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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]

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New post Updated on: 04 Sep 2013, 07:28
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A
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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.

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

Happy Solving!

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Originally posted by Narenn on 03 Sep 2013, 12:16.
Last edited by Bunuel on 04 Sep 2013, 07:28, 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]

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New post 04 Sep 2013, 06:35
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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]

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New post 03 Sep 2013, 22:59
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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%

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

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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New post 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]

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

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New post 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.


% = (75-60) * 100 / 60 = 25% Ans.
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Re: John and Karen begin running at opposite ends of a trail unt  [#permalink]

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New post 01 Oct 2016, 09:13
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]

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New post 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.

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

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New post 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

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

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

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

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New post 03 Aug 2017, 01:51
Smallwonder wrote:
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%

/SM


Dear,

I don't understand this part --> Js*T:(3/2)Js*T = 2:3 = 0.4d : 0.6d
Would you explain what is Js*T? Thank you
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Re: John and Karen begin running at opposite ends of a trail unt  [#permalink]

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New post 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]

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New post 26 Mar 2018, 04:01
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!


Some good solution is provided. Sharing my 2 cents on the topic.

Had john not got a cramp and stopped, time taken to meet, t = \(\frac{distance}{total speed}\) ==> t = \(\frac{d}{2.5v}\)
assuming, john's speed = v, karen's speed = 1.5v and distance = d,

Now, time taken by Karen to reach john, = \(\frac{.75d}{1.5v}\)
Extra time taken by karen to reach john, td = \(\frac{.75d}{1.5v}\)- \(\frac{d}{2.5v}\)
==> td= \(\frac{d}{10v}\)

percent longer would Karen have run = \(\frac{\frac{d}{10v}}{\frac{d}{2.5v}}\) * 100 = 25%

Answer : 25%
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