Title: Tips on cracking Questions Based on Speed & Distance
13 TIPS on cracking Aptitude Questions on
Distance, Speed and Time
2Tip 1 Use the formula Speed Distance/ time
while ensuring that units are same
Question In a flight of 600 km, an aircraft was
slowed down due to bad weather. Its average speed
for the trip was reduced by 200 km/hr and the
time of flight increased by 30 minutes. Calculate
the duration of the flight. Solution Let the
average duration of the flight be t hours.
Distance 600 km. Average Speed (km/ hr) 600/
t Average Speed 200 600 / (t 0.5) (600/
t) 200 600 / (t 0.5) (600 200 t) (t
0.5) 600 t gt 2t2 t - 3 0 gtt 1 hr
Duration of the flight 1.5 hrs Question
Robert is travelling on his cycle and has
calculated to reach point A at 2 P.M. if he
travels at 10 km/hr. But he will reach there at
12 noon if he travels at 15 km/hr. At what speed
must he travel to reach A at 1 P.M.? Solution
According to the question, (D / 10) (D / 15)
2 gtD 60 km. Time taken to reach A at 10
km/hr 60 / 10 6 hours. (Starting time is 8
AM) Therefore, we need to find the speed to cover
60 km should be in 5 hours. Hence, speed 60 / 5
12 km/hr.
3Tip 2 If a person travels same distances with
different speeds, then the average speed is not
the arithmetic mean but the harmonic mean.
If a person covers a distance d first at x km/hr
and then covers the same distance d at y km/hr,
then the average speed is Total distance
travelled/ Total time taken 2d/ (d/x d/y)
2d/ (yd xd)/xy 2dxy/d(xy) 2xy/ xy
(Harmonic mean of x and y) Question A travels
25km at 50 km/hr and then 25km again with
70km/hr. What is As average speed during the
whole journey? Solution Average speed for the
whole journey (2x50x70) / (5070)
58.3km/hr
4Tip 3 If A runs x times faster than B, As
speed is actually 1x the speed of B.
- Question A runs 1? times as fast as B. If A
gives B a start of 80 m, how far must the winning
post be so that A and B might reach it at the
same time? - Solution
- Speed of A, Sa 5/3 x Sb
- Let the distance of the course be d meters
- Time taken by A to cover distance d Time
taken by B to cover distance d-80 - d/5/3 x Sb (d-80)/Sb
- 3d 5d 400
- 2d 640 gt d 200m
- Question A runs 1? times faster than B. If A
gives B a start of 80 m, how far must the winning
post be so that A and B might reach it at the
same time? - Solution
- Speed of A, Sa (1 5/3) x Sb 8/3 x Sb
- Let the distance of the course be d meters
- Time taken by A to cover distance d Time
taken by B to cover distance d-80 - d/8/3 x Sb (d-80)/Sb
- 3d 8d 640
- 5d 640 gt d 128m
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Questions on Distance, Speed Time.