Question 174976
How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 L of a 50% antifreeze solution are needed? 
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active + active = active
0.40x + 0.80(20-x) = 0.50*20
40x + 1600 - 80x = 1000
-40x = -600
x = 15 liters (amount of 40% antifreeze needed in the mixture)
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20-x = 5 liters (amount of 80% antifreeze needed in the mixture)
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Cheers,
Stan H.