# NCERT Solutions Class 8 Maths Chapter 13 Exercise 13.2

One subject that hinders many students is mathematics. Despite their interest in the subject, they may have difficulty understanding the concepts, formulas, and calculations used in it. Due to the fact that mathematics is a purely conceptual subject, students should have a thorough understanding of its curriculum. Chapter 13 of Class 8 Mathematics is Direct and Inverse Proportions. Students may find it difficult to understand the concepts and calculations in this chapter at first. These concepts and calculations can become easier with their own practice and Extramarks’ guidance. The topics that are covered in Chapter 13 of Class 8 Mathematics include introduction, direct proportion and inverse proportion. In direct proportion, two quantities are related in such a way that an increase in one quantity leads to an increase in the other, and their ratio always remains constant. The chapter does not contain many exercises or questions, but it does contain many new concepts. Extramarks provides students with all the essential resources that make their learning experience comprehensive and effortless, and as a result, they can perform very well in their examinations.

NCERT Textbooks do not contain solutions to the questions incorporated in the respective exercises of the listed chapters. The solutions to Class 8 Maths Exercise 13.2 provided by Extramarks help students understand the questions given in the exercise while having a clear solution and approach to each question. NCERT textbooks are the basis for the foundation of the concepts in any chapter of Mathematics. As a result, it is critical to practise Class 8 Maths Chapter 13 Exercise 13.2. Along with the Class 8 Maths Exercise 13.2 Solution, Extramarks provides students with various learning modules so that they can learn comprehensively and perform effectively in their examinations. Extramarks is an e-learning platform that proves that e-learning has various benefits. The K12 study material provided by Extramarks helps students have access to complete and convenient study material, due to which they can score well in their examinations. The in-depth performance reports provided by Extramarks also help them with self-assessment and track their academic progress. Live Doubt Solving Classes provide access to live sessions with expert educators from Extramarks, which also help to clear doubts effectively. Students can also resolve their doubts by going through NCERT Class 8 Maths Chapter 13 Exercise 13.2. Students can practise chapter-wise worksheets, answer practise questions, participate in interactive activities, and more with Extramarks to achieve proficiency in all the subjects. Furthermore, the visual learning journey provided by the Learning App helps students easily grasp the subject’s curriculum and enjoy practising the new concepts. Extramarks make learning enjoyable for students with visuals and animations, so that they do not find their studies monotonous. With Extramarks, students have access to expert teachers who are highly qualified and experienced in their field so that they can receive the best guidance.

**NCERT Class 8 Maths Solutions, Chapter 13: Direct and Inverse Proportions (EX 13.2), Exercise 13.2 **

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**Access NCERT Solutions for Class 8 Chapter 13: Direct and Inverse Proportions**

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**NCERT Solutions for Class 8 Maths, Chapter 13: Direct and Inverse Proportions Exercise 13.2**

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**NCERT Solutions for Class 8**

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**Q.1** Which of the following are in inverse proportion?

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

(ii) The time taken for a journey and the distance travelled in a uniform speed.

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

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

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

**Ans**

(i) It is in inverse proportion because if there are more workers, then it will take less time to complete that job.

(ii) No, these are not in inverse proportion because in more time, we may cover more distance with a uniform speed.

(iii) No, these are not in inverse proportion because in more area, more quantity of crop may be harvested.

(iv) It is in inverse proportion because with more speed, we may complete a certain distance in a lesser time.

(v) It is in inverse proportion because if the population is increasing/decreasing, then the area of the land per person will be decreasing/increasing accordingly.

**Q.2 **In a Television game show, the prize money of ₹ 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?

Number of winners | 1 | 2 | 4 | 5 | 8 | 10 | 20 |

Prize of each winner (in ₹ ) | 1,00,000 | 50,000 | … | … | … | … | … |

**Ans**

Let the unknown prizes be x_{1}, x_{2}, x_{3}, x_{4 }and x_{5.}

The given information is represented in a table below:

Number of winners | 1 | 2 | 4 | 5 | 8 | 10 | 20 |

Prize of each winner (in ₹ ) | 1,00,000 | 50,000 | x1 | x2 | x3 | x4 | x5 |

From the table, we obtain

1 × 1,00,000 = 2 × 50,000 = 1,00,000

Thus, the number of winners and the amount given to each winner are inversely proportional to each other.

Therefore, we obtain

$\begin{array}{l}1\times 1,00,000=4\times {\mathrm{x}}_{1}\\ \Rightarrow {\mathrm{x}}_{1}=\frac{1,00,000}{4}=25,000\\ \\ 1\times 1,00,000=5\times {\mathrm{x}}_{2}\\ \Rightarrow {\mathrm{x}}_{2}=\frac{1,00,000}{5}=20,000\\ 1\times 1,00,000=8\times {\mathrm{x}}_{3}\\ \Rightarrow {\mathrm{x}}_{3}=\frac{1,00,000}{8}=12,500\\ \\ 1\times 1,00,000=10\times {\mathrm{x}}_{4}\\ \Rightarrow {\mathrm{x}}_{4}=\frac{1,00,000}{10}=10,000\\ \\ 1\times 1,00,000=20\times {\mathrm{x}}_{5}\\ \Rightarrow {\mathrm{x}}_{5}=\frac{1,00,000}{20}=5,000\end{array}$

**Q.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.

Number of spokes | 4 | 6 | 8 | 10 | 12 |

Angle betweem a pair of consecutive spokes | 90º | 60º | … | … | … |

(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 15 spokes.

(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?

**Ans**

Let the unknown angles be x_{1}, x_{2} and x_{3}.

\begin{array}{cccccc}\text{Number of spokes}& 4& 6& 8& 10& 12\\ \begin{array}{l}\text{Angle between a pair of}\\ \text{consecutive spokes}\end{array}& {90}^{\circ}& {60}^{\circ}& {x}_{1}& {x}_{2}& {x}_{3}\end{array}

$\begin{array}{l}\text{From the given table},\text{we obtain}\\ \text{4}\times \text{9}0\text{\xb0}=\text{36}0\text{\xb0}=\text{6}\times \text{6}0\text{\xb0}\\ \begin{array}{l}\left(\text{i}\right)\text{Thus},\text{the number of spokes and the angle between a pair of consecutive}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}spokes are inversely proportional to each other}.\end{array}\\ \text{Now},{\text{we will find x}}_{\text{1}},{\text{x}}_{\text{2}}{\text{and x}}_{\text{3}}.\\ \\ 4\times 90\text{\xb0}=8\times {\text{x}}_{1}\\ \Rightarrow {\text{x}}_{1}=\frac{4\times 90\text{\xb0}}{8}={45}^{\circ}.\\ \\ \text{Similarly},{\text{x}}_{2}=\frac{4\times 90\text{\xb0}}{10}={36}^{\circ}{\text{and x}}_{3}=\frac{4\times 90\text{\xb0}}{12}={30}^{\circ}.\\ \text{Thus},\text{the following table is obtained}.\end{array}$

\begin{array}{cccccc}\text{Number of spokes}& 4& 6& 8& 10& 12\\ \begin{array}{l}\text{Angle between a pair of}\\ \text{consecutive spokes}\end{array}& {90}^{\circ}& {60}^{\circ}& {45}^{\circ}& {36}^{\circ}& {30}^{\circ}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\left(\text{ii}\right)\text{Let the angle between a pair of consecutive spokes on a wheel with 15 spokes be}x.\\ \text{Therefore},\text{4}\times \text{9}0\xb0\text{}=\text{15}\times x\\ \Rightarrow x=\frac{4\times 90\xb0}{15}={24}^{\circ}.\\ \text{Hence},\text{the angle between a pair of consecutive spokes of a wheel},\text{which has 15 spokes in it},\text{is 24}\xb0.\end{array}

$\begin{array}{l}\begin{array}{l}\left(\text{iii}\right)\text{Let the number of spokes in a wheel},\text{which has 4}0\xba\text{angles between}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em} a pair of consecutive spokes},\text{be}\mathrm{y}.\end{array}\\ \text{Therefore},\text{4}\times \text{9}0\xb0\text{}=\text{y}\times \text{4}0\xb0\\ \Rightarrow \mathrm{y}=\frac{4\times 90\xb0}{{40}^{\circ}}=9\\ \text{Hence},\text{the number of spokes in such a wheel is 9}.\end{array}$

**Q.4 **If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

**Ans**

Total number of children =24

If the number of the children is reduced by 4, the number of remaining children =24 – 4= 20

Let the number of sweets which each of the 20 students will get be x.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of students}& 24& 20\\ \text{Number of sweets}& 5& \mathrm{x}\end{array}$

If the number of children is reduced, then each student will get more number of sweets. So, this is the case of inverse proportion.

Therefore, we obtain

$\begin{array}{l}24\times 5=20\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{24\times 5}{20}=6\end{array}$

Hence, each student will get 6 sweets.

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

**Ans**

Total number of animals is 20.

If there are 10 more animals, then the total number of animals would be 30.

Let the number of days that the food will last if there were 10 more animals in the cattle be *x*.

The following table is obtained.

$\begin{array}{ccc}\text{Number of animals}& 20& 30\\ \text{Number of days}& 6& \mathrm{x}\end{array}$

The food will last longer if there is less number of animals.

Hence, the number of days the food will last and the number of animals are inversely proportional to each other.

Therefore,

$\begin{array}{l}\text{2}0\times \text{6}=\text{3}0\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{20\times 6}{30}=4\end{array}$

Therefore, the food will last for 4 days.

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

**Ans**

Let the number of days required by 4 persons to complete the job be *x*.

The given information is represented in the following table.

$\begin{array}{ccc}\text{Number of days}& 4& \mathrm{x}\\ \text{Number of persons}& 3& 4\end{array}$

If more persons are employed, it will take lesser time to complete the job.

Hence, the number of days and the number of persons required to complete the job are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 4\times \text{3}=\mathrm{x}\times 4\\ \Rightarrow \mathrm{x}=\frac{4\times 3}{4}=3\end{array}$

Therefore, the number of days required by 4 persons to complete the job is 3.

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

**Ans**

Let the number of boxes filled by using 20 bottles in each box be x.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of bottles}& 12& 20\\ \text{Number of boxes}& 25& \mathrm{x}\end{array}$

If more number of bottles are used, there will be less number of boxes.

Hence, the number of bottles and the number of boxes required to pack these are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 12\times \text{25}=20\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{12\times 25}{20}=15\end{array}$

Hence, the number of boxes required to pack 20 bottles is 15.

**Q.8 **A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Ans

Let the number of machines required to produce the same number of articles in 54 days be x.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of machines}& 42& \mathrm{x}\\ \text{Number of days}& 63& 54\end{array}$

If more number of machines is used, then less number of days it will take to produce the given number of articles.

So, the number of machines and the number of days required to produce the given number of articles are inversely proportional to each other.

$\begin{array}{l}\text{Therefore,}\\ 42\times \text{63}=54\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{42\times \text{63}}{54}=49\end{array}$

Hence, the required number of machines to produce the given number of articles in 54 days is 49.

**Q.9 **A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?

**Ans**

Let the time taken by the car to reach the destination, when it travels at a speed of 80 km/hr, be *x* hours.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Speed(in km/hr)}& 60& 80\\ \text{Tine taken(in hours)}& 2& \mathrm{x}\end{array}$

If the speed of the car is more, then it will take less time to reach the destination.

Hence, the speed of the car and the time taken by the car are inversely proportional to each other.

\begin{array}{l}\end{array} $\begin{array}{l}\text{Therefore},\\ \text{6}0\times \text{2}=\text{8}0\times \mathrm{x}\\ \Rightarrow \mathrm{x}=\frac{\text{6}0\times \text{2}}{80}=\frac{3}{2}\\ \\ \therefore \text{The time required by the car to reach the given}\\ \text{destination is}\frac{3}{2}\text{or 1}\frac{1}{2}\text{hours}.\end{array}$

**Q.10** Two persons could fit new windows in a house in 3 days.

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

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

**Ans**

(i) Since one of the persons fell ill before the work started, so we are left with only one person.

Let the number of days required by 1 man to fit all the windows be x.

The following table is obtained by the given information.

$\begin{array}{ccc}\text{Number of persons}& 2& 1\\ \text{Number of days}& 3& \mathrm{x}\end{array}$

If less number of persons are employed, then it will take more number of days to fit all the windows.

Hence, this is a case of inverse proportion.

Therefore,

2 × 3 = 1× *x*

*x* = 6

Hence, the number of days taken by 1 man to fit all the windows is 6.

(ii) Let the number of persons required to fit all the windows in one day be y.

The given information is represented in the following table:

$\begin{array}{ccc}\text{Number of persons}& 2& \mathrm{y}\\ \text{Number of days}& 3& 1\end{array}$

If less number of days is given, then more people would be required to fit all the windows.

Hence, it is a case of inverse proportion.

Therefore, by given information, we get

2 × 3 = *y *× 1

*y* = 6

Hence, 6 persons are required to fit all the windows in one day.

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

**Ans**

Let the duration of each period, when there are 9 periods a day in the school, be *x* minutes.

The following table is obtained from the given information.

$\begin{array}{ccc}\begin{array}{l}\text{Duration of each}\\ \text{period(in minutes)}\end{array}& 45& \mathrm{x}\\ \text{Number of periods}& 8& 9\end{array}$

If there is more number of periods a day in the school, then the duration of each period will be lesser.

Hence, it is case of inverse proportion.

$\begin{array}{l}\text{Therefore},\\ \text{45}\times \text{8}=\mathrm{x}\times \text{9}\\ \Rightarrow \mathrm{x}=\frac{\text{45}\times \text{8}}{9}=40\end{array}$

Hence, the duration of each period, when there are 9 periods a day in the school, will be 40 minutes.

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