Model 3 Problems on average speed Practice Questions Answers Test with Solutions & More Shortcuts

Question : 1 [SSC CGL Prelim 2007]

A man completes 30 km of a journey at the speed of 6 km/hr and the remaining 40 km of the journey in 5 hours. His average speed for the whole journey is

a) 8 km/hr

b) 7.5 km/hr

c) 7 km/hr

d) 6$4/11$ km/hr

Answer: (c)

Time taken to cover 30km at 6 kmph

= $30/6$ = 5 hours

Time taken to cover 40 km = 5 hours

Average speed

= $\text"Total distance"/ \text"Total time"$

= ${30 + 40}/10 = 70/10 = 7$ kmph

Question : 2 [SSC CHSL 2010]

A man covers the journey from a station A to station B at a uniform speed of 36 km/hr and returns to A with a uniform speed of 45 km/hr. His average speed for the whole journey is :

a) 41 km/hr

b) 42 km/hr

c) 40 km/hr

d) 40.5 km/hr

Answer: (c)

Using Rule 5,

Here same distances are covered at different speeds.

Average speed

= $({2xy}/{x + y})$ kmph

= $[{2 × 36 × 45}/({36 + 45})]$ kmph

= ${2 × 36 × 45}/81$ = 40 kmph

Question : 3 [SSC CGL Tier-II 2016]

A car travels from A to B at the rate of 40 km/h and returns from B to A at the rate of 60 km/ h. Its average speed during the whole journey is

a) 45 km/h

b) 60 km/h

c) 48 km/h

d) 50 km/h

Answer: (c)

Here, distance is same.

Average speed = ${2xy}/{x + y}$

= $({2 × 40 × 60}/{40 + 60})$ kmph.

= $({2 × 40 × 60}/100)$ kmph.

= 48 kmph.

Question : 4 [SSC CPO 2016]

When Alisha goes by car at 50 kmph, she reaches her office 5 minutes late. But when she takes her motorbike, she reaches 3 minutes early. If her office is 25 kms away, what is the approximate average speed at which she rides her motorbike ?

a) 58 kmph

b) 52 kmph

c) 68 kmph

d) 62 kmph

Answer: (c)

Difference of time

= 5 + 3 = 8 minutes

= $8/60$ hour = $2/15$ hour

If the speed of motorbike be x kmph, then

$25/50 - 25/x = 2/15$

$25/x = 1/2 - 2/15$

$25/x = {15 - 4}/30 = 11/30$

11x = 25 × 30

$x = {25 × 30}/11 = 750/11$

= 68.18 kmph ≈ 68 kmph

Question : 5 [SSC CGL Tier-I 2015]

A train travels 500 m in first minute. In the next 4 minutes, it travels in each minute 125 m more than that in the previous minute. The average speed per hour of the train during those 5 minutes will be

a) 50 km/hr

b) 55 km/hr

c) 30 km/hr

d) 45 km/hr

Answer: (d)

Using Rule 2,
If a man travels different distances $d_1,d_2,d_3$,... and so on in different time $t_1,t_2,t_3$ respectively then,Average speed
= $\text"total travelled distance"/\text"total time taken in travelling distance"$
= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Total distance covered by train in 5 minutes

= (500 + 625 + 750 + 875 + 1000) metre

= 3750 metre = 3.75 km.

Time = 5 minutes

= $5/60$ hour = $1/12$ hour

Speed of train = Distance Time

= $(3.75/{1/12})$ kmph

= (3.75 × 12) kmph = 45 kmph

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