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Math Help - Finding the median of a data set

  1. #1
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    Finding the median of a data set

    A data set has 15 elements. The 15 elements in a second data set are obtained by multiplying each element in the first data set by 10. The 15 elements in a third data set are obtained by decreasing each element of the second data set by 20. The median of the third data set is 50. What is the median of the first data set?

    I know the answer is 7 but I don't know how to set up this problem.
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  2. #2
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    50 + 20 = 70

    70 / 10 = 7

    What is the point of this problem? Is there a text book? What section is it in?
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  3. #3
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    It's a practice problem for the ACT, why do we do the steps you posted above? I need more explanation..
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  4. #4
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    Start with the given information: 50
    Work backwards. What decreased by 20 is 50? 70!
    What, when multiplied by 10, is 70? 7!
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  5. #5
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    Instead of working backwards, could you let x = the sum of the 1st data set. Then 10x is the sum of the 2nd? Or am I doing it wrong? I'm not sure how you could express the 3rd data set.
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  6. #6
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    It is the same, whether you wokd backwards deliberately or let the algebra do it for you. You must somehow start with the known information and get the the information you seek. Unique answers don't care how you find them. However, for standardized or competitive exams, there may be an additional element of time or speed. You will need to create solutions, but faster solutions create higher scores.
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  7. #7
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    Re: Finding the median of a data set

    Since the median may not have a nice algebraic expression (i.e., it may land between two numbers and we need to round), it is probably best to work backward from the given median value. You can express the transformations for each dataset algebraically to understand each of the datasets. Let them be denoted by the first capital letters, respectively. Then we start with A having 15 elements.

    B = 10A
    C = B - 20 = 10A - 20 = 10(A - 2)

    Since the median of C = 50, we just follow the backward steps TKHunny pointed out. We can also use the algebra now,

    10(Med(A) - 2) = Med(C) \Rightarrow Med(A) = \frac{Med(C)}{10} + 2 \Rightarrow Med(A) = 7

    Note, this approach is the same as TKHunny's save for the fact I used "10(A - 2)" instead of "10A - 20" to algebraically solve for Med(A). If we used the other form we would have had the same arithmetic as TKHunny's:

    10Med(A) - 20 = Med(C) \Rightarrow Med(A) = \frac{Med(C) + 20}{10} \Rightarrow Med(A) = 7
    Last edited by bryangoodrich; June 13th 2011 at 01:04 PM. Reason: An elaboration
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