SOLUTION: Help!! How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 gallons of a 50% antifreeze solution are needed?

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Question 103123This question is from textbook Intermediate Algebra
: Help!!
How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 gallons of a 50% antifreeze solution are needed? This question is from textbook Intermediate Algebra

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How much of a 40% antifreeze solution must a mechanic mix with an 80% antifreeze solution if 20 gallons of a 50% antifreeze solution are needed?
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Let amt of 80% solution be x gallons: amt of active ingredient= 0.80x gallons
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Amt of 40% solution is "20-x" gals; amt of active is 0.4(20-x)=8-0.4x gallons
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Amt of 50% mixture is 20 gals: amt of active ingredient =0.5*20=10 gallons
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EQUATION:
active + active = active
0.8x + 8 - 0.4x = 10
0.4x = 2
x = 5 gallons (amt of 80% antifreeze in the mixture)
20-x = 15 gallong (amt of 40% antifreeze in the mixture)
Cheers,
Stan H.