Solution

(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.


Thank you to the contributors under CC BY-SA 4.0!

Github:
bookofproofs
non-Github:
@H-Dudeney


References

Project Gutenberg

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

This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at http://www.gutenberg.org. If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.