#### Data Interpretation

• ( 1 ) A train 220 m long is running with a speed of 59 kmph. In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going ?

• 1) 10 sec.
• 2) 12 sec.
• 3) 14 sec.
• 4) 16 sec.
• Discussion in forum
Solution : Speed of train relative to man = (59 + 7) km/hr = 66 km/hr.
= 66 * 5/18 m/sec
= 55/3 m/sec.
Time taken to pass the man = 220 * 3/55 sec = 12 sec.

discussion

• ( 2 ) Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?

• 1) 10
• 2) 12
• 3) 15
• 4) 20
• Discussion in forum
Solution : Speed of the first train = ( 120 /10) m/sec = 12 m/sec.
Speed of the second train = ( 120 /15) m/sec = 8 m/sec.
Relative speed = (12 + 8) = 20 m/sec.
Therefore Required time = [ (120 + 120) /20] sec = 12 sec.

discussion

• ( 3 ) Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is :

• 1) 1:03
• 2) 3:02
• 3) 3:04
• 4) 2:03
• Discussion in forum
Solution : Let the Speed of the two trains be x m/sec and y m/sec respectively. Then, length of the first train = 27x meters, and length of the second train = 17y meters.
(27x + 17y)/(x + y) = 23
=> 27x + 17y = 23x+ 23y
=> 4x = 6y
=> x/y = 3/2.

discussion

• ( 4 ) A train covers the distance 50 m between two cities in 4 hours arriving 2 hours late. What rate would permit the train to arrive on schedule:

• 1) 25 km/hr
• 2) 30 km/hr
• 3) 28 km/hr
• 4) 26 km/hr
• Discussion in forum
Solution : Speed = Distance/Time
Distance = 50 km
Time taken = 4
But train is late by 2 hours
So, the time taken should have been 4 - 2
So, Speed required to be on time = 50/(4 - 2) = 50/2 = 25 km/hr

discussion

• ( 5 ) A train moves with a speed of 108 kmph.Its speed in metres per second is ?

• 1) 10.8 m/sec
• 2) 18m/sec
• 3) 30m/sec
• 4) 38.8m/sec
• Discussion in forum
Solution : Speed = 108 Kmph
= (108 * 5/18)m/sec = 30 m/sec.

discussion

• ( 6 ) A 360 metre long train crosses a platform in 40 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

• 1) 320 m
• 2) 440 m
• 3) 650 m
• 4) 550 m
• Discussion in forum
Solution : Speed = Distance/time = 360/18 = 20 m/sec
Let the length of the platform be x meter
Then Distance = Speed * Time
x + 360 = 20*40
=> x + 360 = 800
=> x = 800 - 360
=> x = 440 meter

discussion

• ( 7 ) How long does a train 110 metres long running at the speed of 72 km / hr take to cross a bridge 132 metres in length ??

• 1) 9.8 sec
• 2) 12.1 sec
• 3) 14.3 sec
• 4) 12.42 sec
• Discussion in forum
Solution : Speed = (72 * 5/18 )m/sec = 20 m/ sec
Total Distance covered = (110 + 132) m = 242 m
Required time = (242 / 60) sec = 12.1 sec

discussion

• ( 8 ) Two stations P and Q are 110 km apart on a straight line. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?

• 1) 9 a.m
• 2) 10 a.m.
• 3) 10.30 a.m.
• 4) 11 a.m
• Discussion in forum
Solution : Suppose they meet x hours after 7 a.m.
Distance covered by P in x hours = 20x km.
Distance covered by Q in (x - 1) hours = 25(x - 1) km.
Therefore 20x + 25(x - 1) = 110
=> 45x = 135
=> x = 3.
So, they meet at 10 a.m.

discussion

• ( 9 ) A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train?

• 1) 70 km/hr
• 2) 72 km/hr
• 3) 79.2 km/hr
• 4) 80 km/hr
• Discussion in forum
Solution : Speed = (132 / 6) m/sec
= (22 * 18 /5)km/hr
= 79.2 km/hr

discussion

• ( 10 ) A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the Length of the other train ?

• 1) 230 m
• 2) 240 m
• 3) 260 m
• 4) 320 m
• Discussion in forum
Solution : Relative Speed = (120 + 80)km/hr
= (200 * 5/18)m/sec
= (500 / 9)m/sec
Let the length of the other train be x metres
Then, x + 270 / 9
x + 270 = 500
x = 230.

discussion