NCERT Solutions for Class 8 Maths Chapter 13 - Direct And Inverse Proportions

Exercise 13.1

1. Following are the car parking charges near a railway station up to  $ \text{4}\ \text{hrs}\ \text{Rs}\ \text{60} $ ,  $ \text{8}\ \text{hrs}\ \text{Rs}\ \text{100} $ ,  $ \text{12}\ \text{hrs}\ \text{Rs}\ \text{140} $ , and  $ \text{24}\ \text{hrs}\ \text{Rs}\ \text{180} $ . Check if the parking charges are in direct proportion to the parking time.

Ans: We know the charges will be in direct proportion to the parking time when each of their ratios will be the same.

We can form a table by using the given information as –

Therefore, the ratio of the parking charges to the time taken for parking will be  $\text{=}\dfrac{\text{charge}}{\text{time taken}}$

Hence,

 $ \dfrac{60}{4}=15 $ ,

 $ \Rightarrow \dfrac{100}{8}=\dfrac{25}{2} $ ,

 $ \Rightarrow \dfrac{140}{12}=\dfrac{35}{3} $ ,

 $ \Rightarrow \dfrac{180}{24}=\dfrac{15}{2} $ .

As, we can observe that the ratios are not the same. Hence, we can conclude that the parking charges are not in direct proportion to the parking time.

2. A mixture of paint is prepared by mixing  $ \text{1} $  part of red pigments with  $ \text{8} $  parts of the base. In the following table, find the parts of the base that need to be added.

Ans:

Given we have a mixture of paint that is prepared by mixing  $ 1 $  part of red pigments with  $ 8 $  parts of base. We can observe that they are in a direct proportion. Therefore,

Let  $ {{a}_{1}},{{a}_{2}},{{a}_{3}}, $  and  $ {{a}_{4}} $  denote the parts of base of the respective parts of red pigment.

Hence, we have –

 $ \dfrac{8}{1}=\dfrac{{{a}_{1}}}{4} $ 

 $ \Rightarrow {{a}_{1}}=32 $ .

Similarly,  $ \dfrac{8}{1}=\dfrac{{{a}_{2}}}{7} $ 

 $ \Rightarrow {{a}_{2}}=56 $ ,

 $ \dfrac{8}{1}=\dfrac{{{a}_{3}}}{12} $ 

 $ \Rightarrow {{a}_{3}}=96 $ , and

 $ \dfrac{8}{1}=\dfrac{{{a}_{4}}}{20} $ 

 $ \Rightarrow {{a}_{4}}=160 $ .

Therefore, now the given table will be as –

3. In Question  $ \text{2} $  above, if  $ \text{1} $  part of a red pigment requires  $ \text{75mL} $  of base, how much red pigment should we mix with  $ \text{1800mL} $  of base?

Ans: Let us assume that  $ a $  denotes the parts of red pigment required. We know that parts of the base are in direct proportion to parts of red pigment.

Therefore,

 $ \dfrac{75}{1}=\dfrac{1800}{a} $ 

 $ \Rightarrow a=\dfrac{1800}{75} $ 

 $ \Rightarrow a=24 $ .

Hence, the table will be as –

4. A machine in a soft drink factory fills  $ \text{840} $  bottles in six hours. How many bottles will it fill in five hours?

Ans: Given we have a soft drink factory that fills  $ 840 $  bottles in six hours. Let us assume that the number of bottles to be filled in five hours will be  $ x $.

Therefore, we can conclude that both the parameters are in direct proportion.

Hence,

 $ \dfrac{840}{6}=\dfrac{x}{5} $ 

 $ \Rightarrow x=\dfrac{840\times 5}{6} $ 

 $ \Rightarrow x=700 $ 

Therefore, the number of bottles to be filled in five hours will be  $ 700 $ .

5. A photograph of a bacteria enlarged  $ \text{50,000} $  times attains a length of  $ \text{5cm} $ . What is the actual length of the bacteria? If the photograph is enlarged  $ \text{20,000} $  times only, what would be its enlarged length?

Ans: Let us assume that the actual length of the bacteria is  $ a\ \text{cm} $  and let us assume that its enlarged length will be  $ b\ \text{cm} $ . The given information implies that the parameters are in direct proportion.

Hence, we can compute a table as –

Therefore, the actual length of the bacteria when the photograph was not enlarged will be –

 $ \dfrac{5}{50000}=\dfrac{a}{1} $ 

 $ \Rightarrow a=\dfrac{1}{10000} $ 

 $ \Rightarrow a={{10}^{-4}}cm $ 

Now, we will calculate the enlarged length of the bacteria when the photograph is enlarged  $ 20,000 $  times.

 $ \dfrac{5}{50000}=\dfrac{b}{20000} $ 

 $ \Rightarrow b=2cm $ .

Therefore, the actual length of the bacteria is  $ 0.0001\ cm $  and the enlarged length is  $ \text{2cm} $ .

6. In a model of a ship, the mast is  $ \text{9}\ \text{cm} $  high, while the mast of the actual ship is  $ \text{12}\ \text{m} $  high. If the length of the ship is  $ \text{28}\ \text{m} $ , how long is the model ship?

Ans:

Let us assume that the length of the model ship is  $ x\ \text{cm} $ . We can observe that the mast and length of the ship are directly proportional to each other. Therefore,

 $ \dfrac{12}{9}=\dfrac{28}{x} $ 

 $ \Rightarrow x=\dfrac{28\times 9}{12} $ 

 $ \Rightarrow x=21\ \text{cm} $ .

Hence, the length of the model ship will be of  $ 21\ \text{cm} $ .

7. Suppose  $ \text{2}\ \text{kg} $  of sugar contains \[\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\] crystals. How many sugar crystals are there in –

Ans:

(i). $ \text{5}\ \text{kg} $  of sugar? 

Let us assume that the  $ 5\ \text{kg} $  of sugar contains  $ x $  number of crystals.

As, weight of sugar is directly proportional to the number of crystals it contains. Therefore,

 $ \dfrac{2}{9\times {{10}^{6}}}=\dfrac{5}{x} $ 

 $ \Rightarrow x=\dfrac{9\times {{10}^{6}}\times 5}{2} $ 

 $ \Rightarrow x=2.25\times {{10}^{7}} $ 

Hence,  $ 5\ \text{kg} $  of sugar contains  $ 2.25\times {{10}^{7}} $  crystals of sugar.

(ii). $ \text{1}\text{.2}\ \text{kg} $  of sugar?

Similarly, let us assume that the  $ 1.2\ \text{kg} $  of sugar contains  $ x $  number of crystals.

As, weight of sugar is directly proportional to the number of crystals it contains. Therefore,

 $ \dfrac{2}{9\times {{10}^{6}}}=\dfrac{1.2}{x} $ 

 $ \Rightarrow x=\dfrac{9\times {{10}^{6}}\times 1.2}{2} $ 

 $ \Rightarrow x=5.4\times {{10}^{6}} $ 

Hence,  $ 1.2\ \text{kg} $  of sugar contains  $ 5.4\times {{10}^{6}} $  crystals of sugar.

8. Rashmi has a road map with a scale of  $ \text{1}\ \text{cm} $  representing \[\text{18}\ \text{km}\]. She drives on a road for \[\text{72}\ \text{km}\]. What would be her distance covered in the map?

Ans:

Given we have a road map with a scale of  $ 1\ cm $  representing \[18\ km\]. Let us assume that the distance covered by her is  $ x\ \text{cm} $ . We can observe that the scale is directly proportional to the distance she covers.

Therefore,

 $ \dfrac{18}{1}=\dfrac{72}{x} $ 

 $ \Rightarrow x=4\ cm $ .

Hence, the distance covered by her represents on the map as  $ 4\ cm $ .

9. A  $ \text{5}\ \text{m}\ \text{60}\ \text{cm} $  high vertical pole casts a shadow  $ \text{3}\ \text{m}\ \text{20}\ \text{cm} $  long. Find at the same time – 

i. The length of the shadow cast by another pole  $ \text{10}\ \text{m}\ \text{50}\ \text{cm} $  high

Ans:

Given we have a vertical pole of height  $ 5\ \text{m}\ \text{60}\ cm $  which casts a shadow of  $ 3\ \text{m}\ 2\text{0}\ cm $ . Let us assume that the length of shadow cast by pole  $ 10\ \text{m}\ 5\text{0}\ cm $  high is  $ x\ \text{m} $ . As we know that the height of the pole is directly proportional to the length of the shadow.

Therefore,

 $ \dfrac{5.60}{3.20}=\dfrac{10.50}{x} $ 

 $ \Rightarrow x=\dfrac{10.50\times 3.20}{5.60} $ 

 $ \Rightarrow x=6 $ 

Hence, the length of the shadow cast by another pole  $ 10\ \text{m}\ 5\text{0}\ cm $  high is  $ 6\ \text{m} $ .

ii. The height of a pole which casts a shadow 5m long.

Ans:

Let us assume that the height of the pole is  $ x\ \text{m} $  which casts a shadow  $ 5\ \text{m} $  long. As we know that the height of the pole is directly proportional to the length of the shadow.

Therefore,

 $ \dfrac{5.60}{3.20}=\dfrac{x}{5} $ 

 $ \Rightarrow x=\dfrac{5\times 5.60}{3.20} $ 

 $ \Rightarrow x=8.75 $ 

Hence, the height of the pole is  $ 8\ \text{m}\ 75\ cm $  which casts a shadow  $ 5\ \text{m} $  long.

10. A loaded truck travels  $ \text{14}\ \text{km} $  in  $ \text{25}\ \text{minutes} $ . If the speed remains the same, how far can it travel in  $ \text{5}\ \text{hours} $ ?

Ans:

Let us assume the distance travelled by truck in  $ 5\ \text{hours} $  is  $ x\ \text{km} $ . As we know that the distance traveled by the loaded truck is directly proportional to the time.

Therefore,

 $ \dfrac{14}{25}=\dfrac{x}{5\times 60} $ 

 $ \Rightarrow x=\dfrac{5\times 60\times 14}{25} $ 

 $ \Rightarrow x=168\ \text{km} $ 

Hence, the distance travelled by the loaded truck in  $ 5\ \text{hours} $  is  $ 168\ \text{km} $ .

Exercise 13.2

1. Which of the following are in inverse proportion? 

(i). The number of workers on a job and the time to complete the job. 

Ans: Since, the number of workers increases, it reduces the time to complete the job. Therefore, we can conclude that this is an inverse proportion.

(ii). The time is taken for a journey and the distance traveled at a uniform speed.

Ans: We know, if we travel at a uniform speed then the distance traveled can be more if we have taken more time. This implies that they are in direct proportion and not in inverse proportion. Hence, we can conclude that this is not in inverse proportion.

(iii). Area of cultivated land and the crop harvested. 

Ans: We know, the more cultivated land we will have, the greater number of crops can be harvested. This implies that they are in direct proportion and not in inverse proportion. Therefore, we can conclude that this is not in inverse proportion.

(iv). The time is taken for a fixed journey and the speed of the vehicle. 

Ans: When we travel with the vehicle at high speed, there is a possibility that we can complete the journey in less time than expected. This implies that this is in inverse proportion.

(v). The population of a country and the area of land per person.

Ans: As the population of a country increases, the area of the land per person will decrease because more people will be requiring the area for themselves. Therefore, this is in inverse proportion.

2. In a Television game show, the prize money of  $ \text{Rs}\ \text{1,00,000} $  is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Ans:

We know, the more the number of winners will be, the lesser the prize money each winner gets. This implies that these are in inverse proportion.

Now, let us assume the prize for each winner as  $ {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}} $  and  $ {{a}_{5}} $ .

Therefore,

 $ 1\times 100000=4\times {{a}_{1}} $ 

 $ \Rightarrow {{a}_{1}}=\dfrac{100000}{4} $ 

 $ \Rightarrow {{a}_{1}}=Rs\ 25000 $ 

Similarly,

 $ 1\times 100000=5\times {{a}_{2}} $ 

 $ \Rightarrow {{a}_{2}}=\dfrac{100000}{5} $ 

 $ \Rightarrow {{a}_{2}}=Rs\ 20000 $ 

Now, we will calculate  $ {{a}_{3}} $ 

 $ 1\times 100000=8\times {{a}_{3}} $ 

 $ \Rightarrow {{a}_{1}}=\dfrac{100000}{8} $ 

 $ \Rightarrow {{a}_{3}}=Rs\ 12500 $ 

Now, we will calculate  $ {{a}_{4}} $ 

 $ 1\times 100000=10\times {{a}_{4}} $ 

 $ \Rightarrow {{a}_{4}}=\dfrac{100000}{10} $ 

 $ \Rightarrow {{a}_{4}}=Rs\ 10000 $ 

Now, we will calculate  $ {{a}_{5}} $ 

 $ 1\times 100000=20\times {{a}_{5}} $ 

 $ \Rightarrow {{a}_{5}}=\dfrac{100000}{20} $ 

 $ \Rightarrow {{a}_{5}}=Rs\ 5000 $ 

Hence, the amount decreases as the number of winners increases.

3. Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.

(i). Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion? 

(ii). Calculate the angle between a pair of consecutive spokes on a wheel with  $ \text{15} $  spokes. 

(iii). How many spokes would be needed, if the angle between a pair of consecutive spokes is  $ \text{40 }\!\!{}^\circ\!\!\text{ } $ ?

Ans:

We know, the more the number of spokes will be, the lesser the angle between two spokes will be. This implies that these are in inverse proportion.

Now, let us assume the angle between two consecutive spokes as  $ {{a}_{1}},{{a}_{2}}, $  and  $ {{a}_{3}} $ .

Therefore,

\[4\times 90{}^\circ =8\times {{a}_{1}}\]

 $ \Rightarrow {{a}_{1}}=\dfrac{4\times 90}{8} $ 

 $ \Rightarrow {{a}_{1}}=45{}^\circ  $ 

Similarly,

 $ 4\times 90{}^\circ =10\times {{a}_{2}} $ 

 $ \Rightarrow {{a}_{2}}=\dfrac{4\times 90}{10} $ 

 $ \Rightarrow {{a}_{2}}=36{}^\circ  $ 

Now, we will calculate  $ {{a}_{3}} $ 

 $ 4\times 90{}^\circ =12\times {{a}_{3}} $ 

 $ \Rightarrow {{a}_{3}}=\dfrac{4\times 90}{12} $ 

 $ \Rightarrow {{a}_{3}}=30{}^\circ  $ 

Hence, table will be as –

(i). Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?

Ans: As the number of spokes increases, we can observe that the angle between the pair of consecutive spokes decreases. Therefore, this implies they are in inverse proportion.

(ii). Calculate the angle between a pair of consecutive spokes on a wheel with  $ \text{15} $  spokes. 

Ans: Let us assume the angle as  $ x{}^\circ  $  of consecutive spokes on a wheel with  $ 15 $  spokes.

Therefore,

\[4\times 90{}^\circ =15\times x\]

 $ \Rightarrow x=\dfrac{4\times 90}{15} $ 

 $ \Rightarrow x=24{}^\circ  $ .

Hence, the angle will be  $ 24{}^\circ  $  between a pair of consecutive spokes on a wheel with  $ 15 $  spokes.

(iii). How many spokes would be needed, if the angle between a pair of consecutive spokes is  $ \text{40 }\!\!{}^\circ\!\!\text{ } $ ?

Ans: Let us assume the number of spokes be  $ y $ .

Therefore,

\[4\times 90{}^\circ =y\times 40{}^\circ \]

 $ \Rightarrow y=\dfrac{4\times 90}{40} $ 

 $ \Rightarrow y=9 $ .

Hence,  $ 9 $  spokes would be needed if the angle between a pair of consecutive spokes is  $ 40{}^\circ  $ .

4. If a box of sweets is divided among  $ \text{24} $  children, they will get  $ \text{5} $  sweets each. How many would each get, if the number of the children is reduced by  $ \text{4} $ ?

Ans: As the number of children is reduced by  $ 4 $ . Hence, the remaining children will be  $ 24-4=20 $  children.

Let us assume that the number of sweets each would get is  $ a $ . We know that the number of sweets each child gets increases whenever the number of children is decreased.

Therefore,

\[24\times 5=20\times a\]

 $ \Rightarrow a=\dfrac{24\times 5}{20} $ 

 $ \Rightarrow a=6 $ .

Hence, each child will get  $ 6 $  sweets.

5. A farmer has enough food to feed  $ \text{20} $  animals in his cattle for  $ \text{6} $  days. How long would the food last if there were  $ \text{10} $  more animals in his cattle?

Ans:  Let us assume that the number of days the food will last be  $ x $  days.

He has  $ 10 $  more animals in his cattle. Hence, total animals will be  $ 20+10=30 $ 

We know, the more animals the food will last for less days. This implies these are in inverse proportion.

Therefore,

 $ 20\times 6=30\times x $ 

 $ \Rightarrow x=\dfrac{20\times 6}{30} $ 

 $ \Rightarrow x=4 $ .

Hence, the food will last for  $ 4 $  days if there were  $ 10 $  more animals in his cattle.

6. A contractor estimates that  $ \text{3} $  persons could rewire Jasminder’s house in  $ \text{4} $  days. If he uses  $ \text{4} $  persons instead of three, how long should they take to complete the job?

Ans: Let us assume the number of days for the persons to complete the job be  $ x $  days.

We know that the number of days and number of persons are in inverse proportion.

Therefore,

 $ 3\times 4=4\times x $ 

 $ \Rightarrow x=\dfrac{3\times 4}{4} $ 

 $ \Rightarrow x=3 $ .

Hence, the number of days will be  $ 3 $  days if he uses  $ 4 $  persons instead of three to complete the job.

7. A batch of bottles was packed in  $ \text{25} $  boxes with  $ \text{12} $  bottles in each box. If the same batch is packed using  $ \text{20} $  bottles in each box, how many boxes would be filled?

Ans: Let us assume the number of boxes needed to be filled is  $ x $ .

We know that the number of boxes and number of bottles in each box are in inverse proportion.

Therefore,

 $ 25\times 12=x\times 20 $ 

 $ \Rightarrow x=\dfrac{25\times 12}{20} $ 

 $ \Rightarrow x=15 $ .

Hence, the number of boxes needed to be filled if the same batch is packed using  $ 20 $  bottles in each box will be  $ 15 $  boxes.

8. A factory required  $ \text{42} $  machines to produce a given number of articles in  $ \text{63} $  days. How many machines would be required to produce the same number of articles in  $ \text{54} $  days?

Ans: Let us assume the number of machines required to produce the articles is  $ x $ .

We know that the number of machines and number of days are in inverse proportion.

Therefore,

 $ 42\times 63=x\times 54 $ 

 $ \Rightarrow x=\dfrac{42\times 63}{54} $ 

 $ \Rightarrow x=49 $ .

Hence, the number of machines required to produce the same number of articles in  $ 54 $  days will be  $ 49 $  machines.

9. A car takes  $ \text{2} $  hours to reach a destination by travelling at the speed of  $ \text{60}\ \text{km/h} $ . how long will it take when the car travels at the speed of  $ \text{80}\ \text{km/h} $ ?

Ans:Let us assume the number of hours the car will take to reach the destination be   $ x\ \text{hrs} $ .

We know that the number of hours and speed of the car are in inverse proportion.

Therefore,

 $ 2\times 60=x\times 80 $ 

 $ \Rightarrow x=\dfrac{2\times 60}{80} $ 

 $ \Rightarrow x=\dfrac{3}{2} $ .

Hence, the time taken when the car travels at the speed of  $ 80\ \text{km/h} $  will be  $ 1.5\ hrs $ .

10. Two people could fit new windows in the house in  $ \text{3} $  days. 

(i). One of the persons fell ill before the work started. How long would the job take now? 

Ans: As, we know that the number of persons and the number of days is in inverse proportion. Let us assume that the number of days for one person to get the job done be  $ x $  days.

Therefore,

 $ 2\times 3=1\times x $ 

 $ \Rightarrow x=\dfrac{2\times 3}{1} $ 

 $ \Rightarrow x=6 $ .

Hence, if one of the persons fell ill before the work started, it will take  $ 6 $  days to fit new windows.

(ii). How many people would be needed to fit the windows in one day?

Ans: As we know that the number of persons and the number of days is in inverse proportion. Let us assume that the number of persons needed to fit the windows is  $ y $ .

Therefore,

 $ 2\times 3=y\times 1 $ 

 $ \Rightarrow y=\dfrac{2\times 3}{1} $ 

 $ \Rightarrow y=6 $ .

Hence,  $ 6 $ person will be needed to fit the windows in one day.

11. A school has  $ \text{8} $  periods a day each of  $ \text{45} $  minutes duration. How long would each period be, if the school has  $ \text{9} $  periods a day, assuming the number of school hours to be the same?

Ans: We know that the time for each period will decrease if we increase the number of periods a day. This implies these are in inverse proportion.

So, let us assume that each period will be of  $ x $  minutes.

Therefore,

 $ 8\times 45=9\times x $ 

 $ \Rightarrow x=\dfrac{8\times 45}{9} $ 

 $ \Rightarrow x=40 $ .
Hence, each period will be of  $ 40 $  minutes if the school has  $ 9 $  periods a day, assuming the number of school hours to be the same.

NCERT Solutions for Class 8 Maths Chapter 13 Free PDF Download

The expert team of Vedantu has made available the Class 8 Maths Direct and Inverse Proportion solutions in PDF format on its official website. You can easily download the NCERT Solutions of Class 8 Maths Chapter 13 PDF from anywhere anytime and save it on your devices for offline access. The solutions are also printable so that you could even have a group study with the hard copy of the PDF.

We Cover all Exercises in the Chapter Given Below:

• Exercise 13.1 - 10 Questions with Solutions.

• Exercise 13.2 - 11 Questions with Solutions.

Chapter 13 - Direct and Inverse Proportions

13.1 Introduction

In the first section of Class 8 Maths Chapter Direct and Inverse Proportion, there are many examples given from daily life where the knowledge of proportions is applied. For example:

• If two students arrange chairs for an assembly in 20 minutes, how much time will 5 students take?

• If you have to make tea for 1 person and you use 1 teaspoon of sugar then how many teaspoons would you need for making tea for 5 people?

From these examples, it can be figured that there is a direct relation in a change of one quantity on another quantity. This can be understood better with the concept of variation. The variation between two objects means that quantities of two objects are related to each other in a way that an increase or decrease in one quantity leads to an increase or decrease in the other quantity.

13.2 Direct Proportion

You will learn the definition of direct proportion in this part of Direct and Inverse Proportion class 8 solutions. Two quantities x, and y, are said to be in direct proportion if they fulfill the following criteria:

• If x increases, y increases too.

• If x decreases, y decreases too.

• x/y = C where C is a constant or, x = C * y. This means that the ratio of the respective values of x and y remain the same despite an increment or decrement in their values.

• From the above rule, we can say that if for the value of x at x1 the value of y is y1 and for the value of x at x2 the value of y is y2 then x1/y1 = x2/y2.

We denote y is proportional to x by the symbol ∝ , i.e. y ∝ x. A real-life example of direct proportion is that if the number of things you purchase increases, there is an increase in expenditure too.

13.2 Inverse Proportion

This part of Direct and Indirect Proportion class 8 will acquaint you with the concept of Inverse Proportion. Two quantities x, and y, are said to be in inverse proportion if they fulfill the following criteria:

• If x increases, there is a decrease in y.

• If x decreases, there is an increase in y.

• x * y = a constant K. This means that the ratio of the respective values of x and y remain the same despite an increment or decrement in their values.

• From the above rule, we can say that if for the value of x at x1 the value of y is y1 and for the value of x at x2 the value of y is y2 then x1 * y1 = x2 * y2= C.

We denote y is inversely proportional to x by the notation y ∝ 1/x. A real-life example of inverse proportion is that if the speed of a car increases, the time to reach a destination will decrease.

Key Features of NCERT Solutions for Class 8 Maths Chapter 13

The solutions provided for Maths chapter Direct and Inverse Proportion of Class 8 by the subject matter experts at Vedantu is apt for the students of class 8 because:

• Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

• Clear and structured presentation for easy comprehension.

• Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

• Visual aids like diagrams and illustrations to simplify complex concepts.

• Additional tips and insights to enhance students' performance.

• Chapter summaries for quick revision.

• Online accessibility and downloadable resources for flexible study and revision.