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17 April, 07:01

The first solution contains 15 % acid, the second contains 25 % acid, and the third contains 70 % acid. She created 48 liters of a 45 % acid mixture, using all three solutions. The number of liters of 70 % solution used is 2 times the number of liters of 25 % solution used. How many liters of each solution should be used?

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  1. 17 April, 08:25
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    First solution = 12 Liters

    Second solution = 12 Liters

    Third solution = 24 Liters

    Step-by-step explanation:

    Let the number of liters of 25% acid = y

    The number of liters of 70 % acid = 2y

    The number of liters of 15 % acid = 48-3y

    Volume of acid in first solution = 0.15 (48-3y)

    Volume of acid in second solution = 0.25y

    Volume of acid in third solution = 0.7 (2y)

    Volume of acid in mixture = 0.45 x48

    = 21.6 liters

    Assuming no loss of volume,

    0.15 (48-3y) + 0.25y + 0.7 (2y) = 21.6

    7.2 - 0.45y + 0.25y + 1.4y = 21.6

    7.2 + 1.2y = 21.6

    1.2y = 21.6 - 7.2

    1.2 y = 14.4

    y = 14.4/1.2

    y = 12 liters

    Substituting the value of y above:

    Volume of first solution = 48-3y

    = 48 - 3 (12)

    = 48-36

    = 12 liters

    Volume of second solution = y

    = 12 liters

    Volume of third solution = 2y

    = 2 (12)

    = 24 liters
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