Question

The statements (S1), (S2), and (S3) pertain to the scores obtained by students in an exam. The maximum possible marks in the exam is 150.
(S1) The highest score is 100.
(S2) The fourth highest score is 76.
(S3) There are at least four students whose scores are within 25 of each other.
Which one of the following options is necessarily correct?

• A. (S1) and (S2) together imply (S3)
• B. (S1) and (S3) together imply (S2)
• C. (S2) and (S3) together imply (S1)
• D. (S1) implies (S3)

Hint:

Work out how close together the top four scores must be once you fix the 1st and 4th highest values.

Answer 1

The Correct Option is A

Step 1: Turn (S1) and (S2) into a range.
If the top score is $100$ and the fourth highest is $76$, then the 2nd and 3rd highest scores are squeezed between them, since the ranking already fixes their order. So the top four scores all sit inside $[76,100]$.

Step 2: Measure that range.
The width of this range is $100-76=24$, which is under $25$. So any two of the top four students differ by at most $24$ marks, and (S3) is automatically satisfied by these four students alone. This makes option (A) always true.

Step 3: Try to break option (B) with an example.
Take a class where the topper scores $100$, and four low scoring students happen to sit close together, say at $30, 32, 34, 36$. This satisfies (S1) and (S3), but the fourth highest score here could easily be something other than $76$, say $85$. So (S1) and (S3) do not force (S2).

Step 4: Try to break option (C) with an example.
Let four students all score exactly $76$, satisfying (S2) (the fourth highest is $76$) and (S3) (these four are within $0$ of each other). Nothing stops the highest score in the class from also being $76$, so (S1), which needs the highest score to be $100$, is not forced.

Step 5: Try to break option (D) with an example.
Let the topper score $100$, but space every other student out by more than $25$ marks from everyone else, for example $100, 74, 48, 22, \ldots$ This keeps (S1) true while making (S3) false, so (S1) alone cannot force (S3).

Step 6: Conclude.
Only option (A) survives every attempt to break it, because the arithmetic in Step 1 and Step 2 shows it must hold in general, not just in one example.
\[ \boxed{\text{(S1) and (S2) together imply (S3)}} \]