Tip #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

=> 2t^{2} + t - 3 = 0 =>t = 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 =>D = 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.

Tip #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(x+y)]

= 2xy/ x+y (Harmonic mean of x and y)

Question: A travels 25km at 50 km/hr and then 25km again with 70km/hr. What is A’s average speed during the whole journey?

Solution:

Average speed for the whole journey = (2x50x70) / (50+70) = 58.3km/hr

Tip #3: If A runs x times faster than B, A’s speed is actually 1+x the speed of B

Question: A runs 1⅔ 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 => d = 200m

Question: A runs 1⅔ 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 => d = 128m

[Note: Here, A 5/3 times faster than B, i.e., A’s speed = B’s speed + 5/3 times B’s speed = 8/3 times B’s speed.]

Are you interested in getting Certificates to boost your Resume? Participate in our
Online Grammar and Aptitude Contests. It only takes 20 mins. All participants get
Participation Certificates while the top 100 winners get Amazon Cash Vouchers every week.
Participate NOW!

As a gesture of support, please follow us on Facebook and YouTube

Become a Campus Ambassador for LearningPundits - Promote our Weekly Online Contests to other students in your Campus via Posters, Facebook, WhatsApp, Email and face to face communication. You will receive a stipend based on your performance and an Internship Certificate to boost your Resume. Email your Resume to
support@learningpundits.com

Speed and distance

Aptitude Test on Speed And Distance: Tips and Aptitude Questions