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attached is the answer. Forgive me for the handwriting and certain assumptions in the explanation. Will provide additional detail if requested.
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Torrance and Harriet run a race along a long, straight path. Torrance runs at a constant speed of 2 miles per hour; Harriet runs at a speed of 8 miles per hour, but whenever Harriet leads Torrance by at least 1 mile, she stops and does not run again until she has fallen 2 miles behind. If both start in the same place and begin running at noon, what time is it when Torrance passes Harriet for the second time?

Considering the first one is not an overtake
First they both start together and Harriot stops when she reaches 1 mile ahead to Torrance= 1/6 *60 = 10mins (Note: distance covered by Harriot =8/6 miles)
Now Torrance has to cover 8/6 + 2 miles time = 20/6 * 60/2 = 100 mins
Again Harriot starts and stops until she covers 1 mile ahead to Torrance = 3/6 *60 = 30 mins (Note total distance covered by Harriot =8/6+ 4 = 32/6 miles,Time =130 mins)
Now Torrance has to cover 32/6 + 2 miles time = 44/6 * 60/2 = 220 mins
Again Harriot starts and this time she just have to overtake Torrance = 2/6 * 60 = 20 mins
Total time now is 220+20 = 240 mins

Considering the first one as an overtake

First they both start together and Harriot stops when she reaches 1 mile ahead to Torrance= 1/6 *60 = 10mins (Note: distance covered by Harriot =8/6 miles)
Now Torrance has to cover 8/6 + 2 miles time = 20/6 * 60/2 = 100 mins
Again Harriot starts and this time she just have to overtake Torrance = 2/6 *60 = 20 mins (Note total distance covered by Harriot =8/6+ 4 = 32/6 miles,Time =120 mins)
Total time now is 100+20 = 120 mins

:? :? :?
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My 2 cents: Here is my solutions!!!
If it was helpful please kudo!!!
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If you examine the question closely you will see that, since Harriet and Torrance start at the same time and run in the same direction along the same path, both H and T cover the same distance in the same time until T catches up with H for the second time, the only difference being that T runs non-stop but H makes two stops: (a) to let T cover the distance of 1 mile between them, pass her and run ahead for another 2 miles and (b) to let T again cover the distance of 1 mile to overtake her. So H is stationery for 2 hours (the time it takes T to run 4 miles). So, if T runs for 't' hrs, H runs for (t-2)hrs. But the distance they cover from the start to meeting for the 2nd time (or for any subsequent passing for that matter) is the same. So, if 'd' is the distance, d=st=2t (in T's case)=(t-2)8 (in H's case). Therefore, 8t-16=2t or t=8/3 hrs or 2hrs 40mins. Hence, answer D.
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Let’s first determine the time (t) it takes for Harriet to be 1 mile ahead of Torrance:

8t - 2t = 1

6t = 1

t = 1/6 hour = 10 minutes

So, Harriet stops after 10 minutes, at which time she has run 8 x 1/6 = 8/6 = 4/3 miles. Torrance has to take (4/3)/2 = 2/3 hour = 40 minutes to catch up with her, and that is the first time he catches up and passes her.

Torrance will continue to run another 2 miles before Harriet starts to run again. Harriet will stop again when she is 1 mile ahead. Thus, Harriet has to run 3 miles more than Torrance when she starts to run again:

8t - 2t = 3

6t = 3

t = 1/2 hour = 30 minutes

So, Harriet stops after another 30 minutes of running, at which time she has run another 8 x 1/2 = 4 miles. Torrance has to take 4/2 = 2 hours to catch up with her, and that is the second time he catches up and passes her.

Since they start at noon and Torrance needs to take 40 min + 2 hrs = 2 hrs 40 min to pass Harriet for the second time, he will pass her at 2:40 p.m.

Answer: D
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sreenu7464
Torrance and Harriet run a race along a long, straight path. Torrance runs at a constant speed of 2 miles per hour; Harriet runs at a speed of 8 miles per hour, but whenever Harriet leads Torrance by at least 1 mile, she stops and does not run again until she has fallen 2 miles behind. If both start in the same place and begin running at noon, what time is it when Torrance passes Harriet for the second time?

Considering the first one is not an overtake
First they both start together and Harriot stops when she reaches 1 mile ahead to Torrance= 1/6 *60 = 10mins (Note: distance covered by Harriot =8/6 miles)
Now Torrance has to cover 8/6 + 2 miles time = 20/6 * 60/2 = 100 mins
Again Harriot starts and stops until she covers 1 mile ahead to Torrance = 3/6 *60 = 30 mins (Note total distance covered by Harriot =8/6+ 4 = 32/6 miles,Time =130 mins)
Now Torrance has to cover 32/6 + 2 miles time = 44/6 * 60/2 = 220 mins
Again Harriot starts and this time she just have to overtake Torrance = 2/6 * 60 = 20 mins
Total time now is 220+20 = 240 mins

Considering the first one as an overtake

First they both start together and Harriot stops when she reaches 1 mile ahead to Torrance= 1/6 *60 = 10mins (Note: distance covered by Harriot =8/6 miles)
Now Torrance has to cover 8/6 + 2 miles time = 20/6 * 60/2 = 100 mins
Again Harriot starts and this time she just have to overtake Torrance = 2/6 *60 = 20 mins (Note total distance covered by Harriot =8/6+ 4 = 32/6 miles,Time =120 mins)
Total time now is 100+20 = 120 mins

:? :? :?



Can anyone confirm which case to consider and why?
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speed * time = distance

Kindly note that the time taken to cover their respective distances will be same when both the runners are running simultaneously.

Therefore 8 * time - 2 * time = 1 or time = 1/6 hours. This means that it takes 1/6 hours for Harriet to be 1 mile ahead of Torrance.

After running for 1/6 hours harriet stops. Torrance will take one and half hours to travel 3 miles @ 2 miles/hour, i.e. she will cover one mile deficit and another two miles before which both start running together.

8 * time - 2 * time = 3 or time = 1/2 hour. After running for 1/2 hours harriet stops. Torrance will take half an hour to travel 1 miles @ 2 miles/hour, i.e. she will cover one mile deficit to pass harriet for the second time.

Therefore total time is equal to 1/6 +3/2+1/2 +1/2 = 2 hours 40 minutes pass noon is 2:40 PM (Option D is the correct answer)
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rate Torrance = 2
rate Harriet = 8

1. Time to get 1 mile ahead of Torrance
1/(8-2) = 1/6 = 10min

2. Harriet pauses and waits for Torrance to get 2 miles ahead. Torrance must first catch up 1 mile, and then go another 2 miles before Harriet starts
3/2 = 3/2 hours = 90min

OR

30min for 1 miles catch
60min for 2 miles to get ahead
90min total Harriet has stood still

3. Harriet now goes ahead of Torrance by a mile. She must catch up 2 miles, and go ahead 1 mile
3/(8-2) = 3/6 = 1/2 = 30min

4. Harriet waits for Torrance to catch up to her a 2nd time. He has to travel 1 mile
1/2 = 1/2 hour = 30min

Total time = 10+90+30+30 = 160min = 2hr 40min
12:00 + 2:40 = 2:40pm
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Approach1 - Start to end.

1.both run until 1 mile gap is created
2.H stops and T runs for 3 miles (1 mile gap + another 2 miles)
3.both run until 1 mile gap is created again. In this case, H runs for 3 miles relative to T, (2 miles gap+ another 1 mile)
4.H stops and T runs for 1 mile

1. 1 mile divided by 6 mph equals 10 minutes.
2. 3 miles divided by 2 mph equals 90 minutes.
3. 3 miles divided by 6 mph equals 30 minutes
4. 1 mile divided by 2 mph equals 30 minutes.

Hence, total equals 2 hours 40 minutes.

———————————————————————

Approach 2 - Comparison at end.

When they meet at second overtake,
distance of T = distance of H
2*T = 8*T’
We need to determine T

In this entire period, H was at rest for {(1 mile + 2 mile) + 1 mile} = 4 miles of T’s run.
Time required for T to run 4 miles equals 4 / 2 =2 hours
Hence, T’ = T -2
2T = 8T -16 or T = 16/6 hours =2 hours 40 minutes.
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