SOLUTION: on a recent trip, a trucker traveled 330 mi at a constant rate. because of road construction, the trucker then had to reduce the speed by 25 mph. an additional 30 mi was traveled
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Question 376163: on a recent trip, a trucker traveled 330 mi at a constant rate. because of road construction, the trucker then had to reduce the speed by 25 mph. an additional 30 mi was traveled at the reduced rate. the total time for the entire trip was 7 h. find the rate of the trucker for the first 330 mi.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let speed for 330 miles be x
next 30 miles x-25 mph.
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time for 330 miles +time for 30 miles = 7
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t=d/r
330/x + 30/(x-25)=7
LCD = x(x-25)
(330(x-25)+30x )/(x(x-25)= 7
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(330x-8250+30x)/(x(x-25)=7
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360x-8250 = 7*(x(x-25)
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360x-8250 = 7x^2-175x
7x^2-360x-175x+8250=0
7x^2-535x+8250=0
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find roots x1,x2 by quadratic formula
discriminant =b^2-4ac= 55225
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x1=(535+sqrt(55225))/14
x1=55 mph
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x2= (535-sqrt(55225))/14
x2=21.42 mph. ( not possible since speed was reduced by 25 mph.)
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So speed = 55 mph for the 330 mile part
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m.ananth@hotmail.ca
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