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Math Help - algebra word problem

  1. #1
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    algebra word problem

    Mike got a score of 56 on a recent 25-question multiple choice test. The scoring for the test was +6 for each correct answer, -2 for each incorrect answer and 0 for each unanswered question. What is the maximum number of questions Mike could have answered correctly?

    My book says the answer is 13 but I'm not sure how to calculate this.
    Last edited by mr fantastic; March 23rd 2010 at 02:29 AM. Reason: Restored deleted question.
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  2. #2
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    well

    the maximum point is 150 (6*25)
    He got 56
    that is 150-94.
    so he had somehow minus 94 points from the maximum.

    Each wrong answer gives you minus 8 points from the maximum points.
    (-6 for not getting it correct and the minus 2 for answering wrong).
    So the question is now.
    How many questions can go wrong and still get 56 points?
    The answer is \frac{94}{8} = 12,\frac{6}{8}
    so 12 questions can go wrong and u can still get 56 points.
    That leaves you with 25-12 = 13

    Understandable?
    Regards Henry
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  3. #3
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    Hello, donnagirl!

    Mike got a score of 56 on a recent 25-question multiple choice test.
    The scoring was: +6 for each correct answer, -2 for each incorrect answer,
    and 0 for each unanswered question.

    What is the maximum number of questions Mike could have answered correctly?

    Let: . \begin{Bmatrix}R &=& \text{number correct} \\ W &=& \text{number wrong} \\ B &=& \text{number blank} \end{Bmatrix}


    We have: . R + W + B \:=\:25 .[1]

    and: . 6R - 2W + 0B \:=\:56 \quad\Rightarrow\quad  3R - W \:=\:28 .[2]


    Add [1] and [2]: . 4R + B \:=\:53  \quad\Rightarrow\quad R \:=\:\frac{53-B}{4}

    Since R is an integer, 53-B must be divisible by 4.

    The first time this happens is when B = 1.

    . . Hence: . R \:=\:\frac{53-1}{4} \:=\:\frac{52}{4} \:=\:13


    There are other values of B\!:\;5,9,14, \hdots
    . . but they produce smaller values of R.


    Therefore, Mike could have had a maximum of 13 right answers.

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