Let the two trains meet at t hours after 9 p.m. Suppose the average speed of the passenger train is v km/hr. Distance moved by the passengertrain when the two trains meet=v×(t+3)=540 ⇒v=
540
t+3
....(1) Distance moved by the express train when the two trains meet =(v+15)×t=540 ⇒v=
540
t
−15.......(2) From (1) and (2), we have
540
t+3
=
540
t
−15 ⇒
540
t
−
540
t+3
=15 ⇒
1620
t(t+3)
=15 ⇒t2+3t−108=0 ⇒(t+12)(t−9)=0 ⇒t=9(t>0) So, the two trains meet 9 hours after 9 p.m., i.e., at 6 a.m.