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 : 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

The correct answers to the above question in:

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.

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

Question : 1

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 : 2

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.

Answer: (c)

Distance = Speed × Time

= (80 × 7) km. = 560 km.

Question : 3

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 : 4

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

Question : 5

The speed of a train going from Nagpur to Allahabad is 100 kmph while its speed is 150 kmph when coming back from Allahabad to Nagpur. Then the average speed during the whole journey is :

a) 140 kmph

b) 135 kmph

c) 120 kmph

d) 125 kmph

Answer: (c)

Using Rule 5,

Here, the distances are equal.

Average speed

= $({2 × 100 × 150}/{100 + 150})$ kmph

= ${2 × 100 × 150}/250$ = 120 kmph

Question : 6

P travels for 6 hours at the rate of 5 km/ hour and for 3 hours at the rate of 6 km/ hour. The average speed of the journey in km/ hour is

a) 1$2/9$

b) 2$2/5$

c) 3$1/5$

d) 5$1/3$

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

= 5 × 6 + 3 × 6

= 30 + 18 = 48 km

Total time = 9 hours

Average speed

= $48/9 = 16/3 = 5{1}/3$ kmph

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