Question 181478: 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 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
Found 3 solutions by Mathtut, stanbon, Alan3354:
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website! let x be the amount drained and added back to the radiator
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we know that the original mixture is 35% water so there is 1.05 liters of water
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we also know that as we drain this, whatever we drain has 35% water. so the drained portion of water(or water lost while draining), would be .35x.
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we also know what ever we do drain we have to add back, that is x
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our end result needs to be 50% of capacity( which is 3 liters)
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so .35(3)-.35x + x =.5(3)
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1.05-.35x + x=1.5
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.65x=.45
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 liters
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The radiator in a car is filled with a solution of 65 per cent antifreeze and 35 per cent water.
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The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50 per cent antifreeze.
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If the capacity of the raditor is 3 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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antifreeze - antifreeze + + antifreeze = antifreeze
0.65*3 - 0.65x + 0*x = 0.50*3
Multiply thru by 100 to get:
65*3 - 65x + 0*x = 50*3
195 = 65x + 0 = 150
-65x = -45
x = 9/13 liters (amt. of liguid that must be replace)
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Cheers,
Stan H.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! 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 liters, how much coolant (in liters) must be drained and replaced with pure water to reduce the antifreeze concentration to 50 per cent?
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At the start, it has 3 liters of 65% antifreeze, so that's 3*0.65 = 1.95 l of AF.
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At the end, we want 3 liters of 50% AF, which will be 1.5 l of AF in the radiator. So 0.45 liters of AF must be removed.
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To drain 0.45 l of a solution that's 65% AF, we must drain 0.45/0.65 liters. See that? Each liter drained removes 0.65 liters of AF. We need to remover 0.45 liters, so we drain 0.45/0.65 liters, = 9/13 liters, apx 692 milliliters.
Then add the water.
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