Speed & Distance - Online Quiz



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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.

Questions and Answers

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

A - 5 m/sec

B - 30 m/sec

C - 40 m/sec

D - 50 m/sec

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:

A - 3 km/hr

B - 4km/hr

C - 4.5 km/hr

D - 9 km/hr

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:

A - 30 min

B - 60 min

C - 75 min

D - 1 hr.30 min

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:

A - 4:1

B - 1:4

C - 2:1

D - 1:2

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?

A - 16.7 kms

B - 18.6 kms

C - 21.5 kms

D - 26.7 kms

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:

A - 990 kms

B - 1200 kms

C - 1320 kms

D - 1440 kms

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:

A - 31.2 kms

B - 38.4 kms

C - 32.8 kms

D - 24 kms

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:

A - 13 min

B - 15 min

C - 19 min

D - 21 min

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:

A - 10 km/hr

B - 20 km/hr

C - 25 km/hr

D - 30 km/hr

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:

A - 1:2

B - 2:1

C - 1:4

D - 4:1

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


aptitude_speed_distance.htm

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