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    • Filling in Multiple Choice Exam

      Meddling with averages

      19 October, 2010

      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.

      Solution link
      Meddling with averages - solution
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      Anonymous

      22 October 2010

      Permalink
      Comment

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

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      Marianne

      25 October 2010

      In reply to Meddling with averages by Anonymous

      Permalink
      Comment

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

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      Anonymous

      30 October 2010

      In reply to The mode is the score that by Marianne

      Permalink
      Comment

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

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      Anonymous

      30 October 2010

      In reply to Meddling with averages by Anonymous

      Permalink
      Comment

      there's no mode in that case

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      Anonymous

      30 October 2010

      In reply to Meddling with averages by Anonymous

      Permalink
      Comment

      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.

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      Anonymous

      29 October 2010

      Permalink
      Comment

      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

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      Anonymous

      3 November 2010

      In reply to Solution - 4 Students by Anonymous

      Permalink
      Comment

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

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      Anonymous

      3 November 2010

      In reply to Solution - 4 Students by Anonymous

      Permalink
      Comment

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

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