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It is necessary to have 40% antifreeze solution in the radiator of a certain car. The radiator now has 50 lites of 20% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
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Answers (1)
let
x = amount to be replaced
50 - x = volume in radiator after x is drained
0.2(50 - x) = amount of 100% antifreeze in what's left in radiator
so
0.2(50 - x) + x = 0.5(50)
x = 18.75 L
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