Let's go for a hike  math puzzle
A hiker walked for two days. On the second day the hiker walked 2 hours longer and at an average speed 1 mph 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 hours walking, what was his average speed on the first day?
Explanation
When t=time and s=speed:
t1 + t2 = 18 and t2 = t1 + 2, from which t1 = 8 and t2 = 10
s2 = s1 + 1 and s1 * t1 + s2 * t2 = 64
so 8*s1 + 10*(s1+1) = 64, from which s1 = 3
t1 + t2 = 18 and t2 = t1 + 2, from which t1 = 8 and t2 = 10
s2 = s1 + 1 and s1 * t1 + s2 * t2 = 64
so 8*s1 + 10*(s1+1) = 64, from which s1 = 3
