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Math Help - Word Problem - Basic Math

  1. #1
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    Word Problem - Basic Math

    Could someone please explain how to solve this? Thanks.
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  2. #2
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    Re: Word Problem - Basic Math

    You need to find out how much is 80% of 10.
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  3. #3
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    Re: Word Problem - Basic Math

    80% = 80/100 = 8/10 = 4/5.

    so what is (4/5)(10/32)?

    one you have that, say as a/b, you need to express this in 32nds.

    if (by luck) there is a number c such that bc = 32, then a/b = ac/32. the "legal limit" of remaining tread is then (10 - ac)/32.
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    Re: Word Problem - Basic Math

    Which question are you referring to?
    8/32 = 1/4.
    Shouldn't the question be: how do you express 1/16 as 32nds, since 2/32 = 1/16?
    2/32 is the answer to both questions btw.
    Last edited by Sprinkledozer; February 1st 2013 at 05:30 PM.
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    Re: Word Problem - Basic Math

    Quote Originally Posted by Deveno View Post
    80% = 80/100 = 8/10 = 4/5.

    so what is (4/5)(10/32)?

    one you have that, say as a/b, you need to express this in 32nds.

    if (by luck) there is a number c such that bc = 32, then a/b = ac/32. the "legal limit" of remaining tread is then (10 - ac)/32.

    Which question are you referring to?
    8/32 = 1/4.
    Shouldn't the question be: how do you express 1/16 as 32nds, since 2/32 = 1/16?
    2/32 is the answer to both questions btw.
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  6. #6
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    Re: Word Problem - Basic Math

    Quote Originally Posted by Sprinkledozer View Post
    Which question are you referring to?
    8/32 = 1/4.
    Shouldn't the question be: how do you express 1/16 as 32nds, since 2/32 = 1/16?
    2/32 is the answer to both questions btw.
    80% of \frac{10}{32} equals \frac{8}{32}

    This can be calculated:

    \frac{10}{32}\cdot 0,80\rightarrow\frac{10\cdot 0,80}{32}=\frac{8}{32}

    Your question states: If 80% of the tread has worn off.

    That means that we subtract 80% from the original number:

    \frac{10-8}{32}=\frac{2}{32} inches

    As for the second part of your problem. It asks:

    How much tread wear remains if the tread depth is \frac{4}{32}?

    We now consider that the legal minimum for tread is \frac{2}{32}
    That means that we have to find the difference between the two:

    \frac{4}{32}-\frac{2}{32}=\frac{2}{32} inches of tread remaining.
    Last edited by Paze; February 2nd 2013 at 06:33 PM.
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