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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
05 Nov 2011, 10:45
4
This post received KUDOS
28
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
45% (04:55) correct
55% (03:44) wrong based on 267 sessions
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?
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 [ 6.44 KiB | Viewed 7134 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 [ 5.48 KiB | Viewed 7127 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
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. ? _________________
D- Day December 30 2011. Hoping for the happiest new year celebrations !
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
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?
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
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
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
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
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
23 Apr 2012, 11:22
Expert's post
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. _________________
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
23 Apr 2012, 23:37
4
This post received KUDOS
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
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
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
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
12 May 2013, 05:06
Expert's post
1
This post was BOOKMARKED
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)
Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim [#permalink]
24 Jul 2014, 18:25
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