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M07 Q24

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M07 Q24 [#permalink] New post 24 Sep 2010, 00:59
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A
B
C
D
E

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Quote:
A plane can fly from Europe to North America in 12.4 hours, but the return trip only takes 10.26 hours. If the average ground speed of a plane is 700 km/h, what is the approximate distance between Europe and North America?

A. 8680
B. 8273
C. 7931
D. 7245
E. 7131



I first solved it in the straightforward way -- average time (11.33 hours) and then multiply by 700km/h.

I was confused after solving this, however, as weighted rates skew off the average. For example, if I walk from home to work at 4mph, but then run home from work at 6mph, my average speed isn't 5mph -- it's 4.8mph, because I spent more time walking the 4mph trip.

Could anybody explain why that situation doesn't apply here?
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Last edited by NewSc2 on 24 Sep 2010, 11:46, edited 1 time in total.
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Re: M07 Q24 [#permalink] New post 24 Sep 2010, 07:52
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps. :)
btw is the answer ~ 9064 km approximately?
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Joined: 17 May 2010
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Location: United States
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GPA: 3.26
WE: Brand Management (Consumer Products)
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Kudos [?]: 27 [0], given: 5

Re: M07 Q24 [#permalink] New post 24 Sep 2010, 11:46
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps. :)
btw is the answer ~ 9064 km approximately?


No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.
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Re: M07 Q24 [#permalink] New post 11 Apr 2011, 22:14
NewSc2 wrote:
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps. :)
btw is the answer ~ 9064 km approximately?


No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.



Can any of experts comment on this problem. I have problem understanding the concept as well. This problem should be solved using the formula Average Distance= Speed * Time, since the time and speed are different for two trips (from A to b and from b to A because there was the speed of Earth which rotates one way), I am trying to use the formula Average Distance= (700+v)*Time, where v is the speed of the Earth's rotation as plane moves from A to B.
However, the OA solution to this problem is very straightforward, taking the average of time and multiplying it by 700.
Any help with this problem would be appreciated. Thank you.
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Re: M07 Q24 [#permalink] New post 11 Apr 2011, 23:45
I got C.

2d / 700 =total time

Or this is the same as d = average speed * average time. = 700 * 11.33
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Re: M07 Q24 [#permalink] New post 11 Apr 2011, 23:49
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mirzohidjon wrote:
NewSc2 wrote:
Pinali wrote:
I don't see any reason why it can't apply here . The only difference here is we know the value of time and unknown is speed. while in the example you said time is unknown factor and we know the speed . I hope this helps. :)
btw is the answer ~ 9064 km approximately?


No, the answer is 700*11.33 = 7931. I went with the assumption that the plane would have spent *more* time flying the slower speed, so the actual distance would be slightly more than 7931, but this was wrong. I've edited my post to show the answer choices.

The only reasoning I can see is that weighted rates skew speed, and not time spent (e.g., if time spent walking to work was 6 hours, and walking home 4 hours -- average is 5 hours), but this question still relies on speed to calculate the distance.



Can any of experts comment on this problem. I have problem understanding the concept as well. This problem should be solved using the formula Average Distance= Speed * Time, since the time and speed are different for two trips (from A to b and from b to A because there was the speed of Earth which rotates one way), I am trying to use the formula Average Distance= (700+v)*Time, where v is the speed of the Earth's rotation as plane moves from A to B.
However, the OA solution to this problem is very straightforward, taking the average of time and multiplying it by 700.
Any help with this problem would be appreciated. Thank you.


Speed_{Average} = \frac{Total \hspace{2} Distance}{Total \hspace{2} Time}

Speed_{Average} = \frac{Total \hspace{2} Distance}{Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}+Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}}

Speed_{Average} = 700

Let the distance between A to B be D.
Total \hspace{2} Distance = D+D=2D

Time \hspace{2} Taken{A \hspace{2} to \hspace{2} B}=12.4

Time \hspace{2} Taken{B \hspace{2} to \hspace{2} A}=10.26

Total \hspace{2} Time = 12.4+10.26=22.66

700=\frac{2D}{22.66}

D=\frac{700*22.66}{2}=7931

Ans: "C"
****************************************

Coming to your question:
"Average Distance= (700+v)*Time". I am not aware of any such formula.
Moreover, distance between 2 points A and B is constant. What is average distance then, I wonder!!!

700= Average Speed of the plane both ways.
Earth's speed = \nu_{e}
Plane's speed = \nu_{p}

A to B: Speed= Plane's speed-Earth's speed=\nu_{p}-\nu_{e}
B to A: Speed= Plane's speed+Earth's speed=\nu_{p}+\nu_{e}

From A to B:
Time=\frac{Distance}{Speed}
12.4=\frac{D}{\nu_{p}-\nu_{e}}

From B to A:
Time=\frac{Distance}{Speed}
10.26=\frac{D}{\nu_{p}+\nu_{e}}

Since we don't know either the speed of plane or the speed of the earth, proceeding this way would prove futile.
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Re: M07 Q24 [#permalink] New post 12 Apr 2011, 00:47
fluke, Thank you for nice explanation. When I meant v, i thought of the speed of Earth. After reading your post, i understood that 700, which is average speed of the plane on the ground already includes that (speed of the Earth). Now, I understand why we use the formula 2D/(t1+t2) to get the average speed.
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Re: M07 Q24   [#permalink] 12 Apr 2011, 00:47
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