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Math Help - mixture problem

  1. #1
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    mixture problem

    I have two cans of paint. Can A has 9 parts of blue paint to one part of yellow paint. Can B is 20% blue and the rest is yellow. How much paint should I use from each can to obtain 4 liters of paint which is half blue and half yellow.

    Answers in liters.
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  2. #2
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    Hello, math34a!

    I have two cans of paint.
    Can A has 9 parts of blue paint to one part of yellow paint.
    Can B is 20% blue and the rest is yellow.
    How much paint should I use from each can to obtain
    4 liters of paint which is half blue and half yellow.
    Answers in liters.
    This is a standard "mixture problem".


    Can A is 90% blue.
    Can B is 20% blue.
    . . We want 4 liters of mixture which contains 2 liters of blue.

    Let x = number of liters from can A.
    . . It contains: . 0.90x liters of blue.

    Then 4-x = number of liters from can B.
    . . It contains: . 0.20(4-x) liters of blue.

    Together, this mixture will contains 2 liters of blue.


    There is our equation: . 0.90x + 0.20(4-x) \:=\:2

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  3. #3
    MHF Contributor
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    Quote Originally Posted by math34a View Post
    I have two cans of paint. Can A has 9 parts of blue paint to one part of yellow paint. Can B is 20% blue and the rest is yellow. How much paint should I use from each can to obtain 4 liters of paint which is half blue and half yellow.

    Answers in liters.
    Here is one way.

    Let A = liters of paint to get from can A.
    And B = liters of paint to get fropm can B.

    For blue parts,
    A(9) +B(2) = 4(5)
    9A +2B = 20 --------------(1)

    For yellow parts,
    A(1) +B(8) = 4(5)
    A +8B = 20 -----------------(2)

    20 = 20,
    9A +2B = A +8B
    9A -A = 8B -2B
    8A = 6B
    A = 6B /8 = (3/4)B ------(i)
    Substitute that into, say, (2),
    (3/4)B +8B = 20
    8.75B = 20
    B = 20/8.75) = 2.286 liters

    So,
    A = (3/4)B = 0.75(2.286) = 1.714 liters

    Check,
    A +B = 4
    1.714 +2.286 =? 4
    4 =? 4
    Yes, so, OK.
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