Model 3 Problems on average speed Practice Questions Answers Test with Solutions & More Shortcuts

Question : 31 [SSC CGL Tier-I (CBE) 2016]

A man covers a total distance of 100 km on bicycle. For the first 2 hours, the speed was 20 km/ hr and for the rest of the journey, it came down to 10 km/hr. The average speed will be

a) 15$1/8$ km/hr

b) 20 km/hr

c) 12$1/2$ km/hr

d) 13 km/hr

Answer: (c)

Distance covered in first 2 hours

= 2 × 20 = 40 km.

Remaining distance

= 100 - 40 = 60 km.

Time taken in covering 60 km at 10 kmph

= $60/10$ = 6 hours

∴ Required average speed

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

= $(100/{2+6})$ kmph = $(100/8)$ kmph

= $25/2$ kmph = 12$1/2$ kmph

Question : 32 [SSC CGL Prelim 1999]

A boy rides his bicycle 10km at an average speed of 12 km/hr and again travels 12 km at an average speed of 10 km/hr. His average speed for the entire trip is approximately :

a) 11.0 km/hr

b) 12.2 km/hr

c) 10.4 km/hr

d) 10.8 km/hr

Answer: (d)

Total distance = 10 + 12 = 22 km

Total time = $10/12 + 12/10 = 244/120$ hours

Required average speed

= $\text"Total distance"/ \text"Total time"$

= $22/{244/120} = 22/244 × 120$ = 10.8 km/hr.

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 = $({d_1 + d_2 + d_3 + ...})/{d_1/S_1 + d_2/S_2 + d_3/S_3 + ...}$

Here, $d_1 = 10, S_1 = 12, d_2 = 12, S_2$ = 10

Average Speed = ${d_1 + d_2}/{d_1/S_1 + d_2/S_2}$

= ${10 + 12}/{10/12 + 12/10} = {22 × 120}/244$ = 10.8 km/hrs.

Question : 33 [SSC CGL Prelim 2007]

A man covers half of his journey at 6km/hr and the remaining half at 3km/hr. His average speed is

a) 4 km/hr

b) 3 km/hr

c) 9 km/hr

d) 4.5 km/hr

Answer: (a)

Using Rule 5,
If a bus travels from A to B with the speed x km/h and returns from B to A with the speed y km/h,then the average speed will be $({2xy}/{x + y})$

If the same distance are covered at different speed of x kmph and y kmph,

the average speed of the whole journey is given by = $({2xy}/{x + y})$ kmph.

Required average speed

= ${2 × 6 × 3}/{6 + 3} = 36/9$ = 4 kmph

Question : 34 [SSC SAS 2010]

A person went from A to B at an average speed of x km/hr and returned from B to A at an average speed of y km/hr. What was his average speed during the total journey ?

a) $2/{x + y}$

b) $1/x + 1/y$

c) ${x + y}/{2xy}$

d) ${2xy}/{x + y}$

Answer: (d)

Using Rule 5,

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

[Since, can be given as corollary If the distance between A and B be z units, then

Average speed = $\text"Totaldistance"/ \text"Time taken"$

= ${z + z}/{z/x + z/y}$

= $2/{1/x + 1/y} = 2/{{x + y}/{xy}} = {2xy}/{x + y}$

Question : 35 [SSC CHSL 2012]

A man goes from Mysore to Bangalore at a uniform speed of 40 km/hr and comes back to Mysore at a uniform speed of 60 km/hr. His average speed for the whole journey is

a) 54 km/hr

b) 55 km/hr

c) 48 km/hr

d) 50 km/hr

Answer: (c)

Using Rule 5,

Average speed

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

= $({2 × 40 × 60}/{40 + 60})$ kmph

= 48 kmph

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