Algebra Help PLZ
posted by Gayle on .
I have tried this several different ways and I am just not satisfied with the answer I keep coming up with.
Trains A and B are traveling at the same direction on parallel tracks. Train A is traveling at 100 mph and train B is traveling at 120 mph. Train A passes a station at 4:10 pm. If train B passes the same station at 4:22 pm, at what time will train B catch up with train A?

Trains A and B are traveling at the same direction on parallel tracks. Train A is traveling at 100 mph and train B is traveling at 120 mph. Train A passes a station at 4:10 pm. If train B passes the same station at 4:22 pm, at what time will train B catch up with train A?
Something is wrong here. If train B is traveling 20 mph faster than A, B should reach the station first. Unless they are traveling on a circular track, B will never "catch up" to A. B is going faster.
Please repost with accurate data. Thanks for asking. 
This is verbatum, and that is why I am having problems with it myself.

The problem does not say the trains started out at the same time so even though B is travelin faster it could still be behind A

Discuss it with your teacher or whomever gave you the problem.
Sorry I can't help you more. 
There is nothing worng with the problem.
When A passes the station B is some distance behind (we don't know how far  we don't care)
When B passes the station A is some distance in front, the distance (Na) being
Na=12(100)/60 miles.
Nb=0 miles
The trains now continue in the same direction.
after time tx the distances are now
Na = 12(100)/60 + tx(100)/60
Nb= tx(120)/60
which are equal when Na=Nb
so
12(100)/60 + tx(100)/60 = tx(120)/60
1200 +100tx = 120tx
1200 = 20tx so Tx=60
So B catches up with A at 5:22, i.e. 60 minutes after B passes the station.
But check my working