Step 1: Start from the average.
There are 100 students spread over 10 standards, so the average number of students per standard is $100/10 = 10$. An average of 10 does not force every standard to have exactly 10, but it does force at least one standard to have 10 or more, because if every standard had fewer than 10 (that is, at most 9), the total could never reach 100.
Step 2: Turn the average argument into a proof by contradiction.
Suppose every one of the 10 standards has 9 or fewer students. Adding up the maximum possible in this case gives $9 \times 10 = 90$ students, which falls short of the actual total of 100 students. Since this is impossible, at least one standard must have 10 or more students. This directly matches option (A): "There are at least 10 students who belong to the same standard."
Step 3: Disprove the remaining options by building extreme, but valid, distributions.
Place all 100 students in the 1st standard and 0 everywhere else; this is a valid split of 100 students, but 9 standards then have zero students, so option (B) fails. Place all 100 students in the 10th standard instead; this is also a valid split, and now the 10th standard has 100 students, far above the "at most 10" claimed in option (C), so option (C) fails. Place all 100 students in the 8th, 9th, and 10th standards only; then the 1st to 5th standards together have 0 students, which is less than the 50 claimed in option (D), so option (D) fails.
Step 4: Conclude.
Since options (B), (C), and (D) can each be broken by a valid distribution of the 100 students, while option (A) is forced to be true no matter how the students are split, option (A) is the one statement that is always correct.