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
_________________

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.

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!