Speed, Time & Distance

ICSE Class 6 Mathematics · Solutions for Class 6 RS Aggarwal ICSE Foundation Mathematics · 38 solved questions

Exercise
Marks
1Exercise 11(A)

A bus travels a distance of 186 km in 2 hours 35 minutes. What is the speed of the bus in km/hr?

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Given,

Distance covered = 186 km

Time taken = 2 hours 35 minutes

= $2\frac{35}{60} \text{ hrs } = 2\frac{7}{12} \text{ hrs } = \frac{31}{12}$ hrs

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{186 \text{ km}}{\frac{31}{12} \text{ hrs}}\\[1em] = 186 \times \frac{12}{31} \text{ km/hr }\\[1em] = 72 \text{ km/hr }\end{aligned}$$

Hence, the speed of the bus is 72 km/hr.

2Exercise 11(A)

A train covers 35 km in 25 minutes. What is the speed of the train in km/hr?

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Given,

Distance covered = 35 km

Time taken = 25 minutes

= $\frac{25}{60} \text{ hr } = \frac{5}{12}$ hr

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{35 \text{ km}}{\frac{5}{12} \text{ hr}}\\[1em] = 35 \times \frac{12}{5} \text{ km/hr }\\[1em] = 84 \text{ km/hr }\end{aligned}$$

Hence, the speed of the train is 84 km/hr.

3Exercise 11(A)

A man walks 8 km in 1 hour 20 minutes. What is the speed of the man in km/hr?

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Given,

Distance covered = 8 km

Time taken = 1 hour 20 minutes

= $1\frac{20}{60} \text{ hrs } = 1\frac{1}{3} \text{ hrs } = \frac{4}{3}$ hrs

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{8 \text{ km}}{\frac{4}{3} \text{ hrs}}\\[1em] = 8 \times \frac{3}{4} \text{ km/hr }\\[1em] = 6 \text{ km/hr }\end{aligned}$$

Hence, the speed of the man is 6 km/hr.

4Exercise 11(A)

A man cycles at 12 km per hour. How much distance does he cover in 35 minutes?

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Given,

Speed = 12 km/hr

Time taken = 35 minutes

= $\frac{35}{60} \text{ hr } = \frac{7}{12}$ hr

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 12 \times \frac{7}{12} \text{ km }\\[1em] = 7 \text{ km }\end{aligned}$$

Hence, the man covers 7 km in 35 minutes.

5Exercise 11(A)

A car is travelling at a speed of 72 km/hr. How far will it travel in 25 minutes?

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Given,

Speed = 72 km/hr

Time taken = 25 minutes

= $\frac{25}{60} \text{ hr } = \frac{5}{12}$ hr

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 72 \times \frac{5}{12} \text{ km }\\[1em] = 30 \text{ km }\end{aligned}$$

Hence, the car will travel 30 km in 25 minutes.

6Exercise 11(A)

A cyclist covers 1.5 km in 5 minutes. Find his speed in km/hr.

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Given,

Distance covered = 1.5 km

Time taken = 5 minutes

= $\frac{5}{60} \text{ hr } = \frac{1}{12}$ hr

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1.5 \text{ km}}{\frac{1}{12} \text{ hr}}\\[1em] = 1.5 \times 12 \text{ km/hr }\\[1em] = 18 \text{ km/hr }\end{aligned}$$

Hence, the speed of the cyclist is 18 km/hr.

7Exercise 11(A)

Sajjal is going on his scooty at a speed of 24 km/hr. How much time will he take to cover 1 km 600 m?

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Given,

Speed = 24 km/hr

Distance = 1 km 600 m

= $1\frac{600}{1000} \text{ km } = 1.6$ km

By formula,

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1.6 \text{ km}}{24 \text{ km/hr}}\\[1em] = \frac{1.6}{24} \text{ hr }\\[1em] = \frac{1}{15} \text{ hr }\\[1em] = \frac{1}{15} \times 60 \text{ min }\\[1em] = 4 \text{ min }\end{aligned}$$

Hence, Sajjal will take 4 minutes to cover 1 km 600 m.

8(i)Exercise 11(A)

Convert the following speed into m/sec:

36 km/hr

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To convert a speed of km/hr into m/sec, we multiply by $\frac{5}{18}$.

$$\begin{aligned}\Rightarrow 36 \text{ km/hr }= \left(36 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{180}{18} \text{ m/sec }\\[1em] = 10 \text{ m/sec }\end{aligned}$$

Hence, 36 km/hr = 10 m/sec.

8(ii)Exercise 11(A)

Convert the following speed into m/sec:

81 km/hr

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To convert a speed of km/hr into m/sec, we multiply by $\frac{5}{18}$.

$$\begin{aligned}\Rightarrow 81 \text{ km/hr }= \left(81 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{405}{18} \text{ m/sec }\\[1em] = \frac{45}{2} \text{ m/sec }\\[1em] = 22\frac{1}{2} \text{ m/sec }\end{aligned}$$

Hence, 81 km/hr = $\mathbf{22\frac{1}{2}}$ m/sec.

8(iii)Exercise 11(A)

Convert the following speed into m/sec:

9 km/hr

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To convert a speed of km/hr into m/sec, we multiply by $\frac{5}{18}$.

$$\begin{aligned}\Rightarrow 9 \text{ km/hr }= \left(9 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{45}{18} \text{ m/sec }\\[1em] = \frac{5}{2} \text{ m/sec }\\[1em] = 2\frac{1}{2} \text{ m/sec }\end{aligned}$$

Hence, 9 km/hr = $\mathbf{2\frac{1}{2}}$ m/sec.

9(i)Exercise 11(A)

Convert the following speed into km/hr:

25 m/sec

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To convert a speed of m/sec into km/hr, we multiply by $\frac{18}{5}$.

$$\begin{aligned}\Rightarrow 25 \text{ m/sec }= \left(25 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{450}{5} \text{ km/hr }\\[1em] = 90 \text{ km/hr }\end{aligned}$$

Hence, 25 m/sec = 90 km/hr.

9(ii)Exercise 11(A)

Convert the following speed into km/hr:

7.5 m/sec

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To convert a speed of m/sec into km/hr, we multiply by $\frac{18}{5}$.

$$\begin{aligned}\Rightarrow 7.5 \text{ m/sec }= \left(7.5 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{135}{5} \text{ km/hr }\\[1em] = 27 \text{ km/hr }\end{aligned}$$

Hence, 7.5 m/sec = 27 km/hr.

9(iii)Exercise 11(A)

Convert the following speed into km/hr:

1 m/sec

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To convert a speed of m/sec into km/hr, we multiply by $\frac{18}{5}$.

$$\begin{aligned}\Rightarrow 1 \text{ m/sec }= \left(1 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{18}{5} \text{ km/hr }\\[1em] = 3\frac{3}{5} \text{ km/hr }\end{aligned}$$

Hence, 1 m/sec = $\mathbf{3\frac{3}{5}}$ km/hr.

10Exercise 11(A)

A train covers 120 m in 5 seconds. What is the speed of the train in km/hr?

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Given,

Distance covered = 120 m

Time taken = 5 seconds

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{120 \text{ m}}{5 \text{ sec}}\\[1em] = 24 \text{ m/sec }\end{aligned}$$

Converting the speed into km/hr :

$$\begin{aligned}\Rightarrow 24 \text{ m/sec } = \left(24 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{432}{5} \text{ km/hr }\\[1em] = 86\frac{2}{5} \text{ km/hr }\end{aligned}$$

Hence, the speed of the train is $\mathbf{86\frac{2}{5}}$ km/hr.

11Exercise 11(A)

A bus is moving at a speed of 84 km/hr. How much distance will it cover in 1 min 45 sec.?

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Given,

Speed = 84 km/hr

Time taken = 1 min 45 sec = (60 + 45) sec = 105 sec

Converting the speed into m/sec :

$$\begin{aligned}84 \text{ km/hr } = \left(84 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{70}{3} \text{ m/sec }\end{aligned}$$

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = \frac{70}{3} \times 105 \text{ m }\\[1em] = 70 \times 35 \text{ m }\\[1em] = 2450 \text{ m }\\[1em] = 2 \text{ km } 450 \text{ m }\end{aligned}$$

Hence, the bus will cover 2 km 450 m.

12Exercise 11(A)

Preeti is going on her moped at a speed of 24 km/hr. How much time will she take to cover 720 metres?

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Given,

Speed = 24 km/hr

Distance = 720 m

Converting the speed into m/sec :

$$\begin{aligned}24 \text{ km/hr } = \left(24 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{20}{3} \text{ m/sec }\end{aligned}$$

By formula,

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{720 \text{ m}}{\frac{20}{3} \text{ m/sec}}\\[1em] = 720 \times \frac{3}{20} \text{ sec }\\[1em] = 108 \text{ sec }\\[1em] = 1 \text{ min } 48 \text{ sec }\end{aligned}$$

Hence, Preeti will take 108 sec = 1 min 48 sec to cover 720 metres.

13Exercise 11(A)

A car, running at 45 km/hr, takes 6 hours to complete a journey. At what speed must it travel to complete the journey in 5 hours?

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Given,

Speed = 45 km/hr

Time taken = 6 hours

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 45 \times 6 \text{ km }\\[1em] = 270 \text{ km }\end{aligned}$$

New time taken = 5 hours

$$\begin{aligned}\text{New speed} = \frac{\text{Distance}}{\text{Time}} = \frac{270 \text{ km}}{5 \text{ hrs}}\\[1em] = 54 \text{ km/hr }\end{aligned}$$

Hence, the car must travel at a speed of 54 km/hr to complete the journey in 5 hours.

14Exercise 11(A)

A car covers a certain distance in 40 minutes, if it runs at 54 km/hr. How much time will it take to cover the same distance at a speed 45 km/hr?

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Given,

Speed = 54 km/hr

Time taken = 40 minutes

= $\frac{40}{60} \text{ hr } = \frac{2}{3}$ hr

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 54 \times \frac{2}{3} \text{ km }\\[1em] = 36 \text{ km }\end{aligned}$$

New speed = 45 km/hr

$$\begin{aligned}\text{New time} = \frac{\text{Distance}}{\text{Speed}} = \frac{36 \text{ km}}{45 \text{ km/hr}}\\[1em] = \frac{4}{5} \text{ hr }\\[1em] = \frac{4}{5} \times 60 \text{ min }\\[1em] = 48 \text{ min }\end{aligned}$$

Hence, the car will take 48 minutes to cover the same distance at a speed of 45 km/hr.

11 Mark · Exercise 11(B) - Multiple Choice Questions

A speed of 15 m/s is the same as:

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To convert a speed of m/sec into km/hr, we multiply by $\frac{18}{5}$.

15 m/s = $\left(15 \times \frac{18}{5}\right)$ km/hr = 54 km/hr

Hence, option 3 is the correct option.

21 Mark · Exercise 11(B) - Multiple Choice Questions

A speed of 90 km/hr is the same as:

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To convert a speed of km/hr into m/s, we multiply by $\frac{5}{18}$.

90 km/hr = $\left(90 \times \frac{5}{18}\right)$ m/s = 25 m/s

Hence, option 1 is the correct option.

31 Mark · Exercise 11(B) - Multiple Choice Questions

A car covers a distance of 7 km in 5 minutes. Its speed is:

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Given,

Distance covered = 7 km

Time taken = 5 minutes

= $\frac{5}{60} \text{ hr } = \frac{1}{12}$ hr

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{7 \text{ km}}{\frac{1}{12} \text{ hr}}\\[1em] = 7 \times 12 \text{ km/hr }\\[1em] = 84 \text{ km/hr }\end{aligned}$$

Hence, option 3 is the correct option.

41 Mark · Exercise 11(B) - Multiple Choice Questions

A motorcyclist travels at 30 m/s. How much distance does he cover in 15 minutes?

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Given,

Speed = 30 m/s

Time taken = 15 min = (15 × 60) sec = 900 sec

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 30 \times 900 \text{ m }\\[1em] = 27000 \text{ m }\\[1em] = \frac{27000}{1000} \text{ km } = 27 \text{ km }\end{aligned}$$

Hence, option 2 is the correct option.

51 Mark · Exercise 11(B) - Multiple Choice Questions

A motorbike covers 84 km in 2 hrs 24 min. The speed of the bike is:

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Given,

Distance covered = 84 km

Time taken = 2 hrs 24 min

= $2\frac{24}{60} \text{ hrs } = 2\frac{2}{5} \text{ hrs } = \frac{12}{5}$ hrs

By formula,

$$\begin{aligned}\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{84 \text{ km}}{\frac{12}{5} \text{ hrs}}\\[1em] = 84 \times \frac{5}{12} \text{ km/hr }\\[1em] = 35 \text{ km/hr }\end{aligned}$$

Hence, option 4 is the correct option.

61 Mark · Exercise 11(B) - Multiple Choice Questions

A car runs at a speed of 90 km/hr. What distance does it cover in 50 minutes?

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Given,

Speed = 90 km/hr

Time taken = 50 minutes

= $\frac{50}{60} \text{ hr } = \frac{5}{6}$ hr

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 90 \times \frac{5}{6} \text{ km }\\[1em] = 75 \text{ km }\end{aligned}$$

Hence, option 2 is the correct option.

71 Mark · Exercise 11(B) - Multiple Choice Questions

A car is running at 20 m/s. How much distance will it cover in 2 hrs 20 min.?

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Given,

Speed = 20 m/s

Time taken = 2 hrs 20 min

= $2\frac{20}{60} \text{ hrs } = 2\frac{1}{3} \text{ hrs } = \frac{7}{3}$ hrs

Converting the speed into km/hr :

20 m/s = $\left(20 \times \frac{18}{5}\right)$ km/hr = 72 km/hr

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 72 \times \frac{7}{3} \text{ km }\\[1em] = 24 \times 7 \text{ km }\\[1em] = 168 \text{ km }\end{aligned}$$

Hence, option 3 is the correct option.

81 Mark · Exercise 11(B) - Multiple Choice Questions

Which of the following is the slowest speed?

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To compare the speeds, we convert each of them into km/hr.

10 m/s = $\left(10 \times \frac{18}{5}\right)$ km/hr = 36 km/hr

0.9 km/min = (0.9 × 60) km/hr = 54 km/hr

Since, 36 km/hr < 45 km/hr < 54 km/hr

⇒ 10 m/s is the slowest speed.

Hence, option 2 is the correct option.

91 Mark · Exercise 11(B) - Multiple Choice Questions

A car travels at 90 km/hr. How much distance does it cover in 8 seconds?

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Given,

Speed = 90 km/hr

Time taken = 8 seconds

Converting the speed into m/sec :

90 km/hr = $\left(90 \times \frac{5}{18}\right)$ m/s = 25 m/s

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 25 \times 8 \text{ m }\\[1em] = 200 \text{ m }\end{aligned}$$

Hence, option 2 is the correct option.

1Exercise 11(B) - Mental Maths

Fill in the blanks:

(i) 12 m/sec = ...... km/hr

(ii) 27 km/hr = ...... m/sec.

(iii) At a speed of 54 km/hr, a car will cover a distance of 600 m in ...... seconds.

(iv) At a speed of 45 km/hr, the distance covered in 18 seconds is ...... m.

(v) At a speed of 5 m/s, a distance of 27 km will be covered in ...... hr ...... min.

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(i) To convert a speed of m/sec into km/hr, we multiply by $\frac{18}{5}$.

$$\begin{aligned}\Rightarrow 12 \text{ m/sec }= \left(12 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{216}{5} \text{ km/hr }\\[1em] = 43\frac{1}{5} \text{ km/hr } = 43.2 \text{ km/hr }\end{aligned}$$

Hence, 12 m/sec = 43.2 km/hr.

(ii) To convert a speed of km/hr into m/sec, we multiply by $\frac{5}{18}$.

$$\begin{aligned}\Rightarrow 27 \text{ km/hr }= \left(27 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{135}{18} \text{ m/sec }\\[1em] = \frac{15}{2} \text{ m/sec } = 7.5 \text{ m/sec }\end{aligned}$$

Hence, 27 km/hr = 7.5 m/sec.

(iii) Given,

Speed = 54 km/hr and distance = 600 m

Converting the speed into m/sec :

$$\begin{aligned}54 \text{ km/hr }= \left(54 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{270}{18} \text{ m/sec }\\[1em] = 15 \text{ m/sec }\end{aligned}$$

By formula,

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{600 \text{ m}}{15 \text{ m/sec}}\\[1em] = 40 \text{ seconds }\end{aligned}$$

Hence, at a speed of 54 km/hr, a car will cover a distance of 600 m in 40 seconds.

(iv) Given,

Speed = 45 km/hr and time taken = 18 seconds

Converting the speed into m/sec :

$$\begin{aligned}45 \text{ km/hr }= \left(45 \times \frac{5}{18}\right) \text{ m/sec }\\[1em] = \frac{225}{18} \text{ m/sec }\\[1em] = 12.5 \text{ m/sec }\end{aligned}$$

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 12.5 \times 18 \text{ m }\\[1em] = 225 \text{ m }\end{aligned}$$

Hence, at a speed of 45 km/hr, the distance covered in 18 seconds is 225 m.

(v) Given,

Speed = 5 m/s and distance = 27 km

Converting the speed into km/hr :

$$\begin{aligned}5 \text{ m/s }= \left(5 \times \frac{18}{5}\right) \text{ km/hr }\\[1em] = \frac{90}{5} \text{ km/hr }\\[1em] = 18 \text{ km/hr }\end{aligned}$$

By formula,

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{27 \text{ km}}{18 \text{ km/hr}}\\[1em] = \frac{3}{2} \text{ hrs } = 1\frac{1}{2} \text{ hrs }\\[1em] = 1 \text{ hr } 30 \text{ min }\end{aligned}$$

Hence, at a speed of 5 m/s, a distance of 27 km will be covered in 1 hr 30 min.

2Exercise 11(B) - Mental Maths

Write (T) for true and (F) for false for each of the following statements:

(i) Distance = (Speed × Time)

(ii) Speed = (Distance × Time)

(iii) 90 km/hr = 25 m/s

(iv) $12\frac{1}{2}$ km/hr = 54 km/hr

(v) A man runs at 18 km/hr. Time taken by him to cover 600 m is 3 minutes.

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(i) Since, Speed = $\frac{\text{Distance}}{\text{Time}}$, we get Distance = (Speed × Time).

Hence, the statement is true (T).

(ii) Speed = $\frac{\text{Distance}}{\text{Time}}$ and not (Distance × Time).

Hence, the statement is false (F).

(iii) 90 km/hr = $\left(90 \times \frac{5}{18}\right)$ m/s = 25 m/s

Hence, the statement is true (T).

(iv) $12\frac{1}{2}$ km/hr = 12.5 km/hr, which is not equal to 54 km/hr.

Hence, the statement is false (F).

(v) Given,

Speed = 18 km/hr and distance = 600 m

Converting the speed into m/sec :

18 km/hr = $\left(18 \times \frac{5}{18}\right)$ m/s = 5 m/s

By formula,

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{600 \text{ m}}{5 \text{ m/s}}\\[1em] = 120 \text{ sec }\\[1em] = 2 \text{ minutes }\end{aligned}$$

Time taken by him to cover 600 m is 2 minutes and not 3 minutes.

Hence, the statement is false (F).

1Exercise 11(B) - Case Study Based Questions

Two stations X and Y are 760 km apart. Two trains A and B start at the same time from X and Y respectively and move towards each other. It is known that A takes 5 hours to travel 620 km, while B covers 810 km in 4.5 hours.

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1. Given,

Distance covered by train A = 620 km and time taken = 5 hours

By formula,

$$\begin{aligned}\text{Speed of train A} = \frac{\text{Distance}}{\text{Time}} = \frac{620 \text{ km}}{5 \text{ hrs}}\\[1em] = 124 \text{ km/hr }\end{aligned}$$

Hence, option (a) is the correct option.

2. Given,

Distance covered by train B = 810 km and time taken = 4.5 hours

By formula,

$$\begin{aligned}\text{Speed of train B} = \frac{\text{Distance}}{\text{Time}} = \frac{810 \text{ km}}{4.5 \text{ hrs}}\\[1em] = 180 \text{ km/hr }\end{aligned}$$

Hence, option (d) is the correct option.

3. As the two trains move towards each other,

Relative speed = 124 km/hr + 180 km/hr = 304 km/hr

$$\begin{aligned}\text{Time taken to meet} = \frac{\text{Distance between X and Y}}{\text{Relative speed}}\\[1em] = \frac{760 \text{ km}}{304 \text{ km/hr}}\\[1em] = \frac{5}{2} \text{ hrs } = 2.5 \text{ hrs }\end{aligned}$$

Distance of the meeting point from station X = Distance covered by train A in 2.5 hrs

= 124 × 2.5 km = 310 km

Hence, option (c) is the correct option.

4. As the two trains move towards each other,

Relative speed = 124 km/hr + 180 km/hr = 304 km/hr

$$\begin{aligned}\text{Time taken to meet} = \frac{760 \text{ km}}{304 \text{ km/hr}}\\[1em] = \frac{5}{2} \text{ hrs }\\[1em] = 2\frac{1}{2} \text{ hrs }\\[1em] = 2 \text{ hrs } 30 \text{ min}\end{aligned}$$

Hence, option (b) is the correct option.

2Exercise 11(B) - Case Study Based Questions

Rajan is travelling from his home to a town 240 km away. For the first quarter of the journey, he takes a bus that travels 100 km in 2.5 hours. For the remaining journey, he takes a cab that covers 252 km in 3.5 hours.

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1. Given,

Distance covered by the bus = 100 km and time taken = 2.5 hours

By formula,

$$\begin{aligned}\text{Speed of the bus} = \frac{\text{Distance}}{\text{Time}} = \frac{100 \text{ km}}{2.5 \text{ hrs}}\\[1em] = 40 \text{ km/hr }\end{aligned}$$

Hence, option (b) is the correct option.

2. Given,

Distance covered by the cab = 252 km and time taken = 3.5 hours

By formula,

$$\begin{aligned}\text{Speed of the cab} = \frac{\text{Distance}}{\text{Time}} = \frac{252 \text{ km}}{3.5 \text{ hrs}}\\[1em] = 72 \text{ km/hr }\end{aligned}$$

Hence, option (a) is the correct option.

3. Distance covered by bus = first quarter of the journey = $\frac{240}{4}$ km = 60 km

$$\begin{aligned}\text{Time taken by bus} = \frac{60 \text{ km}}{40 \text{ km/hr}} = \frac{3}{2} \text{ hrs }\\[1em] = 1.5 \text{ hrs }\end{aligned}$$

Distance covered by cab = 240 km − 60 km = 180 km

$$\begin{aligned}\text{Time taken by cab} = \frac{180 \text{ km}}{72 \text{ km/hr}} = \frac{5}{2} \text{ hrs }\\[1em] = 2.5 \text{ hrs }\end{aligned}$$

Total time taken = 1.5 hrs + 2.5 hrs = 4 hrs

Hence, option (b) is the correct option.

4. Total distance covered = 240 km

Total time taken = 4 hrs

By formula,

$$\begin{aligned}\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}}\\[1em] = \frac{240 \text{ km}}{4 \text{ hrs}}\\[1em] = 60 \text{ km/hr }\end{aligned}$$

Hence, option (c) is the correct option.

1Exercise 11(B) - Assertion–Reason Questions

Assertion (A): A speed of 72 km/h is equivalent to 20 m/sec.

Reason (R): To convert a speed of km/h in m/sec, we multiply by $\frac{5}{18}$.

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Converting the speed,

72 km/h = $\left(72 \times \frac{5}{18}\right)$ m/sec = 20 m/sec

So, Assertion (A) is true.

$$\begin{aligned}1 \text{ km/h } = \frac{1 \times 1000 \text{ m}}{60 \times 60 \text{ sec}}\\[1em] = \frac{1000}{3600} \text{ m/sec }\\[1em] = \frac{5}{18} \text{ m/sec }\end{aligned}$$

So, to convert a speed of km/h into m/sec, we multiply by $\frac{5}{18}$.

So, Reason (R) is true and is the correct explanation of Assertion (A), since the speed of 72 km/h is converted into 20 m/sec by multiplying it by $\frac{5}{18}$.

Hence, option 1 is the correct option.

2Exercise 11(B) - Assertion–Reason Questions

Assertion (A): A car is running at a speed of 43.2 km/h. It will cover 3.5 km in 5 minutes.

Reason (R): Distance = Speed × Time

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Given,

Speed = 43.2 km/h

Time taken = 5 minutes

= $\frac{5}{60} \text{ h } = \frac{1}{12}$ h

By formula,

$$\begin{aligned}\text{Distance} = \text{Speed} \times \text{Time}\\[1em] = 43.2 \times \frac{1}{12} \text{ km }\\[1em] = 3.6 \text{ km }\end{aligned}$$

The car will cover 3.6 km in 5 minutes and not 3.5 km.

So, Assertion (A) is false.

Since, Speed = $\frac{\text{Distance}}{\text{Time}}$, we get Distance = Speed × Time.

So, Reason (R) is true.

Hence, option 4 is the correct option.

1Exercise 11(B) - Competency Focused Questions

A plane flies from City A to City B, covering a distance of 1,200 km in 2 hours. The return trip takes 2 hours and 48 minutes due to strong headwinds. What is the average speed of the plane for the entire journey?

$\left[\text{Average speed} = \frac{\text{total distance covered}}{\text{total time taken}}\right]$

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Given,

Total distance covered = 1200 km + 1200 km = 2400 km

Total time taken = 2 hrs + 2 hrs 48 min

= $2 \text{ hrs} + 2\frac{48}{60} \text{ hrs } = 2 \text{ hrs} + 2\frac{4}{5} \text{ hrs } = 2 \text{ hrs} + 2.8 \text{ hrs } = 4.8$ hrs

By formula,

$$\begin{aligned}\text{Average speed} = \frac{\text{Total distance covered}}{\text{Total time taken}}\\[1em] = \frac{2400 \text{ km}}{4.8 \text{ hrs}}\\[1em] = 500 \text{ km/h }\end{aligned}$$

Hence, option 1 is the correct option.

2Exercise 11(B) - Competency Focused Questions

Two runners, Ramesh and Suresh, start at the same point but run on two different circular tracks.

  • Ramesh completes one round of a 300 m track in 1 minute.

  • Suresh completes one round of a 400 m track in 1.5 minutes.

Who will be ahead after 6 minutes, and by how much distance?

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Given,

Ramesh covers 300 m in 1 minute and Suresh covers 400 m in 1.5 minutes.

By formula,

$$\begin{aligned}\text{Speed of Ramesh} = \frac{\text{Distance}}{\text{Time}} = \frac{300 \text{ m}}{1 \text{ min}}\\[1em] = 300 \text{ m/min }\end{aligned}$$

Distance covered by Ramesh in 6 minutes = 300 × 6 m = 1800 m

$$\begin{aligned}\text{Speed of Suresh} = \frac{\text{Distance}}{\text{Time}} = \frac{400 \text{ m}}{1.5 \text{ min}}\\[1em] = \frac{800}{3} \text{ m/min }\end{aligned}$$

Distance covered by Suresh in 6 minutes = $\frac{800}{3} \times 6$ m = 1600 m

Since, 1600 m < 1800 m, Ramesh will be ahead.

Difference of distances = 1800 m − 1600 m = 200 m

Hence, option 1 is the correct option.

3Exercise 11(B) - Competency Focused Questions

Two friends, Amit and Sunil, start at the same point and walk in opposite directions. Amit walks at 4 km/h, and Sunil walks at 5 km/h. After 3 hours, how far apart will they be?

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Given,

Speed of Amit = 4 km/h, speed of Sunil = 5 km/h and time taken = 3 hours

As Amit and Sunil walk in opposite directions,

Relative speed = 4 km/h + 5 km/h = 9 km/h

By formula,

$$\begin{aligned}\text{Distance between them} = \text{Relative speed} \times \text{Time}\\[1em] = 9 \times 3 \text{ km }\\[1em] = 27 \text{ km }\end{aligned}$$

Hence, option 4 is the correct option.

4Exercise 11(B) - Competency Focused Questions

A person leaves their house at 8:00 am, walks 6 km at 4 km/h, and then cycles another 12 km at 12 km/h to reach their workplace. What time does the person reach their workplace?

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Given,

Distance walked = 6 km at a speed of 4 km/h and distance cycled = 12 km at a speed of 12 km/h

By formula,

$$\begin{aligned}\text{Time taken in walking} = \frac{\text{Distance}}{\text{Speed}} = \frac{6 \text{ km}}{4 \text{ km/h}}\\[1em] = \frac{3}{2} \text{ hrs }\\[1em] = 1 \text{ hr } 30 \text{ min }\end{aligned}$$

$$\begin{aligned}\text{Time taken in cycling} = \frac{\text{Distance}}{\text{Speed}} = \frac{12 \text{ km}}{12 \text{ km/h}}\\[1em] = 1 \text{ hr }\end{aligned}$$

Total time taken = 1 hr 30 min + 1 hr = 2 hrs 30 min

Time of reaching the workplace = 8:00 am + 2 hrs 30 min = 10:30 am

Hence, option 1 is the correct option.

5Exercise 11(B) - Competency Focused Questions

A car completes a certain journey in 8 hours. It covers half the distance at 40 km/h and the rest at 60 km/h. The length of the journey is:

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Let the length of the journey be 2x km, so that each half of the journey is x km.

By formula,

Time taken to cover the first half = $\frac{\text{Distance}}{\text{Speed}} = \frac{x}{40}$ hrs

Time taken to cover the second half = $\frac{\text{Distance}}{\text{Speed}} = \frac{x}{60}$ hrs

Total time taken = 8 hrs

$$\begin{aligned}\Rightarrow \frac{x}{40} + \frac{x}{60} = 8\\[1em] \Rightarrow \frac{3x + 2x}{120} = 8\\[1em] \Rightarrow \frac{5x}{120} = 8\\[1em] \Rightarrow x = \frac{8 \times 120}{5}\\[1em] \Rightarrow x = 192\end{aligned}$$

Length of the journey = 2x km = 2 × 192 km = 384 km

Hence, option 4 is the correct option.

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