Average Speed

If the distance between point A and point B is $d$ , then the amount of time Bob takes on his trip is $\frac{2d}{45}$ . Carl takes $\frac{d}{23}$ hours to get to point A to point B, so the amount of time Carl has left to get from point B to point A is $\frac{2d}{45} - \frac{d}{23}$ .

We find the speed Carl needs to travel at is the distance (which is $d$ ) divided by the time (which is $\frac{2d}{45} - \frac{d}{23}$ ).

$\frac{d}{\frac{2d}{45} - \frac{d}{23}} = \frac{1}{\frac{2}{45}-\frac{1}{23}} = \boxed{1035}$

Comments: This may seem unexpected. We intuitively believe that the answer should be an average, or 67 mph. However, averages need to be weighted the same in order to do this. Carl spends much more time traveling at 23 mph to get from A to B than Bob does at 45 mph. This is why we need to be especially careful when dealing with rate problems and construct tables/do unit analysis.

Another way to see this problem is imagine Carl was going at 30 mph and Bob was going at 60. By the time Bob gets back to A, Carl will have just reached B! The only way Carl can catch up is if he instantly teleports back to A. If he traveled any slower, Bob would have returned before Carl even reaches B!

Let us assume that the distance from point A to point B is 23*45=1035 miles

After 23 hours, Bob will be at point B and after 45 hours, Carl will be at point B and Bob will be at (45-23) 45=22 45 miles away from point B or 45 miles away from point A.

Since it would take Bob an hour to get to point A, Carl must travel the whole distance to go back to point A within an hour, which is 1035 miles and therefore he should go 1035 mph.

Suppose distance from point A to B = 45 miles
So Bob completed entire trip = 2 hrs
Carls completed trip from A to B = 45/23 hrs,
so to catch up Bob, required speed = 45 * 23 = 1035 mph

Because both persons go to the two points, the common sense occurs in this one. The LCD is 1035, so this is the answer.