A train left a station P at 6 am and reached another station Q at 11 a

Assign a distance. Then use the "gap" and relative speed approach

Distance?
Call the trains B and C.
They travel the same distance.

B takes 5 hours (from 6 to 11 a.m.)
C takes 3 hours (from 7 to 10 a.m.)

Let distance between P and Q = 15 miles (LCM of 5 and 3)

Use each train's time to find rates.

(1) Each train's rate?
\(r*t=D\)
B, rate: \((r*5hrs)=15mi\)
B, rate: \(r=\frac{15mi}{5hrs}=3\) mph

C, rate: \((r*3hrs)=15mi\)
C, rate: \(r=\frac{15mi}{3hrs}=5\) mph

(2) Distance "gap" between B and C?
B is at station P
C is at station Q
Initial distance = 15 miles

But B travels alone for 1 hour
-- B covers \((3mph*1hr)=3\) miles
-- B shortens the distance between them from 15 to 12 miles

That 12 miles is a "gap." When B and C at relative speed cover the distance, they close that gap and pass one another.

(3) Relative rate/speed?
Opposite directions (towards or away): ADD rates
Relative rate/speed: \((3+5)=8\) mph

(2) Time required for them to pass? \(R*T=D\), so
\(T=\frac{D}{R}\)

\(T=\frac{12mi}{8mph}=\frac{3}{2}=1.5\) hours

(4) Clock time at which they pass one another?

Calculate clock time from the start time of the second train (then both trains are moving and thus closing the gap together at their relative/combined speed)

Train C left at 7:00 a.m.
1.5 hours later is 8:30 a.m.

Answer C