Jeff, Jill and Skippy, a dog go for a 10 mile walk. Jeff and Jill can
walk at 2 mph and Skippy can run at 4 mph. They have a bike
which only one of them (including the skippy!) can use at a time. When
riding, Jeff and Jill can travel at 12 mph while Skippy can pedal
at 16 mph. What is the shortest time in which all three can complete
the trip?

are we to assume they can leave the bike on the side of the road?

Does someone need to be with the dog at all times?

2 hours 45 min

Assuming they can do whatever they want to get to the end of the walk the fastest....

I think jeff should bike 5.4k, leave the bike, then walk 4.6 mile. He will be home in 2 hour 45 min.
Skippy should run 5.4k, get the bike and go back 0.8k. Leave the bike and carry on running to the end. He will be home in 2 hour 45 min.
Jill should walk 4.6k, pick up the bike that skippy has left and cycle the rest of the way. She will be home in 2 hour 45 min..

However we all know the real answer to this question is 5 hours. Because Jeff gave Jill the bike first and she took off and left him to walk the whole way....

It's miles now, pt. Wake up, they de-metricized the world.

My point of view would be to have one of the humans riding half the way, the other for the other half. That would prolly be the same as pt's answer.

Nope - the little shuffle with the dog in the middle is important.

Skippy could make the journey solo in 2.5 hours, but without him being involved a human riding half way and walking the other half would take 5/2 + 5/12 = 2 11/12 of an hour, which is more than 2 3/4

Having the magic bicycle-riding dog shuttle the bike about means the humans can ride it for a little further than halfway and reduce the journey time.

+1 pt_22 ...............perfect answer. good job

p.s.  I deliberatly did the miles to k thing to try and confuse you yanks!