# Speed & Distance - Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to **Speed & Distance**. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using **Show Answer** button. You can use **Next Quiz** button to check new set of questions in the quiz.

Q 1 - A train is moving with a pace of 180 km/hr. Its rate is:

### Answer : D

### Explanation

180 km/hr = (180*5/18) m/sec = 50 m/sec.

Q 2 - A man covers half of his adventure at 6km/hr and the staying half at 3 km/hr. His normal rate is:

### Answer : B

### Explanation

Average speed = 2xy/(x+y) km/hr = 2*6*3/(6+3)km/hr = 4 km/hr

Q 3 - By strolling at 3/4 of his standard speed, a man achieves his office 20 min. later than Normal. His standard time is:

### Answer : B

### Explanation

At a speed of 3/4 of the usual speed , time taken = 4/3 of usual time ∴ (4/3 of usual time) - (usual time) = 20 min. Let the usual time be x min. then, (4x/3 - x) = 20 ⇒x = 60 min. ∴usual time is 60 min.

Q 4 - A certain separation is secured at a sure speed. On the off chance that half of this separation is secured in twofold the time, the proportion of the two paces is:

### Answer : A

### Explanation

Let x kms be covered in y hrs. then, first speed = x/y km/hr Again, x/2 km is covered in 2y hrs. ∴new speed = (x/2 * 1/2y) km/hr = (x/4y)km/hr Ratio of speeds = x/y : x/4y = 1:1/4 = 4:1

Q 5 - A strolls at a uniform rate of 4 km an hours and 4 hours after his begin, B cycles after him at the uniform rate of 10 km an hours. How a long way from the Beginning stage will B get A?

### Answer : D

### Explanation

Suppose B catches A after x hours. Then, Distance travelled by A in (x+4) hr. =distance travelled by B in x hours 4(x+4) =10x⇒6x=16⇒3x=8⇒x=8/3hrs. Distance travelled by B in 8/3 hrs= (8/3*10)km=80/3km=26.7km.

Q 6 - Two trains begin from stations A and B and travel towards one another at 50 km/hr and 60 km/hr separately. At the season of their meeting, the second prepare has voyage 120 km more than the first. The separation in the middle of A and B is:

### Answer : C

### Explanation

Let the two train meet after x hours . Then, 60x-50x-=120⇒ 10x =120⇒x=12hrs. Distance AB = (distance covered by slower train) + (distance covered by fast train) = [(150*12)+(60*12)]km=(600+720)km=1320km.

Q 7 - A man performs 2/15 of the aggregate adventure by rail, 9/20 by transport and the remaining 10 km on cycle. His aggregate voyage is:

### Answer : D

### Explanation

Let the total journey be x km. then , 2x/15+ 9x/20 +10 = X ⇒ 8x+27 x+600 = 60 x ⇒ 25x = 600 ⇒x = 24 ∴ Total journey = 24 km

Q 8 - If a train keeps running at 40 km/hr, it achieves its destination late by 11 minutes. In any case, in the event that it keeps running at 50 km/hr, it is late by 5 minute just. The right time for the train to cover its trip, is:

### Answer : C

### Explanation

Let the required time be x minutes. Distance covered in (x+11) min at 40 km/hr Distance covered in (x+5) min at 50km/hr ∴(x+11)/60*(x+5)/60*50⇒4(x+11) =5(x+5) ⇒x= (44-25) =19. Hence, the required time is 19 minutes.

Q 9 - A auto covers four progressive 3 km extends at 10 km/hr, 20 km/hr, 30 km/hr and 60 km/hr separately. Its normal rate over this separation is:

### Answer : B

### Explanation

Total distance = (3*4) km = 12 kms Total time taken = (3/10+3/20+3/30+ 3/60) = (36+18+12+6)/120 hrs. = 72/120 hrs = 3/5 hrs. Average speed = 12/ (3/5) km/hr = (12*5)/3 = 20 km/hr

Q 10 - A certain separation is secured by cyclist at a sure speed. On the off chance that a jogger covers a large portion of the separation in twofold the time, the proportion of the rate of the jogger to that of the cyclist is:

### Answer : C

### Explanation

Let distance = d meters and time taken by cyclist = t sec. Speed of the cyclist = d/t m/sec. Again, distance = d/2 meters, time taken by jogger = 2t sec. Speed of the jogger = (d/2)/2t m/sec. = d/4t m/sec. Ratio of speeds of jogger and cyclist = d/4t: d/t = 1/4:1 = 1:4