SOLUTION: How many gallons of a 60% antifreeze solution must be mixed with 60 gallons of 10% antifreeze to get a mixture that is 50% antifreeze?

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Question 1133931: How many gallons of a 60% antifreeze solution must be mixed with 60 gallons of 10% antifreeze to get a mixture that is 50% antifreeze?
Found 2 solutions by josgarithmetic, Boreal:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
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How many gallons of a 60% antifreeze solution must be mixed with 60 gallons of 10% antifreeze to get a mixture that is 50% antifreeze?
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How many gallons x, of a H% antifreeze solution must be mixed with M gallons of L% antifreeze to get a mixture that is T% antifreeze?
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highlight_green%28Hx%2BML=T%28x%2BM%29%29
Solve for x.

Hx%2BML=Tx%2BTM
Hx-Tx=TM-LM
%28H-T%29x=TM-LM
highlight%28x=%28TM-LM%29%2F%28H-T%29%29

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substitute the given values:
x=%2850%2A60-10%2A60%29%2F%2860-50%29
x=%2840%2A60%29%2F10
x=240%2F1
highlight_green%28x=240%29

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x gallons of 60%
60 gallons of 10%
60+x gallons in final
x*.60+60*.10=(60+x)*.50
.60x+6=30+0.5x
0.1x=24
x=240 gallons ANSWER
final solution is 300 gallons of .50% or 150 gallons of pure
Original is 240 gallons of .60 (144 gallons pure) 60 gallons .10 (6 gallons pure), which are 150 gallons