
Re: more "bales of hay" urgh!
Posted:
Nov 9, 1997 9:05 AM


Hello. I'm a recovering mathphobic; 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 289231. So c=58. The combination of c+a must equal the secondlightest total (112), so a=11258=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.
Thanks to all. Colette

