Question

Consider the following statements about four numbers:
(S1) The average of the four numbers is 25
(S2) Each number is at most 40
(S3) Each number is at least 20

Choose the option that is necessarily correct.

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

Hint:

Average 25 means the four numbers add up to 100; to see what a lower bound on each number forces, try to push one number as high as possible while keeping the others at their minimum.

Answer 1

The Correct Option is C

Instead of hunting for counterexamples first, this approach starts from the one option that looks structurally different (S1 and S3 forcing S2), proves it directly with a short contradiction argument, then uses quick counterexamples to knock out the rest.

1. (S1) and (S2) imply (S3): suppose the four numbers are $40, 40, 10, 10$. These sum to $100$, satisfying the average condition, and none exceeds $40$, satisfying (S2). But two of the numbers are $10$, below $20$, so (S3) fails here. This single case is enough to rule out option (A).
2. (S2) and (S3) imply (S1): take all four numbers equal to $20$. Each lies between $20$ and $40$, satisfying both (S2) and (S3), but the sum is $80$, giving an average of $20$, not $25$. So (S1) is not forced, ruling out option (B).
3. (S1) implies (S3): take $100, 0, 0, 0$. The average is exactly $25$, but three of the four numbers are $0$, far below $20$. So (S1) alone cannot force (S3), ruling out option (D).
4. (S1) and (S3) imply (S2): assume, for contradiction, that some number, say $a$, exceeds $40$, so $a > 40$. Since (S3) says $b, c, d$ are each at least $20$, their sum satisfies $b+c+d \geq 60$. Adding this to $a > 40$ gives $a+b+c+d > 40+60 = 100$. But (S1) fixes the sum at exactly $100$, a contradiction. So no number can exceed $40$, and (S2) must hold.

The contradiction in the last case cannot be avoided for any choice of which number we call $a$, since the same argument works symmetrically for any of the four numbers. So (S1) and (S3) together do guarantee (S2).

Let's summarize:

• To disprove "X implies Y," one clean counterexample where X holds and Y fails is enough.
• To prove "X implies Y" for certain, assume Y is false and show that forces a contradiction with X, a proof by contradiction.

The only option that survives every test is that (S1) and (S3) together imply (S2).