An exam was taken by some students. Each problem was worth 1 point for the correct answer, and 0 points for an incorrect one. For each question, at least one student answered it correctly. Also, there are two students with different scores on the exam. Prove that there exists a question for which the following holds: The average score of the students who answered the question correctly is greater than the average score of the students who didn't.
Problem
Source: MEMO 2016 T4
Tags: combinatorics, combinatorics proposed
29.08.2016 09:45
Let Si be the score of person i. Assume otherwise. Average of students who answered this question correctly <= Average of students who didn't is equivalent to Average of students who answered this question correctly <= Average of all people. Let the average be a. So for all problems, we have ∑i solved this problemSinumber of people who solved this problem≤a So ∑i solved this problemSi≤anumber of people who solved this problem Sum this for all problems. LHS - Si appears Si times. ∑ni=1S2i. RHS - Sum of "number of people who solved this problem" is sum of scores. This is a∑ni=1Si. So we have 1n(∑ni=1Si)2≥∑ni=1S2i. Oops. This forces Si=Sj for all i,j. Contradiction since there are two students with different scores.
23.02.2017 14:16
rkm0959 wrote: Average of students who didn't is equivalent to Average of students who answered this question correctly <= Average of all people. why?
23.02.2017 19:06
reveryu wrote: rkm0959 wrote: Average of students who didn't is equivalent to Average of students who answered this question correctly <= Average of all people. why? Let there be two disjoint sets of reals A and B. rkm0959 says that avg(A)<avg(B)⟺avg(A)<avg(A∪B). This is just a consequence of definition of average.
24.02.2017 13:11
@rafayaashary1 thank you