Weighed Two at a Time

Five girls weigh themselves in pairs, trying every one of the ten possible couples. The pair weights, in pounds, come out as $$114, \ 116, \ 118, \ 120, \ 121, \ 122, \ 123, \ 124, \ 125, \ 129.$$ No two girls weigh the same. What is the difference between the heaviest and the lightest?

Solution

Call the five weights $a < b < c < d < e$. Each girl is weighed against the other four, so each weight appears in four of the ten pair-totals. The ten totals therefore add up to $4(a + b + c + d + e)$, and since they sum to $1212$, $$a + b + c + d + e = \frac{1212}{4} = 303.$$ The lightest pairing is the two lightest girls, $a + b = 114$, and the heaviest is the two heaviest, $d + e = 129$. Subtracting these from the total leaves the middle girl, $$c = 303 - 114 - 129 = 60.$$ The second-smallest total is $a + c = 116$, so $a = 56$ and then $b = 58$; the second-largest is $c + e = 125$, so $e = 65$ and then $d = 64$. The five weigh $56, 58, 60, 64, 65$ pounds, and the heaviest exceeds the lightest by $65 - 56 = 9$ pounds.