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

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.

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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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:

RELATED VIDEO

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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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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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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)

Devmitra Sen
GMAT SME

Video solution starts at 29:42. An alternate solution to the same question starts at 47:35

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