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Weight of Each Bale of Hay


Date: 09/26/97 at 10:31:41
From: Anonymous
Subject: Question

You have a question like the one I am going to ask but with different
numbers, so if you could help me ...

You have 5 bales of hay. Instead of being weighed individually, 
they are weighed in all possible combinations of two: bales 1 and 2, 
1 and 3, 1 and 4, 1 and 5, bales 2 and 3, and bales 2 and 4, etc.  

The weight of each of these combinations is written down and the 
weights are arranged in numeric order, without keeping track of which 
weight matches which pair of bales. 

The weights in kilograms are 80, 82, 83, 84, 85, 86, 87, 88, 90, 
and 91. 

How much does each bale weigh? Is there more than one possible 	set of
weights?  Explain your answer.


Date: 09/26/97 at 13:53:34
From: Doctor Rob
Subject: Re: Question

First, notice that no two bales can have the same weight, since not
enough of the numbers appear twice (in fact, no number appears twice).  
If two bales were of equal weight we should see at least three pairs 
of equal numbers.

Second, notice that if you add up all the numbers and divide by four, 
you will get the total weight of the five bales, 214 kilograms.

Third, notice that the largest number is the weight of the two 
heaviest bales, and the smallest number is the weight of the two 
lightest bales.

Now if you subtract the largest and the smallest numbers from 
the total weight, you will get the weight of the middle bale, 43 
kilograms. Since 43 added to each of the other four weights are 
integers, all the weights must be integers.

Then the lightest two bales must have unequal weights at most 42, and 
which sum to 80, so they are either (38,42) or (39,41). Since the 
number 81 is missing, no bale can have weight 81 - 43 = 38, so (38,42) 
is eliminated. Thus the three lightest bales have weights 39, 41, and 
43. This accounts for the numbers 80, 82, and 84. The two heaviest 
bales must have unequal weights that sum to 91 and which are at least 
44, so they are (44,47) or (45,46). Since the number 89 is missing, 
no bale can have weight 89 - 43 = 46, so the last two weights are 
44 and 47.

Now check:

  39 + 41 = 80
  39 + 43 = 82
  39 + 44 = 83
  39 + 47 = 86
  41 + 43 = 84
  41 + 44 = 85
  41 + 47 = 88
  43 + 44 = 87
  43 + 47 = 90
  44 + 47 = 91


-Doctor Rob,  The Math Forum
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