Model 3 Problems on average speed Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 5 EXERCISES
The following question based on time & distance topic of quantitative aptitude
(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 »Answer: (b)
Using Rule 1,Distance = Speed × TimeSpeed = $\text"Distance"/\text"Time"$ , 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 »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 »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 »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 »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 »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
GET time & distance PRACTICE TEST EXERCISES
Model 1 Basic Time & Distance using formula
Model 2 Vehicles in x/y of its usual speed
Model 3 Problems on average speed
Model 4 Time & Distance with Ratios
Model 5 Problems with Races
time & distance Shortcuts and Techniques with Examples
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Model 1 Basic Time & Distance using formula
Defination & Shortcuts … -
Model 2 Vehicles in x/y of its usual speed
Defination & Shortcuts … -
Model 3 Problems on average speed
Defination & Shortcuts … -
Model 4 Time & Distance with Ratios
Defination & Shortcuts … -
Model 5 Problems with Races
Defination & Shortcuts …
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