Model 3 Problems on average speed Practice Questions Answers Test with Solutions & More Shortcuts
time & distance PRACTICE TEST [5 - 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
Question : 16 [SSC CAPFs SI 2013]
On a journey across Kolkata, a taxi averages 50 km per hour for 50% of the distance, 40 km per hour for 40% of it and 20 km per hour for the remaining. The average speed (in km/hour) for the whole journey is :
a) 35
b) 45
c) 42
d) 40
Answer »Answer: (d)
Using Rule 2,
Total distance = 100 km.
Total time = $50/50 + 40/40 + 10/20$
= $1 + 1 + 1/2 = 5/2$ hours
Average speed = ${100 × 2}/5$ = 40 kmph
Question : 17 [SSC CHSL 2012]
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 »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
Question : 18 [SSC CHSL 2012]
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 »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 : 19 [SSC CGL Prelim 2008]
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
Question : 20 [SSC CGL Tier-II 2015]
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.
IMPORTANT quantitative aptitude EXERCISES
Model 3 Problems on average speed Shortcuts »
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Click to Start..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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