document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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?
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Algebra.Com's Answer #367571 by KMST(5328)\"\" \"About 
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).
\n" ); document.write( "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.
\n" ); document.write( "The total volume you need is 70 L.
\n" ); document.write( "The total amount of antifreeze that you need in those 70 L is
\n" ); document.write( "\"0.4%2A70\" L = 28 L
\n" ); document.write( "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.
\n" ); document.write( "The final amount of antifreeze in there will be
\n" ); document.write( "\"0.2%2870-x%29\" L from the \"70-x\" L of 20% antifreeze solution not removed, plus
\n" ); document.write( "\"x\" L that were added, so the total (as a function of \"x\" ) is
\n" ); document.write( "\"0.2%2870-x%29%2Bx=14-0.2x%2Bx=14%2B0.8x\" L
\n" ); document.write( "So we need to find the \"x\" that will make that the needed 28 L of antifreeze calculated above.
\n" ); document.write( "\"0.8x%2B14=28\" --> \"0.8x=14\" --> \"x=14%2F0.8\" --> \"highlight%28x=17.5%29\" L
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