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Math Help - interpreting this word problem

  1. #1
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    interpreting this word problem

    The following conditions are given:
    Hay: Dry density (kg/m^3) = 92
    Straw: Dry density (kg/m^3) = 68
    Straw and manure*: Dry density (kg/m^3) = 171
    Hay and manure*: Dry density (kg/m^3) = 248
    (*Note: Straw or hay was mixed with manure in a 1:2 ratio by dry weight)

    How can we deduce that manure, on its own, would have a dry density that is greater than the dry density of hay?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Hay has density = 92 kg/m^3
    Manure and hay has density = 248 kg/m^3

    Given 1 m^3 of hay, you will have 92 kg of hay.
    Given 1 m^3 of manure and hay, you will have 248 kg of manure and hay, of which, a third is the mass of hay and 2 thirds is the mass of manure, which gives 82.7 kg of hay and 165.3 kg of manure.

    Find the volume of the hay in the manure/hay mixture, using the mass of hay and it's density, and get the volume of manure.

    You will get the density of manure using the values you just got.
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  3. #3
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    I'm still stuck Unknown, what will be the fraction and how do you find it approximately? Does that mean that manure has a greater dry density because 165.3 is bigger than 92 (the original amount)?

    (Also: this was a practice science reasoning ACT multiple choice question with no other info given and the other choices were similiar:
    less than the dry density of straw
    between the dry density of straw and the dry density of hay
    equal to the density of hay)
    Last edited by dannyc; April 2nd 2011 at 12:18 AM.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Ah, if this is a multiple choice question, you will rather use logic instead of calculation. But if you want the 'why', I gave it to you. I will complete it if you want.

    So, from the density of hay, we have:

    92 kg of hay => 1 m^3
    82.7 kg of hay => 0.899 m^3

    The volume of manure is then = 1 - 0.899 = 0.101 m^3

    From this, we get the density of manure = 165.3/0.101 = 163.5 kg/m^3

    which you can see is much higher than the desities of hay and straw.
    ~~~~~~~~~~~


    Now for the logic part.

    Let's assume that manure has the same density as hay. Mixing 2/3 mass (2/3 of any number) of manure with 1/3 mass (1/3 of the number picked before) of hay, the net density will turn out to be the density of hay itself!

    Why? if manure has the same density as hay, then adding the two is the same as adding 2/3 of hay with another 1/3 of hay, giving 1 hay and the density of hay is 92 km/m^3

    But that is not so. The net density of the mixture is much higher than that of hay, which makes us conclude that the density of manure must be higher than that of hay.

    EDIT: Typo: replace 92 km/m^3 of hay by 92 kg/m^3 of hay
    Last edited by Unknown008; April 2nd 2011 at 12:52 AM.
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  5. #5
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    For the calculation part, where did the 0.899 come from? Also, when you divide 165.3/0.101 = 1636.6 which seems unreasonable due to the initial values??
    The logic part makes sense! I guess the same could be said to disprove straw too?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    1. I got 0.899 through proportions.

    92 kg => 1 m^3
    1 kg => 1/92 m^3
    82.7 kg => 1/92 * 82.7 = 0.899 m^3

    2. Yes, that also can be said for the straw, but shorter still, since the hay is denser than the straw and the manure is denser than the hay, then the manure is automatically denser than the hay.
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  7. #7
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    Yep that makes sense. Still not sure why the density answer would be in the thousands though, seems a bit too large.

    How I did this problem originally was to take 248 - 92 = 156 -> the amount of manure but that is twice the amount, I then took 156/2 = 78 and saw it was less Tbh, not sure why that still doesn't work (I mean sure the answer doesn't match of course but I felt my math was right)
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Okay, to be honest, now that I'm seeing that, I'm not sure if the question was properly phrased.

    \rho \left(\dfrac23\right) + 92\left(\dfrac13\right) = 248

    Solving for rho gives you 326 kg/m^3

    which is a better value. But then, this is a 1:2 ratio by volume and not dry mass.

    The way you put it is similar to this one, which is finding through a 1:2 volume ratio.
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  9. #9
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    Hi danny c,
    For smplicity I use gram per cc which 1/1000 0f kg per cumeter
    Hay density 92/1000 =.092 g/cc
    Straw density 68/1000 = .068 g/cc
    theoretical equations for mixtures
    1cc Hay + 2cc manure = 3 cc mix@.248 g/cc m= .326 g/cc
    1cc Straw + 2cc manure = 3cc mix @ .171 g/cc m= .223 g/cc
    In practice with bulk materials 1cc +2 cc does not always equal 3 cc of volume because of voids.Even same materials have differences in densities



    bjh
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