Stage: 5 Challenge Level:
If 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
$$0, 1, 4, 7, \dots, 97, 100$$
Suppose that 4 students take the test and score
$$ 91, 97, 100, 100$$
Sadia chooses mode, Tyler chooses median and Joseph chooses
Suppose that 5 students take the test and score
$$91, 94, 97, 100, 100$$
Sadia chooses mode, Joseph chooses median and Tyler chooses
In these two cases each student is correct.
Proof that there is no choice in
All students cannot score the same mark as then all three averages
are the same, which is inconsistent with Tyler's position.
Sadia cannot then choose arithmetic mean, as this must be less than
the maximum if all scores are not the same.
If Sadia chose the median then the median must be the
maximum. If this is the case then at least half of the values equal
the median, in which case Joseph must choose the arithmetic mean.
However, this means that someone scores the mean, median and mode,
which is inconsistent with Tyler's position. Therefore Sadia cannot
choose the median.
Thus, if all three are to be simultaneously correct Sadia must
always choose the mode, which must equal the highest score.
Thus, neither Joseph nor Tyler can then choose the mode, and they
clearly cannot choose the same average.
Thus we have two cases:
1) Joseph chooses arithmetic mean and Tyler the median.
2) Joseph chooses median and Tyler chooses arithmetic mean.
The median is always an achieved score for an odd number of
Thus, if we have an odd number of students we must select case 2)
to have a chance of satisfying Tyler. In this case the median must
be unique to satisfy Joseph.
If we have an even number of students Joseph cannot choose the
median: the median is only an achieved score for an even number of
students if more than one student scores the median
Thus, if we have an even number of students we must select case 1)
to have a chance of satisfying Joseph. In this case the median must
not be an achieved score to satisfy Tyler.
Note: This does not prove that the conditions can in fact be
simultaneously met for various numbers of students; merely that
there is no choice IF it is met.