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 Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum