A hiker walked for 2 days. On the 2nd day he walked 2 hrs longer and at an avg speed 1 mile per hour faster than he walked on the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hrs walking, what was his avg speed on the 1st day?
Let's say the hiker walked x miles on the first day and it took him t hours to walk those x miles. On the second day, the hiker walked z miles in t+2 hours. From the given information in the problem, we can say that:
x+z=64
t+t+2=18
2t+2=18
t=8
Average speed for first day= total dist./total time= x/t=y--> x=8y (plugged in the value of t)
Average speed for second day= total dist./total time= z/t+2=y+1 z=10y+10 (plugged in the value of t)
Now, we have x=8y and z=10y+10 and we know that x+z=64, so 8y+10y+10=64 and y=3, which represents the hiker's average speed on the first day.
There's probably a much easier solution to this problem, but this is what I could think of.
