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24 April, 09:13

A chemist has three acid solutions. The first solution contains 15% acid, the second contains 35% and the third contains 65%. He wants to use all three solutions to obtain a mixture of 228 liters containing 25% acid, using 2 times as much of the 65% solution as the 35% solution. How many liters of each solution should be used?

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  1. 24 April, 09:34
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    171 liters of 15% acid

    19 liters of 35% acid

    38 liters of 65% acid

    Step-by-step explanation:

    If x is the liters of 15% acid, y is the liters of 35% acid, and z is the liters of 65% acid, then:

    x + y + z = 228

    0.15x + 0.35y + 0.65z = 0.25 (228)

    z = 2y

    Solve the system of equations using elimination or substitution. Using substitution:

    x + y + 2y = 228

    x + 3y = 228

    x = 228 - 3y

    0.15 (228 - 3y) + 0.35y + 0.65 (2y) = 0.25 (228)

    34.2 - 0.45y + 0.35y + 1.3y = 57

    1.2y = 22.8

    y = 19

    x = 228 - 3y = 171

    z = 2y = 38

    The chemist needs 171 liters of 15% acid, 19 liters of 35% acid, and 38 liters of 65% acid.
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