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

From the first race,
Speed (Jerry) = 2000/x m/s
Speed (Jim) = 1800/ (x+30) m/s
From the second race,
Speed(Jerry) = 1000/y m/s
Speed(Jim) = 2000/(180+y)m/s

That means, 2000/x = 1000/y => y=x/2

1800/(x+30) = 4000/(360+y)
2200x = 526000 => x = 239 seconds.

Hence, Jerry will complete 2000m race in 2000/239 = 3.98 ~= 4 minutes.

Speed of Jerry is 1800 / (239+30) = 6.69 m/s => The time required for Jerry to finish 2000m = 2000/6.69 = 298.95 s

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.

Answer (B)

It's a tough question. DO you have the OA? I'm not sure how to work this one out.

Yes, i believe the correct answer is B.

This question was already posted, but I would like to see if someone can offer a faster solution.

time-speed-distance-problem-83740.html

Karishma,

thanks for that wonderful explanation . i was just trying to arrive at the answer in a little diff way .. i.e equating the times
so i get something i like this

1800/s -1/2 =2000/s-3

so im just equating the time taken by Jim across both the races and when i compare it with the equation you have come up with , I realised that i have mafe a mistake and tried to see how it went wrong bu unable to.
Can you please help me with this?

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)

getting -ve ans please help me finding the problem in this approach

According to this equation, time taken by Jim in the first race is same as the time taken by him in the second race. But that is not true. In the second race, he covers half the distance he covered in the first race (since in the same time, Jerry covered half the distance too. Since their speeds are constant in the 2 races, the time in the second race must be half)
Otherwise your equation is the same as mine

1800/s -1/2 = 2(2000/s-3)

Look at the diagrams given above. The distance and time in the two races are different from what you have assumed.
In the first race, Jerry covers 2000 m while we don't know how much distance Jim covers.
In the second race, Jim covers 2000 m in x + 3 mins.

Explained beautifully . A small typing error.

Considering that this question has to be done under 2 mins, I think the fastest way is to Plug-in(The concept has already been taken care of in the posts above)

From the first part, we know that 1800/\(v_{jim}\) - 0.5 = 2000/\(v_{jerry}\)--> 9/10*(2000/\(v_{jim}\))-0.5 = 2000/\(v_{jerry}\)

Or 0.9*\(t_{jim}\)-0.5 = \(t_{jerry}\)

Only B qualifies.

One could also use the second part of the problem and frame an equation : [2000-3*\(v_{jim}\)/\(v_{jim}\)]*\(v_{jerry}\) = (2000-1000)

Or, 2000/\(v_{jim}\) - 3 = 1000/\(v_{jerry}\) --> \(t_{jim}\) -3 = \(t_{jerry}\)/2

B.

Just solved it in 1:08, but I'm not sure about the method.

Well, I created and solved a simply equation: 2,000*(t-30)=1800*t
Where 't' is the time needed by Jim and 't-30' the time needed for his buddy Jerry.

We solve it and we get t=300, 5 min. We look at the answers and we immediately see that only B has 5 min for Jim.

Plz tell me if my way is correct or I got it just for luck !

Posted from GMAT ToolKit

I don't understand the logic behind this equation: 2,000*(t-30)=1800*t
This implies Jerry's Distance * Jerry's Time = Jim's Distance * Jim's Time
Why?

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.

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.

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)

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