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A car travels from Mayville to Rome at an average speed of 3

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A car travels from Mayville to Rome at an average speed of 3 [#permalink]

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New post 18 Jul 2010, 10:53
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A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is closest to the average speed, in miles per hour, for the round trip?

○ 32.0
○ 33.0
○ 34.3
○ 35.5
○ 36.5
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jun 2013, 01:26, edited 1 time in total.
Edited the OA.
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Re: Speed Distance [#permalink]

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EnterMatrix wrote:
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is closest to the average speed, in miles per hour, for the round trip?
○ 32.0
○ 33.0
○ 34.3
○ 35.5
○ 36.5


Choose some smart number for the distance from Mayville to Rome: let the distance be 120 miles (120 is LCM of 30 and 40 and this will make further calculations easier).

\(Average \ speed=\frac{total \ distance}{total \ time}\) --> so the nominator, total distance would be \(2*120=240\);

Total time equals time from Mayville to Rome plus time from Rome to Mayville --> \(\frac{120}{30}+\frac{120}{40}=7\);

\(Average \ speed=\frac{total \ distance}{total \ time}=\frac{240}{7}\approx{34.3}\).

Answer: C.
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Re: Speed Distance [#permalink]

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New post 20 Jul 2010, 06:57
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IMO C

EnterMatrix wrote:
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is closest to the average speed, in miles per hour, for the round trip?
○ 32.0
○ 33.0
○ 34.3
○ 35.5
○ 36.5


Solution:

Average speed for the entire trip = 2uv/(u+v),
where u = speed of going
v = speed of returning


Thus, average speed = 2*40*30/(70) = 34.3 mi/h

Hence, C
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Re: Speed Distance [#permalink]

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New post 20 Jul 2010, 10:20
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EnterMatrix wrote:
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is closest to the average speed, in miles per hour, for the round trip?
○ 32.0
○ 33.0
○ 34.3
○ 35.5
○ 36.5


It is pretty involuntary to come up with 35 miles per hour as the average speed in this case. But, that is wrong. Since, the car has taken more time to travel at 30 mph and less time to travel at 40 mph, the avg speed of the car would be somewhere between 30 and 35(mean) and closer to 35. If I had no time on my hands and had to make a calculated guess, I would have chosen C.

The correct way to find the avg speed is to calculate the total distance and the total time and divide the total distance by total time.

Let the one way distance in this case be D. So the total distance is 2D(travelled over the same route twice)

Total Time = Time taken to travel distance D at 30 mph + time taken to travel distance D at 40 mph
= \((D/30) + (D/40)\)
= \(7D/120\)
Thus avg speed = Total distance/total time
= \((2D)/(7D/120)\)
= 34.285

Hope it helps,
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Re: Speed Distance [#permalink]

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New post 17 Aug 2010, 03:44
Sorry but the OA looks wrong. Answer is C by formula of average speed = 2uv/u+v
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Re: Speed Distance [#permalink]

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New post 17 Aug 2010, 05:34
oldstudent wrote:
Sorry but the OA looks wrong. Answer is C by formula of average speed = 2uv/u+v


By formula of average speed, the answer is 240/7 = 34.2857..... , rounded off to 34.3. Hence, answer choice C.
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Re: Speed Distance [#permalink]

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New post 17 Aug 2010, 14:40
the OA is wrong, answer should be C
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Re: A car travels from Mayville to Rome at an average speed of 3 [#permalink]

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New post 26 Sep 2013, 23:30
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Re: A car travels from Mayville to Rome at an average speed of 3 [#permalink]

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New post 27 Apr 2015, 12:23
Hello from the GMAT Club BumpBot!

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A car travels from Mayville to Rome at an average speed of 3 [#permalink]

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New post 27 Apr 2015, 12:47
We can pick any number for distance that is divisible by both 30 and 40. Since we have rate and distance, by plugging in the rate formula (r=d/t) we can solve for the average rate/speed
    City``` Rate--- Time ---Distance
    M-R.... 30--- 4 ---- 120
    R-M ... 40 --- 3 ---- 120
    Total 34.28 --- 7 ---- 240
34.28 ~ 34.3
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A car travels from Mayville to Rome at an average speed of 3   [#permalink] 27 Apr 2015, 12:47
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