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# Train Problems For SSC CGL

Download SSC CGL Train questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Train on objective questions (MCQ’s) for SSC exams.

Instructions

Question 1: Two trains of length 180 meters with velocity 30 m/s and 60 m/s travel in opposite direction and if the initial distance between them is 0.54 kilometres then what is the time taken for their tails to cross each other ?

a) 5 sec

b) 7 sec

c) 8 sec

d) 10 sec

Question 2: A man travels with a speed of 15 m/s to reach the destination 5 seconds late and travels with 20 m/s to reach the destination 5 seconds early.What is the original distance to be travelled ?

a) 5.85 km

b) 5.15 km

c) 5.25 km

d) 5.35 km

Question 3: A man travels half of the distance with speed ‘v/2’ and one fourth of the distance with ‘v’ and remaining distance with ‘2v’.What is the average speed of the journey ?

a) 8v/13

b) 8v/15

c) 8v/17

d) 8v/19

Question 4: A boat takes 4 hours to travel same distance ‘x’ upstream and downstream.If the speed of the boat is 4 km/hr and stream is 2 km/hr then find the value of x ?

a) 6 km

b) 8 km

c) 4 km

d) 10 km

Question 5: Two trains start from stations A and B which are 280 km apart towards each other with their speeds in the ratio of 2:5.If they meet after 1 hour 20 minutes then what is the difference between the speeds ?

a) 45 km/hr

b) 60 km/hr

c) 75 km/hr

d) 90 km/hr

Instructions

Question 6: Two trains of length 180 meters with velocity 30 m/s and 60 m/s travel in opposite direction and if the initial distance between them is 0.54 kilometres then what is the time taken for their tails to cross each other ?

a) 5 sec

b) 7 sec

c) 8 sec

d) 10 sec

Question 7: A train travelled from A to B at a speed of 60 km/hr and returned from B to A at a speed of 40 km/hr. Find the average speed of the train.

a) 50 km/hr

b) 48 km/hr

c) 42 km/hr

d) 56 km/hr

Question 8: A train travels a certain distance in a certain time. If it covers triple the distance in double the time, then find the ratio between original speed and current speed.

a) 1:3

b) 2:3

c) 3:5

d) 2:5

Question 9: A train crosses a platform of length 204 metres in 36 seconds. It crosses a man standing on the platform in 12 seconds. Find the speed of the train.

a) 102 m/sec

b) 127 m/sec

c) 60 m/sec

d) 168 m/sec

Instructions

Question 10: Two trains start from stations A and B which are 200 km apart towards each other with their speeds in the ratio of 1:2.If they meet after 2 hours 40 minutes then what is the difference between the speeds ?

a) 15 km/hr

b) 20 km/hr

c) 25 km/hr

d) 30 km/hr

Question 11: A train travels from Delhi to Mumbai at a speed of 60 km/hr and returns at a speed of 75 km/hr. Find its average speed.

a) 66.67 km/hr

b) 45 km/hr

c) 60 km/hr

d) 67.5 km/hr

Question 12: A train of length 120 m crossed a platform in 36 seconds. It crosses a man standing on the platform in 24 seconds running at the same speed. Find the length of the platform.

a) 120 m

b) 60 m

c) 104 m

d) 72 m

Question 13: A train crosses a pole in 12 seconds and a platform of length 160 m in 32 seconds running at the same speed. Find the length of the train.

a) 108 m

b) 96 m

c) 120 m

d) 84 m

Question 14: A train travels a distance of 480 km at uniform speed. Due to breakdown, its speed is reduced by 20 km/hr and hence it travels the destination 4 hours late. Find the initial speed of the train.

a) 60 km/hr

b) 40 km/hr

c) 25 km/hr

d) 35 km/hr

Question 15: A train travels at a speed of 125 km/hr without stoppages and at 100 km/hr with stoppages. How many minutes per hour does the train stop?

a) 25 min

b) 20 min

c) 15 min

d) 12 min

Question 16: A train travelled at a speed of 15 km/hr and reached destination 20 minutes late. Had the speed increased to 20 km/hr, it would reach the destination 20 minutes early. Find the distance travelled by the train.

a) 30 km

b) 28 km

c) 40 km

d) 45 km

Question 17: A person walked at a speed of 4 kmph while going to office and returned at a speed of 3 kmph. Find the average speed.

a) 3.6 kmph

b) 3.4 kmph

c) 3.8 kmph

d) 3.5 kmph

Question 18: A train travelled from Chennai to Hyderabad at a speed of 54 kmph and returned to Chennai at a speed of 36 kmph. Find its average speed.

a) 42.5 kmph

b) 43 kmph

c) 43.2 kmph

d) 45 kmph

Question 19: A train travels from Delhi to Agra at a speed of 72 km/hr and returns at a speed of 90 km/hr. Find the average speed.

a) 78 km/hr

b) 81 km/hr

c) 80 km/hr

d) 92 km/hr

Question 20: A train travels a distance of 480 km at uniform speed. Due to breakdown, its speed is reduced by 10 km/hr and hence it travels the destination 8 hours late. Find the initial speed of the train.

a) 30 km/hr

b) 20 km/hr

c) 25 km/hr

d) 35 km/hr

As the both are travelling in opposite direction the relative velocity will be sum of the velocities i.e 30+60=90 m/s
Total distance to be travelled is length of the trains i.e 180+180=360 m and also the distance between them i.e 0.54 km=540 m
Total=540+360=900 m
Time =900/90
=10 seconds

let the original time be t seconds
So we have 15(t+5)=20(t-5)
15t+75=20t-100
5t=175 sec
t=35 sec
Total distance=15*35
=5250 m
=5.25 km

Let the total distance be x
Total time=(x/(v/2))+(x/(4v))+(x/(4*2v)
=19x/8v
Average speed=Total distance/Total time
=x/(19x/8v)
=8v/19

Upstream speed =4-2=2 km/hr
Downstream speed=4+2=6 km/hr
Therefore we have $\frac{x}{6}+\frac{x}{2}$=6
4x/6 =4
x=6 km

let the speeds be 2x and 5x.
As both the trains are moving towards each other their relative velocity will be 2x+5x=7x
Therefore 280/7x =4/3
x=120/4
x=30 km/hr
dIfference=5x-2x
=3x
=90 km/hr

As the both are travelling in opposite direction the relative velocity will be sum of the velocities i.e 30+60=90 m/s
Total distance to be travelled is length of the trains i.e 180+180=360 m and also the distance between them i.e 0.54 km=540 m
Total=540+360=900 m
Time =900/90
=10 seconds

Let the distance between A and B be 120 km each (LCM of 60 and 40).
Time taken by the train to travel 120 km at 60 km/hr = 2 hours
Time taken by the train to travel 120 km at 40 km/hr = 3 hours
Total distance = 120+120 = 240 km
Total time = 2+3 = 5 hours
Average speed $= \dfrac{\text{Total Distance}}{\text{Total Time}} = \dfrac{240}{5} = 48 km/hr$

Let the distance travelled by the train be D km.
Let the time taken by the train to travel D km be T hours.
Original speed = $\dfrac{D}{T}$ km/hr
New Distance = 3D km
New time = 2T hours
New speed = $\dfrac{3D}{2T}$ km/hr
Ratio between original speed and new speed = $\dfrac{D}{T} : \dfrac{3D}{2T} = 1 : \dfrac{3}{2} = 2:3$

Let the length of the train be ‘x’ metres.
Then, Speed required to travel (x+204) metres in 36 seconds = $\dfrac{x+204}{36}$ m\sec
Speed required to travel x metres in 12 seconds = $\dfrac{x}{12}$ m/sec

Here, Speeds are equal.
⇒ $\dfrac{x+204}{36} = \dfrac{x}{12}$

⇒ $x+204 = 3x$
⇒ $2x = 204$
⇒ $x = 102$ metres
Hence, the Length of the train = 102 metres.

let the speeds be x and 2x.
As both the trains are moving towards each other their relative velocity will be x+2x=3x
Therefore 200/3x =8/3
x=200/8
x=25 km/hr
dIfference=2x-x
=x
=25 km/hr

Let the total distance be 300 km (LCM of 60 and 75).
Time required to travel 300 km at 60 km/hr = 300/60 = 5 hours
Time required to travel 300 km at 75 km/hr = 300/75 = 4 hours
Total time to travel 600 km = 9 hours
Total distance = 600 km
Average speed = 600/9 = 200/3 = 66.67 km/hr

Speed of the train to cross the man = 120/24 = 5 m/sec
Speed of the train to cross the platform = (120+P)/36
$\dfrac{120+P}{36} = 5$
⇒ $120+P = 180$
⇒ $P = 60$

Let the length of the train be ‘T’ metres.
Speed of the train to cross the pole in 12 seconds = T/12 m/sec
Speed of the train to cross the platform in 32 seconds = (T+160)/32 m/sec
Here, Speeds are equal.
⇒ $\dfrac{T}{12} = \dfrac{T+160}{32}$

⇒ $8T = 3T + 480$
⇒ $5T = 480$
⇒ $T = 96 m$
Hence, The Length of the train = 96 m

Given that the distance $= 480$ km
Let the speed of the train $= x$ km/hr
Then, time in which it can travel $480$ km $= \dfrac{480}{x}$ hr
Now, Speed is reduced by $20$ km/hr
Reduced speed $= (x-20)$ km/hr
Time in which it can travel $480$ km at reduced speed $= \dfrac{480}{(x-20)}$
Difference in time = 4 hours
⇒ $\dfrac{480}{(x-20)} – \dfrac{480}{x} = 4$
⇒ $\dfrac{480(x-x+20)}{x^2-20x} = 4$
⇒ $9600 = 4(x^2-20x)$
⇒ $x^2-20x-2400 = 0$
⇒ $x^2-60x+40x-2400 = 0$
⇒ $x(x-60)+40(x-60) = 0$
⇒ $(x+40)(x-60) = 0$
x+40 = 0 | x-60 = 0
x = -40 | x = 60
Speeds cannot be negative.
Hence, Initial speed = 60 km/hr

Due to stoppages, train travelled 25 km less in an hour.
Time required to travel 25 km in an hour $= \dfrac{25}{125}\times 60 = 12$ minutes.

Let the distance travelled be ‘D’ km.
Time taken to travel D km at 15 kmph = D/15 hours
Time taken to travel D km at 20 kmph = D/20 hours
Given the difference between time = 40 minutes = $\dfrac{40}{60} = \dfrac{2}{3}$
⇒ $\dfrac{D}{15} – \dfrac{D}{20} = \dfrac{2}{3}$

⇒ $\dfrac{3D-2D}{60} = \dfrac{2}{3}$

⇒ $\dfrac{D}{60} = \dfrac{2}{3}$

⇒ $D = 40 km$

Let the onward distance and return distance be 12 km each (LCM of 4 and 3)
Time taken to travel 12 km at a speed of 4 kmph = 12/4 = 3 hours
Time taken to travel 12 km at a speed of 3 kmph = 12/3 = 4 hours
Total time taken by the train to travel 24 km = 3+4 = 7 hours
Average speed = (Total Distance)/(Total Time) = 24/7 = 3.4 kmph

Let the onward distance and return distance be 108 km each (LCM of 54 and 36).
Time taken to travel 108 km at a speed of 54 kmph = 108/54 = 2 hours
Time taken to travel 108 km at a speed of 36 kmph = 108/36 = 3 hours
Total time taken by the train to travel 216 km = 2+3 = 5 hours
Average speed = (Total Distance)/(Total Time) = 216/5 = 43.2 kmph

Let the distance between Delhi and Agra be 360 km (LCM of 72 and 90)
Time taken to travel from Delhi to Agra at 72 km/hr = 360/72 = 5 hours
Time taken to travel from Agra to Delhi at 90 km/hr = 360/90 = 4 hours
Total time = 5+4 = 9 hours
Total distance = 360+360 = 720 km
Average speed = $= \dfrac{\text{Total distance}}{\text {Total time}} = \dfrac{720}{9} = 80 \text{km/hr}$

Given that the distance $= 480$ km
Let the speed of the train $= x$ km/hr
Then, time in which it can travel $480$ km $= \dfrac{480}{x}$ hr
Now, Speed is reduced by $10$ km/hr
Reduced speed $= (x-10)$ km/hr
Time in which it can travel $480$ km at reduced speed $= \dfrac{480}{(x-10)}$
Difference in time = 8 hours
⇒ $\dfrac{480}{(x-10)} – \dfrac{480}{x} = 8$
⇒ $\dfrac{480(x-x+10)}{x^2-10x} = 8$
⇒ $4800 = 8(x^2-10x)$
⇒ $x^2-10x-600 = 0$
⇒ $x^2-30x+20x-600 = 0$
⇒ $x(x-30)+20(x-30) = 0$
⇒ $(x+20)(x-30) = 0$
x+20 = 0 | x-30 = 0
x = -20 | x = 30
Speeds cannot be negative.
Hence, Initial speed = 30 km/hr

We hope this Problems on Train questions for SSC Exam will be highly useful for your preparation.