Problems on Trains - 03

Explanation:-

$$\eqalign{ & {\text{Average speed of train}} \cr & {\text{ = 40 km/hr}} \cr & {\text{Reach at its destination at on time }} \cr & {\text{New average speed of train}} \cr & {\text{ = 35 km/h}} \cr & {\text{Time = 15 minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr & {\text{Then distance travelled}} \cr & {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = 70}}\,{\text{km}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 9 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr & {\text{Distance covered in crossing a}} \cr & {\text{700 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 700}}{9} \cr & \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ {\because {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \Rightarrow \frac{x}{3} = \frac{{x + 700}}{{10}} \cr & \Rightarrow 10x = 3x + 2100 \cr & \Rightarrow 10x - 3x = 2100 \cr & \Rightarrow 7x = 2100 \cr & \Rightarrow x = \frac{{2100}}{7} = 300{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platfrom}} \cr & \frac{x}{9} = \frac{{x + 800}}{T} \cr & \Rightarrow Tx = 9x + 7200 \cr & \Rightarrow 300T = 2700 + 7200 \cr & \Rightarrow 300T = 9900 \cr & \Rightarrow T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let length of train }} \cr & {\text{ = }}l\,{\text{metre}} \cr & \Rightarrow {\text{Time }} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{relative speed in opposite direction}}}} \cr & \Rightarrow 4\sec \, = \,\frac{{l + 0}}{{\left( {84 + 6} \right) \times \frac{5}{{18}}{\text{m/s}}}} \cr & \Rightarrow 4\, = \frac{l}{{90 \times \frac{5}{{18}}}} \cr & \Rightarrow \,l\, = \,100\,{\text{m}} \cr & \therefore {\text{ length of the train = 100 m}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Relative speed of man \& train}} \cr & {\text{ = }}\frac{{100 \times 5}}{{36}} \times \frac{{18}}{5} \cr & {\text{ = 50km/hr}} \cr & \therefore {\text{speed of train}} \cr & {\text{ = 50}} - {\text{5}} \cr & {\text{ = 45 km/hr}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed of the train}} \cr & {\text{ = }}\frac{{700 + 500}}{{10}} \cr & {\text{ = 120 ft/second}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}} = \left( {\frac{{300}}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50}}{3}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr & {\text{Then}},\left( {\frac{{x + 300}}{{39}}} \right) = \frac{{50}}{3} \cr & \Rightarrow 3\left( {x + 300} \right) = 1950 \cr & \Rightarrow x = 350\,m. \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}} = \left( {78 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{65}}{3}} \right)\,{\text{m/sec}} \cr & {\text{Time = }}\,{\text{1}}\,{\text{minute = 60}}\,{\text{second}}. \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{tunnel}}\,{\text{be}}\,x\,{\text{metres}}. \cr & {\text{Then}},\,\left( {\frac{{800 + x}}{{60}}} \right) = \frac{{65}}{3} \cr & \Rightarrow 3\left( {800 + x} \right) = 3900 \cr & \Rightarrow x = 500 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Total}}\,{\text{distance}}\,{\text{covered}} \cr & = \left( {\frac{7}{2} + \frac{1}{4}} \right)\,{\text{miles}} \cr & = \frac{{15}}{4}\,{\text{miles}} \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = \left( {\frac{{15}}{{4 \times 75}}} \right)\,{\text{hrs}} \cr & = \frac{1}{{20}}\,{\text{hrs}} \cr & = \left( {\frac{1}{{20}} \times 60} \right)\,\min \cr & = 3\,\min \cr}$$

Explanation:-

$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = \left( {60 + 6} \right)\,{\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = \left( {66 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \left( {\frac{{55}}{3}} \right)\,{\text{m/sec}} \cr & \therefore {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr & = \left( {110 \times \frac{3}{{55}}} \right){\text{sec}} = 6\,{\text{sec}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}} = 2x\,{\text{m/sec}} \cr & {\text{Relative}}\,{\text{speed}} = \,\left( {x + 2x} \right)\,{\text{m/sec}} = 3x\,{\text{m/sec}} \cr & \therefore \frac{{\left( {100 + 100} \right)}}{8} = 3x \cr & \Rightarrow 24x = 200 \cr & \Rightarrow x = \frac{{25}}{3} \cr & {\text{So,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}}\, = \frac{{50}}{3}\,{\text{m/sec}} \cr & = \left( {\frac{{50}}{3} \times \frac{{18}}{5}} \right)\,{\text{km/hr}} \cr & = 60\,{\text{km/hr}} \cr}$$