SOLUTION: how do you do this could you explain it to me really slow {{{how many gallons of 50% antifreeze must be mixed with 80 gal of 20% antifreeze to get a mixture that is 40% antifreeze?
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Linear-equations
-> SOLUTION: how do you do this could you explain it to me really slow {{{how many gallons of 50% antifreeze must be mixed with 80 gal of 20% antifreeze to get a mixture that is 40% antifreeze?
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You can put this solution on YOUR website! how many gallons of 50% antifreeze must be mixed with 80 gal of 20% antifreeze to get a mixture that is 40% antifreeze? Gallons of mixture/Rate/Gallons of Antifreeze
X .50 .50x
80 .20 .20(80)
x+80 .40 .40(x+80)
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The 1st column is the amount of liguid
The 2nd column is the percentage of active ingredient in each amount
of liguid
The 3rd column is the actual amount of active ingredient in each amount of liquid
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EQUATION:
active ingredient you start with + active ingredient you add = active ingredient
you end up with.
0.50x + 0.20(80) = 0.40(x+80)
Multiply thru by 100 to get:
50x + 20*80 = 40(x+80)
50x + 1600 = 40x + 3200
10x = 1600
x = 160 (this is the amount of 50% antifreeze that must be added) to the
20% antifreeze to end up with 40% antifreeze.
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Comment:
That is not the way I would set it up but I tried to explain the
setup you have to work with.
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Cheers,
Stan H.
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