to prove the existence of a solution, you need to verify that the equations $(*)$ have a solution and this requires the parity of $r$. For instance, if $r = 1999$, the equation $1900 + 2xy = 1999$ has no integer solution.
Now, to answer your question, you indeed need assumptions on your age and your grandfather's age. If you assume that you are born during the same century as your grandfather, the equations $(*)$ should be changed to
$1900 + 2xy = r$  and $1900 + 2pq = r$, leading immediately to $xy = pq$, which is not possible.
Finally, I invite you to think of similar problems such like this one:

In year $r$, I was as old as three time the number formed by the last two digits of my birth year. Then my grandfather said that the same applied to him also. What is the difference in our ages?

In this case, the equations would be $1900 + 4xy = r$ and $1800 + 4pq = r$, leading to $(1900 - 1800) + 4(xy - pq) = 0$ and finally $pq -xy = 100/4 = 25$, a solution independent of the choice of $r$, but you would need $r-1900$ to be divisible by $4$ (like $1980$) to have a solution.
A: I don't think, such an interesting scenario will occur if the mentioned year is $1999$ or $1899$. To show this, I will tell you a different approach to solve the problem-
Since in $1952$, I was as old as the number formed by last two digits of my birth year, I have to be $\frac{52}{2}=26$ years old in $1952$. If the same holds for my grandfather, he has to be not $26$, but $\left(\frac{100}{2}+26\right)$ (shifting by a century) or $76$ years old. Now, you see why 1999 or 1989 can't satisfy the property- it's because they're odd and you can't divide them by $2$.
Does that help?
Note that I assumed that a man cannot be more than $100$ years old (which is a valid assumption I guess :)). If you assumed infinite age, you can go on shifting centuries and getting more and more solutions.
