Problems on Trains - 02

Explanation:-

$$\eqalign{ & {\text{Speed of the train relative to man}} \cr & {\text{ = }}\left( {\frac{{240}}{{10}}} \right){\text{m/sec}} \cr & {\text{ = 24 m/sec}} \cr & {\text{ = }}\left( {24 \times \frac{{18}}{5}} \right){\text{ km/sec}} \cr & {\text{ = }}\frac{{432}}{5}{\text{km/hr}} \cr & {\text{Let the speed of the train be x kmph}}{\text{.}} \cr & {\text{Then relative speed = }}\left( {x + 3} \right){\text{kmph}} \cr & \therefore x{\text{ + 3 = }}\frac{{432}}{5} \cr & \Rightarrow x = \frac{{432}}{5} - 3 \cr & \Rightarrow x = \frac{{417}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 83.4\,{\text{kmph}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (75 + 100)km/hr}} \cr & {\text{ = 175 km/hr}} \cr & {\text{Time taken to cover 175 km}} \cr & {\text{at relative speed = 1 hr}} \cr & \therefore {\text{T = Time taken to cover 200 km}} \cr & {\text{ = }}\left( {\frac{1}{{175}} \times 200} \right)hr \cr & = \frac{8}{7}\,hr \cr & = 1\frac{1}{7}hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (80 + 55)km/hr}} \cr & {\text{ = 135 km/hr}} \cr & {\text{ = }}\left( {135 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{75}}{2}} \right)m/\sec \cr & {\text{Distance covered}} \cr & {\text{ = (120 + 90 + 90)m}} \cr & {\text{ = 300m}} \cr & {\text{Required time}} \cr & {\text{ = }}\left( {300 \times \frac{2}{{75}}} \right)\sec \cr & = 8\sec \cr}$$

Explanation:-

$$\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (60 + 90) km/hr}} \cr & {\text{ = }}\left( {150 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{125}}{3}} \right){\text{m/sec}} \cr & {\text{Distance coverd}} \cr & {\text{ = (1}}{\text{.10 + 0}}{\text{.9)km}} \cr & {\text{ = 2 km}} \cr & {\text{ = 2000 m}}{\text{.}} \cr & {\text{Required time}} \cr & {\text{ = }}\left( {2000 \times \frac{3}{{15}}} \right)\sec \cr & = 48sec \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed of train relative to man}} \cr & {\text{ = }}\left( {60 + 6} \right){\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = \left( {66 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{55}}{3}} \right)m/\sec \cr & \therefore {\text{Time taken to pass the man}} \cr & \left( {100 \times \frac{3}{{55}}} \right)\sec = 6\,\sec \cr}$$

Explanation:-

$$\eqalign{ & {\text{Length of train}} \cr & {\text{ = 280 m }} \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed = }}\left( {\frac{{150 + 300}}{{40.5}}} \right)m/\sec \cr & = \left( {\frac{{450}}{{40.5}} \times \frac{{18}}{5}} \right)km/hr \cr & = 40km/hr. \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let the length of the train be x metres}}{\text{.}} \cr & {\text{Then, length of the platform = (2}}x{\text{) metres}}{\text{.}} \cr & {\text{Speed of the train}} \cr & {\text{ = }}\left( {90 \times \frac{5}{{18}}} \right)m/\sec \cr & = 25m/sec \cr & \therefore \frac{{x + 2x}}{{25}} = 36 \cr & \Rightarrow 3x = 900 \cr & \Rightarrow x = 300 \cr & {\text{Hence, length of platform}} \cr & {\text{ = }}2x = \left( {2 \times 300} \right){\text{m}} = 600{\text{m}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}} \cr & {\text{ = }}\left( {\frac{{132}}{6}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {22 \times \frac{{18}}{5}} \right){\text{km/sec}} \cr & {\text{ = 79}}{\text{.2 km/hr}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed }} \cr & {\text{ = }}\left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{3}{\text{ m/sec}} \cr & {\text{Total distance covered}} \cr & {\text{ = (100 + 140) m = 240 m}} \cr & \therefore {\text{Required time}} \cr & {\text{ = }}\left( {240 \times \frac{3}{{50}}} \right){\text{sec}} \cr & {\text{ = }}\frac{{72}}{5}{\text{sec}} \cr & {\text{ = 14}}{\text{.4 sec}} \cr}$$