Let the 1st traveler's rate be x mph Let the 2nd traveler's rate be y mph Let's take a look at the moment when they met, which was 9:05. At 9:05 the 1st traveler, who started at 5AM, had been traveling for 4 hours and 5 minutes, orhours or hours or hours. At rate x mph, he had traveled or miles. At 9:05 the 2nd traveler, who started at 7AM, had been traveling for 2 hours and 5 minutes, or hours or hours or hours. At rate y mph, he had traveled or miles. When they met at 9:05, each had the same number of miles yet to go that the other had already been. That is, the 1st traveler had miles yet to go and the 2nd traveler had miles yet to go. We are told that they arrived at the same time. So we will calculate each traveler's time after 9:05 and set them equal. Since , the 1st traveler's time was his distance yet to go, , divided by his rate x, or or or hours till he reached town B. Also the 2nd traveler's time was his distance yet to go, , divided by his rate y, or or or hours till he reached town A. Since they arrived at their respective destinations at the same time, we set these times equal: Multiplying both sides by 12 eliminates the 12's from the denominators: Cross multiplying: Taking positive square roots of both sides Dividing both sides by 5x The 1st traveler's time to reach his destination after 9:05 was which equals which equals or 2 hours and 55 minutes after 9:05. So the 1st traveler reached his destination at 12 noon. It isn't necessary, but, as a check, let's see if the 2nd traveler also reached his destination at 12 noon. The 2nd traveler's time to reach his destination after 9:05 was which equals which equals or 2 hours and 55 minutes after 9:05. So the 2nd traveler also reached his destination at 12 noon. So the 1st traveler started at 5AM and arrived at noon, so his time was 7 hours. The 2nd traveler started at 7AM and arrived at noon, so his time was 5 hours. [Notice that this problem is independent of their speeds and the distance they traveled.] Edwin