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Ambigious _ Question_Veritas Prep

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Ambigious _ Question_Veritas Prep

by prabsahi » Thu Oct 05, 2017 12:09 am
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?

12:40 pm
1:20 pm
1:50 pm
2:40 pm
3:20 pm

Hi Experts,

Is the construction of this question sound?

I had two ambiguities because of its language.

1.reference of she:
Does it refer to Torrance or Harriet.
2.
but whenever Harriet leads Torrance by at least 1 mile, she stops and does not run again until she has fallen 2 miles behind.

Intepretation 1: When H leads T by 1 mile,Harriet has fallen two miles but since there is a lead of 1 mile so total miles behind is 3 miles.
Intepretation 2:When H leads T by 1 mile,Harriet has fallen two miles (1 mile lead plus 1 mile fallen)

Please help.
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by ceilidh.erickson » Sun Oct 08, 2017 4:51 pm
I agree that the wording of this question is bad. The GMAT will never ask you to assume the gender of any person mentioned! a) that would be unfair to anyone unfamiliar with those names, and b) lots of people have gender-non-conforming names. There is also some strange ambiguity with other phrasing, including "at least."

The assumption that this question-writer wants us to make is that Harriet is female and Torrance is male, so every "she" refers to Harriet.

Whenever Harriet gets 1 mile ahead, she stops and does not resume running until she is 2 miles behind. When she begins running again, she runs until she is 1 mile ahead, so she runs a total of 3 miles more than he does in that same amount of time.

To calculate the time when Harriet first stops, create a rate chart. Let t = the time when Harriet gets 1 mile ahead, where d = Torrance's distance at the time.

Image

If we substitute d= 2t, we get:
8t = 2t + 1
6t = 1
t = 1/6
Harriet will first stop after 1/6 of an hour, at 12:10.

It will take Torrance 30 min to catch up to her (to make up a difference of 1 mile, at 2 miles per hour). They will meet at 12:40.

Harriet will wait while Torrance continues running, until he's 2 miles ahead. This will take 1 hr, as he is going 2miles per hour. She will resume running at 1:40.

From this moment, Harriet will run from where she is (2 miles behind) until she is 1 mile ahead. Thus, she will run 3 more miles total than Torrance will. We can set this up in a rate chart again, where T = the new amount of time, D = Torrance's new distance:
Image

Solve for T:
8T = 2T + 3
6T = 3
T = 1/2
Thus, it will take another 30 min for Harriet to outpace Torrance by 1 mile: 2:10.

She will then stop, and wait another 30 min for him to catch up to her again (1 mile at 2 miles per hour). He will pass her for the 2nd time at 2:40.

The answer is D.
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by Matt@VeritasPrep » Wed Oct 11, 2017 5:12 pm
Yup, you're definitely right about the names - we'll edit the question to tidy that up.
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by Jeff@TargetTestPrep » Fri Jan 05, 2018 9:23 am
prabsahi 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?

12:40 pm
1:20 pm
1:50 pm
2:40 pm
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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Head of GMAT Instruction
[email protected]

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