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

This problem is about deciding which pair of conditions pins down the third one for certain, no matter what the actual four numbers are. The cleanest way is to imagine trying to break each implication with the most extreme numbers allowed, and see which attempt always fails.

1. (S1) and (S2) imply (S3): Keep the total fixed at 100 (from S1) and the cap at 40 (from S2), but push two numbers as low as possible: use 40, 40, 20, 0. The total is 100 and no number goes above 40, yet the last number is 0, which breaks the "at least 20" rule. So this pairing cannot guarantee (S3).
2. (S2) and (S3) imply (S1): Take all four numbers equal to their allowed minimum, 20. This satisfies both the cap of 40 and the floor of 20, giving a sum of 80 and an average of 20, not 25. So this pairing cannot guarantee (S1) either.
3. (S1) implies (S3) alone: With only the sum fixed at 100, put almost everything into one number, such as 97, 1, 1, 1. The sum is 100, but three of the numbers are far below 20. So (S1) by itself proves nothing about the floor.
4. (S1) and (S3) imply (S2): Fix the sum at 100 and require every number to be at least 20. If any one number tries to exceed 40, say it becomes $40+k$ for some $k>0$, the other three numbers must still each be at least 20, so they add up to at least 60. The total would then be at least $40+k+60 = 100+k$, which is more than 100 the moment $k>0$. This contradicts the fixed sum of 100, so no number can ever cross 40 while the other three stay at or above 20.

Only the fourth pairing survives every attempt to break it: whenever the sum is 100 and every number is at least 20, the cap of 40 comes out automatically, it cannot be dodged by choosing different numbers.

Let's sum up:

• For the first three options, one concrete set of numbers is enough to show the implication does not always hold.
• For the fourth, the floor of 20 on three numbers plus the fixed sum of 100 mathematically forces the fourth number to stay at or under 40.

So the necessarily correct option is that (S1) and (S3) together imply (S2), option (C).