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Math Help - Challenging Money Math Question

  1. #1
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    Challenging Money Math Question

    How best can I approach the question below:



    If Kwame had 3 cents more he would have twice as much as Ama. If he had 4 cents less, he would have the same amount.
    How many cents does Kwame have?

    a)4


    b)7


    c)11


    d)14



    e)none of the above
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  2. #2
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    Re: Challenging Money Math Question

    Quote Originally Posted by KayPee View Post
    How best can I approach the question below:



    If Kwame had 3 cents more he would have twice as much as Ama. If he had 4 cents less, he would have the same amount.
    How many cents does Kwame have?

    a)4


    b)7


    c)11


    d)14



    e)none of the above
    Let a be the amount that Ama has, now rewrite what you are told in terms of a.

    CB
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  3. #3
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    Re: Challenging Money Math Question

    Let a be what Ama has and b be what Kwame has
    b + 3 = 2a
    b - 4= 2a

    a = b + (3/2) = a

    Substituting a into equation 2
    b-4 = b + 3

    I am stuck and the equation doesn’t sound ok

    Where did I go wrong?
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  4. #4
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    Re: Challenging Money Math Question

    Quote Originally Posted by KayPee View Post
    Let a be what Ama has and b be what Kwame has
    b + 3 = 2a
    b - 4= a
    See the change in red above

    CB
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  5. #5
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    Re: Challenging Money Math Question

    I solved it without using a (number thatAma has) in any equation. Here is how I did it (k being the number of Kwame's cents):

    k + 3 = 2(k-4)
    k + 3 = 2k - 8
    k + 11 = 2k
    11 = k

    So Kwame has 11 cents. Using the simultaneous equations method you were using you can reach the same equation.

    k + 3 = 2a
    k - 4 = a

    Using substituion and subbing the first into the second you can get to k + 3 = 2(k-4), as named earlier. Using elimination you can subtract the second from the first and get to 7 = a. Then sub the value for a into either equation to get k = 11.
    Last edited by Ross137; September 7th 2011 at 04:32 AM.
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  6. #6
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    Re: Challenging Money Math Question

    Quote Originally Posted by Ross137 View Post
    k + 3 = 2a
    k - 4 = a

    Using substituion and subbing the first into the second you can get to k + 3 = 2(k-4), as named earlier. Using elimination you can subtract the second from the first and get to 7 = a. Then sub the value for a into either equation to get k = 11.
    Simply subtract 2nd equation from 1st: k - k + 3 - (-4) = 2a - a
    So 7 = a
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  7. #7
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    Re: Challenging Money Math Question

    Yep - that's what I meant by elimination
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