Just wondering: Should that denominator be the sum of the q_j, rather than of the p_j ? It seems easier to add, and then divide by the sum of, numbers than of problem sets, whose algebraic structure is mysterious. —Dan
On Feb 17, 2016, at 5:23 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Here's my favorite such problem (suggested to me by D G Arthur, with some history before that I can dig out if anyone is interested).
Grading in Concrete Math is based on a series of n problem sets p_1 .. p_n. For each problem set p_i, there are q_i questions; the number of questions a student gets correct is s_i.
At the end of the year, the student may *drop* any k of the n problem sets (k is fixed and given). Let G be the subset of {1..n} remaining (G is of size n-k). The final grade is calculated as
(sum_{j in G} s_j) / (sum_{j in G) p_j)
For instance, if my grades were
6 of 10 (60%), 42 of 60 (70%), 23 of 30 (76.7%)
my best option is not to drop the 6 of 10,which would leave me with 65 of 90 or 72.2%, but to drop the "better" 42 of 60, leaving me with 29 of 40 or 72.5%.
How do you pick which k grades to drop to maximize the final score? The solution has some elegance, though it may not be as simple as many of the puzzles we see.