## Meddling with averages

Submitted by Marianne on October 19, 2010You are just about to sit an exam which has 50 compulsory multiple choice questions. Students score 2 points for a correct answer and -1 point for a wrong answer, with a minimum score of 0 for the whole exam.

Three of your friends, Tyler, Sadia and Joseph, are discussing the possible marks:

Tyler says "Nobody will score the average mark".

Sadia says "Nobody will score higher than the average mark".

Joseph says "I will be the only person to score the average mark ".

Each of the three has their own definition of *average* in mind:

- One of them is thinking of the
*arithmetic mean*, which you get by adding up all the scores and dividing by the total number of students. - Another is thinking of the
*mode*, which is the score that occurs most frequently among all students. - The third is thinking of the
*median*, which is the "middle" number in the list of scores. (List all scores in ascending order. If there are an odd number of scores, the median is the middle number in your list. If the there are an even number of scores, the median is the arithmetic mean of the two middle numbers.)

Suppose there are 4 students all together. Can you create a set of scores and a set of choices of average for Tyler, Sadia and Joseph for which they are all simultaneously correct? What if there are 5 students?

*This puzzle has been adapted from the weekly challenge on our sister site NRICH, which has a wealth of free maths games, challenges, activities and articles. *