Let t be jerry's time to complete the race. In t minutes, jim goes 1800-x meters where x represents how far he goes in 30 seconds.

jim goes x meters in 30 seconds, so proportionally he goes 6x meters when given a head start of 3 minutes. He has 2000-6x meters left to go. During that distance, jerry completed half the race so we can say jim goes 2000-6x meters in a time of t/2 minutes So he goes 4000-12x meters in a time of t minutes. From the first scenario, he also goes 1800-x meters in t minutes. So 4000-12x=1800-x

Solving for x gives 200 meters. So jim goes 200 meters in 30 seconds. Proportionally, he goes 2000 meters in 300 seconds or 5 minutes.

The fastest way to solve would be to realize jim has to go more than half the distance in 3 minutes. By the end the gap is 1000 meters so when jim went for 3 minutes the gap must have been more since jim is slower than jerry.

Posted from my mobile device

In races question i try to use ratios sB/sA=dB/dA finding out for the same time what is the distance travelled.

So for 1 piece of data we know
sB/sA = (2000-200-0.5sB)/2000 (1)
Explanation
For the same time that A goes 2000, B goes 2000-200 (Head Start) - (30 seconds of short distance)

Similarly from the second piece of data we know
sB/sA=(2000-3sB)/2000-1000 (2)

2000-3sB. Because B runs a distance that is 3 minutes short so 3sb = distance covered by B in 3 minutes.

Similarly A runs a distance of 1000 meters short, i.e. 2000-1000 = 1000

Equate 1 and 2. You will get sB.
Plugin sB in either to get sA.
Find tb=d/sb ta=d/sa.

Of all the explanations I found this most useful, don't know why there are no kudos on the most apt explanation (The one that will hit your mind while in exam)

Experts - Please help, I am a bit lost with the next steps.
Bunuel chetansharma VeritasKarishma yashikaaggarwal

Here is my approach.

1. Jerry gives Jim a start of 200m and beats him by 30 seconds - This means he us able to cover 200m in 30seconds.
We can calculate the speed of Jerry and Jim by using D1/D2=S1/S2
So, speed of Jerry = 400m/minute
speed of Jim =360m/minute

2.Similarly, Jerry gives Jim a start of 3mins and is beaten by 1000m - this means total time taken by Jerry is 3minutes(t+3) more than that of Jim(t)
Using D1/D2=T1/T2, I am getting an incorrect value of t.
I tried variations of t/t-3 as well.

Not sure where the issue is. I spent a lot of time over this one, unable to move ahead.
Thanks in advance!

Hi Bunuel,

callmeDP could you please help

Sure, lets see. There are many ways to solve this mate, below is one way

Distance travelled by Jerry= 2000 m
And by jim = 1800

Let time taken by jerry is = T min
Then time taken by jim is = T + 1/2 min

Similarly, for second trip
Distance travelled by Jerry = 1000 m
And by jim = 2000

Time taken by jerry = T1 min
Time taken by jim = T1 + 3 min

Now you know their speed will remain same in both first and second trip.

Speed of Jerry = 1000/T1 = 2000/T

Solving, 2*T1 = T

Speed of jim = 2000/T1 + 3 = 1800/ T + 0.5

Solving, 10/T1+3 = 9/T+0.5

Substitute, 2*T1 = T

10*(2*T1 + 3) = 9*(T1 + 3)

T1 = 2 min

T = 2*T1 = 2*2 = 4 min

Speed of Jerry = 2000/4 = 500 m/min

Jim = 1800/4+0.5 = 400 m/min

Time taken to complete 2000 m

Time taken by jerry = 2000/ 500 = 4 minutes
Time taken by jim = 2000 / 400 = 5 minutes

Ans B

Let me know if this helps, carouselambra

Posted from my mobile device
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You are wrong in the interpretation of your point 1.
Jerry gives Jim a start of 200m and still defeats him by 30secs. Let Jim cover 1800 in say x secs => speed =1800/x mps
Whereas Jerry covers 2000 in x-30 => speed = 2000/(x-30) m/s
You can solve for statement II too on these speeds and get your answer.
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Hi Bunuel,

callmeDP could you please help[/quote]

Sure, lets see. There are many ways to solve this mate, below is one way

Distance travelled by Jerry= 2000 m
And by jim = 1800

Let time taken by jerry is = T min
Then time taken by jim is = T + 1/2 min

Similarly, for second trip
Distance travelled by Jerry = 1000 m
And by jim = 2000

Time taken by jerry = T1 min
Time taken by jim = T1 + 3 min

Now you know their speed will remain same in both first and second trip.

Speed of Jerry = 1000/T1 = 2000/T

Solving, 2*T1 = T

Speed of jim = 2000/T1 + 3 = 1800/ T + 0.5

Solving, 10/T1+3 = 9/T+0.5

Substitute, 2*T1 = T

10*(2*T1 + 3) = 9*(T1 + 3)

T1 = 2 min

T = 2*T1 = 2*2 = 4 min

Speed of Jerry = 2000/4 = 500 m/min

Jim = 1800/4+0.5 = 400 m/min

Time taken to complete 2000 m

Time taken by jerry = 2000/ 500 = 4 minutes
Time taken by jim = 2000 / 400 = 5 minutes

Ans B

Let me know if this helps, carouselambra

Posted from my mobile device[/quote]

Hi,
Liked the approach. However, have a silly question.
As per your calculation,
Time taken by jerry = T1 min
Time taken by jim = T1 + 3 min

But in the question, it is stated that "Jerry gives Jim a start of 3mins and is beaten by 1000m"
Would this not mean that the total time taken by Jerry t+3 because he lost the race? Jerry should be t.
Or, total time taken by Jim= t-3 and Jerry=t

What is the issue with this? Thanks in advance!

Ah, got it. Thanks a ton!

You mean that you try the options this way?

4,5 = 2x + x + 3
x = 0,5
2x = 1 (Jerry)
x+3 = 3,5 (Jim)

What does x even represent here? Some part of the race that Jerry needs for head start to finish at the exact same time as Jim?

Lets try another option:

5,9 = 3x + 3
x = 2,9/3
2x = 5,8/3 (Jerry)
x+3 = 2,9/3 + 3 (Jim)

You can still solve this for any of the options?

Could you please elaborate your trial and error in more detail? Thanks!

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)

Devmitra Sen
GMAT SME
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Video solution starts at 29:42. An alternate solution to the same question starts at 47:35

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Q: Find the time in minutes in which Jerry and Jim can run the race separately?

Units taken in km for distance and min for time

Let the time taken by Jerry and Jim running the race separately be tx and ty respectively
Therefore, tx=2/rx and ty=2/ry
Or, rx=2/tx and ry=2/ty

Now from first race (equate t1, replacing rx with 2/tx and ry by 2/ty)
2/rx=t1-0.5
2tx/2+0.5=t1
also,
1.8/ry=t1
1.8ty/2=t1
Therefore, 1.8ty-2tx=1 ... #1

Similarly equate t2 from second race to get
2ty-tx=6 ... #2

Multiply equation #2 with 2 and subtract equation #1 from it
4ty-2tx=12
1.8ty-2tx=1
----------------
2.2ty=11
ty=5 min
Put ty=5 in either equation #1 or #2 o get tx
tx=4*5-12=4 min

Option B