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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 competitor runs 200 meters race in 24 seconds. His rate is:
Answer : D
Explanation
Speed = 200/24 m/sec = (200/24* 18/5) km/hr = 30 km/hr
Q 2 - The velocities of A and B are in the proportion 3:4. A takes 20 min. more than B to achieve a destination. In what time does A achieve the destination?
Answer : A
Explanation
Let the time taken by A be x hrs. Then, time taken by B = (x-20/60) hrs = (x-1/3) hrs Ratio of speeds = inverse ratio of time taken ∴3:4 =(x- 1/3): x ⇒3x-1/3x = 3/4 ⇒12x- 4 = 9x ⇒3x= 4 ⇒x= 4/3 hrs Required time = 4/3 hrs.
Q 3 - A plane departed 30 min. later than the planned time and with a specific end goal to achieve the destination 1500 km away in time, it needed to build the pace by 250 km/hrs from the standard rate. Its standard rate is:
Answer : D
Explanation
Let the usual speed be x km/hr 1500/x - 1500/(x+250) =1/2 ⇒ 1/x - 1/(x+250) = 1/3000 ⇒ (x+250)- x /x(x+250) = 1/3000 ⇒x(x+250)= 75000 ⇒ x2+250 x-75000=0 ⇒ x2+ 1000x- 750 x- 75000= 0 ⇒x(x+1000)-750(x+1000) = 0 ⇒(x+1000)(x-750)=0 ⇒x = 750 ∴usual speed = 750 km/hr
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 - By what amount of percent must a driver expand his pace so as to lessen the time by 20%, taken to cover a sure separation?
Answer : B
Explanation
Distance= (time*speed) =t*x. Let the required increase in speed be p%. Then, (80%of t)*(100+p)/100=x=t*x ⇒80/100*t*(100+p)/100*x=t*x⇒4 (100+p/500=1⇒4p=100⇒p=25. ∴Required increase in speed=25%.
Q 7 - A auto ventures a separation of 840 km at a uniform pace. On the off chance that the velocity of the auto is 10 km/hr more, it takes 2 hours less to cover the same separation. The first speed of the auto was
Answer : C
Explanation
Let the original speed be x km/hr then, 840/x-840/(x+10) = 2 ⇒ 840(x+10)-840 x = 2x(x+10) ⇒2x2+20 x-8400 = 0 ⇒x2+10x- 4200= 0 ⇒(x+70) (x-60) = 0 ⇒x = 60 ∴ Original speed = 60 km/hr
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 train covers a separation in 50 minutes, in the event that it keeps running at a velocity of 48 km/hr on a normal. The velocity at which the train must rushed to diminish the season of voyage to 40 minutes, will be:
Answer : C
Explanation
Distance = (speed * time) = (48*50/60) km = 40 km New time taken = 40/60 hr = 2/3 hr New speed = (40*3/2) km/hr = 60 km/hr
Q 10 - A train without stoppages goes at the rate of 50 km/hr and with stoppages it goes at 45 km/hr. How long does the train stop on a normal for every hour?
Answer : B
Explanation
Due to stoppage it covers 5 km less per hour. Time taken to cover 5 km= (5/50*60) min. = 6 min. Hence, the train stops on an average 6 min/hr.