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 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

The correct answers to the above question in:

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.

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

Question : 1

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 : 2

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: (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 : 3

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 : 4

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

Question : 5

A train travelled at a speed of 35 km/hr for the first 10 minutes and at a speed of 20 km/hr for the next 5 minutes. The average speed of the train for the total 15 minutes is

a) 31 km/hr

b) 29 km/hr

c) 30 km/hr

d) 23 km/hr

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
= $\text"total travelled distance"/\text"total time taken in travelling distance"$
= ${d_1 + d_2 + d_3 +...}/{t_1 + t_2 + t_3 +...}$

Distance covered

= $(35 × 10/60 + 20 × 5/60)$ km

= $(35/6 + 10/6) = 45/6$ km

Total time = 15 minutes

= $1/4$ hour

Required average speed

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

= $45/6 × 4$ = 30 kmph

Question : 6

With an average speed of 40 km/ hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The total journey is

a) 70 km

b) 80 km

c) 30 km

d) 40 km

Answer: (a)

Let the length of journey be x km, then

$x/35 - x/40 = 15/60 = 1/4$

${8x - 7x}/280 = 1/4$

$x = 280/4 = 70$ km

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