Trains Questions and answers

Let the length of each train be x m.

Relative speed = 46 - 36 = 10 km/h

= \( \Large \frac{10 \times 5}{18} \) m/s = \( \Large \frac{25}{9} \) m/s

\( \Large \frac{Sum \  of \  length \  of \  train}{Relative \  speed \  of \  train} = Time \ taken \)

Therefore, \( \Large \frac{2x}{\frac{25}{9}} = 36 => 2x = \frac{36 \times 25}{9} = 100 \)

Therefore, x = 50 m

Let length of each train be x m.

Then, \( \Large \left(65+85\right) \times \frac{5}{18} = \frac{x+x}{6} \)

\( \Large \frac{150 \times 5 \times 6}{18} = 2x \)

=> 2x = 250

Therefore, x = 125 m

Average speed or train = \( \Large \frac{240}{12} \) = 20 km/h

Average speed of bus

=> \( \Large \frac{3}{4} \times 20 = 3 \times 5 \) =15 km/h

Required distance = \( \Large 15 \times 7 \) = 105 km

If the trains meet after t h.



Relative speed of train = (24-18) = 6 km/h

= Distance = 27

Therefore, t = \( \Large \frac{27}{6} \) = \( \Large \frac{9}{2} \) h

Therefore, QR distance travel by train which is travelling at a speed of 18 km/h

=> 18t = \( \Large 18 \times \frac{9}{2} \)

= 81 km.

Given, a=45 km/h, y=?, \( \Large t_{1} \) = 4h 48 min and \( \Large t_{2} \) = 3 h 20 min

Using y = \( \Large a\sqrt{\frac{t_{1}}{t_{2}}}=45\sqrt{\frac{4 h \  and \  48 \  min}{3 h \  and \  20 \  min}} \)

= \( \Large 45\sqrt{\frac{\frac{24}{5}h}{\frac{10}{3}h}} \)

= \( \Large 45\sqrt{\frac{24 \times 3}{5 \times 10}} \)

= \( \Large 45 \times \sqrt{1.44}=45 \times 12 \) = 54 km/h