WholeLottaLove wrote:
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.
While we know that his average speed was 4 miles/hour, the answer to the question depends on the distance from home to diner. Let's say that the distance was 8. Furthermore we can break it up into two, 4 mile segments.
Time = Distance/Rate
Wendy's Time = 8/3 hours or roughly 160 minutes.
Now Bob traveling at his faster rate would cover 4 miles in an hour. However, Bob would need to travel the 4 miles with Wendy, 4 miles back home, then 8 miles to the diner meaning he covered 16 miles in total. He would have spent 4 hours (240 minutes) walking so she would have spent 240-160 = 80 minutes waiting.
Now lets pretend the diner was 4 miles away.
Time = Distance/Rate
Wendy Time = 4/3 hours or roughly 75 minutes.
Bob would travel 2 miles (the half way point), two miles back home then another 4 miles from home to the diner for a total of 8 miles. At 4 MPH average this would take 2 hours (120 minutes) meaning she waited 120-75 = 35 minutes.
INSUFFICIENT
I would LOVE to see the algebraic explanation if possible!
Thanks!
Just some minor glitch there, not affecting the answer in any way though.
Let the total distance be 6d miles
From F.S 1, we know that time taken by Wendy \(= \frac{6d}{3} = 2d\)hours.
Again, we know that Total time taken by Bob \(= \frac{Total Distance}{Average Speed} = \frac{(3d+3d+6d)}{4} = \frac {12d}{4} = 3d\)hours
Thus, the time for which Wendy waited = 3d-2d = d hours. Clearly, this depends on the initial distance.Insufficient.
From F.S 2, we know that "On his journey, Bob spent 32 more minutes alone than he did walking with Wendy "
Time spent by Bob with Wendy = \(\frac{3d}{3} = d\)hours
Time spent by Bob alone = \(d+\frac{32}{60} = d+\frac{8}{15}\)hours
Total time taken by Bob =\(2d+\frac{8}{15}\)hours
Total time taken by Wendy\(= \frac{6d}{3}= 2d\) hours
Thus, time Wendy waited for Bob \(= 2d-2d+\frac{8}{15} = \frac{8}{15}\) hours
Sufficient.