You 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. *

## Meddling with averages

How is the mode defined if all the scores are different?

## The mode is the score that

The mode is the score that appears most frequently, for example if scores are 22, 55, 43, 43, 10, then the mode is 43.

## But if the scores are all

But if the scores are all different, then the each appear once.

## there's no mode in that case

there's no mode in that case

## If no score appears more than

If no score appears more than once, then there is no mode. This problem can be answered with a mode, that is, with a score that appears more than once.

## Solution - 4 Students

Since Joseph is the only person who gets average marks. His marks cannot be the most frequent marks. With 4 students, median is the arithmetic mean of the second and third numbers. Joseph cannot be the median too as no one else has got same marks as Joseph for the arithmetic mean of Joseph and that guy to be same as Joseph's marks. So, given our conditions, Joseph's average is the arithmetic mean.

Tyler's average score cannot be mode as then no one would get mode marks, which is not possible since mode is the highest frequency number.

So, Tyler's average score is median and Sadia's average score is mode.

So, the three statements are:

No one scores higher than mode

No one scores the median

Joseph is the only person with marks as arithmetic mean

Marks can be of the form 2x - (50-x) = 3x-50

Let the marks be h1>=h2>=h3>=h4

Max frequency is highest marks. So, h1=h2

Now, median is (h3+h1)/2 not equal to h1

So, h3 h4 and is < h1

So, Joseph gets h3 and since arithmetic mean=h3

3*h3=2*h1+h4

So, The marks are

h1, h1, h3, 3*h3-2*h1

So,

Joseph's average is the arithmetic mean

Tyler's average score is median

Sadia's average score is mode

Joseph's score is x

Highest score is say y

Lowest score is 3x-2y

So, the scores are of the form y,y,x,3x-2y (x>y)

Cheers!

Pratik

http://www.pratikpoddarcse.blogspot.com

## I think this is correct but

I think this is correct but in the last sentence it should be x<y.

## I would add that x < y and

I would add that x < y and y>=2/3*y. What if there are 5 students?