John and Karen begin running at opposite ends of a trail unt : Problem Solving (PS)

WholeLottaLove

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

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.

ANSWER: A) 25%

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

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

20 Aug 2015, 20:17

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Karen normally covers 3/5 of the distance.
Now she has to cover 3/4 of the distance.
3/4 / 3/5 = 1.25

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

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%

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

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.

Thus Answer A is the only available option.

Posted from my mobile device

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

14 May 2015, 01:18

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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.

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

20 Nov 2013, 10:11

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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]

23 Aug 2015, 06:45

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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.

% = (75-60) * 100 / 60 = 25% Ans.

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

01 Oct 2016, 09:13

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the wording is a bit confusing. you can change it to 25% of the total distance. i took to be 25% of his normal distance before they met.

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

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

If my post was helpful to you, please kudo :idea:

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

11 Dec 2017, 02:02

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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!

Easy question, but it's so so poorly written.

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

03 May 2020, 10:59

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Js speed= x
Ks speed = 1.5 x
total distance = 100

Relative speed= 2.5x
Time they would take to meet = distance/ relative speed = 100/2.5x = 40/x
distance karen would cover in this time = speed*time = 1.5x*40/x = 60
distance that karen has to cover now because of Johns cramp = 75 (75% = 75 since we took total distance as 100)
% increase= (75-60/60)*100= 25%

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

08 May 2020, 12:38

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Kinshook wrote:

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!

Given:
1. John and Karen begin running at opposite ends of a trail until they meet somewhere in between their starting points.
2. They each run at their respective constant rates until John gets a cramp and stops.

Asked: 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.

vK / vJ = 1.5

Let the speeds of John and Karen be 2x & 3x mph respectively

Let the length of the trail be L miles

Normal Case:
Distance = L miles
Relative speed = 2x + 3x = 5x mph
Time taken to meet = L/5x= .2L/x hours

Abnormal Case : John is only able to cover 25% of the distance before he stops
Time when both were running = (L/4)/5x = L/20x hours
Distance covered by both in L/20x hours = L/20x * (2x + 3x) = L/4

Remaining distance = 3L/4
Time taken by Karen to cover 3L/4 distance = 3L/4 /(3x) = L/4x

Total time run by Karen = L/20x + L/4x = 6L/20x = 3L/10x = .3L/x

Extra time run by Karen = .3L/x - .2L/x = .1L/x

Percent longer would Karen have run ={ (.1L/x)/(.2L/x)}*100% = 50%

IMO B

Bunuel
I think OA should be B
Please check whether OA is correct.

Yes I thought the answer would be 50% too. Assuming the total distance to be 180km, in the normal scenario, Karen would have traveled 108km while John would have traveled 72km when they would have met. So I assumed that John had traveled only 25% of his estimated distance and not the overall. This would've led to Karen having to travel an additional 54km, thereby leading to a 50% increase in time.

Bunuel, chetan2u, Narenn, is the wording of the question ambiguous or have I misinterpreted it?

Thank You!

L

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

09 May 2020, 12:15

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Expert Reply

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%

Let’s assume that John runs at 10 mi/hr, and thus, Karen runs at 10 x 1.5 = 15 mi/hour. If we assume that the distance between them was 100 miles, then when John stopped, he had run 25 miles and Karen had to run 75 miles to meet him, and thus her running time was 75 / 15 = 5 hours.

Had John not stopped, they would have met when he had run 40 miles and she had run 60 miles (ratio of 10 : 15); thus, her running time would have been 60 / 15 = 4 hours.

The percent longer that Karen would have had to run is (50 - 40) / 40 x 100% = 25%.

Answer: A

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

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

Thanks!

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

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

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pclawong

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

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.

Thank you so much.

gmatclubot

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

03 Aug 2017, 00:08

GMAT Problem Solving (PS) Questions

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