Calculating Average Speed: A Problem-Solving Dilemma

[physicsvirgin]

I've got a little trouble here with average speed.

The specific problem is:

A paper delivery boy rides his bike 2 miles out to a farmhouse at 12mph, delivers the paper, then travels back 2 miles at 10mph to get home. What was the delivery boy's average speed during this trip? (Hint: there are two trips here!)
a) 10.9mph b) 11mph c) 11.1mph d) not enough info

Here's my trouble...the problem doesn't give me any information on how long it took the boy to deliver the paper on either trip. That makes me want to choose "D" because average speed is distance covered/time interval. However, the fact that there is a "hint" makes me want to reconsider.

So, I (as a novice in physics) divides 12 and 2 to get 6 minutes on the first trip, and 2 and 10 to get 5 minutes on the second trip. That makes 11 minutes to go 4 miles all together, but the result of that isn't one of the choices when I divide them!

So where am I going wrong?

 

[Doc Al]

You are calculating the time incorrectly. Distance = speed X time. So time = Distance/speed.

 

[David]

Are you sure that it takes 6 minutes on the first trip? Check how you worked out the time taken for the trip. Remember that the time taken is the distance/speed and then remember what units the measurements are in.

However your general approach of working out the total time and the total distance is correct.

 

[physicsvirgin]

Okay, how about 11.1?

I have done as you suggest - divide 2/12 and 2/10, add them together; get 0.36 and then divide 4/0.36 = 11.1 mph?!

 

[Doc Al]

Since the answers are so close, do not round off until the very last step! (Treat the given values as exact.) So take the time as 11/30 of an hour.