SOLUTION: A 10-quart radiator contains a 30% concentration of antifreeze. how much of the sloution must be drained and replaced by 100% antifreeze to bring the solution up to 50%

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Question 419707: A 10-quart radiator contains a 30% concentration of antifreeze. how much of the sloution must be drained and replaced by 100% antifreeze to bring the solution up to 50%
Found 3 solutions by mananth, ikleyn, josgarithmetic:
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
100 ---------------- x
30 ---------------- 10-x
50.00% ---------------- 10
...
100x +30(10 -x) = 50 * 10
100x +300-30 x= 500
100x-30 x=500- 300
70 x - = 200
/70
x=2.86 liters of 100.00%
2.86 liters of 30 % will have to be drained off.

Answer by ikleyn(53418) About Me  (Show Source):
You can put this solution on YOUR website!
.


The answer "x=2.86 liters of 100% antifreeze" in the post by @matanth is incorrect.

The correct answer is "x=2.86 quarts of 100% antifreeze".



Answer by josgarithmetic(39701) About Me  (Show Source):
You can put this solution on YOUR website!
v, to remove and replace.
Total finished contents is to be unchanged.

basic setup %280.3%2A10-0.3v%2B1.0v%29%2F10=0.5

0.3%2A10-0.3v%2Bv=0.5%2A10
v-0.3v=0.5%2A10-0.3%2A10
%281-0.3%29v=%280.5-0.3%2910
v=10%28%280.5-0.3%29%2F%281-0.3%29%29
When this makes sense, just compute v.