A statement is "always correct" only if no possible way of splitting the 100 students among the 10 standards can break it. The fastest way to test each option is to try to build a counterexample, a specific split of students that makes the statement false. If no counterexample can exist, the statement is guaranteed.
- There are at least 10 students who belong to the same standard: Try to break this by spreading students as evenly as possible, giving each of the 10 standards at most 9 students. The largest possible total would then be $9 \times 10 = 90$ students, but we have $100$ students, which is $10$ more than $90$. So some standard must absorb the extra students and reach $10$ or more. No counterexample is possible, this statement always holds.
- There is at least one student in each standard: A counterexample is easy to build, put all $100$ students in the 1st standard and leave the rest empty. The condition breaks, so this is not always true.
- There are at most 10 students in 10th standard: Put all $100$ students in the 10th standard itself. Now the 10th standard has $100$ students, well above $10$, so this statement is broken too.
- The total number of students from 1st to 5th standards is at least 50: Put all $100$ students in standards 6 to 10, leaving standards 1 to 5 with $0$ students combined. This is far below $50$, so the statement fails here as well.
Since options (B), (C), and (D) can each be broken by some valid distribution of the 100 students, they are not guaranteed. Only option (A) survives every possible split, because packing 100 students into 10 standards with 9 or fewer in each is mathematically impossible.
Let's summarize:
- "Always correct" means true for every possible split of students, so a single counterexample is enough to eliminate an option.
- The pigeonhole idea, 100 students cannot fit into 10 standards with at most 9 each, is what forces option (A) to hold.
So the group must always contain at least 10 students in some one standard, confirming option (A).