dune buggie

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dune buggie

by pappueshwar » Thu Mar 22, 2012 9:26 am
Two dune buggies start out accross the desert at the same time. They drive in a straight line at 25 miles per hour for two hours. Dune buggy A then stops for an hour while dune buggie B continues on at the same steady rate. After and hour, buggy A begins to moving again in the same straight line as B, but this time at a constant speed of 30 miles per hour. If both Dune buggies hold their rates and drive in the same line, how long, in hours, will it take for A to catch up to B?

A. 1
B. 2
C. 2.5
D. 4
E. 5

NO OA. pls assist

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by Whitney Garner » Thu Mar 22, 2012 9:44 am
pappueshwar wrote:Two dune buggies start out accross the desert at the same time. They drive in a straight line at 25 miles per hour for two hours. Dune buggy A then stops for an hour while dune buggie B continues on at the same steady rate. After and hour, buggy A begins to moving again in the same straight line as B, but this time at a constant speed of 30 miles per hour. If both Dune buggies hold their rates and drive in the same line, how long, in hours, will it take for A to catch up to B?

A. 1
B. 2
C. 2.5
D. 4
E. 5

NO OA. pls assist
Hi pappueshwar!

There are a TON of different ways to solve D=RT problems, but I'll be honest, my favorite is "Brute Force" - just move the cars through time/space and see what happens.

So the story says they are leaving from the same place going the same direction (or else A would not "catch up" to B). And for the first 2 hours they are side-by-side so we can throw all of that information away. Now, A stops for an hour and then comes back at a faster rate. So let's see what happens. I'm going to call time=0 hours the moment when A starts moving again so that we can calculate the time it takes for A to catch up.

Hrs | Am | Bm
t=0 | 0m | 25m (definitely not caught up)
t=1 |30m | 50m (still not caught up)
t=2 |60m | 75m (not yet)
t=3 |90m |100m (still not yet!)
t=4 |120m|125m (still not caught up, BUT only one choice is over 4 hours)
t=5 |150m|150m (yep, we've caught up)

Is this the fastest method, no, but it works for EVERY GMAT D=RT problem so I love to practice it.

There is a faster way here. If you remember that A is catching up, its only his "relative" speed that matters. Notice that by going 30 mph, A is traveling at 5mph faster than car B. This is the rate by which he will "catch-up". So, he has to make up the 25 mile discrepancy with 5 miles/hour to do it.

D = R*T
25 = 5*t
t = 5 hours

Hope this helps!
:)
Whit
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Math is a lot like love - a simple idea that can easily get complicated :)

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by killer1387 » Thu Mar 22, 2012 9:50 am
pappueshwar wrote:Two dune buggies start out accross the desert at the same time. They drive in a straight line at 25 miles per hour for two hours. Dune buggy A then stops for an hour while dune buggie B continues on at the same steady rate. After and hour, buggy A begins to moving again in the same straight line as B, but this time at a constant speed of 30 miles per hour. If both Dune buggies hold their rates and drive in the same line, how long, in hours, will it take for A to catch up to B?

A. 1
B. 2
C. 2.5
D. 4
E. 5

NO OA. pls assist
after one hour distance traveled by B= 25 miles

let at time t and distance y from B they meet
then 30t=25+y
and 25t=y
hence t=5

E

P.S.: assuming the part in bold i.e. AND to be AN.

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by pappueshwar » Thu Mar 22, 2012 9:52 am
hi,
thats was a grt expln. your force method is a good way of analyzing the problem

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by GMATGuruNY » Thu Mar 22, 2012 11:39 am
pappueshwar wrote:Two dune buggies start out accross the desert at the same time. They drive in a straight line at 25 miles per hour for two hours. Dune buggy A then stops for an hour while dune buggie B continues on at the same steady rate. After and hour, buggy A begins to moving again in the same straight line as B, but this time at a constant speed of 30 miles per hour. If both Dune buggies hold their rates and drive in the same line, how long, in hours, will it take for A to catch up to B?

A. 1
B. 2
C. 2.5
D. 4
E. 5

NO OA. pls assist
Another approach.

After A stops, the distance traveled by B in one hour = 25 miles.
Thus, A has to catch up by 25 miles.
When elements COMPETE, SUBTRACT their rates.
Since A now travels at 30 miles per hour while B still travels at 25 miles per hour, every hour A travels 30-25 = 5 more miles than B.
Thus, time for A to catch up = 25/5 = 5 hours.

The correct answer is E.
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