Speed/Distance Question

Question

Cities A & B are 70 miles apart. A biker leaves town A at the same time that a bus leaves town B. They travel toward each other and meet 84 minutes after their departure at a point between A and B. The bus arrives at City A, stays there for 20 mins, and then heads back to City B. The bus meets the biker again 2 hours and 41 minutes after their first meeting. What are the speeds of the bus and the biker?

Answer 1

Once it becomes apparent that the biker does not reach point B before they meet the second time, it can be worked quickly as a 2 part problem with trial & error. If Vbus = the bus speed and Vbike = tthe bicycle speed

If c is the point where they meet the first time then 1.4 hours = c/Vbike = (70-c)/Vbus

We know that the bus reaches point A and then begins traveiling toward point at time = 70/Vbus + 1/3 hours

So in that time, the bicycle has traveled Vbike{ 70/Vbus + 1/3 } miles. So equating the distances, you can find the time needed for the bus to catch the bicycle again! Adding that time to 2 hours and 41 minutes we find that a Vbus = 35 mph satisfies the conditions. That means that Vbike = 15 mph.