Question 99850: Hello I have two word problems that I needed help on and I need help solving them. Could you please help me on these problems? Thank you very much for your assistance.
Problem 1: You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 mph faster. What is your average speed on the way home?
Problem 2: A car radiator contains 10 liters of a 30% antifreeze solution. How many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?
Thank you,
Kumi
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Problem 1: You commute 56 miles one way to work. The trip to work takes 10 minutes longer than the trip home. Your average speed on the trip home is 8 mph faster. What is your average speed on the way home?
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To work DATA:
distance = 56 miles ; rate = x mph ; time = d/r = 56/(x) hrs
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From work DATA:
disance = 56 miles ; rate = x+8 mph ; time d/r = 56/(x+8) hrs
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EQUATION:
time to = time from + 1/6 hr
56/x = 56/(x+8) + 1/6
Multiply thru by 6x(x+8) to get:
56*6(x+8) = 56*6x + x(x+8)
2688 = x^2+8x
x^2+8x-2688=0
x = [64 +- sqrt(10816)]/2
x = [64 +- 104]/2
Positive answer:
x = 168/2 = 84 mph (rate to work)
x+8 = 84+8 = 92 mph (rate from work)
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Problem 2: A car radiator contains 10 liters of a 30% antifreeze solution. How many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?
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30% antifree DATA:
Amount = 10 L ; amt of active ingredient = 0.10*10 = 1 L
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Extracted antifreeze DATA:
Amount = x L ; amt of active ingredient = 0.10x L
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Replacement Antifree DATA:
Amount = x L ; amt of active ingredient = x L
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50% Mixture DATA:
Amount = 10 L ; amt of active ingredient = 0.50*10 = 5 L
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active - active + active = active
1 - 0.10x + x = 5
0.9x = 4
x = 4.4444... L(amount of 100% antifreeze that must be added)
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Cheers,
Stan H.
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