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Two cities, Kensington MD and Reston VA are 30 km a (m05q14)

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Two cities, Kensington MD and Reston VA are 30 km a (m05q14) [#permalink] New post 09 Sep 2008, 17:21
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Question Stats:

57% (02:12) correct 43% (02:34) wrong based on 341 sessions
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?

(A) 25
(B) 30
(C) 37.5
(D) 45
(E) 60

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Re: problemmmmmm..... too good!!! [#permalink] New post 09 Sep 2008, 17:42
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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?


* 25
* 30
* 37.5
* 45
* 60


Calculate the amount of time it takes for the two hikers to meet:
Each travel at 5 km/hr --> Each travel 15 km
15 km / 5 km/hr = 3 hrs

The fly travels for 3 hrs at 10mph
Total distance = 3hrs * 10mph = 30 miles
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Re: problemmmmmm..... too good!!! [#permalink] New post 09 Sep 2008, 20:48
Question has lots of wordings to make it appear complex. However, what essentially it means that the fly would travel at a constant speed of 10kmph for the time taken by the hiker(s) to travel 15 km.
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Re: problemmmmmm..... too good!!! [#permalink] New post 10 Sep 2008, 14:00
Good One...
I missed that last part of the question...did 30 +15(the remaining journey)
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Re: problemmmmmm..... too good!!! [#permalink] New post 10 Sep 2008, 17:32
sumanamba wrote:
Good One...
I missed that last part of the question...did 30 +15(the remaining journey)



yep! I arrived at 45 as well because I added 30 + 15.

By the time they meet fly would have traveled only 30 miles and continues 15 miles with the MD Hiker.
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Re: problemmmmmm..... too good!!! [#permalink] New post 25 Mar 2010, 14:36
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scthakur wrote:
Question has lots of wordings to make it appear complex. However, what essentially it means that the fly would travel at a constant speed of 10kmph for the time taken by the hiker(s) to travel 15 km.



You cannot assume that the fly flies back to hiker K from hiker R at 10km/hr. If you assume that, then the fly meets hiker K before hiker K and hiker R meet.

The fly meets hiker R after 2hours:
-The fly has traveled 20km and hiker R has traveled 10km toward K.
-Hiker K has also traveled 10 km toward R.
-At this point hiker K and hiker R are 10km apart.

Now the fly goes back to hiker K, if we assume that the fly is flying at 10km/hr toward hiker K, we need to consider that hiker K is traveling at 5km/hr toward the fly. Therefore we need to add their rates to find out how long it takes for them to meet.

distance between / combined rates = time it takes

10 km / (5 km/hr + 10 km/hr) = 2/3 hr.

now add the flight times for the fly, 2hr + 2/3hr = 8/3 hr.
So the fly has flown for 10km/hr * (8/3 hr) = 80/3 km = 26.67 km

Another thing problematic with this question:
The stimulus never states that the fly flew back at 10km/hr, it simply states that it flew back to hiker K at the same time that they met.
If we don't assume that the fly flies back at 10km/hr:
After the fly has met hiker R, the two hikers are 10km apart. If it flies back and lands on hiker K's shoulder at the same time that hiker K and hiker R meet, it has to travel at a slower speed than 10km/hr (it has to travel at 5km/hr -- the same speed as the hikers). Thus it travels another 5km.

Therefore it has travelled 20km (until it met hiker R) then traveled back 5km (at a slower speed) for a total of 25km
So I think the OA is incorrect.

This is a badly worded question in my opinion. It also forces you to make certain assumptions:
1. the fly is flying constantly at the same speed. (even using this assumption it doesn't work out)
2. the fly only flies in a straight line. (I've never seen a fly fly in a straight line -- the fly could actually have covered alot more distance if it didn't fly in a straight line)

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 24 May 2010, 07:47
its B
Time taken by hikers to meet = 30/(5+5) = 3hrs

Distance flown by fly = speed * time
= 10*3
= 30 miles
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 24 May 2010, 10:05
If we go strictly as per the data provided in this word problem , we cannot solve the problem.
"it flies at the speed of 10km /hour", so we don't know if the speed is constant or not.
Therefore it can fly at any speed , had this been a DS problem , result would be information given in the statement is not sufficient.

However if we consider constant speed of 10kph, then we can answer the question easily (30 km).

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Re: problemmmmmm..... too good!!! [#permalink] New post 25 May 2010, 04:47
icandy wrote:
sumanamba wrote:
Good One...
I missed that last part of the question...did 30 +15(the remaining journey)



yep! I arrived at 45 as well because I added 30 + 15.

By the time they meet fly would have traveled only 30 miles and continues 15 miles with the MD Hiker.


The question is "How many kilometers has the fly flown?" not "how many kilometers the fly has travelled". The right answer is 30 km. IMHO.
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 25 May 2010, 06:58
Wow, what an easy method. I calculated three iterations 30/(10+5), then calculated distance traveled, then substract from original distance, then calculated again.

I reached the same conclusion by aproximation, but took a lot of time. I need to solve problems more easily.
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 25 May 2010, 08:59
Well mathematically this is an infinite geometric series

sum = a / (1-r)
a=2/3*30 (fly travels 1st time)
r=1/3

2/3*30 / (1-1/3) = 30

firasath, you are overcomplicating it

1) the fly has infinite accelerate (i.e. it always travels 10 km/hr and can change direction instantly upon contacting one of the travelers)

2) the travelers and the fly are traveling in a straight line. I know this isn't realistic, but this is a math problem!
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 25 May 2010, 12:03
ponderer wrote:
Well mathematically this is an infinite geometric series

sum = a / (1-r)
a=2/3*30 (fly travels 1st time)
r=1/3

2/3*30 / (1-1/3) = 30



Yes, it is. But, I didn't remember how to solve this, that's why I used the iteration process, two to three steps are good enough (without a time limit :) ). It sucks to see the problem one way, but not remember how to solve it.
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 25 May 2010, 19:34
i too got ans : 30 miles

fly travel for 3 hour ,time required to meet hiker.

so 3*10=30

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 26 May 2010, 10:08
Though i got the answer, I was thrown off initially by the wordings of the question.

The sentence says: "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,..." Wordings could make a simple question appear complicating. If the fly left Kensington simultaneously with its hiker, it should be ahead right from the start-off...Otherwise, i would say am confused at the use of the word "simultaneously."

Can someone explain that, please?

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 26 May 2010, 10:34
Can someone explain why 15 km isn't added to the 30km?
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 26 May 2010, 12:38
marcusaurelius wrote:
Can someone explain why 15 km isn't added to the 30km?

If you go through the previous posts i believe your question would have been addressed.
30km is the total distance between K and R; because the two hikers are traveling at the same speed,
they must have met at the the middle - 15km away from starting point.

Time spent by a hiker = time spent by the fly = 3hrs
distance traveled by fly from START = speed (10km/hr) x total time spent (3hrs) = 30km

Hope that helps, otherwise let me know.

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 05 Jun 2010, 10:20
In my openion the total distance flown by the fly would be 30 kms not 45kms, because question asked categorilcally how many kilometers has fly flown not how many kms fly has travelled so no need to add further 15 kms which has been travelled by the fly in shoulder of the hiker.

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 29 Jun 2010, 18:03
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ponderer wrote:
Well mathematically this is an infinite geometric series

sum = a / (1-r)
a=2/3*30 (fly travels 1st time)
r=1/3

2/3*30 / (1-1/3) = 30

firasath, you are overcomplicating it

1) the fly has infinite accelerate (i.e. it always travels 10 km/hr and can change direction instantly upon contacting one of the travelers)

2) the travelers and the fly are traveling in a straight line. I know this isn't realistic, but this is a math problem!


Yup I am over complicating it based on what is stated in the stimulus. It should make it clear that the fly flies at a constant rate and that it continuously flies BACK AND FORTH between the two hikers until they meet.
This same problem was in a sample interview question for a financial Quant position as shown here: (page 23) http://www.math.chalmers.se/Math/Grundu ... age_QF.pdf

If you read the wording of the problem in the link above, it is clear.

The wording of the problem in this thread however is somewhat unclear.

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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 19 Aug 2010, 23:37
Its B.

They meet after 3 hrs (30/5+5). So fly has travelled for 3 hrs at a speed of 10. So distance is 30.
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Re: problemmmmmm..... too good!!! (m05q14) [#permalink] New post 22 Aug 2010, 04:52
Does a GMAT question ever have more information than required to answer a question. I think i came across such a problem in Kaplan and i ended up getting that question wrong. :wink:
Re: problemmmmmm..... too good!!! (m05q14)   [#permalink] 22 Aug 2010, 04:52
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