Question 892117: The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze. If the capacity of the raditor is 3.4 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you have 3.4 liters of solution.
65% of that solution is antifreeze so you have .65 * 3.4 = 2.21 liters of antifreeze.
35% of that solution is water, soyou have .35 * 3.4 = 1.19 liters of water.
2.21 + 1.29 = 3.4
the numbers check out ok.
now you want to have 50% of each.
that turns out to be 1.7 liters of antifreeze and 1.7 liters of water.
2.21 - 1.7 = .51
1.7 - 1.19 = .51
it looks like you have to drain .51 liters of antifreeze and replace it with .51 liters of water.
when you do that, you will get:
2.21 - .51 = 1.7
1.19 + .51 = 1.7
now you have a 50% solution.
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