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Frank144
 
 

Bob and Wendy left home to walk together to a restaurant for

by Frank144 Sun Jul 08, 2007 4:14 am

Hi,

I had a question on the answer for the following question:

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.

(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

(2) SUFFICIENT: To see why this statement is sufficient, it is helpful to think of Bob's journey in two legs: the first leg walking together with Wendy (t1), and the second walking alone (t2). Bob's total travel time tb = t1 + t2. Because Wendy traveled halfway to the restaurant with Bob, her total travel time tw = 2t1. Substituting these expressions for tb - tw,

t1 + t2 - 2t1 = t2 - t1

tb - tw = t2 - t1

Statement (2) tells us that Bob spent 32 more minutes traveling alone than with Wendy. In other words, t2 - t1 = 32. Wendy waited at the restaurant for 32 minutes for Bob to arrive.

The correct answer is B.

My comment:
I understand the algebra in the answer, but was looking for clarity with the logic. I understand that at the halfway point, Bob started walking back and Wendy continued onto the restaurant. I then understand that somewhere between the time Bob went back to the house then finally reached the restaurant, Wendy reached the restaurant first. So, does the 32 minutes of Bob walking alone include the time he went back to the house, while Wendy was still continuing to the restaurant? If that is the case, then it seems to me that Wendy waited less than 32 minutes at the actual restaurant itself, because she still had to walk to get there.

Please let me know if I'm approaching this from the appropriate logical perspective. Thank you!
christiancryan
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by christiancryan Mon Jul 09, 2007 6:49 pm

Subtle question -- but note that statement #2 doesn't say that Bob spent 32 minutes alone. It says that he spent 32 MORE minutes walking alone THAN walking with Wendy. In other words, his time walking alone = 32 minutes PLUS the time he spent walking with Wendy halfway to the restaurant. Let's just call that 10 minutes, for convenience's sake. So he walks alone for 42 minutes, in this scenario. Wendy takes ANOTHER 10 minutes to walk to the restaurant (because she has just walked 1/2 way and now continues onward at the same speed). So she waits for him for 32 minutes exactly. No matter how much time you choose for the first leg of the journey (together), you will get the same answer.

Hope this is helpful!
nithin
 
 

I think it's 22.

by nithin Wed Aug 13, 2008 3:22 am

christiancryan Wrote:Subtle question -- but note that statement #2 doesn't say that Bob spent 32 minutes alone. It says that he spent 32 MORE minutes walking alone THAN walking with Wendy. In other words, his time walking alone = 32 minutes PLUS the time he spent walking with Wendy halfway to the restaurant. Let's just call that 10 minutes, for convenience's sake. So he walks alone for 42 minutes, in this scenario. Wendy takes ANOTHER 10 minutes to walk to the restaurant (because she has just walked 1/2 way and now continues onward at the same speed). So she waits for him for 32 minutes exactly. No matter how much time you choose for the first leg of the journey (together), you will get the same answer.

Hope this is helpful!


Wendy has walked for 10 minutes with bob and 10 minutes alone. So in total of 20 minutes.
Bob has walked 10 minutes with wendy and 32 minutes alone. So in total of 42 minutes.

So wendy waited for 42 - 20 = 22 minutes.
RonPurewal
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Re: I think it's 22.

by RonPurewal Tue Sep 09, 2008 7:13 am

nithin Wrote:Bob has walked 10 minutes with wendy and 32 minutes alone. So in total of 42 minutes.


nope - go back and read chris ryan's post above, which explains why this is wrong. ironically, that post is quoted in your post, directly above your text.

to reiterate, the problem is that bob didn't walk alone for 32 minutes; he walked alone for 32 minutes LONGER than he walked with wendy.
that's 32 + 10, not just 32.
jp.jprasanna
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Re: I think it's 22.

by jp.jprasanna Sun Mar 04, 2012 4:11 pm

RonPurewal Wrote:
nithin Wrote:Bob has walked 10 minutes with wendy and 32 minutes alone. So in total of 42 minutes.


nope - go back and read chris ryan's post above, which explains why this is wrong. ironically, that post is quoted in your post, directly above your text.

to reiterate, the problem is that bob didn't walk alone for 32 minutes; he walked alone for 32 minutes LONGER than he walked with wendy.
that's 32 + 10, not just 32.


Hi I dont see why Statement 1 is insufficient

Assume distance between home to restaurant is 6 miles

CAN WE assume the above??????? IF not WHY?????

so Wendy has taken 2 hrs to reach the restaurant
3 * 1 = 3 (with Bob)
3 * 1 = 3 (alone),

Now Bob has to cover

3 (going back to the home ) + 6 (home to hotel)
So his new speed 4mph

4 * x = 9
x = 9/4 = 2.25 i.e 2 hrs 25 mins
Wendy took 2 hrs
Bob took 2 hrs 25 mins
She waiting for 25 mins.

Please help.

I have posted another question in the below link - very very similar to this one... (same assumptions and also almost same equations)

walk-away-2-tom-linda-stand-at-point-a-t10345.html

Cheers
Jp.
tim
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Re: Bob and Wendy left home to walk together to a restaurant for

by tim Sun Mar 18, 2012 2:52 am

I answered your other post, so please see that post for your answer..
Tim Sanders
Manhattan GMAT Instructor

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