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

Expert's post

00:00

Question Stats:

27% (03:05) correct
72% (03:03) wrong based on 11 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?

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

1

This post received KUDOS

Expert's post

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:

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink]
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

Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim a [#permalink]
30 Jun 2020, 11:59

Expert's post

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]
30 Jun 2020, 11:59