A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
A. 45 m
B. 50 m
C. 54 m
D. 72 m
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr & 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr & {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr & {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr & \therefore 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr & \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr & {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr} $$Join The Discussion
Comments ( 4 )
Related Questions on Problems on Trains
A. 120 metres
B. 180 metres
C. 324 metres
D. 150 metres
A. 45 km/hr
B. 50 km/hr
C. 54 km/hr
D. 55 km/hr
A. 200 m
B. 225 m
C. 245 m
D. 250 m
Any other way to do it? TIA.
22-2 how and how only 9sec why not 10sec
Speeds 2 and 4
Time. 9 and 10
18. and 40
40-18/timediff=22kmph
(22-2)*5/18*9=50mts
super