There are a couple of ways to approach this problem:

The simplest approach is to realize that since Ron runs three times as fast as he walks, it takes him three minutes to walk the same distance he can run in one minute. In other words, every minute he spends running saves him 2 minutes of walking time for the same distance. Since we know he cut ten minutes off his time, he must have ran for five minutes.

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Another (not as elegant) approach is to use algebra. For the walk/run scenario, if we let d_w and t_w be the distance and time walking, and d_r and t_r be the distance and time running; we can specify the conditions of the problem as follows:

d_w + d_r = d
t_w+ t_r = 50

Where d is the overall distance. We know that his walking velocity is v_w = d/60 = d_w/t_w and his running velocity is v_r = d_r/t_r = 3 v_w = d/20. From these we have d_w = t_w d/60 and d_r = t_rd/20. Plugging these values into the first equation gives (with trivial simplification):

t_w+ 3t_r = 60
t_w+ t_r = 50

Solving these gives t_r = 5.