walking towards it at 6 km/hr. Find the length of the train and of the platform.

1) 200 metres 2) 120 metres 3) 220 metres 4) 320 metres 5) None of these

Solution: (5). Distance = length of Train + Length of Platform

Convert all units to metres & Seconds

Platform crossed in 24 seconds

Then Distance = length of train + length of platform

= speed x time

= 30 km X 24 sec

= 30 X 5/18 x 24

Length of Train & Platform = 200

Man crossed in 8 seconds

Distance = length of train

= speed x time (speed = train +man)

= 36 x 5/18 x 8

Length of train = 80 m

Then platform = 200 – 80 = 120 m

2. Two trains are moving in the same direction at 75 km/hr and 48 km/hr. The faster train crosses a

man sitting in the slower in 14 seconds. Find the length of the faster train .

1) 105 metres 2) 115 metres 3) 125 metres 4) 120 metres 5) None of these

Solution: (1).Same direction = relative speed

= faster – slower

= 75 – 48

Relative speed

Time to cross

Length of faster train

= 27 k/hr

= 14 seconds

= speed x time

= 27 km x 14 sec

= 27 x 5/18 x 14

= 105 metre

3. A 175 metre long train crosses a 35 metre platform in 12 seconds. What is the speed of the train in

Km/hr?

1)42 2)64 3)63 4)59 5)None of these

Solution (3). Speed = Distance / Time

[ Distance = Train +platform = 175 + 35 = 210 m ]

= 210 m / 12 sec

= 210/12

x 18/5 km/hr

= 63 km/hr

4. A train covers a distance of 180 km in 4 hours. Another train covers the same distance in one hour

less. What is the difference in the distance covered by these two trains in one hour?

1) 45 km 2) 40 km 3) 15 km 4) 9 km 5) None of these

Solution: (3).Distance covered by the first train in 1 hour

= 180 / 4 = 45 km

Distance covered by the second train in 1 hour = 180/3 = 60km

Therefore required difference = ( 60 - 45)km = 15km

5. A train takes 3 hours to run from one station to another . If it reduces its speed by 12 km/hr, it takes

45 minutes more for the journey . The distance between the station is

1) 220 km 2) 210 km 3) 180 km 4) 160 km 5) None of these

Solution: (3).let the distance between the stations be x km

Original speed = x/3. km/hr

New speed = [ (x/3 ) – 12) km/hr

New time = 3 hrs + 45 minutes = 3 ¾ hrs = 15/4 hrs

Therefore [ (x/3) – 12 ] * 15/4 = x

Or, (5/4) x - 45 =x

Or ,x/4 = 45

Or, x = 180 km

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