SOLUTION: a car radiator containes 10 liters of a 30% antifreeze solution. how many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?

Question 347483: a car radiator containes 10 liters of a 30% antifreeze solution. how many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a car radiator contains 10 liters of a 30% antifreeze solution.
how many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?
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Equation:
active - active + active = active
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0.30*10 - 0.30x + 1.00x = 0.50*10 liters
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Multiply thru by 100 to get:
30*10 - 30x + 100x = 50*10
70x = 20*10
x = 200/70 = 2.8571 liters (amt. to replaced by pure antifreeze)
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Cheers,
Stan H.
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