Speed = \(\left(60\times\frac{5}{18}\right)m/sec =\)  \(\left(\frac{50}{3}\right)m/sec \)     

Length of the train = (Speed x Time) =\(\left(\frac{50}{3}\times9\right)m=150m.\)

Speed of the train relative to man =\(\left(\frac{125}{10}\right)m/sec\)

=\(\left(\frac{25}{2}\right)m/sec\)

=\(\left(\frac{25}{2}\times\frac{18}{5}\right)km/hr.\)

= 45 km/hr.

Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr.

 therefoe x - 5 = 45         x = 50 km/hr.

Speed =\(\left(45\times\frac{5}{18}\right)m/sec= \left(\frac{25}{2}\right)m/sec.\)

Time = 30 sec.

Let the length of bridge be x metres.

Then,\(\frac{130+x}{30}=\frac{25}{2}\)\(\Rightarrow2\left(130+x\right)= 750\)

\(\Rightarrow x =245m.\)

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,

and length of the second train = 17y metres.

\(\therefore \frac{27x+17y}{x+y}=23\)

\(\Rightarrow27x+17y = 23x+23y\)
4x = 6y

\(\frac{x}{y}= \frac{3}{2}\)

Speed = \(\left(54\times\frac{5}{18}\right)m/sec = 15m/sec.\)

Length of the train = (15 x 20)m = 300 m.

Let the length of the platform be x metres.

Then, \(\frac{x+300}{36} = 15\)

x + 300 = 540

x = 240 m.

Speed = \(\left(\frac{240}{24}\right)m/sec = 10m/sec.\)

thairfo Required time = \(\left(\frac{240+650}{10}\right)sec = 89sec.\)

Let the length of each train be x metres.

Then, distance covered = 2x metres.

Relative speed = (46 - 36) km/hr

= \(\left(10\times\frac{5}{18}\right)m/sec\)

=\(\left(\frac{25}{9}\right)m/sec\)

\(\therefore \frac{2x}{36}= \frac{25}{9}\)

2x = 100

x = 50.

Formula for converting from km/hr to m/s:   X km/hr = \(\left(x\times\frac{5}{18}\right)m/s.\)

Therefore, Speed = \(\left(45\times\frac{5}{18}\right)m/sec = \frac{25}{2}sec.\)

Total distance to be covered = (360 + 140) m = 500 m.

Formula for finding Time = \(\left(\frac{Distance}{Speed}\right)\)

Therefor Required time =\(\left(\frac{500\times2}{25}\right)sec = 40sec.\)

Relative speed = (60+ 90) km/hr 

=\(\left(150\times\frac{5}{18}\right)m/sec\)

= \(\left(\frac{125}{3}\right)m/sec.\)

Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.

Required time = \(\left(2000\times\frac{3}{125}\right)sec = 48sec.\)

Speed of train relative to jogger = (45 - 9) km/hr = 36 km/hr. 

   =\(\left(36\times\frac{5}{18}\right)m/sec\)

   = 10 m/sec. 

Distance to be covered = (240 + 120) m = 360 m. 

thairfor Time taken = \(\left(\frac{360}{10}\right)sec = 36 sec. \)