Algebra Help PLZ

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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?

• Algebra Help PLZ -

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.

• Algebra Help PLZ -

This is verbatum, and that is why I am having problems with it myself.

• Algebra Help PLZ -

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

• Algebra Help PLZ -

Discuss it with your teacher or whomever gave you the problem.

Sorry I can't help you more.

• Algebra Help PLZ -

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

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