SOLUTION: Please help me. How do I solve this problem: How many gallons of a 60% antifreeze solution must be mixed with 80 gallons of a 20% antifreeze solution to get a mixture that is 50% a

Question 823471: Please help me. How do I solve this problem: How many gallons of a 60% antifreeze solution must be mixed with 80 gallons of a 20% antifreeze solution to get a mixture that is 50% antifreeze? Use the six-step method.
I haven't a clue how to solve this problem. Thank you so much for your time.
Found 2 solutions by richwmiller, stanbon:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
.6x+.2*80=.5(x+80)
x = 240.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many gallons of a 60% antifreeze solution must be mixed with 80 gallons of a 20% antifreeze solution to get a mixture that is 50% antifreeze? Use the six-step method.
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Equation:
anti + anti = anti
0.60x + 0.20*80 = 0.50(x+80)
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Multiply thru by 100 to get:
60x + 20*80 = 50x + 50*80
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10x = 30*80
x = 3*80
x = 240 (amt. of 60% solution needed)
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Cheers,
Stan H.
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