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20 December, 11:33

If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each solution, to the nearest milliliter, is needed?

A. 167 milliliters of 7% solution and 333 milliliters of 4% solution

B. 110 milliliters of the 10% solution and 210 milliliters of 4% solution

C. 155 milliliters of the 10% solution and 300 milliliters of 4% solution

D. 170 milliliters of the 10% solution and 340 milliliters of 4% solution

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  1. 20 December, 11:52
    0
    The correct answer is A. 167 milliliters of 7% solution and 333 milliliters of 4% solution and here is how:

    If x is the number of milliliters of the 7% saline solution and y is the number of milliliters of the 4% saline solution then add up to 500 milliliters total, so x + y = 500.

    and if we do

    x + y = 500

    x = 500 - y

    0.07x + 0.04y = 25 (substitute 500 - y for x)

    0.07 (500 - y) + 0.04y = 25

    35 - 0.07y + 0.04y = 25

    -0.03y + 35 = 25

    -0.03y = - 10

    y = 333.333 ...

    y = about 333

    x = 500 - y = 500 - 333 = 167

    Then you know why the answer is A.
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