Question

On the first day of his backpacking trip, Andyleft camp at 9:15 A.M. and arrived at his newcamp at 4:45 P.M. His average rate of speed was2 1/2 miles per hour. On the second day he leftthe new camp at 8:45 A.M. and arrived at hisfinal camp at 3:30 P. M. His average rate of speedwas 3 miles per hour. Find the total distancethat he hiked over the two days.

Answer 1

The first step we need to solve this problem is to calculate the time it took to complete each part of the trip. This is done by subtracting the start time from the arrival time plus 12.

`\begin{gathered} \text{trip 1}=4\colon45+12\colon00-9\colon15=16\colon45-9\colon15=7\colon30 \\ \text{trip 2}=3\colon30+12\colon00-8\colon45=15\colon30-8\colon45=6\colon45 \end{gathered}`

Now we need to multiply the times by the average speed:

`\begin{gathered} d1=7.5\cdot2.5=18.75\text{ miles} \\ d2=6.75\cdot3=20.25\text{ miles} \end{gathered}`

Now we need to add the two distances:

`\text{total}=18.75+20.25=39`

The total distance is 39 miles.