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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 05 Nov 2011, 10:45
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?

A. 8,10
B. 4,5
C. 5,9
D. 6,9
E. 7,10
[Reveal] Spoiler: OA
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Re: RTD Problem [#permalink] New post 05 Nov 2011, 19:05
It's a tough question. DO you have the OA? I'm not sure how to work this one out.
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Re: RTD Problem [#permalink] New post 06 Nov 2011, 15:55
Mindreko wrote:
It's a tough question. DO you have the OA? I'm not sure how to work this one out.


Yes, i believe the correct answer is B.

This question was already posted, but I would like to see if someone can offer a faster solution.

time-speed-distance-problem-83740.html
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Re: RTD Problem [#permalink] New post 06 Nov 2011, 23:25
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SonyGmat wrote:
Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?

1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10

I find it rather challenging, any inputs??


Make diagrams in races. They help you understand the question better.
Attachment:
Ques3.jpg
Ques3.jpg [ 6.44 KiB | Viewed 4017 times ]

Jerry gives Jim a head start of 200 m so Jim starts not from the starting point but from 200 m ahead. Jerry still beats him by 30 sec which means that Jerry completes the race while Jim takes another 30 sec to complete it.
In this race, Jerry covers 2000m. In the same time, Jim covers the distance shown by the red line. Since Jim needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance which is (1/2)*s where s is the speed of Jim. The distance Jim has actually covered in the same time as Jerry is [1800 - (1/2)*s] shown by the red line.

Attachment:
Ques4.jpg
Ques4.jpg [ 5.48 KiB | Viewed 4016 times ]

Jerry gives Jim a start of 3 mins means Jim starts running first while Jerry sits at the starting point. After 3 mins, Jerry starts running too. Now, Jim beats Jerry by 1000 m which means that Jim reaches the end point while Jerry is still 1000 m away from the end.

In this race, Jerry covers a distance of 1000 m only. In that time, Jim covers the distance shown by the red line (the distance before that was covered by Jim in his first 3 mins). This distance shown by the red line is given by 2000 - 3s (3s is the distance covered by Jim in 3 minutes)

Now you see that in the first race, Jerry covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, Jim would also have covered half the previous distance i.e. the second red line will be half the first red line.

First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

Answer (B)
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Last edited by VeritasPrepKarishma on 07 Nov 2011, 22:31, edited 1 time in total.
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Re: RTD Problem [#permalink] New post 07 Nov 2011, 08:35
VeritasPrepKarishma wrote:
First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

Answer (B)


Thanks so much!! +1

I tried to solve it by writing down the tables for the 2 races and tried to find relationships between the R,T,D, but it took way much time.
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Re: RTD Problem [#permalink] New post 07 Nov 2011, 19:53
VeritasPrepKarishma wrote:
SonyGmat wrote:
Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?

1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10

I find it rather challenging, any inputs??


Make diagrams in races. They help you understand the question better.
Attachment:
Ques3.jpg

Attachment:
Ques4.jpg

Now you see that Jerry covers half the distance in the second race shown by the blue line.
Therefore, in the same time, Jim will also cover half the previous distance i.e. the second red line will be half the first red line.
First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

Answer (B)



Hey karishma,

Can you explain it more elaborately. ?
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Re: RTD Problem [#permalink] New post 07 Nov 2011, 22:32
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ksp wrote:

Hey karishma,

Can you explain it more elaborately. ?


I have given more details next to the relevant diagrams in the post above.
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Re: RTD Problem [#permalink] New post 07 Nov 2011, 22:43
Hey karishma, Thanks for the editing. !
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 21 Apr 2012, 06:25
Quote:
First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min


Karishma,

thanks for that wonderful explanation . i was just trying to arrive at the answer in a little diff way .. i.e equating the times
so i get something i like this

1800/s -1/2 =2000/s-3

so im just equating the time taken by Jim across both the races and when i compare it with the equation you have come up with , I realised that i have mafe a mistake and tried to see how it went wrong bu unable to.
Can you please help me with this?
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 22 Apr 2012, 08:56
Hi,

I am trying to solve this problem in following way but not able to solve it, there s some mistake in the process but couldnot find it if anyone can help.
1st case
distance time
jerry 2000m y-0.5
jim 1800 y
2nd case
jerry 1000 x
jim 2000 x-3

now
equate the speed in both the case
2000/(y-5)= 1000/x
1800/y= 2000/(x-3)

getting -ve ans please help me finding the problem in this approach
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 23 Apr 2012, 11:17
Expert's post
shankar245 wrote:
Quote:
First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min


Karishma,

thanks for that wonderful explanation . i was just trying to arrive at the answer in a little diff way .. i.e equating the times
so i get something i like this

1800/s -1/2 =2000/s-3

so im just equating the time taken by Jim across both the races and when i compare it with the equation you have come up with , I realised that i have mafe a mistake and tried to see how it went wrong bu unable to.
Can you please help me with this?


According to this equation, time taken by Jim in the first race is same as the time taken by him in the second race. But that is not true. In the second race, he covers half the distance he covered in the first race (since in the same time, Jerry covered half the distance too. Since their speeds are constant in the 2 races, the time in the second race must be half)
Otherwise your equation is the same as mine

1800/s -1/2 = 2(2000/s-3)
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 23 Apr 2012, 11:22
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ruprocks wrote:
Hi,

I am trying to solve this problem in following way but not able to solve it, there s some mistake in the process but couldnot find it if anyone can help.
1st case
distance time
jerry 2000m y-0.5
jim 1800 y
2nd case
jerry 1000 x
jim 2000 x-3

now
equate the speed in both the case
2000/(y-5)= 1000/x
1800/y= 2000/(x-3)

getting -ve ans please help me finding the problem in this approach


Look at the diagrams given above. The distance and time in the two races are different from what you have assumed.
In the first race, Jerry covers 2000 m while we don't know how much distance Jim covers.
In the second race, Jim covers 2000 m in x + 3 mins.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 23 Apr 2012, 15:40
Wow tough problem, still trying to grasp it.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 23 Apr 2012, 23:37
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From the first race,
Speed (Jerry) = 2000/x m/s
Speed (Jim) = 1800/ (x+30) m/s
From the second race,
Speed(Jerry) = 1000/y m/s
Speed(Jim) = 2000/(180+y)m/s

That means, 2000/x = 1000/y => y=x/2

1800/(x+30) = 4000/(360+y)
2200x = 526000 => x = 239 seconds.

Hence, Jerry will complete 2000m race in 2000/239 = 3.98 ~= 4 minutes.

Speed of Jerry is 1800 / (239+30) = 6.69 m/s => The time required for Jerry to finish 2000m = 2000/6.69 = 298.95 s

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.

Answer (B)
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 30 Dec 2012, 07:41
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This question is too tough to figure and solve under 2 minutes in the conventional method.

See below an intuitive method to do the same.

On reading the question carefully, you can comprehend,

In the second instance, Jerry covers only half the distance (1000 m) while Jim covers the full distance (2000m) with 3 minutes extra

Let the time taken for Jerry to cover half the distance be x, and therefore full will be 2x
Hence Jim covers the full distance in x + 3

Therefore your answer choice must be of the form (2x, x+3)

B is the only choice that satisfies this.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 12 May 2013, 00:57
riyazv2 wrote:
From the first race,
Speed (Jerry) = 2000/x m/s
Speed (Jim) = 1800/ (x+30) m/s
From the second race,
Speed(Jerry) = 1000/y m/s
Speed(Jim) = 2000/(180+y)m/s

That means, 2000/x = 1000/y => y=x/2

1800/(x+30) = 4000/(360+X)
2200x = 526000 => x = 239 seconds.

Hence, Jerry will complete 2000m race in 2000/239 = 3.98 ~= 4 minutes.

Speed of Jerry is 1800 / (239+30) = 6.69 m/s => The time required for Jerry to finish 2000m = 2000/6.69 = 298.95 s

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.

Answer (B)

Explained beautifully . A small typing error.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 12 May 2013, 05:06
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SonyGmat wrote:
Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a start of 200m and beats him by 30 seconds. Next, Jerry gives Jim a start of 3mins and is beaten by 1000m. Find the time in minutes in which Jerry and Jim can run the race seperately?

A. 8,10
B. 4,5
C. 5,9
D. 6,9
E. 7,10


Considering that this question has to be done under 2 mins, I think the fastest way is to Plug-in(The concept has already been taken care of in the posts above)

From the first part, we know that 1800/v_{jim} - 0.5 = 2000/v_{jerry}--> 9/10*(2000/v_{jim})-0.5 = 2000/v_{jerry}

Or 0.9*t_{jim}-0.5 = t_{jerry}

Only B qualifies.

One could also use the second part of the problem and frame an equation : [2000-3*v_{jim}/v_{jim}]*v_{jerry} = (2000-1000)

Or, 2000/v_{jim} - 3 = 1000/v_{jerry} --> t_{jim} -3 = t_{jerry}/2

B.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink] New post 24 Jul 2014, 18:25
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim   [#permalink] 24 Jul 2014, 18:25
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