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    lp107's Avatar
    lp107 Posts: 7, Reputation: 1
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    #1

    Jul 17, 2007, 02:49 PM
    Algebra Inequality
    A Jockey must weigh in at less than 130 pounds. This includes the jockey's weight plus the weight of the blanket, saddle, and bridle. If the combined weight of the equipment is 1/8 of the jockey's weight, what is the maximum a jockey can weigh?
    Curlyben's Avatar
    Curlyben Posts: 18,514, Reputation: 1860
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    #2

    Jul 17, 2007, 02:52 PM
    Please refer to This Announcement
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    Capuchin Posts: 5,255, Reputation: 656
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    #3

    Jul 17, 2007, 02:53 PM
    Well, that proportion of the 130 pounds can he weigh, if the equipment is 1/8, how much does that leave for the jockey?
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    #4

    Jul 17, 2007, 02:56 PM
    All right... I have tried this (as per your first post) and I have come up with 65... but that seems way to low... I have also gotten 115.5?? I don't know if that second one is right though
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    Capuchin Posts: 5,255, Reputation: 656
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    #5

    Jul 17, 2007, 03:05 PM
    You didn't answer my question: if the equipment is 1/8, how much does that leave for the jockey?

    Just as a proportion.
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    lp107 Posts: 7, Reputation: 1
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    #6

    Jul 17, 2007, 03:08 PM
    well... um x+1/8x<130? Is that what you are asking?
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    #7

    Jul 17, 2007, 03:15 PM
    Okay, say you have a cake. But 1/8 of that cake is your brother's. How much of the cake is yours?

    Now say you have a certain weight on a horse. But 1/8 of that weight is the equipment. How much of the weight is left for the jockey?

    I hope the answer is obvious now...
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    #8

    Jul 17, 2007, 03:16 PM
    7/8 is left for the jockey... I think the answer is 115.5
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    #9

    Jul 17, 2007, 03:19 PM
    130*7/8 is not 115.5
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    lp107 Posts: 7, Reputation: 1
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    #10

    Jul 17, 2007, 03:22 PM
    I know... but 115.5 as an answer is closer to 130 than 113.75... and if you do 130 *7/8... where do the x's come in?? That is a really confusing way to do it... but the answer doesn't make sense
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    #11

    Jul 17, 2007, 03:24 PM
    What xs? I don't see any xs. How did you get 115.5?
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    #12

    Jul 17, 2007, 03:27 PM
    x's an x is what you use in algebra when you don't know a number... so x+1/8x<130... and then I came but with 115.5 because 8(x+1/8x)<130 and then 9x<1040 divided by nine, so x is 115.5
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    #13

    Jul 17, 2007, 03:33 PM
    The way you worked it out there calculates 8/9 of 130. Do you not think that 7/8 makes more sense?
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    #14

    Jul 17, 2007, 03:35 PM
    Thanks for your help! Night!

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