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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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New post Updated on: 25 Nov 2012, 06:14
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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...

Originally posted by cv3t3l1na on 25 Nov 2012, 05:57.
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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New post 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.

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

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New post 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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New post 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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New post 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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New post 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
Answer is A

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

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

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New post 20 Jul 2017, 00:07
Total distant covered by Carol after his mother starts driving upto next 3 hours is 15(30mins)+30(1hr)+60(2hr)+60(3hr)=105 miles. Divide that by 3 hours (total time taken) gives 35.
Note: Mother has to drive 15 miles extra to compensate the half hour delay as stated in Problem statement.
Option A.
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Re: Carol started from home on trip averaging 30 miles per hour.  [#permalink]

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New post 27 Sep 2017, 23:54
Can someone let me know why is this method incorrect. Carol: speed is 30 mph , Time =3 hrs and the distance is 30*3=90 miles
Mother: Speed=x ,Time=2.5 hrs and distance is a constant = 90 miles
Therefore the speed is 36 mph
Why is this method incorrect??
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Carol started from home on trip averaging 30 miles per hour.  [#permalink]

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New post 28 Sep 2017, 16:16
longhaul123 wrote:
Can someone let me know why is this method incorrect. Carol: speed is 30 mph , Time =3 hrs and the distance is 30*3=90 miles
Mother: Speed=x ,Time=2.5 hrs and distance is a constant = 90 miles
Therefore the speed is 36 mph
Why is this method incorrect??

longhaul123 , think again about distance. Carol drives for 3 hours and 30 minutes.

And if you're going to calculate distance this way, look at the post by FB2017 -- directly above yours, or the one from rajathpanta .

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

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New post 03 Oct 2017, 10:06
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


We know that Carol’s rate is 30, and let’s let her mom’s rate = r. We know that her mom will have driven for 3 hours, and Carol will have driven for (3 + ½), or 3.5, hours when her mom catches up. Using the formula distance = rate x time, and knowing that both Carol and her mom have driven equal distances, we have:

30(3.5) = 3r

105 = 3r

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

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New post 03 Oct 2017, 22:36
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...


Speed of Carol = 30 mph
In 30 minutes Carol travels = 30 * (30/60) = 15 miles

Let the speed of carol's mother = V
So relative speed of Carol's mother w.r.t carol = V-30

Relative distance to be covered by Carol's mother in 3 hours = 15 miles with a relative speed of (V-30) mph

So, 15/(V-30) = 3
V = 35 mph

Answer A
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Re: Carol started from home on trip averaging 30 miles per hour. &nbs [#permalink] 03 Oct 2017, 22:36
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