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# A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 mi

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Re: A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 mi [#permalink]
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gmatt14 wrote:
A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 minutes later the cop tears off after him averaging a speed of 120mph. At this rate how long will it take for our friendly copper to catch the speeder?

BTW: What would be the least known variable? The cop's distance?

Solution :
in 2 mis, the motorcyclist goes : 3 miles. [ 90/60 * 2]
relative speed = 120 - 90 = 30 mph
At 30 mph rate, 3 miles will be covered in 60/30 * 3 = 6 minutes.
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Re: A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 mi [#permalink]
bike 90 mph = 1.5 miles/ minute
cop 120 mph = 2 miles/ minute

they both go the same distance (d) and bike gets 2 minute head start (t+2) and we'll call cop's time (t)
bike's equation: 1.5(t+2)=d this is R*T=D
cop's equation: 2t=d

we can set their d's equal since that's the point where they meet and we're trying to solve for t

1.5(t+2)=2t
t=6

I know other people explained it. I just wanted to boost my GMAT confidence and give it my own shot.
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Re: A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 mi [#permalink]
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Re: A cop clocks a motorcyclist speeding down the highway at 90 mph. 2 mi [#permalink]
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