Model 3 Problems on average speed Section-Wise Topic Notes With Detailed Explanation And Example Questions

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The following question based on time & distance topic of quantitative aptitude

Questions : At an average of 80 km/hr Shatabdi Express reaches Ranchi from Kolkata in 7 hrs. The distance between Kolkata and Ranchi is

(a) 560 m.

(b) 650 m.

(c) 560 km.

(d) 506 km.

The correct answers to the above question in:

Answer: (c)

Distance = Speed × Time

= (80 × 7) km. = 560 km.

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Read more problems on average speed Based Quantitative Aptitude Questions and Answers

Question : 1

A train runs at an average speed of 75 km/hr. If the distance to be covered is 1050 kms, how long will the train take to cover it ?

a) 15 hrs

b) 14 hrs

c) 13 hrs

d) 12 hrs

Answer: (b)

Using Rule 1,
Distance = Speed × Time
Speed = $\text"Distance"/\text"Time"$ , Time = $\text"Distance"/\text"Speed"$
1 m/s = $18/5$ km/h, 1 km/h = $5/18$ m/s

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

= $1050/75$ = 14 hours

Question : 2

A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/ hr. His average speed for the whole journey in km/hr is

a) 35

b) 40

c) 37.5

d) 37

Answer: (c)

Using Rule 5,

Average speed of whole journey

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

= ${2 × 50 × 30}/{50 + 30}$

= ${2 × 50 × 30}/80$ = 37.5 kmph

Question : 3

Durga walks 5 km from her home to school in 60 minutes, then bicycles back to home along the same route at 15 km per hour. Her sister Smriti makes the same round trip, but does so at half of Durga's average speed. How much time does Smriti spend on her round trip ?

a) 160 minutes

b) 80 minutes

c) 120 minutes

d) 40 minutes

Answer: (a)

Durga's average speed

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

= $({2 × 5 × 15}/{5 + 15})$ kmph.

= $({2 × 5 × 15}/20)$ kmph = $15/2$ kmph

Distance of School = 5 km.

Smriti's speed = $15/4$ kmph

∴ Required time = $2(5/{15/4})$ hours

= $({2 × 5 × 4}/15) = 8/3$ hours

= $(8/3 × 60)$ minutes = 160 minutes

Question : 4

One third of a certain journey is covered at the rate of 25 km/ hour, one-fourth at the rate of 30 km/hour and the rest at 50 km/ hour. The average speed for the whole journey is

a) 30 km/hour

b) 37$1/12$ km/hour

c) 35 km/hour

d) 33$1/3$ km/hour

Answer: (d)

Let the total distance be x km.

Total time = ${x/3}/25 + {x/4}/30 +{{5x}/12}/50$

= $x/75 + x/120 + x/120$

= $x/75 + x/60 = {4x + 5x}/300 = {3x}/100$ hours

Average speed

= $\text"Total distance"/\text"Time taken"$

= $x/{{3x}/100} = 100/3 = 33{1}/3$ kmph

Using Rule 18,If a man covers $1/x$ part of Journey at u km/h,$1/y$ part at v km/h and $1/z$ part at w km/hr and so on, then his average speed for the whole journey will be $1/{1/{xu} + 1/{yv} + 1/{zw} +...}$

Here, x = 3, u = 25

y = 4, v = 30

z = $12/5$ , w = 50

Average Speed= $1/{1/{xu} + 1/{yv} + 1/{zw}}$

= $1/{1/{3 × 25} + 1/{4 × 30} + 1/{12/5 × 50}}$

= $1/{1/75 + 1/120 + 1/120}$

= $1/{1/75 + 1/60} = 1/{{4 + 5}/300}$

= $300/9 = 100/3$

= 33$1/3$ km/hr.

Question : 5

A man walks from his house at an average speed of 5 km per hour and reaches his office 6 minutes late. If he walks at an average speed of 6 km/h he reaches 2 minutes early. The distance of the office from his house is

a) 12 km

b) 4 km

c) 6 km

d) 9 km

Answer: (b)

Required distance of office from house = x km. (let)

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

According to the question,

$x/5 - x/6 = {6 + 2}/60 = 2/15$

${6x - 5x}/30 = 2/15$

$x/30 = 2/15$

$x = 2/15$ × 30 = 4 km.

Using Rule 10,
If a man travels at the speed of $s_1$, he reaches his destination $t_1$ late while he reaches $t_2$ before when he travels at $s_2$ speed, then the distance between the two places is D = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

Here, $S_1 = 5, t_1 = 6, S_2 = 6, t_2$ = 2

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${(5 × 6)(6 + 2)}/{6 - 5}$

= $30 × 8/60$ = 4 km.

Question : 6

A train covers a distance of 3584 km in 2 days 8 hours. If it covers 1440 km on the first day and 1608 km on the second day, by how much does the average speed of the train for the remaining part of the journey differ from that for the entire journey ?

a) 4 km/hour more

b) 5 km/hour less

c) 3 km/hour more

d) 3 km/hour less

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

Remaining distance

= (3584 - 1440 - 1608) km

= 536 km.

This distance is covered at the rate of $536/8$ = 67 kmph.

Average speed of whole journey

= $3584/56$ = 64 kmph

Required difference in speed

= (67 - 64) kmph i.e. = 3 kmph more

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