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 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

The correct answers to the above question in:

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

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

Question : 1

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: (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 : 2

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: (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 : 3

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: (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

Question : 4

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: (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 : 5

A car completed a journey of 400 km in 12$1/2$ hrs. The first $3/4$th of the journey was done at 30 km/hr. Calculate the speed for the rest of the journey.

a) 40 km/hr

b) 30 km/hr

c) 45 km/hr

d) 25 km/hr

Answer: (a)

Total distance covered = 400 km.

Total time = $25/2$ hours

$3/4$th of total journey

= $3/4$ × 400 = 300 km.

Time taken = $\text"Distance"/ \text"Speed"$

= $300/30$ = 10 hours

Remaining time = $25/2$ –10

= ${25 - 20}/2 = 5/2$ hours

Remaining distance = 100 km.

∴ Required speed of car

= $100/{5/2} = {100 × 2}/5$ = 40 kmph.

Question : 6

A man goes from A to B at a uniform speed of 12 kmph and returns with a uniform speed of 4 kmph His average speed (in kmph) for the whole journey is :

a) 6

b) 4.5

c) 8

d) 7.5

Answer: (a)

Using Rule 5,

If two equal distances are covered at two unequal speed of x kmph and y kmph,

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

= ${2 × 12 × 4}/{12 + 4} = 96/16$ = 6 kmph

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