Model 4 Train Vs Pole/Signal Post/Man Section-Wise Topic Notes With Detailed Explanation And Example Questions
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The following question based on trains topic of quantitative aptitude
(a) 200m
(b) 300m
(c) 250m
(d) 400m
The correct answers to the above question in:
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
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Read more crossing pole signal post man Based Quantitative Aptitude Questions and Answers
Question : 1
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 »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 : 2
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 »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 : 3
In what time will a 100 metre long train running with a speed of 50 km/hr cross a pillar ?
a) 72 seconds
b) 7.2 seconds
c) 70 seconds
d) 7.0 seconds
Answer »Answer: (b)
Speed of train = 50 kmph
= $({50 × 5}/18)$ m./sec. = $125/9$ m./sec.
Required time = $({100/125}/9)$ seconds
= $({100 × 9}/125)$ = 7.2 seconds
Question : 4
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
Answer »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
Question : 5
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 »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 : 6
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 »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
trains Shortcuts and Techniques with Examples
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Model 1 Train Vs Train in same direction
Defination & Shortcuts … -
Model 2 Train Vs Train in opposite direction
Defination & Shortcuts … -
Model 3 Train Vs Bridge/Platform
Defination & Shortcuts … -
Model 4 Train Vs Pole/Signal Post/Man
Defination & Shortcuts … -
Model 5 Train Vs Both platform and a man/a pole
Defination & Shortcuts … -
Model 6 Change in speed Vs Change with time travel
Defination & Shortcuts …
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