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Math Help - (SAT) Proportionality

  1. #1
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    (SAT) Proportionality

    If 500 pounds of mush will feed 20 pigs for a week, for how many days will 200 pounds of mush feed 14 pigs?

    Inversely proportional.

    \displaystyle 500\ \times\ \frac{20}{7} = \frac{10 000}{7}
    \displaystyle = 1428\frac{4}{7} \implies 1428\frac{4}{7} pig-days

    \displaystyle 200\ \times\ y\ days = 1428\frac{4}{7}
    \displaystyle y = \frac{1428\frac{4}{7}}{200} = 7\frac{1}{7}
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  2. #2
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    n_{pigs} \cdot n_{days} \cdot n_{lb/(day \cdot pig)} = n_{lb}

    Since the amount of mush eaten per pig per day is assumed to be constant then 20 \cdot 7 \cdot  n_{lb/(day \cdot pig)} = 500

    n_{lb/(day \cdot pig)} = \dfrac{500}{20 \cdot 7} = \dfrac{25}{7}

    Using this info:

    14 \cdot n_{days} \cdot \dfrac{25}{7} = 200


    Solve for n. I get 4 days which is around what we'd expect from intuition (not like 0.4, 40 and 25 which I got when trying to solve!)
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  3. #3
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    First, you need to know how much a pig will eat each day.

    If 500 lbs feeds 20 pigs for 7 days, then if we set up the ratio mush : day

    \displaystyle 500 : 7

    \displaystyle \frac{500}{7} : 1.


    So 20 pigs will eat \displaystyle \frac{500}{7} lbs of mush per day.

    Now we can set up another ratio of mush : pigs

    \displaystyle \frac{500}{7} : 20

    \displaystyle \frac{25}{7} : 1

    \displaystyle 50:14.

    So 14 pigs will eat 50 lbs of mush per day.


    Finally, setting up the ratio mush : days

    \displaystyle 50 : 1

    \displaystyle 200 : 4.


    So 14 pigs will eat 200 lbs of mush in 4 days.
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  4. #4
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    This is a tough question for the SAT. I would do it with a ratio (direct proportion) followed by an inverse proportion as follows:

    First I'll answer this question: "If 500 pounds of mush will feed 20 pigs for a week, then 200 pounds of mush will feed how many pigs for a week?"

    \rm{mush}\ \       500   \   200

    \rm{pigs}   \ \ \  \      20 \ \    \   x

    So \frac{500}{20}=\frac{200}{x}

    500x=4000

    x=8

    Now I'll answer this question: "If you can feed 8 pigs for 7 days, then for how many days can you feed 14 pigs?"

    8\cdot 7=14x

    56=14x

    x=4

    The problem is particularly confusing because there are three active variables here. I don't believe I've ever seen this in an actual SAT problem (if it ever did occur you could probably get it wrong and still get an 800). I have seen an additional inactive variable by which I mean the third variable wasn't needed to solve the problem (it was just there to confuse you). I'm curious what book this problem came from.
    Last edited by DrSteve; January 11th 2011 at 05:54 AM.
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