Model 4 Train Vs Pole/Signal Post/Man Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

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The following question based on trains topic of quantitative aptitude

Questions : A train 100 metre long is running at a speed of 120 km/hr. The time taken to pass a person standing near the line is

(a) 3 seconds

(b) 5 seconds

(c) 7 seconds

(d) 1 second

The correct answers to the above question in:

Answer: (a)

Speed of train = 120 kmph.

= $({120 × 5}/18)$ m./sec. = $100/3$ m./sec.

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

= $(100/{100/3})$ seconds

= $(100/100 × 3)$ seconds = 3 seconds

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Read more crossing pole signal post man Based Quantitative Aptitude Questions and Answers

Question : 1

A train 240 metres in length crosses a telegraph post in 16 seconds. The speed of the train is

a) 52 km/hr

b) 54 km/hr

c) 56 km/hr

d) 50 km/hr

Answer: (b)

When a train crosses a pole it travels a distance equal to its length.

Speed of train

= $240/16$ = 15 m./sec.

= $(15 × 18/5)$ kmph = 54 kmph.

Question : 2

A train is 125 m long. If the train takes 30 seconds to cross a tree by the railway line, then the speed of the train is :

a) 15 km/hr

b) 16 km/hr

c) 12 km/hr

d) 14 km/hr

Answer: (a)

Using Rule 1,

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

=$125/30$ = 41.6 m/s

4.16 m/s = 4.16 × $18/5$ = 15 km/hr

Question : 3

A train 300 m long is running with a speed of 54 km/hr. In what time will it cross a telephone pole?

a) 15 seconds

b) 17 seconds

c) 18 seconds

d) 20 seconds

Answer: (d)

Using Rule 1,

Speed of train = 54 kmph

= $({54 × 5}/18)$ m/sec = 15 m/sec

Required time = $\text"Length of trains"/ \text"Speedof train"$

= $300/15$ = 20 seconds

Question : 4

A train 150m long passes a km stone in 30 seconds and another train of the same length travelling in opposite direction in 10 seconds. The speed ot the second train is :

a) 125 km/hr

b) 25 km/hr

c) 75 km/hr

d) 90 km/hr

Answer: (d)

Speed of train A

= $150/30$ = 5 m/sec.

Speed of train B = x m/sec.

Relative speed = (5+x) m/sec.

Length of both trains

= Relative speed × Time

300 = (5 + x) × 10

5 + x = $300/10$ = 30

x = 30 - 5 = 25 m/sec.

= $({25 × 18}/5)$ kmph. = 90 kmph.

Question : 5

A train 100 metres long meets a man going in opposite direction at 5 km/hr and passes him in 7$1/5$ seconds. What is the speed of the train (in km/hr) ?

a) 60 km/hr

b) 55 km/hr

c) 50 km/hr

d) 45 km/hr

Answer: (d)

Using Rule 6,

Let speed of train be x kmph

Relative speed = (x + 5) kmph

Length of train

= $100/1000$ km = $1/10$ km

${1/10}/{x + 5} = 36/{5 × 60 × 60}$

$1/{10(x + 5)} = 1/500$

x + 5 = 50 ⇒ x = 45 kmph

Question : 6

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

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GET trains PRACTICE TEST EXERCISES

Model 1 Train Vs Train in same direction

Model 2 Train Vs Train in opposite direction

Model 3 Train Vs Bridge/Platform

Model 4 Train Vs Pole/Signal Post/Man

Model 5 Train Vs Both platform and a man/a pole

Model 6 Change in speed Vs Change with time travel
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