A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively. The speed of the train is
A. 28 kmph
B. 27 kmph
C. 25 kmph
D. 24 kmph
Answer: Option C
Solution(By Examveda Team)
1st method: Let the speed of the train be S. And length of the train be x. When a train crosses a man, its travels its own distance. $$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$Join The Discussion
Comments ( 5 )
Related Questions on Speed Time and Distance
A. 48 min.
B. 60 min.
C. 42 min.
D. 62 min.
E. 66 min.
A. 262.4 km
B. 260 km
C. 283.33 km
D. 275 km
E. None of these
A. 4 hours
B. 4 hours 30 min.
C. 4 hours 45 min.
D. 5 hours
3*10=30
5*11=55
Actual speed=55-30=25
time =1
Speed=25
Train 1=3*10=30
Train2=5*11=55
Actual speed->55-30=25
The speed of train=25kmph
3_ _ _ _ _ 5
10_ _ _ _ _11
3*10=30_ _ _5*11=55
time diff=11-10=1
speed=55-30=25
How do you get 5/18 in both the equations
18x=55s-275. 18x=50s-150. Then equate two equations... We will get the speed of train