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M05-14

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M05-14  [#permalink]

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New post 16 Sep 2014, 00:25
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  75% (hard)

Question Stats:

56% (02:11) correct 44% (02:43) wrong based on 93 sessions

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Two cities, Kensington, MD and Reston, VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington. It flies at the speed of 10 km/hour and passes the hiker from its city. When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker. It keeps doing so until the hikers meet. If the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston at the moment the two hikers meet, how many kilometers has the fly flown?

A. 25
B. 30
C. 37.5
D. 45
E. 60

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New post 16 Sep 2014, 00:25
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Official Solution:

Two cities, Kensington, MD and Reston, VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington. It flies at the speed of 10 km/hour and passes the hiker from its city. When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker. It keeps doing so until the hikers meet. If the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston at the moment the two hikers meet, how many kilometers has the fly flown?

A. 25
B. 30
C. 37.5
D. 45
E. 60


Calculate how long the fly was flying. We know that the fly flew until the two hikers met and they did so after \(\frac{30}{5+5} = 3\) hours. Thus it was flying for 3 hours at 10 km/h and covered a distance of 30 km.


Answer: B
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Re: M05-14  [#permalink]

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New post 25 Dec 2016, 14:44
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Damn!!! I feel so stupid seeing this solution, (I literally was calculating to and fro distance covered by the fly) Sometimes it just doesn't strike you in the test environment.
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Re: M05-14  [#permalink]

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New post 26 Jan 2017, 06:58
The answer is 45. On the second half the fly travels with 5Km/h.
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Re: M05-14  [#permalink]

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New post 26 Jan 2017, 07:03
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Re: M05-14  [#permalink]

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New post 06 Mar 2017, 10:21
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Bunuel, that is an amazing shortcut. For those who didn't see the shortcut (like me), this is also the sum of a geometric series, with the first term a = 20, and the common ratio r= 1/3. 20*(1/(1-1/3)) = 30.
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Re M05-14  [#permalink]

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New post 05 Sep 2017, 17:45
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. ...But the explanation is very poor. What is the most efficient way to calculate the answer within 2 minutes? I had to draw this out and took way longer.
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Re: M05-14  [#permalink]

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New post 05 Sep 2017, 20:38
nealpatel89 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. ...But the explanation is very poor. What is the most efficient way to calculate the answer within 2 minutes? I had to draw this out and took way longer.


The solution given IS in fact shortest and most efficient way to get the answer. It should take no more than 30 seconds. So, please re-read the solution and try to understand the logic behind it.
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Re: M05-14  [#permalink]

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New post 29 Oct 2017, 04:45
hi bunnel,

in the first 2 hrs the fly travels 20 kms and at this point it reaches the man from Reston who has travelled 10 kms in this time.After this when it travels back along with the man from Reston,both men meet at another 5 kms.
So the fly has travelled 25 kms when it lands on the shoulder of the man from Kensington.

Where am i going wrong?
Pls reply
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Re: M05-14  [#permalink]

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New post 29 Oct 2017, 06:50
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yamini.mv wrote:
hi bunnel,

in the first 2 hrs the fly travels 20 kms and at this point it reaches the man from Reston who has travelled 10 kms in this time.After this when it travels back along with the man from Reston,both men meet at another 5 kms.
So the fly has travelled 25 kms when it lands on the shoulder of the man from Kensington.

Where am i going wrong?
Pls reply



Hi..
Where you are going wrong is Taking speed of fly also as 5km/h after it meets the other hiker

So the fly travels 20km and meets second hiker.
Then it turns back and moves with 10km/h to meet first hiker, and it will not meet him at 5km /h but they travel distance of 10km in combined speed of 10+5..... Time taken =10/15, so fly will travel 10*10/15 in this time..
And again turn back and move..

Best method is to find after how much time these hikers meet as nothing is hampering their move and then see how much that fly will travel in that time.
So distance 30 combined speed 5+5=10, so time taken =20/10=3h..
In 3h fly will travel 10*3=30
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Re: M05-14  [#permalink]

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New post 20 Dec 2017, 23:11
This my first reply and let me tell the method given in the explanation is the best way to get the answer. I went into deep calculations and even though I got the answer I wouldn't have even gone down that road in the real test.
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Re: M05-14  [#permalink]

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New post 06 Aug 2019, 12:02
Quote:
Two cities, Kensington, MD and Reston, VA are 30 km apart. From both of these cities, simultaneously, two hikers start their journeys towards each other. They are walking at a constant speed of 5 km/hour each. Simultaneously, a fly leaves the city of Kensington. It flies at the speed of 10 km/hour and passes the hiker from its city. When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker. It keeps doing so until the hikers meet. If the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston at the moment the two hikers meet, how many kilometers has the fly flown?


I believe there is a flaw in the logic of the task.
When the fly meets the second hiker (from Reston), it flew 20km already. (\(2*10=20\)). It returns back to the first hiker (from Kensington). But the first hiker does not stop at the point of 10km, he moves forward with the speed of 5km/h. So does the second hiker. Thus, when the hikers meet, the fly cannot be in their meeting point, as its speed was 10km/h, not 5km/h, and it would be at the point of 10km, not 15km which is the meeting point for two hikers!
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Re: M05-14   [#permalink] 06 Aug 2019, 12:02
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