After a cyclist has gone 2/3 of his route he gets a flat tire.
Finishing on foot he spends twice as long walking as he
did riding. If his walking and riding rates are constant how
much faster does he ride th[a]n walk?
The first part of his journey can be expressed by the
equation d = s1 * t
We are comparing speed, so s1 = d / t
The second part is defined by 1/2 d (1/3 vs 2/3) and 2t.
The equation is now 1/2 d = s2 * 2t
We will equalize the distance by multiplying the second
equation by 2
2 * (1/2 d) = 2 * (s2 * 2t)
d = s2 * 4t
Still focusing on speed, 4 * s2 = d / t
We have d / t = s1 while cycling,
and d / t = 4 * s2 while walking.
Thus, s1 = 4 * s2, i.e., his cycling speed was
four times his walking speed.