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

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by netigen » Wed Jun 11, 2008 12:36 pm
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.
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Source: — Data Sufficiency |

by roy_priya » Wed Jun 11, 2008 12:49 pm
Answer - B

The time bob spent walking with wendy, is same as the time wendy spent walking to the restaurant. Substracting that from the total time bob spent walking back to the house and then to the restaurant will give how much time Wendy had to wait.
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by sm612 » Wed Jun 11, 2008 8:15 pm
I think both the conditions are required to get the answer.
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by durgesh79 » Wed Jun 11, 2008 9:44 pm
I think its B.

the distance is 2d.

W's time = 2d/3
B's Time = d/3 + 3d/x ------- x is the new speed of Bob,

we have to find out = d/3 + 3d/x - 2d/3 = 3d/x - d/3 ------- (1)

Statement 1 - we can find out x, but not d, insuff.
statement 2
B's time with windy = d/3
B's time alone = 3d/x
difference = 3d/x - d/3

This is what we want. suff.
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