Speed & Distance Aptitude Questions

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:

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 - 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?

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 - By strolling at 3/4 of his standard speed, a man achieves his office 20 min. later than Normal. His standard time is:

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 train secured a separation at a uniform velocity. On the off chance that the train had been 6 km/hr speedier, it would have taken 4 hours not exactly the booked time, and if the train were slower by 6 km/hr, the train would have taken 6 hours more than the planned time. The length of the trip is?

Let the required distance be x km and uniform speed by y km/hr
x/y - x/(y+6) = 4 ...(a) x/(y-6) - x/y = 6 ...(b)
⇒xy+6x-xy = 4y (y+6) and xy - xy +6x = 6y (y-6)
⇒4y2+24 y - 6x = 0 and 6y2- 36y - 6x = 0
⇒2y2- 60 y = 0
⇒ 2y (y-30) = 0 ⇒y = 30
∴x/30 - x/36 = 4 ⇒6x- 5x = 720 ⇒x = 720 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?

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?

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?

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?

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 man on visit ventures initial 160 km at 64 km/hr and the following 160 km at 80 km/hr. The normal rate for the entire excursion is:

Average speed = 2xy/(x+y) km/hr = (2*64*80)/ (64+80) km/hr
= (2*64*80)/144 km/hr = 640/9 km/hr = 71.11 km/hr

Q 10 - A train voyaged separations of 10 km, 20 km and 30 km at rates of 50 km/hr, 60 km/hr and 90 km/hr separately. The normal rate of the train was

Total distance covered = (10+20+30) km = 60 km
Total time taken = (10/50 + 20/60+ 30/90) hr = (1/5+1/3+1/3) hr = 13/15 hr.
Average speed = (60* 15/13) km/hr = (900/13) km/hr = 69.23 km/hr