Yes, if I had only 20 secs to read and answer the question, I would guess (B) (given these options) and move on. Had the options been different, I might have been able to eliminate some and then guess out of remaining.

The options which give time taken by Jim as some value between 3 mins and 6 mins (excluding) could work.

The reason is this:
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?

So first, Jerry gives Jim a head start of 1/10th of the race but still beats him. This means Jim is certainly quite a bit slower than Jerry.
Next Jerry gives Jim a start of 3 mins and Jim beats him by 1000 m i.e. half of the race. What does this imply? It implies that Jim ran more than half the race in 3 mins. Look - Say Jim covers x meters in 3 mins. Once Jerry starts running, he starts reducing the distance between them since he is covering more distance every second than Jim because he is faster. At the end, the distance between them is still 1000 m. This means the initial distance that Jim created between them by running for 3 mins was certainly more than 1000 m. So total time Jim would take to finish the race will be less than 2*3 min i.e. less than 6 mins. Jerry must be even faster and hence would take even lesser time. Hence only option (B) will satisfy here.
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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

let t=Jerry's time in race 1
Jim's speed in race 1 is 1800/(t+1/2)
Jim's speed in race 2 is 2000/(t/2+3)
as Jim's speed is constant,
1800/(t+1/2)=2000/(t/2+3)
t=4 minutes
Jerry runs race 1 (2000 m) in 4 minutes
Jim runs race 2 (2000 m) in t/2+3=5 minutes
4,5
B

For those still struggling I have tried to present the solution in a neater and more eatable form.

STEP 1
I find the RTD chart suggested by mgmat quite handy for organization of info for distance/work problems. See the attachment with the RTD for the first case and for the second one. As you can see by organizing everything into the matrix boxes we also have compiled the necessary algebraic equations.

STEP 2
Here we should decide through which variable of the RTD formula we should present the relationship in the equations. YOU WILL VERY OFTEN COME ACROSS SUCH CHOICE ON THE DISTANCE PROBLEMS INVOLVING ALGEBRAIC SOLUTIONS - SO CONSIDER WHICH OF THE TREE COMPONENTS OF THE RTD FORMULA TO TAKE USE OF IN FURTHER MANIPULATIONS.

So, Time is varying across all the four equations - out. Distance? Well from the equations in their current form I believe I can come to solution but this gonna take more sweat and additional steps (e.g. 1800 needs to be balanced up to 2000 or down to 1000, same with 1000 in the second case). Let us try with rate - the speeds are constant and we can express the equations through the speed variable, this promises to be a shorter path because the equations will look neat and somewhat compact, and no need for additional steps. Let us try.

STEP 3
*I will start with A: a = \(\frac{2000}{t}\) and a = \(\frac{1000}{q-180}\), hence equate the two \(\frac{2000}{t}\) = \(\frac{1000}{q-180}\). Manipulate and get t = 2q-360.

*Equate the speed of B in the 2nd case in the same manner as above: \(\frac{1800}{t+30}\) = \(\frac{2000}{q}\). And plug the value of t derived from above into this equation and find q = 300 seconds = 5 min. Only answer B contains a 5 min option for B (Jim).

Final tips
The above approach and the habit of ogranization should save u time and nerves. However I would like to highlight ocne again this brilliant solution suggested by eaakbari above jerry-and-jim-run-a-race-of-2000-m-first-jerry-gives-jim-122757.html

The problem is hard because first you need to form algebraic equations (and smart numbers cannot be used here) correctly, and there are 4 of them. Second you need to figure out how to play with them. And finally you need to be accurate in calculations. So I don't think this is solvable in 2 mins under exam conditions. Having organized the info accurately one can get the aha moment quite soon if sticks to the above reasoning. But I like the ingenuity and brevity of eaakbari solution, and I think this is what GMAT tests.

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KUDO me plenty

hi

the problem is in the second part



jim starts 3 min early, so time taken will be x+3 and not x-3.. now do it you will get the answer
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Dear Experts, could you , please, evaluate my solution? I cannot find a flaw in it but not sure that my logic is correct. In my humble opinion, the solution is attractive because it's easier to understand.

I use Karishma's diagrams. ( in attachment below )


step 1 - finding Jim speed ;
1st Race :
Let say \(S_m\)- Jim speed ;
Because 200 meters was given to Jim, we know that overall at the 1st race he made 2000 - 200 = 1800 meters before Jerry had finished .
Because Jim had lost the 1st race by 30 second, this means that if additional 30 second were given to Jim, he finished the 1st race simultaneously with Jerry.

To sum up all this facts we have :
(2000meters - 200meters ) + \(S_m*\frac{1}{2}\)minute = 2000 meters
1800 + \(\frac{1}{2}min\)* \(S_m\) = 2000
\(S_m\)= 400 meters per min ;



****
****
****

step 2 - finding Jerry speed;
2nd Race :
\(S_m\)- Jim speed ;
\(S_r\) - Jerry speed ;
\(t_r\) - Jerry time from second race before Jim finished the race ;

We didn't know how many minutes it takes Jim to finish the 2d race. However, we know that he ran 3 minutes plus time that Jerry ran his 1000 meters before Jim ended the race.
So we have \(t_m = t_r + 3min\)


System of equations for 2nd Race :
(1) \(S_r*t_r = 1000\)
(2)\(S_m*(t_r + 3min ) = 2000\)
(3) \(S_m\)= 400 mpm ;

First solution :
(2) \(400 *t_r + 1200meters = 2000meters\)
\(t_r\)= \(\frac{2000 - 1200}{400} = 2 minutes\);
put \(t_r\) in (1) and find \(S_r\)
\(S_r = 500\) mpm

Second solution (alternative) :
From race 1, we can find that Jim can run 2000 in 5 minutes => \(\frac{2000 meters}{400 mpm}= 5 min\)
So 5 min - 3 min = 2 min
\(t_r\)= 2 minutes;
put \(t_r\) in (1) and find \(S_r\)
\(S_r = 500\) mpm




step 3 - finding the time in minutes in which Jerry and Jim can run the race separately

Jim: \(\frac{2000 meters}{400 mpm}= 5 min\)
Jerry: \(\frac{2000 meters}{500 mpm} = 4 min\)

Answer B

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Please, press the +1 KUDOS button , if you find this post helpful

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

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.

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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Brent Hanneson – GMATPrepNow.com

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I know my question may not be clear, but I am new to this forum and I am still learning. I coloured the problematic expressions to clarify my question. Aren't the purple-coloured expressions equivalent to each other? Also aren't the maroon-coloured expressions equivalent to each other? Because they give different answers. Thanks in advance.

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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