THE TRUSSES OF HAY.
(
Money Puzzles)
Farmer Tompkins had five trusses of hay, which he told his man Hodge to
weigh before delivering them to a customer. The stupid fellow weighed
them two at a time in all possible ways, and informed his master that
the weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120,
and 121. Now, how was Farmer Tompkins to find out from these figures how
much every one of the five trusses weighed singly? The reader may at
first think that he ought to be told "which pair is which pair," or
something of that sort, but it is quite unnecessary. Can you give the
five correct weights?
Answer:
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.
