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Math Help - Pre-Algebra Word Problem - PLEASE HELP

  1. #1
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    Question Pre-Algebra Word Problem - PLEASE HELP

    I have the hardest word problem I have ever come across. It is called The Haybailer Problem here is how it goes....
    The Situation:
    You have five bales of hay.
    For some reason, instead of being weighed individually they were weighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1 and 5, bales 2 and 3, bales 2 and 4, and so on.
    The weight of each of these combinations were written down and arranged in numerical order, without keeping track of which weight matched which pair of bales. The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90 and 91.
    The Task:
    The initial task is to find out how much each bale weighs. In particular, you should determine if there is more than one possible set of weights, and explain how you know.
    Once you are done looking for solutions, look back over the problem to see if you can find some easier or more efficient way to find the weights.

    Please Help Me!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GodsDsipl View Post
    I have the hardest word problem I have ever come across. It is called The Haybailer Problem here is how it goes....
    The Situation:
    You have five bales of hay.
    For some reason, instead of being weighed individually they were weighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1 and 5, bales 2 and 3, bales 2 and 4, and so on.
    The weight of each of these combinations were written down and arranged in numerical order, without keeping track of which weight matched which pair of bales. The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90 and 91.
    The Task:
    The initial task is to find out how much each bale weighs. In particular, you should determine if there is more than one possible set of weights, and explain how you know.
    Once you are done looking for solutions, look back over the problem to see if you can find some easier or more efficient way to find the weights.

    Please Help Me!!!
    here is one solution.

    let the masses of bales 1,2,3,4, and 5 be a,b,c,d, and e respectively.

    (for no real reason) let

    a + b = 80 .................(1)
    a + c = 82 .................(2) ==> c = b + 2
    a + d = 83 .................(3) ==> d = b + 3
    a + e = 84 .................(4) ==> e = b + 4
    b + c = 85 .................(5) ==> b = 83/2

    .
    .
    .
    let's stop there, we have enough info now. back-substituting we get:

    a = 77/2, c = 87/2, d = 89/2 and e = 91/2

    i hope you picked up the pattern i was using. and also how i solved for b and hence the rest

    EDIT: (the pattern was, i let the masses represent, respectively, bales 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3, 2 and 4, 2 and 5, 3 and 4, 3 and 5, 4 and 5)

    try to see if you can come up with another way to do the problem, maybe get another solution. using my approach as a hint of how to think about it. what other patterns could you use to set up equations?
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  3. #3
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    I don't understand how much each bale weighs. .... Also I was told that there can be no fractions or decimals ....
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by GodsDsipl View Post
    I don't understand how much each bale weighs. .... Also I was told that there can be no fractions or decimals ....
    i don't know about that. my solution works.

    what is your experience with solving simultaneous equations? what exactly is confusing you?
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  5. #5
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    I dont have any experience. I am new to this and this was our homework due tomorrow. I have been racking my brain all night
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  6. #6
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    b + c = 85 .................(5) ==> b = 83/2

    .
    .
    .
    let's stop there, we have enough info now. back-substituting we get:

    a = 77/2, c = 87/2, d = 89/2 and e = 91/2
    No pair of bails will together weigh 91 in this solution. The reason that the method failed was that b+c may weigh less than a+d

    If this is really at pre-algebra level I have no idea how you are expected to solve it, but I did it like this:

    For convenience I will number the bails in ascending order of weight and call them B1, B2, B3, B4 and B5 in the hope that someone wittier than me will make a joke about bananas in pajamas.

    We know that no bales weigh the same amount because no sums are the same.

    We know that B1+B2 = 80, B1+B3 = 82, B4+B5=91 and B3+B5 = 90

    Notice also that there are 6 even weights in the list and 4 odd ones. An odd sum will be produced when an odd and an even are added. This will occur 4 times if and only if there are 4 odd and one even weight or 4 even and one odd weight.
    Since B4+B5=91, one of B4 and B5 are the different ones. Since B3+B5 is even, B5 is not the different one, so B4 is and all odd sums must contain B4.

    The next smallest sum is 83 and logically must be either B1+B4 or B2+B3. But it is odd and so is B1+B4 as it must contain B4

    The next largest sum is 88 and logically must be either B2+B5 or B3+B4. Since it is even, it must be B2+B5. So B2+B5 = 88

    The next largest sum is is 87 and so must include B4. This means it is B3+B4

    So we have:
    B1+B2 = 80
    B1+B3 = 82
    B4+B5 = 91
    B3+B5 = 90
    B1+B4 = 83
    B2+B5 = 88
    B3+B4 = 87

    You can now solve these simultaneously to find all of the weights. I got:
    B5 = 47
    B4 = 44
    B3 = 43
    B2 = 41
    B1 = 39

    It is necessary to check the list of sums produced from the solution against the original list of sums to ensure that a solution is possible; what we have discovered here is what the solution will be if it exists.

    The sums calculated from the solution are: 80, 82, 83, 84, 85, 86, 87, 88, 90, 91. This is the same as the starting list so we have our solution. Wheee
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