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# Time and Distance for SSC-CGL Set-3 PDF

Download SSC CGLTime and Distance Questions with answers  PDF based on previous papers very useful for SSC CGL exams. Very Important Time and Distance Questions for SSC exams

Question 1: A 360 m long train running at a uniform speed, crosses a platform in 55 seconds and a man standing on the platform in 24 seconds. What is the length (in metre) of the platform?

a) 465

b) 445

c) 480

d) 410

Question 2: Excluding stoppages, the speed of a bus is 70 kmph and including stoppages, it is 56 kmph. For how many minutes does the bus stop per hour?

a) 10 min

b) 8 min

c) 15 min

d) 12 min

Question 3: Mohit starts moving from a place A and reaches the place B in 12 hours. He covers $\frac{1}{5^{th}}$ part of the total distance at the speed of 6 km/hr and covers the remaining distance at the speed of 8 km/hr. What is the distance between A and B?

a) 90 km

b) 95 km

c) 60 km

d) 75 km

Question 4: The ratio between the speeds of two cars is 6 : 5. If the second car runs 600 km in 6 hours, then the speed of the first car is:

a) 100 km/hr

b) 120 km/hr

c) 110 km/hr

d) 90 km/hr

Question 5: A car covers a certain distance at a speed of 48 km/hr in 14 hours. How much time it will take to cover the same distance at the speed of 84 km/hr?

a) 6 hr

b) 8 hr

c) 9 hr

d) 12 hr

Question 6: Two stations P and Q are 570 km apart on a straight line. One train starts from P at 5 a.m. and travels towards Q at 90 km/h. Another train starts from Q at 6 a.m. and travels towards P at a speed of 70 km/h. At what time will they meet?

a) 11 a.m.

b) 9 a.m.

c) 1 p.m.

d) 3 p.m.

Question 7: The length of the bridge, which a train 400 metres long and travelling at 60 km/hr can cross in 36 seconds, is:

a) 450 m

b) 250 m

c) 200 m

d) 500 m

Question 8: A train passes a platform in 42 seconds and a man standing on the platform in 25 seconds. If the speed of the train is 72 km/hr, what is the length of the platform?

a) 300 m

b) 270 m

c) 340 m

d) 370 m

Question 9: Two trains 85 m and 155 m long, run at the speeds of 62 km/h and 82 km/h respectively, in opposite directions on parallel tracks. The time which they take to cross each other is:

a) 4 seconds

b) 5 seconds

c) 6 seconds

d) 8 seconds

Question 10: Smith travels along four sides of a square at speeds of 10, 12, 15 and 20 km/hr. The average speed of the Smith is:

a) $13\frac{1}{3}$ km/hr

b) $12\frac{1}{2}$ km/hr

c) $14\frac{1}{4}$ km/hr

d) $10$ km/hr

Question 11: A car travels a certain distance at a speed of 120km/hr in 4 hours. What should be the speed of the car if the car covers the same distance in 3 hours 20 minutes.

a) 142 km/hr

b) 140 km/hr

c) 124 km/hr

d) 144 km/hr

Question 12: A car travels 60km in the first hour , 65km in the second hour , 70km in the third hour, then the average speed of the car is

a) 66 km/hr

b) 62 km/hr

c) 56 km/hr

d) 65 km/hr

Question 13: Two trains start at the same time, P from A to B and Q from B to A.If they arrive at B and A,respectively, $2\frac{1}{2}$ hours and 10 hours after they passed each other, and the speed of P is 90 km/hr, then the speed of Q in kin/hr is?

a) 80

b) 75

c) 45

d) 60

Question 14: A person covered a distance of 250 km at a certain speed. If his speed is 20% less, he would have covered the same distance in $1\frac{1}{4}$ hours more time. His speed (in km/hr), initially was:

a) 45

b) 50

c) $52\frac{1}{2}$

d) $47\frac{1}{2}$

Question 15: A train is moving with a uniform speed. Train crosses a bridge of length 243 meters in 30 seconds and a bridge of length 343 meters in 36 seconds. Whatis the speed of the train?

a) 60 km/hr

b) 72 km/hr

c) 64 km/hr

d) 65 km/hr

The train with length 360 m passes by a man standing at the platform in 24 seconds .The speed of the train is

V= 360/24= 15 m/s

Let the length of the platform be L

(L+ 360)/ 55 = 15

L+360= 15*55

L+360=825

L= 825-360

L= 465 m.

Due to stoppages, bus travelled 14 km less in one hour.
Time taken to travel 14 km without stoppages = $\dfrac{14}{70}\times60 = 12$ min

Let the total distance be x km
Given, Total time taken to reach x km = 12 hours
Speed for $\dfrac{x}{5}$ km = 6 km/hr
Time taken to travel $\dfrac{x}{5}$ km = $\dfrac{x}{5\times6} = \dfrac{x}{30}$ hours
Speed for remaining $\dfrac{4x}{5}$ km = 8 km/hr
Time taken to travel $\dfrac{4x}{5}$ km = $\dfrac{4x}{5\times8} = \dfrac{x}{10}$ hours
Total time = $\dfrac{x}{30}+\dfrac{x}{10} = \dfrac{4x}{30}$ hours
Given, $\dfrac{4x}{30} = 12$ => x = 90
Therefore, Total distance = 90 km

Let the speeds of the two cars be 6x km/hr and 5x km/hr.
Given, Speed of second car = 600km/6 hours = 100 km/hr
5x = 100 => x = 20
Therefore, Speed of the first car = 6x = 6*20 = 120 km/hr

Given, Speed = 48 km/hr
Time = 14 hours
Then, Distance = $48 \times 14 = 672 km$
New speed = 84 km/hr
Then, Time taken = 672/84 = 8 hours

Distance between P and Q = 570 km
Speed of first train = 90 km/hr
Speed of second train = 70 km/hr
Second train starts after 1 hour of start of first train.
Then, First train travels 90 km in that 1 hour.
Remaining distance = 570 km – 90 km = 480 km
Relative speed = 90+70 = 160 km/hr
Time taken to meet = 480/160 = 3 hours.
Hence, They will meet 3 hours after 6am = 9am

Let the length of the bridge = L m
Speed of the train = 60 km/hr = $60\times\dfrac{5}{18} = \dfrac{50}{3} m/sec$
$\dfrac{400+L}{36} = \dfrac{50}{3}$
=> 400+L = 600
=> L = 200
Therefore, Length of the bridge = 200 m

Given, Speed of the train = 72 km/hr = $72\times\dfrac{5}{18} = 20 m/sec$
Time taken to cross a man on the platform = 25 seconds
Let the length of the train be T m and length of the platform be P m.
Given, $\dfrac{T}{25} = 20 => T = 500$
Hence, The length of the train = 500 m
Given, $\dfrac{500+P}{42} = 20 => 500+P = 840 => P = 340$
Therefore, The length of the platform = 340 m

As both are travelling in opposite directions relative velocity=62+82=144 km/hr
144*5/18 =40 m/s
Total distance=155+95
=240 m
Time taken=240/40
=6 sec

Average Speed=Total distance/Total time
Let the side of the square be ‘s’ and so total distance=4s
Total time =(s/10)+(s/12)+(s/15)+(s/20)
=18s/60
Average speed=4s/(18s/60)
=40/3 km/hr
=$13\frac{1}{3}$ km/hr

Distance=Speed*time
=120*4
=480 km
Speed=distance/time
=480/(10/3)
=144 km/hr

Average speed=Total distance/Total time
Distance 1=60*1=60 km
Distance 2=65*1=65 km
Distance 3=70*1=70 km
Total distance=195 km
Total time=3 hours
Average speed=195/3
=65 km/hr

Let the distance between A and B be ‘d’
speed of Q be q
‘t’ be the time taken for them to meet
d/(90+q) =t
d/(t+10)=q
d/(t+(5/2))=90
d=90t+225
d=90t+qt
d=qt+10q
qt=225
q=9t
$t^{2}$=25
t=5 hours
q=9*5=45 km/hr

Let the original speed be ‘V’ and time taken be ‘t’ hours
Therefore 250/V =t
Now speed=0.8V
Therefore 250/0.8V =t+5/4
250/V =0.8(t+5/4)
t=0.8t+1
0.2t=1
t=5 hours
Original speed V=250/5
=50 km/hr

Speed of the train = $\dfrac{T+243}{30} = \dfrac{T+343}{36}$
=> $\dfrac{T+243}{5} = \dfrac{T+343}{6}$
=> $6T+1458 = 5T+1715$
Therefore, Speed of the train $= \dfrac{243+257}{30} = \dfrac{500}{30} = \dfrac{50}{3} m/sec = \dfrac{50}{3} \times \dfrac{18}{5} = 60 km/hr$