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 - In what time can Somali spread a separation of 400m, in the event that she keeps running at a rate of 20 km/hr?
Answer : A
Explanation
20 km/hr = (20*5/18) m/sec =50/9m/sec. Time taken to cover 400 m = (400*9/50) sec= 72 sec. = 6/5 min.
Q 2 - A auto covers a separation of 715 km at a steady speed. On the off chance that the pace of the auto would have been 10 km/hr all the more, then it would have taken 2 hours less to cover the same separation. What is the first speed of the auto?
Answer : C
Explanation
Let the constant speed be x km/hr. Then, 715/x-715/(x+10) =2⇒1/x-1/(x+10) =2/715 ⇒(x+10)-x/x(x+10) =2/715⇒x(x+10) =3575 ⇒x2+10x-3575=0⇒x2+65x-55x-3575=0 ⇒x(x+65)-55(x+65)=0 ⇒(x+65)(x-55)=0 ⇒x=55. ∴Original speed of the car is 55km/hr.
Q 3 - A auto going with 5/7 of its typical rate covers 42 km in 1 hr. 40 min. 48sec.What is the typical pace of the auto?
Answer : D
Explanation
Let the usual speed be x km/hr. 42/(5x/7)= 126/75 ⇒ 42*7/5x = 42/25 ⇒5x = (25*7) ⇒x = (5*7) = 35
Q 4 - A and B begin all the while from a sure point in North and South bearings on engine cycles. The velocity of A is 80 km/hr and that Of B is 65 km/hr. What is the separation in the middle of A and B following 12 minutes?
Answer : B
Explanation
Required distance = sum of distance covered by A and B
= {(80*12/60) + (65*12/60)} = (16+13) = 29 km
Q 5 - Sunil spreads a separation by strolling for 6 hours .While giving back his pace, diminishes by 1 km/hr and he takes 9 hr. to cover the same distance. What was his velocity consequently traveled?
Answer : A
Explanation
Let the speed in return journey be x km/hr. then 6(x+1) = 9x ⇒3x= 6 ⇒ x= 2 Hence, the speed in return journey is 2 km/hr
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 bullock truck needs to cover a separation of 80 km in 10 hours. On the off chance that it covers half of the excursion in 3/5 th of the time, what ought to be its velocity to cover the remaining separation in the time left?
Answer : C
Explanation
Distance left = (1/2 *80) km = 40 km
Time left = {(1-3/5)*10} hrs = (2/5*10)= 4hrs.
Speed required = 40/4 km/hr = 10 km/hr
Q 8 - A train leaves Meerut at 6 am and achieves Delhi at 10 am. Another train leaves Delhi at 8am and ranges Meerut at 11.30 am. At what time do the two trains cross one another?
Answer : D
Explanation
Let the distance between Meerut and Delhi be x km. Average speed of train from Meerut = x/4 km/hr Suppose they meet y hrs. After 6 am. Then, (X/4*y)+2x/7 * (y-2) = x ⇒ y/4+ 2(y-2)/7 = 1 ⇒7y+8(y-2) = 28 ⇒15 y= 44 ⇒ y = 44/15 hrs = 2 hrs. 56 min. So, the trains meet at 8.56 am
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
