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A train moving at 108 km/hr crosses a platform in 30 seconds. Then it crosses a man running at 12 km/hr in the same direction to train in 9 seconds. What is the length of train and platform in m?
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Solution
Speed of train $= 108 imes frac{5}{18} = 30$ m/s
Speed of man $= 12 imes frac{5}{18} = frac{10}{3}$ m/s
Relative speed $= 30 - frac{10}{3} = frac{80}{3}$ m/s
Length of train $= frac{80}{3} imes 9 = 240$ m
Train length + platform length $= 30 imes 30 = 900$ m
Platform length $= 900 - 240 = 660$ m
Hence, the answer is 240 and 660.
Speed of man $= 12 imes frac{5}{18} = frac{10}{3}$ m/s
Relative speed $= 30 - frac{10}{3} = frac{80}{3}$ m/s
Length of train $= frac{80}{3} imes 9 = 240$ m
Train length + platform length $= 30 imes 30 = 900$ m
Platform length $= 900 - 240 = 660$ m
Hence, the answer is 240 and 660.
Two trains are moving towards each other with speeds 50 km/hr and 55 km/hr from different stations R and S. While they meet, the second train from station S has covered 20 km more distance than the first train which starts from station R. What is the distance between the two stations?
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Solution
Distances are proportional to speeds.
Ratio $= 50:55 = 10:11$
Difference $= 1$ part $= 20$ km
Total distance $= 21 imes 20 = 420$ km
Hence, the distance is 420 km.
Ratio $= 50:55 = 10:11$
Difference $= 1$ part $= 20$ km
Total distance $= 21 imes 20 = 420$ km
Hence, the distance is 420 km.
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