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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 30 Jan 2017, 22:15
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Zoser
Quote:

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.

Can you explain why the second red line is half the first red line? As I did not get it
Show more


Jim and Jerry run at their own constant speeds.
So the ratio of the distance covered by them will be the same in same time.

i.e. if Jim's speed is say 8 mph and Jerry's speed is 5 mph, in 2 hrs, Jim will cover a distance of 16 miles and Jerry will cover 10 miles.
In half the time, i.e. in 1 hr, Jim will cover half the distance i.e. 8 miles. Jerry will also cover half the distance i.e. 5 miles.

In the second case, Jerry runs half the distance (1000 m) so he runs for half the time. In the same time, Jim runs the distance of two red lines. So in the second case, Jim's distance covered will be half his distance covered in the first case. So second red line will be half the first red line.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 31 Jan 2017, 07:08
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Quote:
Jim and Jerry run at their own constant speeds.
So the ratio of the distance covered by them will be the same in same time.

i.e. if Jim's speed is say 8 mph and Jerry's speed is 5 mph, in 2 hrs, Jim will cover a distance of 16 miles and Jerry will cover 10 miles.
In half the time, i.e. in 1 hr, Jim will cover half the distance i.e. 8 miles. Jerry will also cover half the distance i.e. 5 miles.

In the second case, Jerry runs half the distance (1000 m) so he runs for half the time. In the same time, Jim runs the distance of two red lines. So in the second case, Jim's distance covered will be half his distance covered in the first case. So second red line will be half the first red line.
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Thanks for your reply. But can you tell me why you assumed that they are running at the same speed in both races. The question did not say anything about that.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 31 Jan 2017, 21:24
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Zoser
Quote:
Jim and Jerry run at their own constant speeds.
So the ratio of the distance covered by them will be the same in same time.

i.e. if Jim's speed is say 8 mph and Jerry's speed is 5 mph, in 2 hrs, Jim will cover a distance of 16 miles and Jerry will cover 10 miles.
In half the time, i.e. in 1 hr, Jim will cover half the distance i.e. 8 miles. Jerry will also cover half the distance i.e. 5 miles.

In the second case, Jerry runs half the distance (1000 m) so he runs for half the time. In the same time, Jim runs the distance of two red lines. So in the second case, Jim's distance covered will be half his distance covered in the first case. So second red line will be half the first red line.

Thanks for your reply. But can you tell me why you assumed that they are running at the same speed in both races. The question did not say anything about that.
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The question implies that both have their own constant speeds.

"Find the time in minutes in which Jerry and Jim can run the race seperately?"

Otherwise this question makes no sense. The time in which Jerry runs the race has to be constant and the time in which Jim runs the race has to be constant too.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim Updated on: 27 Dec 2019, 07:30
Originally posted by BrentGMATPrepNow on 05 May 2017, 12:28.
Last edited by BrentGMATPrepNow on 27 Dec 2019, 07:30, edited 1 time in total.
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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 separately?

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

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:
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B

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 27 May 2017, 21:54
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SonyGmat
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
Show more
Let speed of Jerry be s1 and that of Jim be s2
case 1: s1=2000/t1 , s2= 1800/(t1+30)
case 2: s1=1000/t2, s2=2000/(t2+180)
Solving the equations, we get t1=240, t2=120
So Time taken by Jerry for full distance is t1=240sec or 4 min
Time taken by Jim for the full distance is t2+180=300sec or 5 min
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 01 Jun 2017, 09:49
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SonyGmat
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
Show more

Let’s denote Jerry’s rate as v meters/min and Jim’s rate as w meters/min.

Since time = distance/rate, the time it takes to finish the race is 2000/v minutes for Jerry and 2000/w minutes for Jim.

Since Jerry can beat Jim by 30 seconds = 1/2 minute when he gives Jim a head start of 200 meters, the time it takes Jerry to run 2000 meters is 1/2 minute less than the time it takes Jim to run 1800 meters; hence:

2000/v = 1800/w - 1/2

Jerry is beaten by 1000 meters when he gives Jim a head start of 3 minutes, so when Jim finishes the 2000 meters, Jerry has run only 1000 meters. This means that the time it takes Jerry to run 1000 meters is 3 minutes less than the time it takes Jim to run 2000 meters; therefore:

1000/v = 2000/w - 3

Let’s multiply the first equation by 2vw and the second by vw:

4000w = 3600v - vw

1000w = 2000v - 3vw

Multiplying the equation 4000w = 3600v - vw by 3, we get:

12000w = 10800v - 3vw

Isolating 3vw in each of the equations, we get:

3vw = 10800v - 12000w

3vw = 2000v - 1000w

Since we have two expressions equal to 3vw, we can set these equations equal to each other:

10800v - 12000w = 2000v - 1000w

8800v = 11000w

88v = 110w
8v = 10w

4v = 5w

w = 4v/5

Let’s substitute w = 4v/5 in the equation 1000/v = 2000/w - 3:

1000/v = 2000/(4v/5) - 3

1000/v = 10000/4v - 3

Let’s multiply each side by 4v:

4000 = 10000 - 12v

12v = 6000

v = 500

Jerry’s rate is 500 meters/minute; thus, he can complete the race in 2000/500 = 4 minutes.

Since w = 4v/5, Jim’s rate is 4(500)/5 = 400 meters/minute; thus, he can complete the race in 2000/400 = 5 minutes.

Answer: B
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 22 Apr 2021, 04:38
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Solution:

Concept: The question tests on the application of Rates(Time-Speed-Distance) in a Race question

1st Race

Let Jerry take x sec

=>Jim would take (x+30) sec to complete the same race

=>Jerry’s speed = Distance of the race/His speed

= 2000 / x m/s

Jim’s speed= Distance of the race/His speed

=1800 / (x+30) m/s

2nd Race

Let Jerry take y sec to cover the race

Jerry has not covered the 2000m; instead he is beaten by 1000m

=>He has covered 2000-1000 = 1000m in y sec

=> Jerry’s speed here = 1000/y m/s

Jim gets a start of 3min or 180sec

=> He gets 180 sec additionally for the race and he also completes the 2000 m race

Speed of Jim = 2000 / (180+y)m/s ----------(1)


That means, 2000/x = 1000/y (Equating Jerry’s speed from both races)

=> y=x/2

=> 1800 / (x+30) = 4000 / (360 + y ) (Substitute y=x/2 in eq(1))

=>2200x = 526000

=> x = 239 seconds.

Hence, Jerry will complete 2000m race in 2000/239

= 3.98 ~= 4 minutes. (You may not need to solve ahead as the option b starts for Jerry with 4)

Speed of Jim is 1800 / (239+30)

= 6.69 m/s

=> The time required for Jim to finish 2000m = 2000/6.69 = 298.95 s

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.(option b)

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 24 Dec 2021, 06:35
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Video solution starts at 29:42. An alternate solution to the same question starts at 47:35
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 09 Jul 2024, 09:48
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E: Jerry's speed
M: Jim's speed

First, Jerry gives Jim a start of 200m and beats him by 30 seconds

Start                                                                     Finish line
A------------------B-----------------------C-------------------Z
 [___200m____]                             [__Jim's 30s___] 
 [___________________2000m_________________]

(i) Jim starts at B
(ii) By the time Jerry at Z, Jim is at C and must run 30s more to reach Z

\(\frac{2000}{E} = \frac{1800}{M} - 30\)

Next, Jerry gives Jim a start of 3mins and is beaten by 1000m

Start                                                                       Finish line
A--------------------H-----------K-------------------------------Z
 [__Jim's 180s__]                 [________1000m______] 
 [_____________________2000m_________________]

By the time Jim at Z
+) Distance run by Jerry: 1000m
+) Distance run by Jim: 2000m

However, Jim runs before Jerry 180s, Jerry is at K

\(\frac{1000}{E} = \frac{2000}{M}- 180\)


So we have

\(\frac{2000}{E} = \frac{1800}{M} - 30\)

­\(\frac{1000}{E} = \frac{2000}{M}- 180\)­

=> \(\frac{900}{M} - 15 = \frac{2000}{M} - 180\)

=> \(\frac{1100}{M} = 165\)

=> \(\frac{100}{M} = 15\)

=> \(M = \frac{100}{15} = \frac{20}{3} (m/s)\)

\(\frac{20m}{3s} = \frac{20 m}{3 s} * \frac{60 s}{1 min} = \frac{400 m}{ 1 min}\)

=> M = 400 m/s
=> time for M to finish the race: \(\frac{2000}{400} = 5\)min
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 09 Jul 2024, 13:06
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­Take the second instance 

Jim covers 2000m in say y mins
Then Jerry covers 1000m in y-3 mins
Therefore Jerry covers 2*1000m in 2*(y-3) mins 

Plug option A. Here Jim's time ie y= 10 
Then Jerry's time = 2*(10-3)= 14 mins. Reject. 

Plug option B. Here Jim's time ie y= 5
Then Jerry's time = 2(5-3) = 4 mins. Satisfies. 

Plug option C. Here Jim's time ie y= 9 
Then Jerry's time = 2*(9-3)= 12 mins. Reject. 

Plug option D. Here Jim's time ie y= 9 
Then Jerry's time = 2*(9-3)= 12 mins. Reject. 

Plug option E. Here Jim's time ie y= 10
Then Jerry's time = 2*(10-3)= 14 mins. Reject. 

Answer: B
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 10 Jul 2024, 01:26
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We don't even need the 2nd part of the information in the question & can use plugins to get our answer effectively

For the first part of the question, since Jerry gives a 200 metre headstart to Jim at the start of the race & beats him by 30 seconds, we can assume that Jim has to cover 1800 m & Jerry has to cover 2000 m.

If we look at option B, which states the individual time taken by each to cover the distance of 2000 metres, we find that Jerry can cover that distance in 4 minutes or 240 seconds.

Now Jim has to cover 1800m & since in option B it states that he can cover 2000m in 5 seconds, we can use the equivalence formula for uniform speed to understand that Jim will take 1800/(2000/5) minutes to cover that distance, which is equal to 270 seconds.

Since there is a difference of 30 seconds between Jim & Jerry's option as per the information given, we can formulate the answer as B

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 06 Aug 2025, 13:10
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Let's Keep it Simple....

Race 1.
Jerry:
Distance - 2000
time - x
Speed - 2000/x

Jim
Distance - 1800
time = x + 0.5 (30 seconds is 0.5 mins)
Speed= 1800/(x + 0.5)

Race 2
Jerry
Distance - 1000
time = y - 3 (We will assume Jim as y)
Speed = 1000/(y-3)

Jim
Distance - 2000
time = y
Speed = 2000/y

So
Jerry's speed in race 1 = Jerry's Speed in race 2
which means 1000/(y-3) = 2000/x

Jim's speed in race 1 = Jim's speed in race 2
1800/(x+0.5) = 2000/y


Now Input values...B fits...
This was the simplest way i could figure...
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 11 Jun 2026, 11:23
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Two algebric methods for solving this question. The first method is a bit time consuming, the 2nd one is faster and preferred.

Method 1 ---> taking speed as x m/s and y m/s for jerry and Jim
The following equation would hold

EQ1: 2000/x = 1800/y - 30 | Please take note of the 1800 and MINUS 30
EQ2: 2000/x = 2000/y - 180 + 1000/x

If you solve you ll get y = 20/3 and x = 25/3 ---> This is speed, use 2000 as distance to find the time which is 4 and 5 respect for x and y respectively


Method 2 (preferred since less calculations) ---> Take x seconds as time for 2000 meters for Jerry
y seconds as time for 2000 meters for Jim


EQ1: x = y - 200/2000y - 30 ....i

EQ2: x = y - 180 + 1/2x ....ii

Solve simultaneously you ll get x as 240 seconds and y as 300 seconds which is 4 mins and 5 mins repsectively Answer B
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 11 Jun 2026, 19:34
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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 separately?

Let the time in minutes in which Jerry and Jim can run the race separately be x & y minutes respectively

Jerry's speed = 2000/60x = 100/3x m/s
Jim's speed = 2000/60y = 100/3y m/s

-----200m----| Jim------------ t+30seconds----
---------Jerry--------------t-----------------------
First: -
Time taken by Jerry t = (2000)/(100/3x) = 2000*3x/100 = 60x seconds
Time taken by Jim = t+30 = 60x + 30
Distance travelled by Jim = (60x+30)*100/3y + 200 = 2000
60x + 30 = 54y
y = (60x + 30)/54 = 10x/9 + 10/18

------3m = 180 seconds |---------Jim----------------
-------------------Jerry--t----|-----1000m------------
Second: -
Time taken by Jerry = t= 1000/(100/3x) = 30x seconds
Time taken by Jim = 30x + 3*60 = (30x + 180) seconds
Distance travelled by Jim = (30x + 180)*100/3y = 2000
30x + 180 = 60y
y = x/2 + 3

y = x/2 + 3 = 10x/9 + 10/18
x(10/9 - 1/2) = 3 - 10/18 = 44/18
x/18 (20 - 9) = 44/18
11x/18 = 44/18
x = 4 minutes

y = x/2 + 3 = 4/2 + 3 = 2 + 3 = 5 minutes

IMO B
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 12 Jun 2026, 11:02
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Let Jim's speed = (s) m/min.
The trick is to compare only the periods during which Jerry is actually running.
--------------------------------------------------------------------------------

RACE 1
Jim gets a 200 m head start.
So Jim only needs to run: 1800 meters.
Jerry finishes first.
Jim still needs another 30 seconds = 0.5 minute to finish.
In that extra 0.5 minute, Jim covers:
0.5s
meters.
Therefore, when Jerry finishes, Jim has covered:
1800-0.5s
meters.
So during the time Jerry runs his race:
Jerry covers:
2000
Jim covers:
1800-0.5s
Call this Jim's "Race-1 red distance".
--------------------------------------

RACE 2
Jim gets a 3-minute head start.
Before Jerry even starts, Jim covers:
3s
meters.
When Jim reaches the finish line, Jerry is still 1000 m away.
Therefore Jerry has covered only:
1000
meters.
During the period when Jerry is actually running:
Jim covers:
2000-3s
meters.
Call this Jim's "Race-2 red distance".
So during Jerry's running time:
Jerry covers:
1000
Jim covers:
2000-3s
-------

THE KEY INSIGHT
Compare Jerry's distances.
Race 1:
2000
Race 2:
1000
Jerry's speed is constant.
Distance = Speed × Time.
Since
1000 = (1/2)(2000)
Jerry must have run for exactly half as long in Race 2 as in Race 1.
This is the step that often confuses people.
Suppose Jerry runs at 500 m/min.
If he covers 2000 m, time = 4 min.
If he covers 1000 m, time = 2 min.
The second time is exactly half the first.
So:
T2 = (1/2)T1
where (T1) and (T2) are Jerry's running times in the two races.
Now Jim's speed is also constant.
If the time is halved, Jim's distance must also be halved.
Therefore:
Jim's Race-2 red distance = (1/2)(Jim's Race-1 red distance)
So:
2000 - 3s = (1/2)(1800 - 0.5s)
Multiply by 2:
1800 - 0.5s = 2(2000 - 3s)
--------------------------

SOLVE
1800 - 0.5s = 4000 - 6s
5.5s = 2200
s = 400
So Jim's speed is:
400 m/min
---------

FIND JIM'S TIME
2000/400 = 5
minutes.
--------

FIND JERRY'S TIME
Go back to Race 1.
Jim has to run only 1800 m.
At 400 m/min:
1800/400 = 4.5
minutes.
Jerry beats him by 30 seconds = 0.5 minute.
Therefore Jerry's time is:
4.5 - 0.5 = 4
minutes.
--------

ANSWER
Jerry: 4 minutes
Jim: 5 minutes
(4, 5)
Quick takeaway:
The entire shortcut comes from noticing:
(Jerry's Race-2 distance)/(Jerry's Race-1 distance) = 1000/2000 = 1/2
Since Jerry's speed is unchanged, the corresponding comparison times are also in the ratio (1:2). Once you see that, Jim's comparison distances must also be in the ratio (1:2), giving a one-variable equation.
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