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Speed and Distance

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Speed and Distance

by MI3 » Tue May 31, 2011 4:03 am
Q. 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.

Thanks,
M
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Source: — Data Sufficiency |
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by cans » Tue May 31, 2011 5:55 am
if distance till restaurant is xkm,
bob traveled 2x and windy traveled x.
windy speed = 3, thus time taken = x/3
a)
bob speed = 4, thus time taken by bob = 2x/4 = x/2
Thus windy waited for x/2-x/3 hrs = x/6 but as we don't know x, insufficient.
b) suppose t time traveling with windy, thus t+32 time alone.
thus bob spent total of 2t+32 hrs.
also windy spent t hrs with bob and as she continued at same pace, thus she spent t time alone
Total of 2t hrs.
thus she had to wait for 2t+32-2t = 32 hrs
sufficient.
IMO B
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by Frankenstein » Tue May 31, 2011 5:59 am
Hi,
Let the house, restaurant and mid point be denoted by A,B,C and the distance between A and B be '2d'
and the time taken for them to reach midpoint be 't'. Let the tiem taken by Bob to go back from C to A and then A to B be t1.
Wendy's speed is 3mph . So, (2d)/(2t)=3 =>d=3t
Bob's avg speed = (4d)/(t+t1)
We need to find (t+t1)-2t =t1-t
From(1): 4d/(t+t1) = 4 =>d= t+t1 =>t1= d-t = 3t-3 = 2t
So, t1-t = t. We need to know the value of t
Not sufficient
From(2): t1-t = (32/60)hours.
Sufficient

Hence, B

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