## Meddling with averages - solution

Submitted by mf344 on October 19, 2010If the scores are 2 points for a correct answer and −1 point for a wrong answer, then the score is 100 for 50 correct, 97 for 49 correct and so on, giving possible scores of

First note that not all students can have equal scores because both Tyler and Joseph's statements would be wrong in that case. Sadia says that no-one scores higher than the average. Since not all scores are equal, both the median and the mean are smaller than the highest score, so Sadia must be thinking of the mode. This tells us that the highest score is the most common score.

Now suppose there are four students in total and that the highest score among them is 100. This score must occur at least twice because it is the mode. If it occurred three times, then Joseph’s average would have to be the smallest number in the list of scores (since it only occurs once), but this is impossible for both mean and median. Therefore, the score 100 can only occur twice. Suppose the list of scores is with . The median lies half-way between and 100, and therefore does not occur as a score. This means that Joseph must be thinking of the arithmetic mean. This gives

so

Setting gives 91, 97, 100, 100 as the list of scores. The mode is 100, the arithmetic mean is 97 and the median is 98.5, so all three students are correct.

Now suppose that there are five students in total. As before, assume that the highest score is 100 and that it occurs at least twice (because it is the mode). Suppose the list of scores is with . Then is the median, so Tyler must be thinking of the arithmetic mean. A possible configuration of scores would be , , . This gives 100 as the mode, 97 as the median and 96.4 as the arithmetic mean.