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

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

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17 Jan 2015, 04:10
2
VeritasPrepKarishma wrote:
SonyGmat 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?

1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10

I find it rather challenging, any inputs??

Attachment:
Ques3.jpg

Jerry gives Jim a head start of 200 m so Jim starts not from the starting point but from 200 m ahead. Jerry still beats him by 30 sec which means that Jerry completes the race while Jim takes another 30 sec to complete it.
In this race, Jerry covers 2000m. In the same time, Jim covers the distance shown by the red line. Since Jim needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance which is (1/2)*s where s is the speed of Jim. The distance Jim has actually covered in the same time as Jerry is [1800 - (1/2)*s] shown by the red line.

Attachment:
Ques4.jpg

Jerry gives Jim a start of 3 mins means Jim starts running first while Jerry sits at the starting point. After 3 mins, Jerry starts running too. Now, Jim beats Jerry by 1000 m which means that Jim reaches the end point while Jerry is still 1000 m away from the end.

In this race, Jerry covers a distance of 1000 m only. In that time, Jim covers the distance shown by the red line (the distance before that was covered by Jim in his first 3 mins). This distance shown by the red line is given by 2000 - 3s (3s is the distance covered by Jim in 3 minutes)

Now you see that in the first race, Jerry covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, Jim would also have covered half the previous distance i.e. the second red line will be half the first red line.

First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

Hi Karisma,

Thank you for detailed explanation. Can one finish this kind of problem with the analysis you have given in 2 min during the test? Is there any faster way to do it? Anyway to guesstimate? or would it be more strategic to skip?

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

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18 Jan 2015, 21:50
2
thorinoakenshield wrote:
VeritasPrepKarishma wrote:
SonyGmat 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?

1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10

I find it rather challenging, any inputs??

Attachment:
Ques3.jpg

Jerry gives Jim a head start of 200 m so Jim starts not from the starting point but from 200 m ahead. Jerry still beats him by 30 sec which means that Jerry completes the race while Jim takes another 30 sec to complete it.
In this race, Jerry covers 2000m. In the same time, Jim covers the distance shown by the red line. Since Jim needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance which is (1/2)*s where s is the speed of Jim. The distance Jim has actually covered in the same time as Jerry is [1800 - (1/2)*s] shown by the red line.

Attachment:
Ques4.jpg

Jerry gives Jim a start of 3 mins means Jim starts running first while Jerry sits at the starting point. After 3 mins, Jerry starts running too. Now, Jim beats Jerry by 1000 m which means that Jim reaches the end point while Jerry is still 1000 m away from the end.

In this race, Jerry covers a distance of 1000 m only. In that time, Jim covers the distance shown by the red line (the distance before that was covered by Jim in his first 3 mins). This distance shown by the red line is given by 2000 - 3s (3s is the distance covered by Jim in 3 minutes)

Now you see that in the first race, Jerry covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, Jim would also have covered half the previous distance i.e. the second red line will be half the first red line.

First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

Hi Karisma,

Thank you for detailed explanation. Can one finish this kind of problem with the analysis you have given in 2 min during the test? Is there any faster way to do it? Anyway to guesstimate? or would it be more strategic to skip?

TO

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

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17 Sep 2015, 12:08
SonyGmat 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

Race #1:
$$t_{Jim}=200m+t_{Jerry}+30s$$
$$t_{Jim}=\frac{1}{10}*t_{Jim}+t_{Jerry}+\frac{1}{2}$$
$$\frac{9}{10}*t_{Jim}=t_{Jerry}+\frac{1}{2}$$
$$t_{Jerry}=\frac{9}{10}*t_{Jim}-\frac{1}{2}$$

Race #2:
Intuitively if Jerry started 3 minutes after, and had 1000 meters left when Jim finished, he had been running for half his total required time.
$$t_{Jim}=3+\frac{1}{2}*t_{Jerry}$$
$$t_{Jim}-3=\frac{1}{2}*t_{Jerry}$$
$$t_{Jerry}=2*t_{Jim}-6$$

$$t_{Jerry}=\frac{9}{10}*t_{Jim}-\frac{1}{2}$$
$$t_{Jerry}=2*t_{Jim}-6$$
$$\frac{9}{10}*t_{Jim}-\frac{1}{2}=2*t_{Jim}-6$$
$$6-\frac{1}{2}=\frac{11}{10}*t_{Jim}$$
$$55=11*t_{Jim}$$
$$5=t_{Jim}$$

$$t_{Jerry}=2*t_{Jim}-6$$
$$t_{Jerry}=2*5-6$$
$$t_{Jerry}=10-6=4$$

$$t_{Jim}=5$$
$$t_{Jerry}=4$$
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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Updated on: 31 Jan 2017, 13:51
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

Originally posted by gracie on 17 Sep 2015, 18:55.
Last edited by gracie on 31 Jan 2017, 13:51, edited 1 time in total.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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02 Oct 2015, 02:04
We need to find time difference between two swimmers

for 200m handicap ----> 30 sec. overtake. For 0m. handicap ----- 1min overtake. That means that time difference should be 1 min for whole distance

B fits

P.S. First, I have spent 4 min. to create algebraic expression but could not (time confused with rates etc.). If it is true GMAT question, as it is written, they must have meant solution like above
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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Updated on: 15 Mar 2016, 22:18
2
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.
Attachments

rtd ab.jpg [ 24.99 KiB | Viewed 1868 times ]

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Originally posted by shasadou on 11 Mar 2016, 11:54.
Last edited by shasadou on 15 Mar 2016, 22:18, edited 1 time in total.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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15 Mar 2016, 20:37
ruprocks wrote:
Hi,

I am trying to solve this problem in following way but not able to solve it, there s some mistake in the process but couldnot find it if anyone can help.
1st case
distance time
jerry 2000m y-0.5
jim 1800 y
2nd case
jerry 1000 x
jim 2000 x-3

now
equate the speed in both the case
2000/(y-5)= 1000/x
1800/y= 2000/(x-3)

I tried doing the same, not sure what the problem is.
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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15 Mar 2016, 20:49
MeghaP wrote:
ruprocks wrote:
Hi,

I am trying to solve this problem in following way but not able to solve it, there s some mistake in the process but couldnot find it if anyone can help.
1st case
distance time
jerry 2000m y-0.5
jim 1800 y
2nd case
jerry 1000 x
jim 2000 x-3

now
equate the speed in both the case
2000/(y-5)= 1000/x
1800/y= 2000/(x-3)

I tried doing the same, not sure what the problem is.

hi

the problem is in the second part

Quote:
2nd case
jerry 1000 x
jim 2000 x-3

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

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16 Mar 2016, 01:33
SonyGmat 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

It is better to go from the options.
Plugging in A) 8, 10
Speed of A is 250 and B is 200
In 30 sec B can cover 100 m; implies B covered 1700 m in 8 min, which implies a speed more than 200 -incorrect
Plugging in B) 4,5
Speed of A is 500 and B is 400
In 30 sec B can cover 200 m; implies B covered 1600 m in 4 sec, which implies a speed of 400 which is correct by our supposition.
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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31 Mar 2016, 06:45
VeritasPrepKarishma wrote:
SonyGmat 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?

1. 8,10
2. 4,5
3. 5,9
4. 6,9
5. 7,10

I find it rather challenging, any inputs??

Attachment:
The attachment Ques3.jpg is no longer available

Jerry gives Jim a head start of 200 m so Jim starts not from the starting point but from 200 m ahead. Jerry still beats him by 30 sec which means that Jerry completes the race while Jim takes another 30 sec to complete it.
In this race, Jerry covers 2000m. In the same time, Jim covers the distance shown by the red line. Since Jim needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance which is (1/2)*s where s is the speed of Jim. The distance Jim has actually covered in the same time as Jerry is [1800 - (1/2)*s] shown by the red line.

Attachment:
The attachment Ques4.jpg is no longer available

Jerry gives Jim a start of 3 mins means Jim starts running first while Jerry sits at the starting point. After 3 mins, Jerry starts running too. Now, Jim beats Jerry by 1000 m which means that Jim reaches the end point while Jerry is still 1000 m away from the end.

In this race, Jerry covers a distance of 1000 m only. In that time, Jim covers the distance shown by the red line (the distance before that was covered by Jim in his first 3 mins). This distance shown by the red line is given by 2000 - 3s (3s is the distance covered by Jim in 3 minutes)

Now you see that in the first race, Jerry covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, Jim would also have covered half the previous distance i.e. the second red line will be half the first red line.

First red line = 2*second red line
1800 - (1/2)s = 2*(2000 - 3s) (where s is the speed of Jim in m/min)
s = 400 m/min

Time taken by Jim to run a 2000 m race = 2000/400 = 5 min

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: $$\frac{2000 meters}{500 mpm} = 4 min$$

Attachments

File comment: Race2

race 2.jpg [ 5.48 KiB | Viewed 1749 times ]

File comment: Race1

race 1.jpg [ 6.44 KiB | Viewed 1748 times ]

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

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18 Nov 2016, 00:20
SonyGmat 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

Here's a somewhat alternative approach..

The ratio of speeds is equal to the distances traveled in the same time.
Let Jerry's speed = a m/s
Jim's speed = b m/s

In the first case, when Jerry covers 2000 m, Jim covers?... 2000 - 200 - 30b = 1800 - 30b in the same time.

In the second case, when Jerry covers 1000 m, Jim covers?... 2000 - 180b in the same time. We have subtracted the distance that Jim covers during the head start(3 mins) to make the time of their travel equal.

Therefore,

$$\frac{2000}{1800-30b} = \frac{1000}{2000-180b}$$

Solving gives $$b = \frac{20}{3} m/s$$

Thus time to cover the total distance for b is 5 mins. (B)

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

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30 Jan 2017, 20:30
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
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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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30 Jan 2017, 22:15
Zoser wrote:
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

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

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31 Jan 2017, 07:08
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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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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31 Jan 2017, 21:24
Zoser wrote:
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.

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

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05 May 2017, 12:28
Top Contributor
1
SonyGmat 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 separately?

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.
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

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

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Updated on: 28 May 2017, 08:35
GMATPrepNow wrote:
SonyGmat 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 separately?

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.
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

Dear Experts,

It is quite straight forward that we have to equate Jerry's speed in the first and second race, and do the same with Jim to evaluate the solution with 2 equations & 2 unknowns. But I was hoping anyone could clear out the problem in the following:

Jerry gives Jim a start of 200m and beats him by 30 seconds:-

So we can call t1 = Jim's Time <---> t1 - 0.5 = Jerry's Time || Jerry's speed = 2000 / (t1 - 0.5) <---> Jim's speed = 1800 / t1
OR we can call t1 = Jerry's Time <---> t1+ 0.5 = Jim's Time || Jerry's speed = 2000 / t1 <---> Jim's speed = 1800 / (t1 + 0.5)

Therefore, both of these scenarios imply that Jerry finished in 30 seconds less than Jim which is equivalent to Jim finished in 30 seconds more than Jerry

Jerry gives Jim a start of 3 mins and is beaten by 1000m:-

So we can call t2 = Jim's time <---> t2 - 3 = Jerry's time || Jim's speed = 2000 / t2 <---> Jerry's speed = 1000 / (t2 - 3)
OR we can call t2 = Jerry's time <---> t2 + 3 = Jim's time || Jim's speed = 2000 / (t2 + 3) <---> Jerry's speed = 1000 / t2

Therefore, both of these scenarios imply that Jim ran for 3 minutes more than Jerry which is also equivalent to Jerry ran for 3 minutes less than Jim

Now the issue lies in that I have 4 possibilities of combinations when equating Jerry's speed in first and second race, and equating Jim's speed in first and second race. Each possibility gives me a different answer where the right answer is given by:

Jerry speed equations: 2000 / t1 = 1000 / (t2 - 3)
Jim speed equations: 1800 / (t1 + 0.5) = 2000 / t2

How should we think to choose the right possibility? Why aren't all the word problem interpretations given equivalent resulting in same answer?

*Typo in red fixed

Originally posted by rosmann on 24 May 2017, 08:04.
Last edited by rosmann on 28 May 2017, 08:35, edited 1 time in total.
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Posts: 12
Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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Updated on: 28 May 2017, 08:34
rosmann wrote:
GMATPrepNow wrote:
SonyGmat 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 separately?

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.
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

Dear Experts,

It is quite straight forward that we have to equate Jerry's speed in the first and second race, and do the same with Jim to evaluate the solution with 2 equations & 2 unknowns. But I was hoping anyone could clear out the problem in the following:

Jerry gives Jim a start of 200m and beats him by 30 seconds:-

So we can call t1 = Jim's Time <---> t1 - 0.5 = Jerry's Time || Jerry's speed = 2000 / (t1 - 0.5) <---> Jim's speed = 1800 / t1
OR we can call t1 = Jerry's Time <---> t1+ 0.5 = Jim's Time || Jerry's speed = 2000 / t1 <---> Jim's speed = 1800 / (t1 + 0.5)

Therefore, both of these scenarios imply that Jerry finished in 30 seconds less than Jim which is equivalent to Jim finished in 30 seconds more than Jerry

Jerry gives Jim a start of 3 mins and is beaten by 1000m:-

So we can call t2 = Jim's time <---> t2 - 3 = Jerry's time || Jim's speed = 2000 / t2 <---> Jerry's speed = 1000 / (t2 - 3)
OR we can call t2 = Jerry's time <---> t2 + 3 = Jim's time || Jim's speed = 2000 / (t2 + 3) <---> Jerry's speed = 1000 / t2

Therefore, both of these scenarios imply that Jim ran for 3 minutes more than Jerry which is also equivalent to Jerry ran for 3 minutes less than Jim

Now the issue lies in that I have 4 possibilities of combinations when equating Jerry's speed in first and second race, and equating Jim's speed in first and second race. Each possibility gives me a different answer where the right answer is given by:

Jerry speed equations: 2000 / t1 = 1000 / (t2 - 3)
Jim speed equations: 1800 / (t1 + 0.5) = 2000 / t2

How should we think to choose the right possibility? Why aren't all the word problem interpretations given equivalent resulting in same answer?

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.

Originally posted by rosmann on 26 May 2017, 12:05.
Last edited by rosmann on 28 May 2017, 08:34, edited 1 time in total.
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Posts: 630
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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27 May 2017, 21:54
SonyGmat 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 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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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim  [#permalink]

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01 Jun 2017, 09:49
SonyGmat 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 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.

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Re: Jerry and Jim run a race of 2000 m. First, Jerry gives Jim   [#permalink] 01 Jun 2017, 09:49

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