Question 178308
If the slower train was bhind the faster train, 
the slower train would never catch up.
This problem puts the faster train behind the slower
one, so it has a chance to catch up.
7:32 minus 7:20 is 12 minutes. That's the elapsed
time between the slower train passing the station
and the faster train passing it
How far did the 1st train go in that 12 min?
12 min is 12/60 of an hour
{{{d = r*t}}}
{{{d = 80*(12/60)}}}
{{{d = 16}}}mi
So, train A has a 16 mi headstart over B
If I have a stopwatch, and I start it when train B
passes the station, and I'm also able to stop it
when B catches A, They'll both be travelling for the
same length of time, except A is 16 mi closer to
where they're going to meet.
For B:
(1) {{{d = 90t}}}
For A:
{{{d - 16 = 80t}}} (notice {{{t}}} is the same for both)
(2) {{{d = 80t + 16}}}
Since both are equal to {{{d}}}, I'll set the 
right sides equal to eachother
{{{90t = 80t + 16}}}
{{{10t = 16}}}
{{{t = 1.6}}}hr
This is 1 hr + {{{.6*60 = 36}}}min
This is the elapsed time since 7:32 when B passed the station 
7:32 + 1 = 8:32 and 8:32 + 36 min = 9:08
Train B catches up to train A at 9:08
I'll check my answer:
What's the distance from the station to where they meet?
(1) {{{d = 90t}}}
{{{d = 90*1.6}}}
{{{d = 144}}}mi
(2) {{{d = 80t + 16}}}
(2) {{{144 = 80t + 16}}}
{{{80t = 144 - 16}}}
{{{80t = 128}}}
{{{t = 1.6}}}hr
OK