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Quantitative Aptitude – Practice – Time Speed and Distance

 

average speed poster

 

PSA_LEVEL-UP_QNR29. Rahul travels a distance 10 km on his bike with average speed of 6km/h. He further covers a distance of 20 km and this time with an average speed of 15 km/h. What is his bike’s average speed during the entire journey?

a) 12 km/h
b) 10 km/h
c) 13 km/h
d) 10.5 km/h
e) None of these

PSA_LEVEL-UP_QNR30. Sumeet goes half of his distance with average speed 20 km/h and the rest of the distance with speed 30 km/h. What is the average speed during the entire journey?

a) 22 km/h
b) 24 km/h
c) 26 km/h
d) 25 km/h
e) None of these

Ans:

PSA_LEVEL-UP_QNR29)

Average speed = Total distance covered/Total time taken

Time taken to cover 10 km= 10/6 hr
Time taken to cover 20 km= 20/15 hr

Average speed= (10+20)/[10/6 + 20/15]

= 30*30/[10*5+ 20*2]
= 900/90
=10 km/h. Option b.

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PSA_LEVEL-UP_QNR30)

Average speed = Total distance covered/Total time taken

Since, the distance given is same, we can use the below expression to find out the total average speed.

S= 2*A*B /[ A + B]

Here, A= 20km/h, B = 30 km/h

Average speed= 2*20*30/[20+30]

= 1200/50

= 24 km/h. Option b.

P.S. In both the cases you must notice that average speed is not (speed1+speed2)/2. In first case average speed comes out to be 10 km/h which is not same as (6+15)/2 or 10.5. Similarly in the second case average speed is 24 km/h and not (20+30)/2 or 25. However, both the wrong answers are present in options.

(Speed1+Speed2)/2 is applicable only when the contributing journeys are covered in equal times. For example, If Sumeet covers some ‘x’ distance in 1 hour at speed ‘s1’ and some ‘y’ distance in 1 hour at speed ‘s2’,

average speed= total distance/time taken

= (s1*1 + s2*1)/[1+1] = (s1+s2)/2.

** For more quantitative aptitude questions visit link1, link2:

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