Practice Crossing pole signal post man - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A train is 250m long. If the train takes 50 seconds to cross a tree by the railway line, then the speed of the train in km/hr is :
(a)
(b)
(c)
(d)
Using Rule 1,
Speed of train = $\text"Length of train"/ \text"Time taken in crossing"$
= $250/50$ = 5 m/sec.
= $(5 × 18/5)$ kmph = 18 kmph
Q-2) A man observed that a train 120 m long crossed him in 9 seconds. The speed (in km/hr) of the train was
(a)
(b)
(c)
(d)
Using Rule 1,
Speed of train
= $120/9 × 18/5$ = 48 kmph
Q-3) If a train, with a speed of 60 km/ hr, crosses a pole in 30 seconds, the length of the train (in metres) is :
(a)
(b)
(c)
(d)
Using Rule 1,
Speed of train = 60 kmph
= 60 × $5/18 = 50/3$ m/sec
Length of train = Speed × Time
= $50/3 × 30$ = 500 m
Q-4) 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)
(b)
(c)
(d)
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
Q-5) 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)
(b)
(c)
(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.
Q-6) 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)
(b)
(c)
(d)
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
Q-7) A train 100m long is running at the speed of 30 km/hr. The time (in second) in which it will pass a man standing near the railway line is :
(a)
(b)
(c)
(d)
Using Rule 1,
If a train crosses an electric pole, a sitting/standing man, km or milestone etc. then distance = Length of the train.
Then, Length of train = Speed × Time And Time = $\text"Length of train"/\text"Speed"$
and Speed = $\text"Length of train"/\text"Time"$
In this situation, the train covers it length.
Required time = $100/{30 × 1000}$ hr.
= ${100 × 60 × 60}/{30 × 1000}$ = 12 seconds
Q-8) A train 150m long passes a telegraphic post in 12 seconds. Find the speed of the train.(in km/hr)
(a)
(b)
(c)
(d)
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
Q-9) 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)
(b)
(c)
(d)
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
Q-10) A train is running at 36 km/hr. If it crosses a pole in 25 seconds, its length is
(a)
(b)
(c)
(d)
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