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Peter and Tom shared the driving on a certain trip. If Peter
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Updated on: 27 Oct 2013, 06:20
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Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but Peter only drove 2/5 of the total distance, what was the ratio of Peter's average speed to Tom's average speed? A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3
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Originally posted by goodglory on 04 Dec 2007, 19:07.
Last edited by Bunuel on 27 Oct 2013, 06:20, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



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Re: Peter and Tom PLEASE HELP
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04 Dec 2007, 19:41
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but peter on drove 2/5 of the total distance, what was the ration of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3
welcome to gmatclub
peter speed (2d/5)/t
tom speed d/t
ratio = peter/tom = 2/5 => 2:5



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Re: Peter and Tom PLEASE HELP
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04 Dec 2007, 22:04
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but peter on drove 2/5 of the total distance, what was the ration of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3
since the time take by each is equal, it is straight respective distances: 2:5



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The answer is suppose to be E. 2:3. It's from the GMATPrep



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I got E.
Peter
R1 * T = 2/5D
Tom
R2 * T = 3/5D
Using the easist LCM, 5, and easiest T, 1.
2/3



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goodglory wrote: The answer is suppose to be E. 2:3. It's from the GMATPrep
This is what happens when we don't read the question correctly .
A 500 level question ..and here I am stumped...



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i purposely omitted the correct answer in my first post. I don't quite understand what you mean "easier LCM is 5". How would you calculate this? Where are you plugging in your numbers. It's obvious but what is the concept here? Thank you



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goodglory wrote: i purposely omitted the correct answer in my first post. I don't quite understand what you mean "easier LCM is 5". How would you calculate this? Where are you plugging in your numbers. It's obvious but what is the concept here? Thank you
2/5 of the distance + 3/5 of the distance = total distance
2/5 of the distance + 3/5 of the distance = 5/5
set the total distance to 5 and the respective distances are 2 and 3.
set the total distance to 10 and the respective distances are 4 and 6
etc....



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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27 Oct 2013, 04:54
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but peter on drove 2/5 of the total distance, what was the ration of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3 As time is constant for both Distance is directly proportional to speed pete distance = 2/5 tom distance = 3/5 Pete distance/tom distance = pete speed/tom speed =(2/5)/(3/5) pete speed/tom speed= 2:3



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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28 Oct 2013, 03:04
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Re: Peter and Tom shared the driving on a certain trip. If Peter
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22 Nov 2013, 05:29
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but Peter only drove 2/5 of the total distance, what was the ratio of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3 If distance and time are constant then the ratio of the speeds must be 2:3 Cheers! J Kudos Rain!



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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14 May 2014, 13:27
Let Total Distance= 5 Peter drove= 2 mile and Tom Drove= 3 mile. Let, Each took 1 hour Their speed = 2/1 : 3/1 = 2: 3



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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07 Feb 2015, 06:34
Let total distance be D.
Tom's covered distance is = 2D/5 Tom's rate= 2D/5T
Peter's covered distance = D2D/5 = 3D/5 Peter's rate= 3D/5T
So Tom: Peter = 2D/5T x 5T/3D = 2 : 3



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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15 Apr 2018, 11:30
Let Peter's rate= r1= (2d/5)/t => 2d/5t= r tom's rate= r2= (3d/5)/t => 3d/5t We're asked for the ratio of their speeds Hence: r1/r2=(2d/5t)/(3d/5t) r1:r2= 2:3



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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03 Jul 2018, 03:53
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but Peter only drove 2/5 of the total distance, what was the ratio of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3 hello niks18 hope you are well ! can you pls advise if my solution is correct ? i took total distance 100 so peter did the distance 2/5 *100 = 40 km hence Tom did 60 km ratio of peters speed to toms speed is 40/60 = 2/3 Answer E
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Re: Peter and Tom shared the driving on a certain trip. If Peter
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03 Jul 2018, 05:09
dave13 wrote: goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but Peter only drove 2/5 of the total distance, what was the ratio of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3 hello niks18 hope you are well ! can you pls advise if my solution is correct ? i took total distance 100 so peter did the distance 2/5 *100 = 40 km hence Tom did 60 km ratio of peters speed to toms speed is 40/60 = 2/3 Answer E Hi dave13yes your approach is completely fine here we can directly take the ratio of distance here because both drove for the "same amount of time". if the time driven had been different then you would have to find out their speed and then arrive at the ratio.



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Re: Peter and Tom shared the driving on a certain trip. If Peter
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03 Jul 2018, 05:17
goodglory wrote: Peter and Tom shared the driving on a certain trip. If Peter and Tom both drove for the same amount of time, but Peter only drove 2/5 of the total distance, what was the ratio of Peter's average speed to Tom's average speed?
A. 1:5 B. 2:5 C. 1:2 D. 3:5 E. 2:3 Let the total distance = 5 miles, implying that Peter's distance = \(\frac{2}{5} * 5 = 2\) miles and that Tom's distance = \(52 = 3\) miles. If Peter and Tom each drive for 1 hour, then Peter's speed : Tom's speed = 2 mph : 3 mph.
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