Step 1: Try to disprove each option by building a counterexample.
A fast way to solve "which statement is always true" questions is to hunt for one valid arrangement of the 100 students that breaks each option. If you find even a single arrangement where a statement fails, that option is eliminated, since "always correct" cannot allow any exception.
Step 2: Break option (B).
Put all 100 students in the 1st standard, with 0 students in standards 2 to 10. The total is still $100$, so this is allowed, but standards 2 through 10 now have no students at all. This contradicts "at least one student in each standard," so (B) fails.
Step 3: Break option (C).
Put all 100 students in the 10th standard, with 0 elsewhere. The total is still $100$, but now the 10th standard alone has 100 students, far more than "at most 10." So (C) fails.
Step 4: Break option (D).
Put all 100 students in standards 6 to 10 only, with 0 in standards 1 to 5. This still totals $100$, but the count from 1st to 5th standards is $0$, which is nowhere close to "at least 50." So (D) fails.
Step 5: Try to break option (A), and see why you cannot.
Suppose, for the sake of argument, that every one of the 10 standards has 9 or fewer students. The largest possible total in that case would be $10 \times 9 = 90$ students. But we are told there are 100 students, and $100 > 90$, so this assumption is impossible. There is no way to keep every standard at 9 or below and still reach a total of 100.
Step 6: Conclude that option (A) cannot be broken.
Since assuming "every standard has fewer than 10 students" leads to a contradiction (a maximum total of only 90, not 100), it must be true that at least one standard has 10 or more students, in every possible arrangement.
Final Answer:
Every attempt to break option (A) fails, while (B), (C), and (D) can each be broken by a simple example, so option (A) is the statement that is always correct.
\[ \boxed{\text{Option (A)}} \]