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Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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09 Dec 2010, 12:18

I think the OA is wrong since "At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston." Suggests the flies speed reduces to 5 km/hr after the hiker meet..... so it should be around 20+15+6.33(something).... Please put some light on this...
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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26 May 2011, 05:33

I would propose a different way of solving this question.. Key point for me is

"When it reaches the Reston hiker, it turns around and flies back to the Kensington hiker"

Time to reach Reston from Kensington = 02 hours which is calculated by following method.

fly and Reston moving to each other ...> commutative speed = 15km/h Total distance between them = 30km. .... > Time to meet ( Fly and Reston ) = 30/15 = 2hours

In 02 hours , Fly travels a distant = 2h * 10km/h = 20 Km ----------------- (a)

Mean while in two hours , Kensington has traveled = 5km/h * 2 = 10 km.

Distance between Fly and Kensington = 10 km.

Time in which Kensington and Fly meet = {10} km /{(10+5) km/h} = 2/3 hours

Second Distance traveled by Fly to Meet Kensington = 10Km/h * 2/3H = 20 /3 = 6.66 km ----- (b)

Fly Returns back after meeting Kensington , so there is a 3rd distance covered here

we know that the point where two hikers meet is = 15 km from each city.

So Remaining Distance = 15 - 10 - 3.333 = 1.666

3rd distance covered by fly = 10 * 1.666 /5 = 3.333 km ------------- (c)

Adding a , b & C ... > Total distance covered by Fly = 30 km .

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. At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston. How many kilometers has the fly flown?

First of all I'd remove the "the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston" part.

The question basically asks: how many kilometers is covered by fly by the time two hikers meet.

Notice that the fly travels all the time till the hikers meet, so for 30/(5+5)=3 hours, which means that the fly covers Rate*Time=Distance --> 10*3=30 kilometers.

Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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30 May 2012, 05:45

The relative speed (velocity) of the two hikers is 5 + 5 = 10 km/hr. The two hirers put together covered a distance of 30 kms. T = D / R => T = 30 / 10 = 3 Hours

In the meantime, the flier continues to fly at the rate of 10 km/hr. So, in three hours the flier will travel: D = RT = 10 * 3 = 30 km.

B is the correct answer. Excellent Problem!!

If you like my response, please do not forget to click on Kudos +1 !! Cheers!!!

Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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30 May 2012, 10:04

Bunuel wrote:

arjtryarjtry wrote:

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. At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston. How many kilometers has the fly flown?

First of all I'd remove the "the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston" part.

The question basically asks: how many kilometers is covered by fly by the time two hikers meet.

Notice that the fly travels all the time till the hikers meet, so for 30/(5+5)=3 hours, which means that the fly covers Rate*Time=Distance --> 10*3=30 kilometers.

Answer: B.

Hope it's clear.

Bunuel we can reason in a similar way: rate constant for each and also distance. R= 6 ---> Fly D = r*t ---> 10*6 = 60.

Considering that the fly meets runner in the middle (where they meet each other) we have 30. It is correct or prone of error ??'

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. At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston. How many kilometers has the fly flown?

First of all I'd remove the "the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston" part.

The question basically asks: how many kilometers is covered by fly by the time two hikers meet.

Notice that the fly travels all the time till the hikers meet, so for 30/(5+5)=3 hours, which means that the fly covers Rate*Time=Distance --> 10*3=30 kilometers.

Answer: B.

Hope it's clear.

Bunuel we can reason in a similar way: rate constant for each and also distance. R= 6 ---> Fly D = r*t ---> 10*6 = 60.

Considering that the fly meets runner in the middle (where they meet each other) we have 30. It is correct or prone of error ??'

thanks

Sorry, but I don't understand your approach at all.
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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31 May 2012, 05:52

Sorry

Ok I try to explain.

R*T = D

K - 5 * t = 30

V - 5 * t = 30

K Time = 6 V Time = 6

Fly 10*6 = 60 (distance) but from this wording problem (a lot convoluted) the fly flight for half of distance (is the only logic thing to think, also because the hikers in their time and distance are constant). Thanks Gmatclub!!
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink]

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30 May 2013, 20:17

arjtryarjtry wrote:

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. At the moment they do meet, the fly lands on the shoulder of the Kensington hiker as he continues his journey to Reston. How many kilometers has the fly flown?

And, awkward wording aside, seems like a great question. I started out calculating it segmentally, then realized what the question boils down to, which cost me a good minute of useless calculations.