Question 147501This question is from textbook Intermediate Algebra
: A mixture of water and antifreeze in a car is 10% antifreeze. In colder climates this mixture should contain 50% antifreeze. If the radiator contains 4 gallons of fluid, how many gallons of radiator fluid should be drained and replaced with a mixture containing 80% antifreeze?
This question is from textbook Intermediate Algebra
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A mixture of water and antifreeze in a car is 10% antifreeze.
In colder climates this mixture should contain 50% antifreeze.
If the radiator contains 4 gallons of fluid, how many gallons of radiator fluid should be drained and replaced with a mixture containing 80% antifreeze?
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Start: 4 gallons liquid ; 0.1*4 = 0.4 gallons active ingredient
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After draining x gallons:
4-x gallons ; 0.1(4-x)= 0.4-0.1x gallons active ingredient
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Adding 80% antifreeze:
Add x gallons ; 0.8x gallons of active ingredient.
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Final Mixture:
4 gallons; 0.5*4 = 2 gallons active ingredient
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EQUATION:
active + active = active
0.4-0.1x+0.8x = 2
0.7x + 0.4 = 2
0.7x = 1.6
x = 2.2857 gallons (number of gallons that should be drained and replaced)
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Cheers,
Stan H.
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