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 passenger train 150m long is travelling with a speed of 36 km/ hr. If a man is cycling in the direction of train at 9 km/hr., the time taken by the train to pass the man is

(a) 15 sec

(b) 18 sec

(c) 20 sec

(d) 10 sec

The correct answers to the above question in:

Answer: (c)

Using Rule 5,
Let 'a' metre long train is running with the speed 'x' m/s. A man is running in same direction and with the speed 'y' m/s, then time taken by the train to cross the man = $a/{(x - y)}$seconds. And a = (x - y)t

Relative speed of train

= (36 - 9) kmph = 27 kmph

= ${27 × 5}/18$ m/sec = $15/2$ m/sec

Required time

= $\text"Length of the train"/ \text"Relative speed"$

= ${150 × 2}/15 = 20$ seconds

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

Question : 1

A train passes an electrical pole in 20 seconds and passes a platform 250 m long in 45 seconds. Find the length of the train.

a) 200m

b) 300m

c) 250m

d) 400m

Answer: (a)

Using Rule 1,

If the length of train be x metre, then speed of train

= $x/20 = {x + 250}/45$

$x/4 = {x + 250}/9$

9x = 4x + 1000

9x - 4x = 1000

5x = 1000

x = $1000/5$ = 200 metre

Question : 2

A train 180 m long moving at the speed of 20 m/sec. over-takes a man moving at a speed of 10m/ sec in the same direction. The train passes the man in :

a) 9 sec

b) 18 sec

c) 27 sec

d) 6 sec

Answer: (b)

Using Rule 5,
Let 'a' metre long train is running with the speed 'x' m/s. A man is running in same direction and with the speed 'y' m/s, then time taken by the train to cross the man = $a/{(x - y)}$seconds. And a = (x - y)t

Relative speed of man and train

= 20 - 10 = 10m/sec.

Required time = $180/10$ = 18 seconds

Question : 3

The time taken by a train 160 m long, running at 72 km/hr, in crossing an electric pole is

a) 9 seconds

b) 6 seconds

c) 4 seconds

d) 8 seconds

Answer: (d)

Distance covered in crossing a pole

= Length of train Speed of train = 72 kmph

= $({72 × 5}/18)$ m./sec. = 20 m./sec.

Required time = $160/20$ = 8 seconds

Question : 4

A 75 metre long train is moving at 20 kmph. It will cross a man standing on the platform in

a) 14 seconds

b) 13.5 seconds

c) 15.5 seconds

d) 12 seconds

Answer: (b)

Using Rule 1,

Speed of train (in m/s)

= 20 × $5/18 = 50/9$ m/sec

Required time = $75/50 × 9$

= 13.5 seconds

Question : 5

A train, 240 m long crosses a man walking along the line in opposite direction at the rate of 3 kmph in 10 seconds. The speed of the train is

a) 75 kmph

b) 83.4 kmph

c) 86.4 kmph

d) 63 kmph

Answer: (b)

Using Rule 6,

Let 'a' metre long train is running with the speed 'x' m/s. A man is running in the opposite direction of train with the speed of 'y' m/s. Then, time taken by the train to cross the man = $(a/{(x + y)})$seconds.

If the speed of train be x kmph then,

Its relative speed = (x + 3) kmph

Time = $\text"Length of the train"/ \text"Relative speed"$

$10/3600 = {240/1000}/{(x + 3)} = 240/{1000(x + 3)}$

x + 3 = 86.4 ⇒ x = 83.4 kmph

Question : 6

If a man walks at the rate of 5 km/hour, he misses a train by 7 minutes. However if he walks at the rate of 6 km/hour, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is

a) 7 km

b) 6.25 km

c) 4 km

d) 6 km

Answer: (d)

Let the required distance be x km.

Difference of time = 7 + 5 = 12

minutes = $1/5$ hour

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

According to the question,

$x/5 - x/6 = 1/5$

${6x - 5x}/30 = 1/5$

$x/30 = 1/5$ ⇒ $x = 30/5$ = 6 km.

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