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

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   24 Jun 2020, 09:23
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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
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   24 Jun 2020, 09:36
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Carcass 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


Let's start with some word equations

Jerry gives Jim a start of 200m and beats him by 30 seconds.
So, Jerry runs 2000 meters, and Jim runs 1800 meters.
Also, Jerry runs the race 30 seconds FASTER than Jim.
In other words, Jerry's travel time is 30 seconds (0.5 minutes) LESS THAN Jim's travel time

So, we can write: (Jerry's travel time + 0.5 minutes) = (Jim's travel time)
Let R = Jerry's speed in meters per minute
Let M = Jim's speed in meters per minute
time = distance/speed, so we can write: 2000/R + 0.5 = 1800/M
To eliminate the fractions, multiply both sides by MR to get: 2000M + 0.5MR = 1800R
Rearrange to get: 2000M = 1800R - 0.5MR


Jerry gives Jim a start of 3mins and is beaten by 1000m
This time Jerry runs 1000 meters, and Jim runs 2000 meters.
Also, Jerry's travel time is 3 minutes LESS THAN Jim's travel time

So, we can write: (Jerry's travel time + 3 minutes) = (Jim's travel time)
time = distance/speed, so we can write: 1000/R + 3 = 2000/M
To eliminate the fractions, multiply both sides by MR to get: 1000M + 3MR = 2000R
Rearrange to get: 1000M = 2000R - 3MR

We now have two equations:
2000M = 1800R - 0.5MR
1000M = 2000R - 3MR

Take the bottom equation and multiply both sides by 2 to get:
2000M = 1800R - 0.5MR
2000M = 4000R - 6MR

Since both equations are set equal to 2000M, we can now write: 1800R - 0.5MR = 4000R - 6MR
Rearrange to get: 5.5MR = 2200R
Divide both sides by R to get: 5.5M = 2200
Solve to get: M = 400 meters per minutes (this is Jim's speed)

Time = distance/speed
So, time for Jim to run 2000 meters = 2000/400 = 5 minutes
Check the answer choices....only one answer choice has 5 minutes as Jim's running time
Answer: B
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   24 Jun 2020, 09:24
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   24 Jun 2020, 10:10
Carcass 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



do such questions appear on gre?:P its taken me around 5 minutes and half i think i did a long way

can anyone provide simpler?

basically we have 2 cases:

1) 200m gap
t (tom) = t (Jerry) + 30
(2000/Vtom) = (1800/VJ) +30
thats equation 1

2) 1000/Vtom = (2000-180Vjerry)/Vjerry

and solve for these

is there a simpler easier way?
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   30 Jun 2020, 11:59
georgekaterji wrote:
do such questions appear on gre?:P its taken me around 5 minutes and half i think i did a long way

can anyone provide simpler?

basically we have 2 cases:

1) 200m gap
t (tom) = t (Jerry) + 30
(2000/Vtom) = (1800/VJ) +30
thats equation 1

2) 1000/Vtom = (2000-180Vjerry)/Vjerry

and solve for these

is there a simpler easier way?


This is an official GMAT question.
Since the two tests have virtually the same math syllabus, it's conceivable that a similar question could appear on the GRE (although it would most definitely be a 168-170 level question)
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   04 Feb 2021, 01:00
Carcass 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


I don't know if this helps!
But let me try

Let,
Jim's speed be X m/min
Jerry's speed be Y m/min

Statement 1: Jerry gives Jim a start of 200m and beats him by 30 seconds
Let us write an equation w.r.t time of both;
1800/X - 2000/Y = 0.5
1800Y - 2000X = 0.5XY
3600Y - 4000X = XY .....(1)

Statement 2: Jerry gives Jim a start of 3mins and is beaten by 1000m
Again equation w.r.t time;
2000/X - 1000/Y = 3
2000Y - 1000X = 3XY .....(2)

Solving both the equations;
10800Y - 12000X = 2000Y - 1000X
108Y - 120X = 20Y - 10X
88Y = 110X
8Y = 10X
X/Y = 4/5

Since, the speeds of Jim and Jerry are in 4:5, the ratio of their time has to be 5:4
i.e. Jim would take 5 mins and Jerry would take 4 mins

Hence, option B
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   04 Mar 2021, 00:18
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I think best way to go here is simple try answer choices
Jim's time minus 3 minutes should be equal to half jerry's time hence b is answear and you can solve it very fast
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   21 Sep 2021, 10:14
georgekaterji wrote:
Carcass 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



do such questions appear on gre?:P its taken me around 5 minutes and half i think i did a long way

can anyone provide simpler?

basically we have 2 cases:

1) 200m gap
t (tom) = t (Jerry) + 30
(2000/Vtom) = (1800/VJ) +30
thats equation 1

2) 1000/Vtom = (2000-180Vjerry)/Vjerry

and solve for these

is there a simpler easier way?


I also have doubt that ets would give such a question on GRE. The GRE quantitative questions are simpler in my opinion.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   16 Jan 2022, 04:03
Carcass 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


Jerry gives Jim a start of 200m and beats him by 30 seconds.
When Jerry starts to run the entire 2000-meter race, Jim is 200 meters ahead thus starts to run the remaining 1800 meters of the race.
Jerry -------------------------------------------------------> 2000 meters
_____200-meter head start______Jim---------------> 1800 meters
Since Jim finishes 30 seconds behind Jerry, it takes Jim 30 seconds longer to travel 1800 meters than it takes Jim to travel 2000 meters.

We can PLUG IN THE ANSWERS, which represent the two times to travel 2000 meters.
Given that all of the values in the prompt are INTEGERS, the two times are almost certain to be values that divide cleanly into 2000, making the correct answer either A or B

B: Jerry's time for 2000 meters = 4 minutes, Jim's time for 2000 meters = 5 minutes
Jim's rate \(= \frac{distance}{time} = \frac{2000}{5} = 400\) meters per minute
Time for Jim to run 1800 meters \(= \frac{distance}{rate} = \frac{1800}{400} = 4.5\) minutes
Success!
Jim's time for 1800 meters (4.5 minutes) is 30 seconds longer than Jerry's time for 2000 meters (4 minutes).

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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   31 Jan 2022, 22:51
I don't understand how we got the second equation with 1000 & 2000?

GreenlightTestPrep wrote:
Carcass 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


Let's start with some word equations

Jerry gives Jim a start of 200m and beats him by 30 seconds.
So, Jerry runs 2000 meters, and Jim runs 1800 meters.
Also, Jerry runs the race 30 seconds FASTER than Jim.
In other words, Jerry's travel time is 30 seconds (0.5 minutes) LESS THAN Jim's travel time

So, we can write: (Jerry's travel time + 0.5 minutes) = (Jim's travel time)
Let R = Jerry's speed in meters per minute
Let M = Jim's speed in meters per minute
time = distance/speed, so we can write: 2000/R + 0.5 = 1800/M
To eliminate the fractions, multiply both sides by MR to get: 2000M + 0.5MR = 1800R
Rearrange to get: 2000M = 1800R - 0.5MR


Jerry gives Jim a start of 3mins and is beaten by 1000m
This time Jerry runs 1000 meters, and Jim runs 2000 meters.
Also, Jerry's travel time is 3 minutes LESS THAN Jim's travel time

So, we can write: (Jerry's travel time + 3 minutes) = (Jim's travel time)
time = distance/speed, so we can write: 1000/R + 3 = 2000/M
To eliminate the fractions, multiply both sides by MR to get: 1000M + 3MR = 2000R
Rearrange to get: 1000M = 2000R - 3MR

We now have two equations:
2000M = 1800R - 0.5MR
1000M = 2000R - 3MR

Take the bottom equation and multiply both sides by 2 to get:
2000M = 1800R - 0.5MR
2000M = 4000R - 6MR

Since both equations are set equal to 2000M, we can now write: 1800R - 0.5MR = 4000R - 6MR
Rearrange to get: 5.5MR = 2200R
Divide both sides by R to get: 5.5M = 2200
Solve to get: M = 400 meters per minutes (this is Jim's speed)

Time = distance/speed
So, time for Jim to run 2000 meters = 2000/400 = 5 minutes
Check the answer choices....only one answer choice has 5 minutes as Jim's running time
Answer: B
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   01 Feb 2022, 06:56
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Chaithraln2499 wrote:
I don't understand how we got the second equation with 1000 & 2000?


I will add some annotations to that part of my solution....

Jerry gives Jim a start of 3mins and is beaten by 1000m
This time Jerry runs 1000 meters, and Jim runs 2000 meters.
Also, Jerry's travel time is 3 minutes LESS THAN Jim's travel time.

If Jerry's travel time is 3 minutes LESS THAN Jim's travel time, we can create a word equation for this information,
Since Jerry's travel time is 3 minutes LESS THAN Jim's travel time, we can make their travel times equal by adding 3 minutes to Jerry's travel time.

So, we can write: (Jerry's travel time + 3 minutes) = (Jim's travel time)
time = distance/speed, so we can write: 1000/R + 3 = 2000/M [Jerry traveled 1000 meters at a speed of R meters per minute, so his travel time is 1000/R. Similarly, Jim traveled 2000 meters at a speed of M meters per minute, so his travel time is 2000/M.]
To eliminate the fractions, multiply both sides by MR to get: 1000M + 3MR = 2000R
Rearrange to get: 1000M = 2000R - 3MR

Does that help?
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink] New post   31 Dec 2023, 10:35
GreenlightTestPrep
You used the time difference in your solution, I tried to use speeds.
My solution failed. Where is my mistake?

Information about the first run:

0M_____200M______________________2,000M

Jerry:
Distance - 2,000
Time - t
Speed - 2,000\t

Jim:
Distance - 1,800
Time - (t + 0.5)
Speed - 1,800\(t + 0.5)



Information about the second run:

0M______________1,000M_______________2,000M

Jerry:
Distance - 2,000
Time - t
Speed - 2,000\t

Jim:
Distance - 1,000
Time - (t + 3)
Speed - 1,000\(t + 3)



The speed that Jim is running in the first and the second run are same (I assume so...),
so let's create and solve the equation:


1,800\(t + 0.5) = 1,000\(t + 3)

Divide by 100
18\(t + 0.5) = 10\(t + 3)

Solve it
18t + 48 = 10t +5

The result is wrong, but why?
8t = -43
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