SOLUTION: I do not understand how to work these problems, I am pretty good at algebra but i do not understand what formula to use or how to work through them, any help is appreciated. The

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Question 570085: I do not understand how to work these problems, I am pretty good at algebra but i do not understand what formula to use or how to work through them, any help is appreciated.
The radiator in a certain make of car needs to contain 70 liters of 40% antifreeze. the radiator now contains 70 liters of 20% antifreeze. How many liters of this solvent must be drained and replaced with 00% antifreeze to get the desired strength?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the radiator has 20% antifreeze now and it needs 40%, you may need to replace some of the liquid with 100% antifreeze (not 00%, which I think is some kind of typo).
This kind of problem requires that you do some accounting of both, the total volume, and the amount of the substance of interest that is mixed in.
The total volume you need is 70 L.
The total amount of antifreeze that you need in those 70 L is
0.4%2A70 L = 28 L
If you remove x L of what is in the radiator now, and replace them with x L of 100% antifreeze, you expect to have 70 L again in the end.
The final amount of antifreeze in there will be
0.2%2870-x%29 L from the 70-x L of 20% antifreeze solution not removed, plus
x L that were added, so the total (as a function of x ) is
0.2%2870-x%29%2Bx=14-0.2x%2Bx=14%2B0.8x L
So we need to find the x that will make that the needed 28 L of antifreeze calculated above.
0.8x%2B14=28 --> 0.8x=14 --> x=14%2F0.8 --> highlight%28x=17.5%29 L