Question 520554: I'm having trouble setting the equation up correctly. It reads - on a recent trip, a trucker traveled 330 miles at a constant rate. Because of road construction, the trucker then had to reduce the speed by 25mph. An additional 30 miles was traveled at the reduced rate. The total time for the entire trip was 7 hours. Find the rate of the trucker for the first 330 miles.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! on a recent trip, a trucker traveled 330 miles at a constant rate.
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Because of road construction, the trucker then had to reduce the speed by 25mph.
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An additional 30 miles was traveled at the reduced rate. The total time for the entire trip was 7 hours.
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Find the rate of the trucker for the first 330 miles.
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Faster leg DATA:
distance = 330 miles ; rate = r mph ; time = d/r = 330/r hrs.
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Slower leg DATA:
distance = 30 miles ; rate = r-25 mph ; time = d/r = 30/(r-25) hrs
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Equation:
time + time = 7 hrs.
330/r + 30/(r-25) = 7 hrs.
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Multiply thru by r(r-25) to get:
330(r-25) + 30r = 7r(r-25)
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330r - 330*25 + 30r = 7r^2 - 175r
7r^2 - 535r + 330*25 = 0
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r = 55 mph (rate for the 1st 330 miles)
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Cheers,
Stan H.
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