Practice Trains problems - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) A goods train starts running from a place at 1 P.M. at the rate of 18 km/hour. Another goods train starts from the same place at 3 P.M. in the same direction and overtakes the first train at 9 P.M. The speed of the second train in km/hr is
(a)
(b)
(c)
(d)
Distance covered by the first goods train in 8 hours = Distance covered by the second goods train in 6 hours.
18 × 8 = 6 × x
$x = {18 × 8}/6 = 24$ kmph
Q-2) A train 'B' speeding with 100 kmph crosses another train C, running in the same direction, in 2 minutes. If the length of the train B and C be 150 metre and 250 metre respectively, what is the speed of the train C (in kmph)?
(a)
(b)
(c)
(d)
Let the speed of train C be x kmph.
Relative speed of B
= (100 – x ) kmph.
Time taken in crossing
= $\text"Length of both trains"/ \text"Relative speed"$
$2/60 = {({150 + 250}/1000)}/{100 – x}$
$1/30 = 2/{5(100 – x)}$
$1/6 = 2/{100 – x}$
100 – x = 12
x = 100 – 12 = 88 kmph.
Q-3) Two trains travel in the same direction at the speed of 56 km/h and 29 km/h respectively. The faster train passes a man in the slower train in 10 seconds. The length of the faster train (in metres) is
(a)
(b)
(c)
(d)
Relative speed
= 56 – 29 = 27 kmph
= $27 × 5/18 = 15/2$ m/sec
Distance covered in 10 seconds
= $15/2$ × 10 = 75 m
Hence, length of train = 75 m.
Q-4) Two trains, 80 metres and 120 metres long, are running at the speed of 25 km/hr and 35 km/hr respectively in the same direction on parallel tracks. How many seconds will they take to pass each other ?
(a)
(b)
(c)
(d)
Relative speed
= 35 – 25 = 10 kmph
= ${10 × 5}/18$ m/sec.
Total length = 80 + 120 = 200 metres
Required time
= $\text"Sum of the length of trains"/ \text"Relative speed"$
= $200/{{10 × 5}/18} = {200 × 18}/{10 × 5}$
= 72 seconds
Q-5) A boy started from his house by bicycle at 10 a.m. at a speed of 12 km per hour. His elder brother started after 1 hr 15 mins by scooter along the same path and caught him at 1.30 p.m. The speed of the scooter will be (in km/hr)
(a)
(b)
(c)
(d)
Let the speed of Scooter be x Distance covered by cycling in 3$1/2$ hours
= Distance covered by scooter in 2$1/4$ hours
$12 × 7/2 = x × 9/4$
$x = {12 × 7 × 2}/9$
= $56/3 = 18{2}/3$ kmph
Q-6) Two trains are running 40 km/hr and 20 km/hr respectively in the same direction. The fast train completely passes a man sitting in the slow train in 5 seconds. The length of the fast train is
(a)
(b)
(c)
(d)
Relative speed
= 40 – 20 = 20 km/hour = ${20 × 5}/18$ m/sec.
Length of the faster train
= ${20 × 5}/18 × 5$ metres
= $250/9 = 27{7}/9$ metres
Q-7) Two trains of equal length are running on parallel lines in the same direction at 46 km/hour and 36 km/hour. The faster train passes the slower train in 36 seconds. The length of each train is
(a)
(b)
(c)
(d)
Let the length of each train be x metre.
Relative speed
= (46 – 36) kmph = 10 kmph
= $({10 × 5}/18)$ m./sec.
= $25/9$ m./sec.
${2x}/{25/9} = 36$
$2x = 36 × 25/9$ = 100
$x = 100/2$ = 50 metre
Q-8) A train passes two bridges of lengths 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is :
(a)
(b)
(c)
(d)
Using Rule 1,
When a train crosses a bridge it covers the distance equal to length of Bridge & its own length
Let the length of the train be = x
Speed of the train
= ${x + 800}/100$ m/s
Since train passes the 800 m bridge in 100 seconds.
Again, train passes the 400 m bridge in 60 seconds.
${400 + x}/{{x + 800}/100} = 60$
${(400 + x) × 100}/{x + 800}$ = 60
40000 + 100x = 60x + 48000
100x - 60x = 48000 - 40000
40x = 8000 ⇒ x = $8000/40$ = 200m
Q-9) 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-10) 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-11) Two trains start from stations A and B and travel towards each other at speeds of 50 kmph and 60 kmph respectively. At the time of their meeting, the second train has travelled 120 km more than the first. The distance between A and B is
(a)
(b)
(c)
(d)
Let both trains meet after t hours.
Distance = speed × time
60t – 50t = 120
10t = 120 ⇒ t = 12 hours
Required distance
= 60t + 50t
= 110t = 110 × 12 = 1320 km
Q-12) Two trains start at the same time from A and B and proceed toward each other at the speed of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have travelled 175 km more than the other. Find the distance between A and B.
(a)
(b)
(c)
(d)
Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t - 50t = 175
25t = 175 ⇒ t = $175/25$ = 7 hours
Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.
Using Rule 13,
Here, a = 75, b = 50, d = 175
Required distance
= $({a + b}/{a - b}) × d$
= $({75 + 50}/{75 - 50}) × 175$
= $125/25 × 175$ = 125 × 7 = 875 km
Q-13) Two trains start from stations A and B and travel towards each other at speed of 50 km/hour and 60 km/hour respectively. At the time of their meeting, the second train has travelled 120 km more than the first. The distance between A and B is :
(a)
(b)
(c)
(d)
Let train A start from station A and B from station B.
Let the trains A and B meet after t hours.
Distance covered by train A in t hours = 50t
Distance covered by train B in t hours = 60t km
According to the question,
60t - 50t = 120
t = $120/10$ = 12 hours.
Distance AB = 50 × 12 + 60 × 12
= 600 + 720 = 1320 km
Using Rule 13,From stations A and B, two trains start travelling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as$({a + b}/{a - b}) × d$
Here, a = 60, b = 50, d = 120
Distance between A and B = $({a + b}/{a - b}) × d$
= $({60 + 50}/{60 - 50}) × 120$
= $110/10 × 120$ = 1320 km
Q-14) Two towns A and B are 500 km. apart. A train starts at 8 AM from A towards B at a speed of 70 km/ hr. At 10 AM, another train starts from B towards A at a speed of 110 km/hr. When will the two trains meet ?
(a)
(b)
(c)
(d)
Let two trains meet after t hours when the train from town A leaves at 8 AM.
Distance covered in t hours at 70 kmph
+ Distance covered in (t - 2) hours at 110 kmph = 500km
70t + 110 (t - 2) = 500
70t + 110t - 220 = 500
180 t = 500 + 220 = 720
t = $720/180$ = 4 hours
Hence, the trains will meet at 12 noon.
Q-15) A train, 150 m long, takes 30 seconds to cross a bridge 500 m long. How much time will the train take to cross a platform 370 m long ?
(a)
(b)
(c)
(d)
Using Rule 10,
When a train croses a bridge, distance covered
= length of (bridge + train).
Speed of train = ${150 + 500}/30$
=$650/30 = 65/3$ m/sec.
Time taken to cross the 370m long platform
= ${370 + 150}/{65/3}$
= ${520 × 3}/65$ = 24 seconds
Q-16) In what time will a train 100 metres long cross an electric pole, if its speed be 144 km/hour ?
(a)
(b)
(c)
(d)
Using Rule 1,
Speed of the train = 144 kmph
= $144 × 5/18$ = 40 m/s
When a train crosses a pole, it covers a distance equal to its own length.
The required time
= $100/40 = 5/2$ = 2.5 seconds.
Q-17) A moving train passes a platform 50 metres long in 14 seconds and a lamp-post in 10 seconds. The speed of the train is
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Suppose length of train be x
According to question
${x + 50}/14 = x/10$
14x = 10x + 500
4x = 500
$x = 500/4 = 125 m$
Therefore, speed
=$125/10 × 18/5 = 45$ kmph
Q-18) A train 110 metre long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going ?
(a)
(b)
(c)
(d)
Relative speed of train
= (60 + 6) kmph.
=$({66 × 5}/18)$ m/sec.
= $55/3$ m/sec.
Length of train = 110 metre
∴ Required time = $(110/{55/3})$ seconds
= $({110 × 3}/55)$ seconds = 6 seconds
Q-19) A train moving at a rate of 36 km/hr. crosses a standing man in 10 seconds. It will cross a platform 55 metres long, in :
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Speed of train = 36 kmph
= $36 × 5/18$ = 10 m/sec
Length of train
= 10 × 10 = 100 metres
Required time = ${100 + 55}/10$
= 15$5/10 = 15{1}/2$ second
= 15.5 seconds
Q-20) A train passes a platform 90 metre long in 30 seconds and a man standing on the platform in 15 seconds. The speed of the train is :
(a)
(b)
(c)
(d)
Rule 10 and Rule 1,
Let the length of the train be x
According to the question,
Speed of the train
= ${x + 90}/30 = 15$
x + 90 = 2x ⇒ x = 90 m
Speed of train = $90/15$
= 6 m/s = $6 × 18/5$ kmph
= 21.6 kmph