SOLUTION: I m a 4 digit no. My ones digit is three, My thousands digit is twice the value of one digit, my haundreds digit is two less than the thousands digit, sum of my digit is 25

Let's put 4 spaces for the digits.
_ _ _ _

So we put a 3 last:

_ _ _ 3

So the first two digits are either 9&7, 8&6, 7&5, 6&4, 5&3, 4&2,
3&1 or 2&0

So the number is either

9 7 _ 3
8 6 _ 3
7 5 _ 3
6 4 _ 3
5 3 _ 3
4 2 _ 3
3 1 _ 3
2 0 _ 3

We go through them:

9 7 _ 3, since 9+7+3=19, missing digit would have to be 25-19=6  
8 6 _ 3, since 8+6+3=17, missing digit would have to be 25-17=8
7 5 _ 3, since 7+5+3=15, missing digit would have to be 25-15=10,
         impossible, for 9 is the largest digit.  So the others
         require an even larger tens digit.  So they are all 
         ruled out.

So it's either

9 7 6 3 

or

8 6 8 3

But the thousands digit of 9763 is 9 which is not twice 
any one digit

And the thousands digit of 8683 is 8 which is not twice 
any one digit
either.

So there is no such 4-digit number.  Tell your teacher 
that he or she has made a mistake and given you an
impossible problem.

Edwin