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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.

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?

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 - The proportion between the paces of two trains is 7:8. On the off chance that the second prepare keeps running in 5 hours 400 km, the pace of the first prepare is :

Let the speed of first train be 7x km/hr. Then the speed of the second train is 8x km/hr. But speed of the second train=400/5km/hr=80 km/hr ∴8x=80⇒x=10. Hence the speed of first train is (7*10) km/hr=70 km/hr.

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:

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 ⇒ x^{2}+250 x-75000=0 ⇒ x^{2}+ 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 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?

Required distance = sum of distance covered by A and B = {(80*12/60) + (65*12/60)} = (16+13) = 29 km

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?

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 - A man strolling at 3 km/hr crosses a square field corner to corner in 2 minutes. The zone of the field is:

Speed =(3*5/18)m/sec. = 5/6 m/sec Distance covered in 2 min. = (5/6* 2* 60) m = 100 m Length of the diagonal of the square field = 100 m Area = 1/2 * (diagonal)^{2}= (1/2 *100 *100 )m^{2}= 5000 m^{2}= 5000/100 ares = 50 ares {1 are= 100 m^{2}}

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

Let the original speed be x km/hr then, 840/x-840/(x+10) = 2 ⇒ 840(x+10)-840 x = 2x(x+10) ⇒2x^{2}+20 x-8400 = 0 ⇒x^{2}+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:

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:

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 - 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?

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 ⇒ x^{2}+5x - 750 = 0 ⇒x^{2}+30x- 25x -750 = 0 ⇒x (x+30)-25(x+30) =0 ⇒(x+30) (x-25) = 0 ⇒ x= 25 ∴ Normal speed = 25 km/hr

aptitude_speed_distance.htm

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