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11 November, 09:12

A metallurgist has one alloy containing 21% titanium and another containing 50% titanium. How many pounds of each alloy must he use to make 44 pounds of a third alloy containing 37% titanium? (Round to two decimal places if necessary.)

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  1. 11 November, 10:41
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    x = amount of 21% alloy

    y = amount of 50% alloy

    The metallurgist wants a combination weighing a total of 44 lb, so

    x + y = 44

    Each pound of either alloy contributes either 0.21 or 0.5 pound of titanium. The final product needs to be comprised of 37% titanium; weighing at 44 lb, this means it should contain 0.37 * 44 = 16.28 lb. So

    0.21x + 0.5y = 16.28

    From the first equation,

    x + y = 44 = => y = 44 - x

    Substitute this into the second equation and solve for x:

    0.21x + 0.5 (44 - x) = 16.28

    -0.29x = - 5.72 = => x = 19.72

    Substitute this into the first equation to solve for y:

    19.72 + y = 44 = => y = 24.28
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