I thought the answer would simply be $2 \times 9 \times 8 \times 7 = 1008$, since the last digit must be 0 or 5, so we have two choices. Then after picking the last digit, we have 9 choices for the 1st digit, 8 for the 2nd digit, and 7 for the 3rd digit.
However, the correct answer is 952. Can anyone tell me where am I over counting, and how I could solve it?
Your count includes the number $0435$, but this is not a four-digit numeral because numerals do not begin with zero.
One way to do this problem is to subtract the sequences that begin with 0. They all end in 5, and that makes $1\times 8\times 7\times 1 = 56$ of them. $1008-56=952$.
Another way is to count the $xyz5$ and the $xyz0$ numerals separately. There are $8\times 8\times 7\times 1 = 448$ of the first kind and $9\times 8\times 7\times 1 = 504$ of the second kind; $448+504=952$.