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

Question : 6 [SSC CGL Tier-I 2016]

Gautam goes to office at a speed of 12 kmph and returns home at 10 kmph. His average speed is :

a) 10.9 kmph

b) 12.5 kmph

c) 11 kmph

d) 22 kmph

Answer: (a)

Here distances are same.

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

= $({2 × 12 × 10}/{12 + 10})$ kmph

= $(240/22)$ kmph = 10.9 kmph

Question : 7 [SSC MTS 2013]

With an average speed of 40 km/ hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The total journey is

a) 70 km

b) 80 km

c) 30 km

d) 40 km

Answer: (a)

Let the length of journey be x km, then

$x/35 - x/40 = 15/60 = 1/4$

${8x - 7x}/280 = 1/4$

$x = 280/4 = 70$ km

Question : 8 [SSC Constable (GD) 2013]

A train travelled at a speed of 35 km/hr for the first 10 minutes and at a speed of 20 km/hr for the next 5 minutes. The average speed of the train for the total 15 minutes is

a) 31 km/hr

b) 29 km/hr

c) 30 km/hr

d) 23 km/hr

Answer: (c)

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
= $\text"total travelled distance"/\text"total time taken in travelling distance"$
= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Distance covered

= $(35 × 10/60 + 20 × 5/60)$ km

= $(35/6 + 10/6) = 45/6$ km

Total time = 15 minutes

= $1/4$ hour

Required average speed

= $\text"Distance covered"/ \text"Time taken"$

= $45/6 × 4$ = 30 kmph

Question : 9 [SSC CPO S.I.2007]

A man goes from A to B at a uniform speed of 12 kmph and returns with a uniform speed of 4 kmph His average speed (in kmph) for the whole journey is :

a) 6

b) 4.5

c) 8

d) 7.5

Answer: (a)

Using Rule 5,

If two equal distances are covered at two unequal speed of x kmph and y kmph,

then average speed = $({2xy}/{x + y})$

= ${2 × 12 × 4}/{12 + 4} = 96/16$ = 6 kmph

Question : 10 [SSC CAPFs (CPO) SI 2016]

A car completed a journey of 400 km in 12$1/2$ hrs. The first $3/4$th of the journey was done at 30 km/hr. Calculate the speed for the rest of the journey.

a) 40 km/hr

b) 30 km/hr

c) 45 km/hr

d) 25 km/hr

Answer: (a)

Total distance covered = 400 km.

Total time = $25/2$ hours

$3/4$th of total journey

= $3/4$ × 400 = 300 km.

Time taken = $\text"Distance"/ \text"Speed"$

= $300/30$ = 10 hours

Remaining time = $25/2$ –10

= ${25 - 20}/2 = 5/2$ hours

Remaining distance = 100 km.

∴ Required speed of car

= $100/{5/2} = {100 × 2}/5$ = 40 kmph.

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