Question 526663: How many gallons of 70% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture of that is 60% antifreeze?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With mixture problems, you need to keep track of how much 'pure' stuff you need.
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You need to end up a mixture that is 60% antifreeze.
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You have 90 gallons of 25% antifreeze, which means it has .25*90 = 22.5 gallons of pure antifreeze mixed with a solvent (probably water).
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You will add 'x' gallons of 70% antifreeze to make the 60% mixture.
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So, at the conclusion, you will have (90+x) gallons @ 60% 'pure' antifreeze. That can be shown with the equation:
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.25*90 + .7*x = .6*(90+x)
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Multiply by 100 to eliminate fractions.
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25*90 + 70x = 60(90 +x)
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2250 + 70x = 5400 + 60x
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10x = 5400 -2250
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10x = 3150
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x = 315 gallons of 70% antifreeze.
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How much will you end up having on hand?
90 + 315 = 405 gallons
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If it is 60% antifreeze, how much 'pure' antifreeze will be in the mixture?
.6*405 = 243
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In the original 90 gallons, you have 22.5 gallons of pure antifreeze.
In the additional 315 gallons, you have .7*315 = 220.5 gallons.
22.5+220.5 = 243 gallons
Correct.
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Answer: Add 315 gallons of 70% antifreeze to 90 gallons of 25% antifreeze to make a 60% mixture.
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Done.
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