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Math Help - Annoying Puzzle

  1. #1
    Senior Member Mukilab's Avatar
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    Annoying Puzzle

    What should be the value of the fourth column?


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  2. #2
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    Quote Originally Posted by Mukilab View Post
    What should be the value of the fourth column?


    Write a system of equations using each Blue and Red as a variables and the 2nd and 3rd columns.

    Then use the first column to find yellow.

    Then the final answer is R+2Y+B=41
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  3. #3
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    Write a system of equations using each Blue and Red as a variables and the 2nd and 3rd columns.

    Then use the first column to find yellow.

    Then the final answer is R+2Y+B=41

    That's exactly what I did but I ended up scrawling all over my notepad. I'll try again and see if I get your answer.
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  4. #4
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    Quote Originally Posted by Mukilab View Post
    What should be the value of the fourth column?


    call red is R, yellow is Y and blue is B
    so on the first column we have 2R+Y+B=43
    second column we have Y+3B=55
    last column we have 3R+B=45
    so you have
    2R+Y+B=43
    Y+3B=55
    3R+B=45
    then you can find R, B, and Y by yourself. try this and talk your answer.
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  5. #5
    Senior Member Mukilab's Avatar
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    Sorry, I don't know how I got muddled up. I think it's because i named them a, b and c then got a mixed with c and so on...

    Anyways I got

    Y+2=R
    39-3Y=B

    SUB INTO: R+2Y+B=?

    Y+2+2Y+39-3Y

    41=?


    Thanks
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  6. #6
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    Hello, Mukilab!

    What should be the value of the fourth column?

    We have a system of equations:

    . . \begin{array}{ccccc}\text{Column 1:} & 2R + Y + B &=& 43 \\ \text{Column 2:} & R \qquad+ 3B &=& 55 \\ \text{Colukn 3:} & 3R \qquad + B &=& 45 \end{array}

    Solution: . R = 10,\;Y = 8,\;B = 15


    Columkn 4: . R + Y + B + Y \;=\;10+8+15+8\;=\;41

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  7. #7
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by Mukilab View Post
    Sorry, I don't know how I got muddled up. I think it's because i named them a, b and c then got a mixed with c and so on...

    Anyways I got

    Y+2=R
    39-3Y=B

    SUB INTO: R+2Y+B=?

    Y+2+2Y+39-3Y

    41=?


    Thanks
    Quote Originally Posted by Soroban View Post
    Hello, Mukilab!

    We have a system of equations:

    . . \begin{array}{ccccc}\text{Column 1:} & 2R + Y + B &=& 43 \\ \text{Column 2:} & R \qquad+ 3B &=& 55 \\ \text{Colukn 3:} & 3R \qquad + B &=& 45 \end{array}

    Solution: . R = 10,\;Y = 8,\;B = 15


    Columkn 4: . R + Y + B + Y \;=\;10+8+15+8\;=\;41


    Sorry for the misunderstanding, I thought my previous post showed I got it. Thanks for the contribution though!
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