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ruprocks
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 22 Apr 2012, 09:56
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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
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 04 Dec 2014, 03:42
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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 !

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 30 Jan 2017, 20:30
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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.
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Can you explain why the second red line is half the first red line? As I did not get it
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 31 Jan 2017, 07:08
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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.
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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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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 31 Jan 2017, 21:24
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Zoser
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.
Show more


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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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 27 May 2017, 21:54
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SonyGmat
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
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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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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 01 Jun 2017, 09:49
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SonyGmat
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
Show more

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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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 22 Apr 2021, 04:38
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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)

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 09 Jul 2024, 09:48
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E: Jerry's speed
M: Jim's speed

First, Jerry gives Jim a start of 200m and beats him by 30 seconds

Start                                                                     Finish line
A------------------B-----------------------C-------------------Z
 [___200m____]                             [__Jim's 30s___] 
 [___________________2000m_________________]

(i) Jim starts at B
(ii) By the time Jerry at Z, Jim is at C and must run 30s more to reach Z

\(\frac{2000}{E} = \frac{1800}{M} - 30\)

Next, Jerry gives Jim a start of 3mins and is beaten by 1000m

Start                                                                       Finish line
A--------------------H-----------K-------------------------------Z
 [__Jim's 180s__]                 [________1000m______] 
 [_____________________2000m_________________]

By the time Jim at Z
+) Distance run by Jerry: 1000m
+) Distance run by Jim: 2000m

However, Jim runs before Jerry 180s, Jerry is at K

\(\frac{1000}{E} = \frac{2000}{M}- 180\)


So we have

\(\frac{2000}{E} = \frac{1800}{M} - 30\)

­\(\frac{1000}{E} = \frac{2000}{M}- 180\)­

=> \(\frac{900}{M} - 15 = \frac{2000}{M} - 180\)

=> \(\frac{1100}{M} = 165\)

=> \(\frac{100}{M} = 15\)

=> \(M = \frac{100}{15} = \frac{20}{3} (m/s)\)

\(\frac{20m}{3s} = \frac{20 m}{3 s} * \frac{60 s}{1 min} = \frac{400 m}{ 1 min}\)

=> M = 400 m/s
=> time for M to finish the race: \(\frac{2000}{400} = 5\)min
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 09 Jul 2024, 13:06
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­Take the second instance 

Jim covers 2000m in say y mins
Then Jerry covers 1000m in y-3 mins
Therefore Jerry covers 2*1000m in 2*(y-3) mins 

Plug option A. Here Jim's time ie y= 10 
Then Jerry's time = 2*(10-3)= 14 mins. Reject. 

Plug option B. Here Jim's time ie y= 5
Then Jerry's time = 2(5-3) = 4 mins. Satisfies. 

Plug option C. Here Jim's time ie y= 9 
Then Jerry's time = 2*(9-3)= 12 mins. Reject. 

Plug option D. Here Jim's time ie y= 9 
Then Jerry's time = 2*(9-3)= 12 mins. Reject. 

Plug option E. Here Jim's time ie y= 10
Then Jerry's time = 2*(10-3)= 14 mins. Reject. 

Answer: B
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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 10 Jul 2024, 01:26
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We don't even need the 2nd part of the information in the question & can use plugins to get our answer effectively

For the first part of the question, since Jerry gives a 200 metre headstart to Jim at the start of the race & beats him by 30 seconds, we can assume that Jim has to cover 1800 m & Jerry has to cover 2000 m.

If we look at option B, which states the individual time taken by each to cover the distance of 2000 metres, we find that Jerry can cover that distance in 4 minutes or 240 seconds.

Now Jim has to cover 1800m & since in option B it states that he can cover 2000m in 5 seconds, we can use the equivalence formula for uniform speed to understand that Jim will take 1800/(2000/5) minutes to cover that distance, which is equal to 270 seconds.

Since there is a difference of 30 seconds between Jim & Jerry's option as per the information given, we can formulate the answer as B

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Jerry and Jim run a race of 2000 m. First, Jerry gives Jim 06 Aug 2025, 13:10
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Let's Keep it Simple....

Race 1.
Jerry:
Distance - 2000
time - x
Speed - 2000/x

Jim
Distance - 1800
time = x + 0.5 (30 seconds is 0.5 mins)
Speed= 1800/(x + 0.5)

Race 2
Jerry
Distance - 1000
time = y - 3 (We will assume Jim as y)
Speed = 1000/(y-3)

Jim
Distance - 2000
time = y
Speed = 2000/y

So
Jerry's speed in race 1 = Jerry's Speed in race 2
which means 1000/(y-3) = 2000/x

Jim's speed in race 1 = Jim's speed in race 2
1800/(x+0.5) = 2000/y


Now Input values...B fits...
This was the simplest way i could figure...
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