A Weighty Problem

Strategies and Hints

The girls weigh 92, 96, 100, 107, and 112 pounds. Since every combination is different, no girls weigh the same. The girls must have gotten on in 10 possible combinations. This can only be done by A-B, A-C, A-D, A-E, B-C, B-D, B-E, C-D,C-E and D-E. This means each girl was weighed 4 times. By summing the weights and dividing by 4 you can get the weight of the 5 girls together being or 507 pounds (A + B + C + D + E = 507). If we continue by placing the girls in order from the lightest to the heaviest (A, B, C, D, E) we also know that the two lightest must together weigh 188 (A + B = 188), and the two heaviest must together weigh 219 (D + E = 219). Substituting and solving, we get C = 100. The second lightest weight of 192 must be girls A and C; therefore A + C = 192, and since C = 100, A = 92.

Similarly, the second heaviest weight must be girls C and E, so C + E = 212. Substituting for C, we get E = 112. We know the combined weights of A and B, also D and E, so we can solve for the other missing weights.