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Math Help - Stumped

  1. #1
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    Question Stumped

    Please forgive me if I have this in the wrong forum, but I am 'stuck' my 12yr old daughter came to me with this & I want to run & hide..

    Avg. of 4#'s is 27
    Avg. of 3#'s is 31
    What is the 4th #?

    Any help seriously appreciated
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  2. #2
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    Quote Originally Posted by alpina17 View Post
    Please forgive me if I have this in the wrong forum, but I am 'stuck' my 12yr old daughter came to me with this & I want to run & hide..

    Avg. of 4#'s is 27
    Avg. of 3#'s is 31
    What is the 4th #?

    Any help seriously appreciated
    Dear alpina17,

    I don't quite understand your question. What do you mean by "Avg. of 4#'s". Is it average of four numbers in a series?
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  3. #3
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    call your 4 numbers:  x1, x2, x3, x4

    the average of these numbers is \frac{(x1 + x2 + x3 + x4)}{4}

    but you have the answer to this

    i.e. \frac{(x1 + x2 + x3 + x4)}{4} = 27

    => multiplying across by the  4 you get:  x1 + x2 + x3 + x4 = 108

    if it's the same set of numbers,  x1, x2, x3

    then the average of these is

    \frac{(x1+ x2+ x3)}{3}

    you also have the answer to this

    i.e.  \frac{(x1+ x2+ x3)}{3} = 31

    => multiplying across by the  3 you get:  x1 + x2 + x3 = 93

    so therefore  x4 must be:  108 -93 = 15
    Last edited by Tekken; April 14th 2010 at 09:06 AM. Reason: latex
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  4. #4
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    Excellent! believe it or not that's what I did; 108 - 93
    but being honest I had no idea why I was submitting 15,
    Now that you have explained, thank you very very much
    maybe my daughter will not look at me as a 'd'oh...(well probably asking too much there!)
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