SOLUTION: A mechanic needs an antifreeze-water mixture that 60% antifreeze to fill a radiator. The mechanic has antifreeze that is a 50% concentration and antifreeze that is 100% concentrati

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Question 1168702: A mechanic needs an antifreeze-water mixture that 60% antifreeze to fill a radiator. The mechanic has antifreeze that is a 50% concentration and antifreeze that is 100% concentration. How many liters of each concentration of antifreeze must be mixed to produce 10L of 60% antifreeze?

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Target is 60% antifreeze.
60% is closer to 50% and farther from 100%.
MOST of the antifreeze should come from the lower concentration 50% antifreeze.


%28100-60%29%2F%28100-50%29 of the 10L is of the 50% antifreeze.
4%2F5 of 10L is of the 50% antifreeze.

8L of 50% and 2L of 100%.

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If you would do the solution with algebra and hold off on all computation until the very end, you would see that same fraction (or 1 minus it).

v, how much 100%
10-v, how much 50%
100v%2B50%2810-v%29=10%2A60----to account for pure antifreeze itself
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100v%2B50%2A10-50v=10%2A60
100v-50v=10%2A60-10%2A50
%28100-50%29v=10%2860-50%29
v=10%28%2860-50%29%2F%28100-50%29%29------see the one-minus-the-other fraction form; depends on which antifreeze the variable is assigned.

%2860-50%29%2F%28100-50%29=10%2F50=1%2F5