Step 1: Understand what "always correct" means.
A statement is always correct only if it holds true for every possible way of splitting 100 students across the 10 standards, and it fails if even a single valid distribution breaks it.
Step 2: Try to break options (B), (C) and (D) using one extreme distribution.
Suppose all 100 students are placed in the 10th standard, and every other standard (1st to 9th) has 0 students. This is a valid distribution since it uses exactly 100 students across 10 standards. Here, standard 1 has no students, so "there is at least one student in each standard" (option B) is false. The 10th standard has 100 students, which is far more than 10, so "at most 10 students in 10th standard" (option C) is false. The 1st to 5th standards together have 0 students, which is less than 50, so option (D) is false as well.
Step 3: Check whether option (A) survives this extreme case.
In this same distribution, the 10th standard has 100 students, and \(100 \geq 10\), so "there are at least 10 students who belong to the same standard" is true here.
Step 4: Prove option (A) cannot fail for any distribution, using an averaging argument.
The average number of students per standard is $100/10 = 10$. If every one of the 10 standards had strictly fewer than 10 students, that is at most 9 each, the total across all standards could be at most $9 \times 10 = 90$ students, which contradicts the given total of 100. So it is impossible for every standard to have fewer than 10 students, meaning at least one standard must always have 10 or more, no matter how the 100 students are distributed.
Step 5: Conclude.
Since option (A) holds both in the extreme case and by the general averaging argument, while options (B), (C) and (D) all fail in the extreme case, option (A) is the only statement that is always correct.
\[ \boxed{\text{Option (A)}} \]