Model 4 Train Vs Pole/Signal Post/Man Practice Questions Answers Test with Solutions & More Shortcuts

Question : 26 [SSC CHSL 2012]

A train is running at a speed of 90 km/hr. If it crosses a signal in 10 sec., the length of the train (in metres) is

a) 324

b) 900

c) 250

d) 150

Answer: (c)

Using Rule 1,

Speed of train = 90 kmph

= $({90 × 5}/18)$ metre/second

= 25 metre/second

If the length of the train be x then,

Speed of train

= $\text"Length of train"/ \text"Time taken in crossing the signal"$

25 = $x/10$ ⇒ x = 250 metre

Question : 27 [SSC CGL Prelim 2003]

In what time will a train 100 metres long cross an electric pole, if its speed be 144 km/hour ?

a) 5 seconds

b) 12.5 seconds

c) 3$5/4$ seconds

d) 2.5 seconds

Answer: (d)

Using Rule 1,

Speed of the train = 144 kmph

= $144 × 5/18$ = 40 m/s

When a train crosses a pole, it covers a distance equal to its own length.

The required time

= $100/40 = 5/2$ = 2.5 seconds.

Question : 28 [SSC Constable 2013]

A train, 120 m long, takes 6 seconds to pass a telegraph post; the speed of train is

a) 62 km/hr

b) 55 km/hr

c) 85 km/hr

d) 72 km/hr

Answer: (d)

Using Rule 1,

Speed of train

= $\text"Length of train"/ \text"Time taken in crossing the pole"$

= $120/6$ = 20 m/sec

= 20 × $18/5$ = 72 kmph

Question : 29 [SSC CGL Tier-I 2010]

Buses start from a bus terminal with a speed of 20 km/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes?

a) 4 km/hr

b) 5 km/hr

c) 7 km/hr

d) 3 km/hr

Answer: (b)

Distance covered in 10 minutes at 20kmph

= distance covered in 8 minutes at (20 + x ) kmph

20 × $10/60 = 8/60 (20 + x)$

200 = 160 + 8x

8x = 40 ⇒ $x = 40/8$ = 5 kmph

Question : 30 [SSC CGL Tier-I 2016]

A train 150m long passes a telegraphic post in 12 seconds. Find the speed of the train.(in km/hr)

a) 12.5

b) 25

c) 45

d) 50

Answer: (c)

Distance covered by train in crossing a telegraphic post

= length of train

Speed of train = $\text"Distance"/ \text"Time"$

= $(150/12)$ m./sec.

= $(150/12 × 18/5)$ kmph = 45 kmph

IMPORTANT quantitative aptitude EXERCISES

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