Question 507816: A driver averaged a speed of 20 kph more on a trip from A to B than on the return trip. The return trip took one and a half times as long. What was the average speed of the trip from a) point A to point B and b) point B to point A?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! Always start with d=rt, the fundamental distance equation.
In this case, d = d because the driver goes from A to B and then from B to A.
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A-to-B trip:
r = x
t = y
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B-to-A trip:
r = x-20
t = 3/2*y
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d=d
so
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x*y = (x-20)*(3/2*y)
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multiply both sides by 2 to eliminate fraction
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2xy =(x-20)(3y)
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2xy = 3xy -60y
xy -60y = 0
y(x-60) = 0
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x = 60
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Recall 'x' is the average speed going from A to B.
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So, x-20 = 40 is the average speed going from B to A.
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Note that we do not know 't' or 'd', but we have answered the question about 'r'.
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Done.
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