Question 1207753: The cooling system of a certain foreign-made car has a capacity of 15 liters. If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze?
Let me see.
Mixture 1 + mixture 2 = total mixture
15(0.40) + x = 0.60(x + 15)
Is this the correct set up?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39617)   (Show Source):
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
The cooling system of a certain foreign-made car has a capacity of 15 liters.
If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained
and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze?
Let me see.
Mixture 1 + mixture 2 = total mixture
15(0.40) + x = 0.60(x + 15)
Is this the correct set up?
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Your setup equation is INCORRECT.
This problem is a standard and typical mixture problem.
To see many similar problems solved with detailed explanations, look into the lessons
- Word problems on mixtures for antifreeze solutions
in this site. Learn the subject from there.
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Here I will show you very uncommon method of solution,
which allows to solve the problem practically MENTALLY.
The amount of the pure antifreeze in 15 liters of the 40% mixture
is 0.4*15 = 6 liters.
The amount of the pure antifreeze in 15 liters of the 60% mixture
is 0.6*15 = 9 liters.
So, the original mixture contained 6 liters of the pure antifreeze;
the final mixture contains 9 liters of the pure antifreeze.
Now, Let V be the volume of the mixture to replace.
When we drain V liters of the 40% mixture, 0.4*V liters of the pure antifreeze go out.
We replace it with V liters of the pure antifreeze.
So, V should be 3 liters more than 0.4V, which means
V - 0.4V = 3, or 0.6V = 3, hence V =  = 5 liters.
At this point, the solution is complete, and we get the
ANSWER. 5 liters of the of the original 40% mixture should be drained
and replaced by 5 liters of pure antifreeze.
Solved in full and explained completely.
------------------
Re-phrasing O'Henry, the famous American writer,
The blind begin to walk and the dumb begin to see
when they receive such beautiful solutions to their problems.
There is nothing here except a mind game.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Your setup is NOT correct.
Your setup says you are mixing 15 liters of 40% antifreeze with an unknown amount of pure (100%) antifreeze to get (15+x) liters of 60% antifreeze.
The capacity of the cooling system is 15 liters, so that can't be right.
The x liters of pure antifreeze can only be put into the cooling system after x liters of the 40% antifreeze is drained out. So you are mixing (15-x) liters of the 40% antifreeze with x liters of pure antifreeze to get 15 liters of 60% antifreeze:
(15-x)(.40)+x(1.00)=15(.60)
Solve as shown in the response from the other tutor.
If formal algebra is not required, here is a solution using a quick and easy informal method that can be used to solve any 2-part mixture problem like this.
You are mixing 40% antifreeze with 100% antifreeze to get 60% antifreeze.
Look at the three percentages (on a number line, if it helps) and observe that 60% is 1/3 of the way from 40% to 100%.
That means 1/3 of the mixture needs to be the higher percentage antifreeze.
1/3 of 15 liters is 5 liters, so the mixture needs to use 5 liters of pure antifreeze. That of course means 5 liters of the 40% antifreeze must be drained to make room for the pure antifreeze.
ANSWER: 5 liters
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