Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 23 Oct 2011
Posts: 82

Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
05 Nov 2011, 11:45
Question Stats:
39% (03:10) correct 61% (03:00) wrong based on 592 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India

Re: RTD Problem
[#permalink]
Show Tags
Updated on: 07 Nov 2011, 23:31
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 [ 6.44 KiB  Viewed 16174 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 16164 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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Manager
Joined: 24 Mar 2010
Posts: 70

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
30 Dec 2012, 08:41
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.
_________________
 Stay Hungry, stay Foolish 




Manager
Status: livin on a prayer!!
Joined: 12 May 2011
Posts: 110
Location: Australia

Re: RTD Problem
[#permalink]
Show Tags
05 Nov 2011, 20:05
It's a tough question. DO you have the OA? I'm not sure how to work this one out.
_________________
Aim for the sky! (800 in this case) If you like my post, please give me Kudos



Manager
Joined: 23 Oct 2011
Posts: 82

Re: RTD Problem
[#permalink]
Show Tags
06 Nov 2011, 16:55
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. timespeeddistanceproblem83740.html



Manager
Joined: 23 Oct 2011
Posts: 82

Re: RTD Problem
[#permalink]
Show Tags
07 Nov 2011, 09:35
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.



Manager
Status: SC SC SC SC SC.... Concentrating on SC alone.
Joined: 20 Dec 2010
Posts: 189
Location: India
Concentration: General Management
GMAT Date: 12302011

Re: RTD Problem
[#permalink]
Show Tags
07 Nov 2011, 20:53
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. ?
_________________
D Day December 30 2011. Hoping for the happiest new year celebrations !
Aiming for 700+
Kudo me if the post is worth it



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India

Re: RTD Problem
[#permalink]
Show Tags
07 Nov 2011, 23:32
ksp wrote: Hey karishma,
Can you explain it more elaborately. ?
I have given more details next to the relevant diagrams in the post above.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Status: SC SC SC SC SC.... Concentrating on SC alone.
Joined: 20 Dec 2010
Posts: 189
Location: India
Concentration: General Management
GMAT Date: 12302011

Re: RTD Problem
[#permalink]
Show Tags
07 Nov 2011, 23:43
Hey karishma, Thanks for the editing. !
_________________
D Day December 30 2011. Hoping for the happiest new year celebrations !
Aiming for 700+
Kudo me if the post is worth it



Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 88
Concentration: Strategy, General Management
GPA: 3.6
WE: Consulting (Computer Software)

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
21 Apr 2012, 07:25
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/s3 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?



Intern
Joined: 25 Jul 2008
Posts: 3

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
22 Apr 2012, 09:56
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 y0.5 jim 1800 y 2nd case jerry 1000 x jim 2000 x3
now equate the speed in both the case 2000/(y5)= 1000/x 1800/y= 2000/(x3)
getting ve ans please help me finding the problem in this approach



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
23 Apr 2012, 12:17
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/s3 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/s3)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
23 Apr 2012, 12:22
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 y0.5 jim 1800 y 2nd case jerry 1000 x jim 2000 x3
now equate the speed in both the case 2000/(y5)= 1000/x 1800/y= 2000/(x3)
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.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Senior Manager
Joined: 01 Apr 2010
Posts: 280
Location: Kuwait
GPA: 3.2
WE: Information Technology (Consulting)

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
23 Apr 2012, 16:40
Wow tough problem, still trying to grasp it.



Manager
Status: Sky is the limit
Affiliations: CIPS
Joined: 01 Apr 2012
Posts: 68
Location: United Arab Emirates
Concentration: General Management, Strategy
WE: Supply Chain Management (Energy and Utilities)

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
24 Apr 2012, 00:37
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)



Intern
Joined: 20 Apr 2013
Posts: 20
Concentration: Finance, Finance
GMAT Date: 06032013
GPA: 3.3
WE: Accounting (Accounting)

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
12 May 2013, 01:57
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.



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 613

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
12 May 2013, 06:06
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 Plugin(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 : [20003*\(v_{jim}\)/\(v_{jim}\)]*\(v_{jerry}\) = (20001000) Or, 2000/\(v_{jim}\)  3 = 1000/\(v_{jerry}\) > \(t_{jim}\) 3 = \(t_{jerry}\)/2 B.
_________________
All that is equal and notDeep Dive Inequality
Hit and Trial for Integral Solutions



Intern
Joined: 17 May 2012
Posts: 39

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
16 Nov 2014, 23:24
Hi Bunuel, Can you add your spin to this question and advise on how to about solving it? Thanks, AJ



Intern
Joined: 13 Oct 2014
Posts: 4

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
04 Dec 2014, 03:42
Just solved it in 1:08, but I'm not sure about the method. Well, I created and solved a simply equation: 2,000*(t30)=1800*t Where 't' is the time needed by Jim and 't30' the time needed for his buddy Jerry. We solve it and we get t=300, 5 min. We look at the answers and we immediately see that only B has 5 min for Jim. Plz tell me if my way is correct or I got it just for luck ! Posted from GMAT ToolKit



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8418
Location: Pune, India

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim
[#permalink]
Show Tags
04 Dec 2014, 21:00
LinusITA wrote: Just solved it in 1:08, but I'm not sure about the method. Well, I created and solved a simply equation: 2,000*(t30)=1800*t Where 't' is the time needed by Jim and 't30' the time needed for his buddy Jerry. We solve it and we get t=300, 5 min. We look at the answers and we immediately see that only B has 5 min for Jim. Plz tell me if my way is correct or I got it just for luck ! Posted from GMAT ToolKitI don't understand the logic behind this equation: 2,000*(t30)=1800*t This implies Jerry's Distance * Jerry's Time = Jim's Distance * Jim's Time Why?
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim &nbs
[#permalink]
04 Dec 2014, 21:00



Go to page
1 2 3
Next
[ 42 posts ]



