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 : 11 [SSC Investigator 2010]
A man goes from a place A to B at a speed of 12 km/hr and returns from B to A at a speed of 18 km/hr. The average speed for the whole journey is
a) 15$1/2$ km/hr
b) 16 km/hr
c) 14$2/5$ km/hr
d) 15 km/hr
Answer »Answer: (c)
Using Rule 5,
Average speed
= $({2xy}/{x + y})$ kmph
= $({2 × 12 × 18}/{12 + 18})$ kmph
= $({2 × 12 × 18}/30)$ kmph
= 14$2/5$ kmph
Question : 12 [SSC SO 2005]
A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train:
a) 48 kmph
b) 30 kmph
c) 37.5 kmph
d) 36 kmph
Answer »Answer: (d)
Using Rule 3,
Average speed
= $\text"Total distance"/\text"time taken"$
= ${30 × 12/60 + 45 × 8/60}/{12/60 + 8/60}$
= 12 × 3 = 36 kmph
Question : 13 [SSC CGL Tier-I 2014]
A and B are 20 km apart. A can walk at an average speed of 4 km/ hour and B at 6 km/hr. If they start walking towards each other at 7 a.m., when they will meet ?
a) 9.00 a.m.
b) 10.00 a.m.
c) 8.00 a.m.
d) 8.30 a.m.
Answer »Answer: (a)
Using Rule 11,Time taken by 1st man to reach B after meeting 2nd man at C is '$t_1$' and time taken by 2nd man to reach A after meeting 1st man at C is '$t_2$' then:${\text"Speed of 1st man"(s_1)}/{\text"Speed of 2nd man"(s_2)} = √{t_2/t_1}$Distance from A to B = $s_1t_1 + S_2t_2$
If A and B meet after t hours, then
4 t + 6 t = 20
10 t = 20 ⇒ t = $20/10$ = 2 hours.
Hence, both will meet at 9 a.m.
Question : 14 [SSC CGL Tier-II 2016]
A bus travels 150 km in 3 hours and then travels next 2 hours at 60 km/hr. Then the average speed of the bus will be
a) 50 km/hr.
b) 60 km/hr.
c) 55 km/hr.
d) 54 km/hr.
Answer »Answer: (d)
Total distance covered by the bus
= 150 km. + 2 × 60 km.
= (150 + 120) km.
= 270 km.
∴ Average speed = $\text"Total distance"/ \text"Time taken"$
= $270/5$ = 54 kmph.
Question : 15 [SSC MTS 2013]
A bus covers four successive 3 km stretches at speed of 10 km/ hr, 20 km/hr, 30 km/hr and 60 km/hr respectively. Its average speed over this distance is
a) 20 km/hr
b) 10 km/hr
c) 30 km/hr
d) 25 km/hr
Answer »Answer: (a)
Using Rule 3,If a man travels different distances $d_1,d_2,d_3$, and so on with different speeds $s_1,s_2,s_3$, respectively then,Average speed = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$
Average speed
= $\text"Total distance"/ \text"Time taken"$
= $12/{3/10 + 3/20 + 3/30 + 3/60}$
= $12/{3({6 + 3 + 2 + 1}/60)}$
= ${12 × 60}/{3 × 12} = 20$ kmph
IMPORTANT quantitative aptitude EXERCISES
Model 3 Problems on average speed Shortcuts »
Click to Read...Model 3 Problems on average speed Online Quiz
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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