(related to Problem: The Trusses of Hay)

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.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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