Posted by Reid on Thursday, January 17, 2013 at 2:52pm.
at time t on the first day, they were at distance (t-8)/8 of the way there.
at time t on the way back, they were (t-11)/8 of the way back, or 1-(t-11)/8 of the way out.
(t-8)/8 = 1 - (t-11)/8
t = 27/2 = 13:30
so, at 13:30 on the way out, they were 5 1/2 hours into the trip, with 2 1/2 hours to go.
on the way back, they were 2 1/2 hours into the trip, with 5 1/2 hours to go, so they were at the same place.
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