Let $S_i$ 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 $\sum_{i \text{ solved this problem}} \frac{S_i}{\text{number of people who solved this problem}} \le a$ So $\sum_{i\text{ solved this problem}} S_i \le a \text{number of people who solved this problem}$ Sum this for all problems. LHS - $S_i$ appears $S_i$ times. $\sum_{i=1}^n S^2_i$. RHS - Sum of "number of people who solved this problem" is sum of scores. This is $a \sum_{i=1}^n S_i$. So we have $\frac{1}{n} (\sum_{i=1}^n S_i)^2 \ge \sum_{i=1}^n S^2_i$. Oops. This forces $S_i=S_j$ for all $i, j$. Contradiction since there are two students with different scores.

rkm0959 wrote: Average of students who didn't is equivalent to Average of students who answered this question correctly <= Average of all people. why?

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 $\text{avg}(A)<\text{avg}(B)\iff \text{avg}(A)< \text{avg}(A\cup B)$. This is just a consequence of definition of average.

@rafayaashary1 thank you