Dear Readers, Here We are providing a quant Quiz of 15 question in accordance with the syllabus of SSC CGL. These Questions cover the topic of Speed, Distance and Time
Q1. A train is travelling at the rate of 45 km/hr. How many seconds it will take to cover a distance of 4/5 km?
(a) 36 sec.
(b) 64 sec.
(c) 90 sec.
(d) 120 sec.
Q2. An aeroplane covers a certain distance at a speed of 240 km/hour in 5 hours. To cover the same distance in 1 (2/3 )hours it must travel at a speed of:
(a) 30 km./hr.
(b) 360 km./hr.
(c) 600 km./hr.
(d) 720 km./hr.
Q3. A train 100 m long is running at the speed of 30 km/hr. The time (in second) in which it passes a man standing near the railway line is:
(a) 10
(b) 11
(c) 12
(d) 15
Q4. If a man walks 20 km at 5 km/hr. he will be late by 40 minutes. If he walks at 8km/hr. how early from the fixed time will he reach?
(a) 15 minutes
(b) 25 minutes
(c) 50 minutes
(d) 1 (1/2) hours
Q5. A man walking at the rate of 5 km/hr. crosses a bridge in 15 minutes. The length of the bridge (in metres) is:
(a) 600
(b) 750
(c) 1000
(d) 1250
Q6. A man crosses a road 250 metres wide in 75 seconds. His speed in km/hr is:
(a) 10
(b) 12
(c) 12.5
(d) 15
Q7. The length of a train and that of a platform are equal. If with a speed of 90 km/hr the train crosses the platform in one minute. Then the length of the train (in metres) is:
(a) 500
(b) 600
(c) 750
(d) 900
Q8. A train passes a 50 metres long platform in 14 seconds and a man standing on the platform in 10 seconds. The speed of the train is:
(a) 24 km/hr
(b) 36km/hr
(c) 40km/hr
(d) 45 km/hr
Q9. An athlete runs 200 metres race in 24 seconds. His speed (in km/hr.) is:
(a) 20
(b) 24
(c) 28.5
(d) 30
Q10. A car goes 10 metres in a second. Find its speed in km/hour.
(a) 40
(b) 32
(c) 48
(d) 36
Q11. A man riding his bicycle covers 150 metres in 25 second. What is his speed in km per hour?
(a) 25
(b) 21.6
(c) 23
(d) 20
Q12. 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) 80 m
(b) 90 m
(c) 200 m
(d) 150 m
Q13. A train is 125 m long. If the train takes 30 seconds to cross a tree by the railway line, then the speed of the train is:
(a) 14 km/hr.
(b) 15 km/hr.
(c) 16 km/hr.
(d) 12 km/hr.
Q14.A 120 m long train takes 10 seconds to cross a man standing on a platform. What is the speed of the train?
(a) 12 m/sec
(b) 10 m/sec
(c) 15 m/sec
(d) 20 m/sec
Q15. A 75 metre long train is moving at 20 kmph. It will cross a man standing on the platform in
(a) 12 seconds
(b) 14 seconds
(c) 13.5 seconds
(d) 15.5 seconds
SOLUTIONS
S1. Ans.(b)
Sol. Time =Distance/Speed
∴ Time =4/(5/45)=4/225 hrs
Time (sec.) =4/225×3600 = 64 sec
S2. Ans.(d)
Sol. Distance = Constant
So, speed α 1/Time
Ratio of time =5∶5/3
Ratio of time = 3 : 1
Ratioof speed = 1 : 3
1 unit → 240 km/hr
3 units →240 × 3 = 720 km/hr
So, i.e. Required speed = 720 km/hr
S3. Ans.(c)
Sol.Speed = 30 km/hr=30×5/18 m/sec
=25/3 m/sec
So, Time =D/S=100/(25/3)=12 sec
S4. Ans.(c)
Sol. Time taken at 5 km/hr=20/5
= 4 hrs
Actual time =(4-2/3)=10/3 hrs
Time taken at 8 km/hr =20/8
=5/2 hrs
Time difference =10/3-5/2=5/6 hrs
= 50 min. required time
S5. Ans.(d)
Sol. 15 min =1/4 hrs
1 hrs = 5 kms
1/4 hrs. =5/4 kms
So, length of bridge =5/4 kms. = 1250 mt.
S6. Ans.(b)
Sol. S=D/T=250/75 m/sec =250/75×18/5
= 12 km/hr
S8. Ans.(d)
Sol. Distance travelled in 14 sec. = 50 + l
Distance travelled in 10 sec = l
So, speed of train
=50/(14-10) m/sec
=50/4×18/5km/hr = 45 km/hr
S9. Ans.(d)
Sol. Speed =D/T=200/24 m/sec
=200/24×18/5km/hr
= 30 km/hr
S10. Ans.(d)
Sol. Speed = 10 m/sec =10×18/5km/hr
= 36 km/hr
S11. Ans.(b)
Sol. Speed of =150/25 m/sec =6×18/5km/hr
= 21.6 km/hr
S12. Ans.(c)
Sol. Distance covered in 100 sec = 800 + l
Distance covered in 60 sec = 400 + l
So, Distance covered in 40 sec
= (800 + l) – (400 + l) = 400 mtr.
Speed → 400/40 m/sec. = 10 m/s
Distance covered in 60 sec = 10 × 60 = 600 meter
So, 400 + l = 600
⇒ l = 200 meter
S13. Ans.(b)
Sol. Speed =D/T=125/30 m/sec
=125/30×18/5km/hr
= 15 km/hr
S14. Ans.(a)
Sol. Speed =120/10 m/sec = 12 m/sec
S15. Ans.(c)
Sol. T=D/S=(75 × 18)/(20 ×5 )=27/2 sec
= 13.5 sec
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