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John and Karen begin running at opposite ends of a trail unt

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John and Karen begin running at opposite ends of a trail unt [#permalink] New post 03 Sep 2013, 11:16
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Question Stats:

54% (03:37) correct 46% (02:13) wrong based on 78 sessions
A Nice Question from VERITAS.
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John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.

A) 25%
B) 50%
C) 75%
D) 100%
E) 200%

Happy Solving!
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Sep 2013, 06:28, edited 1 time in total.
Added the OA.
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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 03 Sep 2013, 19:55
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if john speed is S then Karen is 1.5S
D = 1.5ST + ST
D= 2.5ST
where t is the time taken by each runner in normal case.
Now john only travel .25D so Karen has to travel .75D to meet him

.75D/1.5D = 1.25t
so she needs to travel 25% more time than usual time.
A
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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 03 Sep 2013, 20:54
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Narenn wrote:
A Nice Question from VERITAS.
OA and OE will be posted after few responses. Brief and Correct explanations will be rewarded with a Kudo.



John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.

A) 25%
B) 50%
C) 75%
D) 100%
E) 200%

Happy Solving!


Took some time to understand the Q.

Here is my Solution.

Let distance be John and Karen by 90 Kms
John's speed: 10 km/hr ---Time taken: 9 hrs
Karen's speed: 15Km/hr, time taken : 6 hrs

Now if both are running at their constant speed then they will meet in 3 hrs 36 minutes (See below)

at 0 hrs ---Distance between the 2 is 90 kms
after 1 hrs: 65 Km
After 2 hrs : 40 kms
After 3 : 15 kms


In 1 hr distance covered by the 2 jointly is 25 kms so 15 kms will be covered in 15/25*60----> 36 minutes

At the meeting point Distance covered by John : 36 Kms and by Karen: 54 Kms
Now John covered on D/4 distance ie. 22.5 Km distance and thus Karen would have to travel the extra distance of 13.5 kms.
To cover 13.5 kms----> Karen would need 54 mins (13.5/15*60---- 54 minutes)
So Total time taken by Karen : 3 hrs 36 minutes+ 54 minutes----> 4.5 hrs or 9/2 hrs
Usual time: 3hr 36 minutes -----> 18/5 hrs

Hence % More ((9/2-18/5) / 18/5 )*100-----> 25%

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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 03 Sep 2013, 21:59
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Consider d1+d2 =d ( d1 = distance run by John and d2 = distance run by Karen)

d1:d2 = Js*T:(3/2)Js*T = 2:3 = 0.4d : 0.6d

Since the John stops after 25% run d1:d2 = 0.25d : 0.75d

Hence the % longer K needs to run is = 0.75-0.6/0.6 = 25%

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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 04 Sep 2013, 05:35
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Thank You so much for your responses and explanations.

Here is the Official Explanation from the VERITAS

Choice A.

If John covered 25% of the course before stopping, that means that Karen covered 75% of it. But she should have only had to run 60% of it – that’s because she runs 3 miles for every 2 that John runs (covering 50% more distance in the same amount of time), so she should have covered 3/5 of the territory. So then you can use the Percent Change calculation:

\frac{(75-60)}{60} = \frac{1}{4} = 25%

A is therefore the correct answer.

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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 06 Sep 2013, 08:22
AMITAGARWAL2 wrote:
if john speed is S then Karen is 1.5S
D = 1.5ST + ST
D= 2.5ST
where t is the time taken by each runner in normal case.
Now john only travel .25D so Karen has to travel .75D to meet him

.75D/1.5D = 1.25t
so she needs to travel 25% more time than usual time.
A


Could someone explain in more detail this approach?

I don't understand its last division: .75D/1.5D = 1.25t
Where did he or she get 1.5D as divisor? :s

Thanks!
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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 09 Sep 2013, 03:17
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John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points. They each run at their respective constant rates until John gets a cramp and stops. If Karen runs 50% faster than John, who is only able to cover 25% of the distance before he stops, what percent longer would Karen have run than she would have had John been able to maintain his constant rate until they met.

Lets say the distance of the trail is 100 miles. Lets also say that J rate = 10 miles/hour and K rate = 15 miles/hour.

If John stops at the 25% mark that means he travels 25 miles in 2.5 hours. It would take Karen t=d/r t=75/15 = 5 hours to reach john. If John had not stopped, their combined rate would 10+15 = 25 miles/hour meaning they would have met in 4 hours. Therefore, she ran one hour longer (25%) longer than she would have needed to if John ran for the entire time.

ANSWER: A) 25%
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Re: John and Karen begin running at opposite ends of a trail unt [#permalink] New post 20 Nov 2013, 09:11
Let john's speed=2 then karen's speed=(3/2)*2=3
Let total distance=40 then johns distance=(1/4)*40=10. Karens distance=30

Karens time=30/3=10hours
If everything were ok, then these guys would have met in 8 hours
40/(3+2)=8h

So, (10-8)/8*100%=25%

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Re: John and Karen begin running at opposite ends of a trail unt   [#permalink] 20 Nov 2013, 09:11
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