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

Step 1: Convert the average into a sum.
Statement (S1) says the average of four numbers is \(25\). Since average is the sum divided by the count,
\[ \frac{a+b+c+d}{4} = 25 \implies a+b+c+d = 100 \]
This is the fact we will use throughout.

Step 2: Test option (A): do (S1) and (S2) force (S3)?
(S2) says every number is at most \(40\). Try the numbers \(40, 40, 10, 10\). Their sum is \(100\) (so S1 holds) and none exceeds \(40\) (so S2 holds), but \(10\) is less than \(20\), so (S3) fails. Since we found a valid case where (S1) and (S2) hold but (S3) does not, option (A) is false.

Step 3: Test option (B): do (S2) and (S3) force (S1)?
(S2) and (S3) together only say each number lies between \(20\) and \(40\). Try \(20, 20, 20, 20\): every number is between \(20\) and \(40\), but the sum is \(80\), giving an average of \(20\), not \(25\). So (S1) fails here even though (S2) and (S3) hold, making option (B) false.

Step 4: Test option (C): do (S1) and (S3) force (S2)?
We know \(a+b+c+d=100\) and each of \(a,b,c,d\) is at least \(20\) (statement S3). Rearranging for one number, \(a = 100-(b+c+d)\). Since \(b,c,d\) are each at least \(20\), their sum \(b+c+d\) is at least \(60\), so
\[ a = 100-(b+c+d) \leq 100-60 = 40 \]
The same argument applies to \(b\), \(c\), and \(d\) individually, since any one of the four numbers can be isolated the same way. So no number can ever exceed \(40\): (S1) and (S3) together force (S2) to hold, in every possible case. Option (C) is true.

Step 5: Test option (D): does (S1) alone force (S3)?
Try \(100, 0, 0, 0\). The average is \(25\) (S1 holds), but three of the numbers are \(0\), well below \(20\), so (S3) fails. Option (D) is false.

Step 6: Final Answer.
Only option (C), "(S1) and (S3) together imply (S2)," holds in every possible case.
\[ \boxed{\text{Option (C)}} \]