Hello. I'm a recovering math-phobic; I finished secondary school in 1974 with unwarranted high grades in math. Now I'm a student teacher, and last year I took a wonderful course at the U. of Toronto called "Elementary Concepts in Math", which helped me to develop some mathematical thinking ability. In light of that, here's how I solved this problem. There may well be more elegant ways, which I would be delighted to know about, although I'm quite pleased with this one.
The problem: suppose five bales of hay are weighed two at a time in all possible ways. The weights in lbs are 110, 112, 113,114,115, 116,117,118,120 and 121. How much does each bale weigh?
I called the individual bales a, b, c, d and e. I totalled the combined weights: 1156 I reckoned that a+b+c+d+e = 1156 divided by 4 (since each bale of hay is weighed with four others), or 289. I extrapolated that a+b (the 2 lightest) =110, and d+e (the 2 heaviest) = 121, so a+b+d+e=231, and c must equal 289-231. So c=58. The combination of c+a must equal the second-lightest total (112), so a=112-58=54.
And so on, extrapolating the other values in the same way.
I want to add that I'm sending this along not because I think it's a brilliant solution - as I said above, there are very likely more elegant ones - but simply because I'm delighted to be able to solve this problem at all, and so pleased to have found this site to visit from time to time.