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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.

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03 Sep 2013, 20:55

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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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03 Sep 2013, 21:54

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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%
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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.
_________________

Re: John and Karen begin running at opposite ends of a trail unt [#permalink]

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06 Sep 2013, 09:22

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

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09 Sep 2013, 04:17

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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.

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21 Nov 2014, 09:51

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Re: John and Karen begin running at opposite ends of a trail unt [#permalink]

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14 May 2015, 01:18

WholeLottaLove wrote:

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%

Very nice explanation. Thanks a lot.Its easy to assume values.

Concentration: General Management, Entrepreneurship

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John and Karen begin running at opposite ends of a trail unt [#permalink]

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23 Aug 2015, 06:45

Let's say total distance between Johm & Karen is 100 miles when they start. If John would not have stopped then John & Karen would have approached each other at S+1.5S = 2.5S speed and met each other after time T hours and total distance by both have them would have covered = 100 miles.

Hence T = 100/2.5S

In Time T, distance covered by Karen at 1.5S speed is : 1.5S * 100/2.5S = 60 miles. Hence Karen would have covered 60 miles to meet John if John would not have stopped.

Given that John stopped after covering 25 miles means Karen covered 75 miles.

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30 Sep 2016, 05:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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02 Oct 2016, 21:07

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Skip the math. 1st recognize that the answer choices are far apart and thus, this is a probaby a perfect problem to estimate.

*IF* Karen and John ran at the same rate, then they would have met at 50% mark. But since John is a wimp who can't push through his cramps Karen now has to run 50% farther (to reach John who's sitting at 25%)...

BUT remember that Karen actually runs faster than wimpy old John, so they would have met somewhere closer to Johns starting point had John not quit. But he did quit, so we know that Karen is going to have to run <50% farther to reach crying John.

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26 Jan 2017, 13:14

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My 2 cents, Karen’s speed = 1.5 * John’s speed = 3/2 * John’s speed. Therefore, Sk = 3 mph and Sj = 2 mph. Total distance = 60 miles (3*2*10) Sk/Sj = dk/dj = 3/2 => dk/D = 3/5 & dj/D = 2/5 Therefore, dk = 60 * 3/5 = 36 miles. Total time taken T (would be sum of speeds of BOTH Karen and John); T = D/(Sk + Sj) = 60/5 = 12 hrs.

Now, as John stops at 25% of distance due to cramps; Karen has to cover 75% of distance = ¾ * 60 = 45 miles. Therefore, % increase of distance covered by Karen = (New - Old)/Old * 100 = (45 – 36)/36 * 100 = 9/36* 100 = ¼ * 100 = 25% | A

John and Karen begin running at opposite ends of a trail unt [#permalink]

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02 Aug 2017, 16:32

I think the trick of this question is instantly plugging in made up values.

Also knowing you can add rates together. For instance if I am driving at 50KM an hour and you're driving at 50 and we're both driving at eachother the actual speed is 100KM of the gap we're closing.

So if there is 100KM between us and we're both moving at 50KM we would meat eachother in 1 hour as our total speed is 100KM. D=RT 100 = 100 T. So this knowledge helps

Re: John and Karen begin running at opposite ends of a trail unt [#permalink]

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03 Aug 2017, 00:08

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

Dear, Why do you divided 0.75D by 1.5 D? I don't understand the 1.5 D part.

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