Question 930125: Two automobiles each travel 60 km at steady rates. One car goes 6 kph faster than
the other, thereby taking 20 minutes less time for the trip. Find the rate of the
slower car.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! x = speed of slow car
y = speed of fast car
t = time of slow car
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y = x + 6
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s = d/t
t = d/s
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time of slow car:
t = 60/x
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time of fast car:
t - 20/60 = 60/y
t - 20/60 = 60/(x + 6)
t = 60/(x + 6) + 20/60
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equate times:
60/x = 60/(x + 6) + 20/60
60/x - 60/(x + 6) = 20/60
60(x + 6)/x(x + 6) - 60x/x(x + 6) = 20/60
60(x + 6) - 60x = (20/60)x(x + 6)
60x + 360 - 60x = (20/60)xx + 6*(20/60)x
(1/3)xx + 2x - 360 = 0
0.333333333xx + 2x - 360 = 0
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the above quadratic equation is in standard form, with a=0.333333333, b=2 and c=-360
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
0.333333333 2 -360
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
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x = 30
x = -36
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the negative root doesn't fit the problem statement, so use the positive root:
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x = 30
y = x + 6
y = 36
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answer:
x = speed of slow car = 30 kph
y = speed of fast car = 36 kph
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