<PRE><B>Question 53186</B>
One solution contains 10% alcohol while another contains 30% alcohol. How many liters of each should be mixed to give 10 L of a solution which is 25% alcohol?
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We know that the two mixture amts add up to 10L;
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Let x = 10% amt; then (10-x) = 30% amt
:
Write a percent equation:
:
.10x + .30(10-x) = .25(10)
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.10x + 3.0 - .3x  = 2.5
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  .1x - .3x = 2.5 - 3.0
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       -.2x = -.5
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          x = -.5/-.2
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          x = + 2.5 liters of 10% mixture
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10 - 2.5 = 7.5 liters of the 30% mixture
:
:
Check our solutions:
.10(2.5) + .30(7.5) = .25(10)
    .25  +   2.25   = 2.5
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