Pow 6

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  • Topic: Hay
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  • Published : December 13, 2012
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POW:6 The Haybaler Problem
Problem Statement: You bought 5 bales of hay but they weighed them in pairs not individually like they used to. The bales of hay could be matched up in any combination like, 1 and 2, 1 and 3, 1 and 4, and so on. The salesperson did not keep track of the weight of each bale of hay. Your job is to find out the weight of each bale of hay using combinations, 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. Remember there are 5 punkins and you can only make combinations of 2. Process: First I thought of every combination of bales. I ended up getting ten different combinations. The weights in Kg are listed in order of numbers not in order of weight in pairs of bales. So I knew I needed to figure out the heaviest and the lightest bales.The smallest combination of bales is 80 so I will divide that by 2 and I get 40 so I know that both numbers will be around 40. I thought that no 2 hay bales can be the same weight so I decided to use the numbers 39 and 41 Next I divided 82 by 2 and got 41 so the 2 numbers I decided to use were: since I already used 41 I will use the first weight 39 and subtract 39 from 82 to get 43. So far I have 2 of the combinations finished and 3 out of the 5 hay bale weights. Hay bale 1- 39 kg

Hay bale 2- 41 kg
Hay bale 3- 43 kg
The next Combination on the list is 83, 83 divided by 2 is around 41-42 so if we can use one of the weights of the 3 bales of hay we already have. I choose the first hay bale to use, so 83-39= 44. I don’t know if that is the weight for hay bale 4 but lets assume it is. Hay bale 4- 44 kg

The next combination is 84. 84/2=42. We can use the weights of the 2nd and 3rd hay bale to equal this combination 41+43=84. Now lets move on to the next combination which is 85. 85/2= around 42 to 43. So I can use the weight of hay bales 2 and 4 (41+44) to equal the combination 85. The next combination is 86. 86/2= 43. We can use only one weight this time, because none of the weights we have found already match...
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