SOLUTION: The radiator of a car is filled with 3.8L of a solution of 45% antifreeze. How much of the solution should be drained and replaced with antifreeze to increase the concentration of
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Question 694592: The radiator of a car is filled with 3.8L of a solution of 45% antifreeze. How much of the solution should be drained and replaced with antifreeze to increase the concentration of antifreeze to 60%? Round to the nearest tenth of liter.
I tried the problem and got it wrong. I did:
3.8L/1=45%/60% = 2.9L Found 2 solutions by ankor@dixie-net.com, josmiceli:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The radiator of a car is filled with 3.8L of a solution of 45% antifreeze.
How much of the solution should be drained and replaced with antifreeze to increase the concentration of antifreeze to 60%?
Round to the nearest tenth of liter.
:
Use this equation:
:
Let x = amt to be removed and then added
.45(3.8-x) + 1x = .60(3.8)
You can put this solution on YOUR website! You start with 3.8 liters and end with 3.8 liters
This is one of the keys. liters of pure antifreeze to start with
Let = liters to be drained off and replaced
with pure antifreeze = liters of pure antifreeze drained off.
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1.036 liters of the solution should be drained and
replaced with pure antifreeze
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check:
Did I get it right?
