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:

Try to break each implication with an extreme example first; only the (S1)+(S3) implies (S2) pairing survives every attempt.

Answer 1

The Correct Option is C

Step 1: Translate the statements into algebra.
Let the four numbers be \(a, b, c, d\).
(S1): \(\dfrac{a+b+c+d}{4} = 25\), so \(a+b+c+d = 100\).
(S2): \(a,b,c,d \le 40\).
(S3): \(a,b,c,d \ge 20\).

Step 2: Test option (A), whether (S1) and (S2) force (S3).
Try \(a=40, b=40, c=19, d=1\).
Sum \(= 40+40+19+1 = 100\), so (S1) holds, and every number is at most 40, so (S2) holds.
But \(c=19\) and \(d=1\) are both less than 20, so (S3) fails.
So (S1) and (S2) do not force (S3), and option (A) is wrong.

Step 3: Test option (B), whether (S2) and (S3) force (S1).
Try \(a=b=c=d=20\).
Every number is between 20 and 40, so (S2) and (S3) both hold.
But the sum is \(80\), giving an average of \(20\), not \(25\), so (S1) fails.
So option (B) is also wrong.

Step 4: Test option (D), whether (S1) alone forces (S3).
Try \(a=0, b=0, c=0, d=100\).
The sum is \(100\), so the average is \(25\) and (S1) holds.
But \(a=b=c=0\) are all less than 20, so (S3) fails.
So option (D) is wrong too.

Step 5: Prove option (C), that (S1) and (S3) together force (S2).
Suppose, for contradiction, that some number, say \(a\), is greater than 40.
By (S3), each of \(b, c, d\) is at least 20, so \(b+c+d \ge 60\).
Then \(a+b+c+d > 40+60 = 100\), which contradicts (S1), \(a+b+c+d=100\).
So no number can exceed 40, meaning every number is at most 40, which is exactly (S2).
The same argument applies no matter which of the four numbers we assume exceeds 40, by symmetry.

Final Answer:
(S1) and (S3) together necessarily imply (S2), so the correct option is (C). \[ \boxed{\text{Option (C)}} \]