Add together the ten weights and divide by 4, and we get 289 lbs. as the
weight of the five trusses together. If we call the five trusses in the
order of weight A, B, C, D, and E, the lightest being A and the heaviest
E, then the lightest, no lbs., must be the weight of A and B; and the
next lightest, 112 lbs., must be the weight of A and C. Then the two
heaviest, D and E, must weigh 121 lbs., and C and E must weigh 120 lbs.
We thus know that A, B, D, and E weigh together 231 lbs., which,
deducted from 289 lbs. (the weight of the five trusses), gives us the
weight of C as 58 lbs. Now, by mere subtraction, we find the weight of
each of the five trusses--54 lbs., 56 lbs., 58 lbs., 59 lbs., and 62
lbs. respectively.