A goods train left the town M for the town N at 5 a.m. An hour and a half later a passenger train left M, whose speed was $5$ km/h higher than that of the goods train. At 9.30 p.m. of the same day the distance between the trains was $21$ km. Find the speed of the goods train.
Let the speed of the goods train $= u$ kmph
Speed of passenger train $= \left(u + 5\right)$ kmph
Goods train leaves town M at 5.00 am
Passenger train leaves town M an hour and a half later i.e. at 6.30 am
At 9.30 pm, time for which goods train is traveling $= 16.5$ hrs
At 9.30pm, time for which passenger train is traveling $= 15$ hrs
Distance covered by goods train in $16.5$ hrs $= d_g = u \times 16.5$ km
Distance covered by passenger train in $15$ hrs $= d_p = \left(u + 5\right) \times 15 = \left(15 u + 75\right)$ km
Given: Distance between the trains at 9.30 pm is $21$ km
i.e. $\;$ $d_p - d_g = 21$ $\;$ or $\;$ $d_g - d_p = 21 $
i.e. $\;$ $15u + 75 - 16.5 u = 21$ $\;$ or $\;$ $16.5 u - 15 u - 75 = 21$
i.e. $\;$ $1.5 u = 54$ $\;$ or $\;$ $1.5 u = 96$
i.e. $\;$ $u = \dfrac{54}{1.5} = 36$ kmph $\;$ or $\;$ $u = \dfrac{96}{1.5} = 64$ kmph
Therefore, speed of the goods train $= 36$ kmph $\;$ or $\;$ $64$ kmph