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

Question : 11 [SSC CGL Prelim 2002

A 120 m long train takes 10 seconds to cross a man standing on a platform. What is the speed of the train ?

a) 10 m/sec.

b) 15 m/sec.

c) 20 m/sec.

d) 12 m/sec.

Answer: (d)

Using Rule 1,

In crossing a man standing on platform, train crosses its own length.

Speed of train = $120/10 = 12$m/s

Question : 12 [SSC CGL Tier-I 2015]

A train 180 metres long is running at a speed of 90 km/h. How long will it take to pass a post ?

a) 7.8 secs

b) 8 secs

c) 7.2 secs

d) 8.2 secs

Answer: (c)

Using Rule 1,

Speed of train = 90 kmph

= $(90 × 5/18)$ m/sec. = 25 m/sec.

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

Required time = $\text"Distance"/ \text"Speed"$

= $180/25$ = 7.2 seconds

Question : 13 [SSC CGL Tier-I 2016]

In what time will a train, 60 metre long, running at the rate of 36 km/hr pass a telegraph post ?

a) 8 seconds

b) 7 seconds

c) 6 seconds

d) 9 seconds

Answer: (c)

Speed of train = 36 kmph

= $({36 × 5}/18)$ m./sec. = 10 m./sec.

Required time = $\text"Length of train"/ \text"Speed of train"$

= $60/10$ = 6 seconds

Question : 14 [SSC CPO S.I.2006]

A train passes two persons walking in the same direction at a speed of 3 km/hour and 5km/ hour respectively in 10 seconds and 11 seconds respectively. The speed of the train is

a) 27 km/hour

b) 25 km/hour

c) 24 km/hour

d) 28 km/hour

Answer: (b)

Let the speed of train be x kmph

and its length be y km.

When the train crosses a man, it covers its own length

According to he question,

$y/{(x - 3) × 5/18}$ = 10

18y = 10 × 5(x –3)

18y = 50x –150 ..... (i)

and, $y/{(x - 5) × 5/18}$ = 11

18y = 55(x–5)

18y = 55x –275 .... (ii)

From equations (i) and (ii),

55x –275 = 50x–150

55x –50x = 275 - 150

5x = 125 ⇒ x = $125/5$ = 25

Speed of the train = 25 kmph

Using Rule 7,
A train crosses two men in $t_1$ seconds and $t_2$ seconds running in the same direction with the speed $s_1$ and $s_2$ then the speed of train is = ${t_1S_1 - t_2S_2}/{t_1 - t_2}$and length of train is l = $(S_1 - S_2)({t_1 - t_2}/{t_1 - t_2})$

Here, $S_1 = 3, S_2 = 5, t_1 = 10/3600 , t_2 = 11/3600$

Speed of train = ${t_1S_1 - t_2S_2}/{t_1 - t_2}$

= ${{3 × 10}/3600 - {5 × 11}/3600}/{10/3600 - 11/3600}$

= ${-25}/3600 × 3600/{- 1}$ = 25 m/sec

Question : 15 [SSC Investigator 2010]

A train is running at 36 km/hr. If it crosses a pole in 25 seconds, its length is

a) 250 m

b) 255 m

c) 260 m

d) 248 m

Answer: (a)

Using Rule 1,

Speed of train = 36 kmph

= $({36 × 5}/18)$ m/sec = 10 m/sec.

Length of train = Speed × time

= 10 × 25 = 250 metre

IMPORTANT quantitative aptitude EXERCISES

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