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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 man finishes 30 km of a voyage at 6km/hr and the staying 40km of the venture in 5 hr.His normal pace for the entire voyage is:
Answer : B
Explanation
Total distance = (30+40)km= 70 km Total time taken = (30/6 +5) hrs =10 hrs Average speed = 70/10 km/hr = 7 km/hr
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 - Two train approach one another at 30 km/hr and 27 km/hr from two spot 342 km separated. After how long will they meet?
Answer : B
Explanation
Suppose the two trains meet after x hours. Then, 30x+27 x= 342 ⇒ 57 x = 342 ⇒ x = 342/57 = 6. So the two trains will meet after 6 hours.
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 taken 2 hr. more than B to walk d km. If A twofold his rate then he can make it in 1 hours less than B. How much time does B require for strolling d km?
Answer : C
Explanation
Suppose B takes x hours to walk d km. Then, A takes (x+2) hours to walk d km. With double of the speed, A will take 1/2 (x+2) hours. ∴ x- 1/2 (x+2) = 1 ⇒2x- (x+2) = 2 ⇒ x = 4 Hence B takes 4 hours to walk d km.
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 - X and y are two stations 500 km separated. A train begins from X and moves towards Y at 20 km/hr. Another train begins from Y in the meantime and moves towards X at 30 km/hr .How a long way from X will they cross one another?
Answer : B
Explanation
Suppose they meet x km from X. then, X/20 = (500-x)/30 = 30 x = 10000- 20 x ⇒50x = 10000 ⇒x = 200 So, they meet 200 km from X.
Q 9 - A auto finishes a sure excursion in 8 hours. It covers a large portion of the separation at 40 km/hr and the rest at 60 km/hr. The length of the adventure is:
Answer : C
Explanation
Let the total journey be x km. then, (x/2 * 1/40)+ (x/2* 1/60) = 8 ⇒ x/80 + x/120 = 8 ⇒3x+ 2x = 1920 ⇒5x= 1920 ⇒ x= 384 ∴ Total journey = 384 km
Q 10 - If the velocity of a railroad train is expanded by 5 km/hr from its ordinary pace, then it would have taken 2 hours less for a journey of 300 km. What is its ordinary velocity?
Answer : B
Explanation
Let the normal speed be x km/hr. Then 300/x - 300/(x+5) = 2 ⇒1/x - 1/(x+5) = 1/150 ⇒(x+5)-x/ x(x+5) = 1/150 ⇒ x2+5x - 750 = 0 ⇒x2+30x- 25x -750 = 0 ⇒x (x+30)-25(x+30) =0 ⇒(x+30) (x-25) = 0 ⇒ x= 25 ∴ Normal speed = 25 km/hr