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# Carol started from home on trip averaging 30 miles per hour.

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Carol started from home on trip averaging 30 miles per hour. [#permalink]

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25 Nov 2012, 05:57
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5% (low)

Question Stats:

80% (02:19) correct 20% (01:25) wrong based on 272 sessions

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Carol started from home on trip averaging 30 miles per hour. How fast must her mother drive to catch up her in 3 hours if she leaves 30 minutes after Carol?

A. 35 m.p.h
B. 39 m.p.h
C. 40 m.p.h
D. 55 m.p.h
E. 60 m.p.h

It is a simple one, but the answer i get is 36. Maybe my calculation is wrong...
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Nov 2012, 06:14, edited 1 time in total.
Edited the question, tags and added the OA.
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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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25 Nov 2012, 06:18
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cv3t3l1na wrote:
Carol started from home on trip averaging 30 miles per hour. How fast must her mother drive to catch up her in 3 hours if she leaves 30 minutes after Carol?

A. 35 m.p.h
B. 39 m.p.h
C. 40 m.p.h
D. 55 m.p.h
E. 60 m.p.h

It is a simple one, but the answer i get is 36. Maybe my calculation is wrong...

In 30 minutes at 30 miles per hour Carol covers 15 miles.

So, we need her mother to compensate 15 miles in 3 hours. Thus relative speed must be 15/3=5 miles per hour, which means that the speed of Carol's mother should be 30+5=35 miles per hour.

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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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25 Nov 2012, 08:40
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Another approach

Before her Mom starts Carol has already travelled 15 miles(30/2)

Now in st hr her mom travels at

A. 35 miles/hr

1.Carol completes 45 miles in 1st hr after her mom whereas her mom travels 35 miles.

2nd hr-75(45+30) and 70(35+35)

3rd hr-105(75+30) and 105(70+35)
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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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26 Nov 2012, 18:33
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I know why TC got 36mph.
It comes from the understanding "catch up in three hours".
If we assume that Carol and Mom should be equal in 3 hours after Carol leaves - than answer is 36 mph. 30*3=X*2.5
However the question says that 3-hours timer starts when Mom leaves the house - answer is 35 mph. 30*3.5=X*3
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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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27 Nov 2012, 03:24
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In 3 hours carol covers 90 miles. In 3hrs 30 mins carol covers 105 miles. To cover the same distance her mother should drive at 105/3 miles/hr=35
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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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26 Nov 2013, 10:48
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cv3t3l1na wrote:
Carol started from home on trip averaging 30 miles per hour. How fast must her mother drive to catch up her in 3 hours if she leaves 30 minutes after Carol?

A. 35 m.p.h
B. 39 m.p.h
C. 40 m.p.h
D. 55 m.p.h
E. 60 m.p.h

It is a simple one, but the answer i get is 36. Maybe my calculation is wrong...

In 30 minutes (1/2hr) Carol drives 15 miles.
Relative speed is x-30, x being the rate of mother.

So (x-30)(3) = 15

X =35

Cheers
Let the kudos rain begin!!
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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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25 Mar 2015, 23:58
Hello from the GMAT Club BumpBot!

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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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31 Jul 2016, 07:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Carol started from home on trip averaging 30 miles per hour. [#permalink]

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01 Sep 2016, 21:38
The point were they will meet will be the same distace and D= Time x Speed
Carol T=3 V= 30 so D= 10
Mom T=3.5 V= x D= 10 so X = (3.5)(10) = 35
Re: Carol started from home on trip averaging 30 miles per hour.   [#permalink] 01 Sep 2016, 21:38
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