Algebra
posted by David .
Train a is traveling @ 60 miles per hour. Train b is traveling @ 80 miles per hour. A arrives @ 9:25 am and b arrives @ 9:40 am what ime will train b catch up to train a?

Do you not mean departs, not arrives? If so:
a travels 4025 = 15 minutes longer than b
if b travels for t minutes, then a travels for t+15 minutes
a speed = 60/60 = 1 miles/min
b speed = 80/60 = 4/3 miles/min
distances are the same so
1(t+15) = (4/3) t
3 t + 45 = 4 t
t = 45 minutes
so time = 9:40 + 45 = 10:25 am 
9:40am  9:25am = 15min.
15min/60min/h = 0.25h.
When train "B" catches up,the distance
traveled will be the same for each train:
d1 = d2.
60(t+0.25) = 80t,
60t + 15 = 80t,
t = 0.75h = 45min = Time required for
"B" to catchup.
Time = 9:40 + 45 = 10:25am.
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