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Torrance and Harriet run a race along a long, straight path. Torrance
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24 Apr 2017, 04:23
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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? A. 12:40 pm B. 1:20 pm C. 1:50 pm D. 2:40 pm E. 3:20 pm
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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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24 Apr 2017, 12:09
Bunuel wrote: 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?
A. 12:40 pm B. 1:20 pm C. 1:50 pm D. 2:40 pm E. 3:20 pm time they both run for a distance gap of 1 mile between them = 1/(82) =1/6 hrs = 10 minutesfor covering 1 mile @ 2mph torrace needs another 30 minutesfor the next 2 miles gap ,until harriet restarts , torrance taken a time of 1 hrs @ 2mph again time taken to cover 2 miles by harriet = 2/(82) =1/3 hrs = 20 minutes.and further 10 minutes for a gap of 1 miles between them. thus torrance need another 30 minutes for meeting again harriet(rest position) thus total time = 10min + 30 min + 1hrs + 20 min + 10 min + 30 min = 2 hrs 40 mints since they started Noon ie 12:00 torrance meet harrier 2nd time = 12:00 + 2:40 = 14:40 hrs = 2:40 pm Ans D




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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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24 Apr 2017, 16:14
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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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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01 May 2017, 06:16
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 overtakeFirst 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 overtakeFirst 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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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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08 Oct 2017, 11:06
My 2 cents: Here is my solutions!!! If it was helpful please kudo!!!
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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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11 Oct 2017, 04:06
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 nonstop 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 (t2)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)=(t2)8 (in H's case). Therefore, 8t16=2t or t=8/3 hrs or 2hrs 40mins. Hence, answer D.



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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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13 Oct 2017, 09:20
Bunuel wrote: 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?
A. 12:40 pm B. 1:20 pm C. 1:50 pm D. 2:40 pm E. 3:20 pm 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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Re: Torrance and Harriet run a race along a long, straight path. Torrance
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13 Oct 2017, 10:00
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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Torrance and Harriet run a race along a long, straight path. Torrance
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12 May 2018, 00:21
sreenu7464 wrote: 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 overtakeFirst 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 overtakeFirst 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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Torrance and Harriet run a race along a long, straight path. Torrance
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19 Oct 2018, 05:09
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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Torrance and Harriet run a race along a long, straight path. Torrance
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