# NCERT Solutions Class 8 Maths Chapter 13

## NCERT Solution for Class 8 Mathematics Chapter 13 – Direct and Inverse Proportions

Mathematics has constantly been gaining importance for years due to its wide range of applications. Many students find Mathematics to be hard, but with planned study, Mathematics can turn out to be the most interesting subject for them.

The chapter’s direct and inverse proportions are based on the proportionality of various conditions or applications directly or indirectly. This would help students to interpret conditions in different types of word problems and jump to conclusions in their higher classes. Students will be able to do situational analysis and write equations accurately after completing this chapter.

You will find a detailed guide for Class 8 Mathematics Chapter 13 in the NCERT Solution for Class 8 Mathematics Chapter 13. We have suggested certain steps to analyse any given word problem followed by writing the correct equations that will be helpful in almost every section of Algebra in higher classes of Mathematics.

Extramarks’ believe in high-quality academic content. Hence, all our NCERT Solutions for Class 8 Mathematics Chapter 13 are prepared by Mathematics teachers with decades of experience. You can avail of NCERT study material, NCERT solutions, CBSE revision notes, important set of questions with solved answers, and all other resources from our Extramarks’ website.

After studying and revising the chapter from our NCERT study solutions, students will get rid of their fears of solving word problems. As a result, they would approach their school as well as competitive examinations with more confidence.

## Key Topics Covered In NCERT Solution for Class 8 Mathematics Chapter 13

Class 8 Mathematics Chapter 13 helps in formulating the concepts of forming an equation in different conditional applications, which is largely used in the study of Algebra.

Our academics team has structured the study notes in our NCERT Solution for Class 8 Mathematics Chapter 13 in a way that students will gain full understanding about the chapter Direct and Inverse Proportions.  We have also provided a step by step guide to solving any given problem within a stipulated time frame. You can get access to our Class 8 Mathematics Chapter 13 solutions from our Extramarks’website.

### Introduction

In this chapter, we will cover proportionality. To explain proportionality, we are giving some examples below that will help you in understanding what proportionality is.

For example,

• If we increase the number of articles, its cost also increases.
• If more money is deposited in the bank, more interest is earned.
• If the speed of a vehicle increases, the time taken to cover the same distance will decrease.
• In office work, the more the workers, the less time is taken to cover the work will be less.

For more examples to understand this concept, students can refer to our NCERT Solution for Class 8 Mathematics Chapter 13. The NCERT solutions are designed to meet the demands of students in order to help them prepare for exams and score good marks.

### Direct Proportions

This section talks about the direct proportions. Direct proportions is the direct relationship between two variables.

Say, there are two variables X and Y. If there is an increment in X, there will be an increment in Y too i.e if there is an increment in an object, then there will be an increment in other objects too.

The two variables X and Y are said to be in direct proportions

If  X/Y = k or X = k.Y,, then the variables X and Y are in direct proportions, where k is the proportionality constant

Also, the direct proportions are given by X1.X2 = Y1.Y2

To learn more about the direct proportionality in detail, please go through our study material on our Extramarks’ website and refer to NCERT Solution for Class 8 Mathematics Chapter 13. All the concepts will be cleared through our NCERT solutions. As a result, it will help in building up their confidence.

### Inverse Proportions

In this section, we will learn about the inverse proportions. Inverse Proportion is the inverse relationship between two variables. Say there are two variables X and Y. If there is an increment in X, then there will be decrement in Y i.e. if there is an increment in one object, then there will be a decrement in other object.

The two variables X and Y are said to be in inverse proportions,

If X1 .Y1 = X2 .Y2 then the variables X and Y are in inverse proportions.

It is also given by X1/X2 = Y2/Y1 or X.Y= k , where k is the proportionality constant.

To get more detailed knowledge about the inverse proportions, refer to our NCERT Solution for Class 8 Mathematics Chapter 13 available on Extramarks’ website. The NCERT solutions provide detailed chapter-end questions which help them to comprehend the concepts. Hence, students learn to manage their time efficiently during their preparation.

## NCERT Solution for Class 8 Mathematics Chapter 13 Exercise & Solutions

We design NCERT solutions in the best way for the students. NCERT Solution for Class 8 Mathematics Chapter 13 includes all the exercises and solutions from the NCERT textbook. It is updated as per the latest CBSE syllabus. All the solutions are checked and analyzed by the subject matter experts before handing them over to students. Hence, you can trust the resources and use them to the fullest.

You can find for exercise specific questions and solutions for NCERT Class 8 Mathematics Chapter 13 by referring to the following links:

 Chapter 13 – Direct and Inverse Proportions Exercise 13.1 Question and answers Exercise 13.2 Question and answers

Along with NCERT Solution for Class 8 Mathematics Chapter 13, students can explore NCERT solutions on our Extramarks’ website for all primary and secondary classes.

## NCERT Exemplar for Class 8 Mathematics

NCERT Exemplar Class 8 Mathematics is a perfect guide for all the students who want to be a step ahead in their preparation. It has questions designed by a team of experts. Hence, one can trust its content and give credibility to all its resources.

The questions are NCERT based advanced level questions. Once students become capable of solving questions from these books, they can easily solve their NCERT Book level questions and hence will start scoring good marks in their examinations.

Students are advised to include NCERT Exemplar books in their study material by the experts because of the quality of questions. Students should study concepts from the NCERT Solution for Class 8 Mathematics Chapter 13 and practice problems from NCERT Exemplar. Once they do this, they will see a drastic change in their learning pattern.

### Key Features of NCERT Solution for Class 8 Mathematics Chapter 13

In order to excel in examinations, the conceptual understanding must be strong. Hence, NCERT Solution for Class 8 Mathematics Chapter 13 deeply emphasises on clarifying all your concepts. The key features are as follows:

• You will find all the academic notes being presented in a way that will help in quick understanding of each and every concept.
• You will also grasp the interlinked concepts in an efficient way, thus improving scores.
• After completing the NCERT Solution for Class 8 Mathematics Chapter 13, you will become an expert in solving word problems based on questions

Q.1

$\begin{array}{l}\begin{array}{l}\text{A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base.}\\ \text{In the following table, find the parts of base that need to be added.}\end{array}\\ \begin{array}{cccccc}\text{Parts of red pigment}& \text{1}& \text{4}& \text{7}& \text{12}& \text{20}\\ \text{Parts of base}& \text{8}& \dots & \dots & \dots & \dots \end{array}\end{array}$

Ans

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. Since the ratio of red pigment and the base is same every time, therefore, the parts of red pigments and the parts of base are in direct proportion.

Let us take the unknown values as x1, x2, x3 and x4.
The given information in the form of a table is as follows.

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& {x}_{1}& {x}_{2}& {x}_{3}& {x}_{4}\end{array}$ $\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{8}=\frac{4}{{\mathrm{x}}_{1}}⇒{\mathrm{x}}_{1}=8×4⇒{\mathrm{x}}_{1}=32\\ \frac{1}{8}=\frac{7}{{\mathrm{x}}_{2}}⇒{\mathrm{x}}_{2}=8×7⇒{\mathrm{x}}_{2}=56\\ \frac{1}{8}=\frac{12}{{\mathrm{x}}_{3}}⇒{\mathrm{x}}_{3}=8×12⇒{\mathrm{x}}_{3}=96\\ \frac{1}{8}=\frac{20}{{\mathrm{x}}_{4}}⇒{\mathrm{x}}_{4}=8×20⇒{\mathrm{x}}_{4}=160\end{array}$

Therefore, the parts of a base to be added are shown as follows:

$\begin{array}{cccccc}\text{Parts of red pigment}& 1& 4& 7& 12& 20\\ \text{Parts of base}& 8& 32& 56& 96& 160\end{array}$

Q.2 In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Ans

Let the parts of red pigment required to mix with 1800 mL of base be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Parts of red pigment}& 1& \mathrm{x}\\ \text{Parts of base(in mL)}& 75& 1800\end{array}$

The parts of red pigment and the parts of base are in direct proportion.
Therefore, we obtain

$\begin{array}{l}\text{According to direct proportion},\\ \frac{1}{75}=\frac{\mathrm{x}}{1800}⇒75\mathrm{x}=1800⇒\mathrm{x}=24\end{array}$

Thus, 24 parts of red pigments should be mixed with 1800 mL of base.

Q.3 A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Ans

Let the number of bottles filled by the machine in five hours be x.

The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Number of bottles}& 840& \mathrm{x}\\ \text{Time taken(in hours)}& 6& 5\end{array}$

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

$\begin{array}{l}\frac{840}{6}=\frac{\mathrm{x}}{5}⇒6\mathrm{x}=840×5\\ ⇒6\mathrm{x}=4200\\ ⇒\mathrm{x}=700\end{array}$ 

Thus, 700 bottles will be filled in 5 hours.

Q.4 A photograph of a bacterium enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacterium? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Ans

Let the actual length of the bacterium be x cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& x\\ \begin{array}{l}\text{Number of times photograph of}\\ \text{the bacterium was enlarged}\end{array}& 50000& 1\end{array}$

The number of times the photograph of the bacterium was enlarged and the lengths of the bacterium are in direct proportion.
Therefore, we obtain

$\begin{array}{l}\frac{5}{50000}=\frac{x}{1}\\ ⇒50000x=5\\ ⇒x=\frac{1}{10000}\\ ⇒x={10}^{-4}\\ \\ \text{Hence},\text{the actual length of the bacterium is 1}{0}^{-\text{4}}\text{cm}.\end{array}$

Now, we have to find its enlarged length if the photograph is enlarged 20,000 times only.

Let the enlarged length of the bacterium be y cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Length of the bacterium (in cm)}& 5& \mathrm{y}\\ \begin{array}{l}\text{Number of times photograph}\\ \text{of the bacterium was enlarged}\end{array}& 50000& 20000\end{array}$

The number of times the photograph of the bacterium was enlarged and the lengths of bacterium are in direct proportion.

Therefore, we obtain

$\begin{array}{l}\frac{5}{50000}=\frac{\mathrm{y}}{20000}\\ ⇒100000=50000\mathrm{y}\\ ⇒\mathrm{y}=\frac{100000}{50000}\\ ⇒\mathrm{y}=2\\ \text{Hence},\text{the enlarged length of the bacterium is 2 cm}.\end{array}$

Q.5 In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Ans

Let the length of the mast of the model ship be x cm.
The given information in the form of a table is as follows:

$\begin{array}{ccc}& \text{Height of mast}& \text{Length of ship}\\ \text{Model of ship}& 9& \mathrm{x}\\ \text{Actual ship}& 12& 28\end{array}$

The dimensions of the actual ship and the model ship are directly proportional to each other.

Therefore, we obtain:

$\begin{array}{l}\frac{9}{12}=\frac{\mathrm{x}}{28}\\ ⇒9×28=12\mathrm{x}\\ ⇒\mathrm{x}=\frac{252}{12}\\ ⇒\mathrm{x}=21\\ \end{array}$

Thus, the length of the model ship is 21 cm.

Q.6 Suppose 2 kg of sugar contains 9 × 106 crystals. How many sugar crystals are there in
(i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Ans

(i) Let the number of sugar crystals in 5 kg of sugar be x.
The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 5\\ \text{Number of crystals}& 9×{10}^{6}& \mathrm{x}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{2}{9×{10}^{6}}=\frac{5}{\mathrm{x}}\\ ⇒2\mathrm{x}=5×9×{10}^{6}\\ ⇒\mathrm{x}=\frac{5×9×{10}^{6}}{2}\\ ⇒\mathrm{x}=22.5×{10}^{6}\\ ⇒\mathrm{x}=2.25×{10}^{7}\\ \end{array}$

Hence, the number of sugar crystals is 2.25 × 107.

(ii) Let the number of sugar crystals in 1.2 kg of sugar be y.

The given information in the form of a table is as follows:

$\begin{array}{ccc}\text{Amount of sugar(in kg)}& 2& 1.2\\ \text{Number of crystals}& 9×{10}^{6}& \mathrm{y}\end{array}$

The amount of sugar and the number of crystals it contains are directly proportional to each other

.

$\begin{array}{l}\frac{2}{9×{10}^{6}}=\frac{1.2}{\mathrm{y}}\\ ⇒2\mathrm{y}=1.2×9×{10}^{6}\\ ⇒\mathrm{x}=\frac{1.2×9×{10}^{6}}{2}\\ ⇒\mathrm{x}=5.4×{10}^{6}\\ \\ \text{Hence},\text{the number of sugar crystals is}5.4×{10}^{6}.\end{array}$

Therefore, we obtain

Q.7 Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Ans

Let the distance represented on the map be x cm.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance covered on map(in cm)}& 1& \mathrm{x}\\ \text{Distance covered on road(in km)}& 18& 72\end{array}$

The distances covered on road and represented on map are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{1}{18}=\frac{\mathrm{x}}{72}\\ ⇒72=18\mathrm{x}\\ ⇒\mathrm{x}=\frac{72}{18}\\ ⇒\mathrm{x}=4\end{array}$

Hence, the distance represented on the map is 4 cm.

Q.8 A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5m long.

Ans

Let the length of the shadow of the other pole be x m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& 10.50\\ \text{Length of the shadow(in m)}& 3.20& \mathrm{x}\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{10.50}{\mathrm{x}}\\ ⇒5.60\mathrm{x}=10.50×3.20\\ ⇒\mathrm{x}=\frac{33.6}{5.60}\\ ⇒\mathrm{x}=6\end{array}$

Hence, the length of the shadow will be 6 m.

(ii) Let the height of the pole be y m.
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Height of the pole(in m)}& 5.60& \mathrm{y}\\ \text{Length of the shadow(in m)}& 3.20& 5\end{array}$

The height of an object and length of its shadow are directly proportional to each other.
Therefore, we obtain

$\begin{array}{l}\frac{5.60}{3.20}=\frac{\mathrm{y}}{5}\\ ⇒5.60×5=3.20\mathrm{y}\\ ⇒\mathrm{y}=\frac{5.60×5}{3.20}\\ ⇒\mathrm{y}=8.75\end{array}$

Thus, the height of the pole is 8.75 m or 8 m 75 cm.

Q.9 A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Ans

Let the distance travelled by the truck in 5 hours be x km.
We know, 1 hour = 60 minutes
∴ 5 hours = (5 × 60) minutes = 300 minutes
The given information in the form of a table is as follows.

$\begin{array}{ccc}\text{Distance travelled(in km)}& 14& \mathrm{x}\\ \text{Time(in min)}& 25& 300\end{array}$

The distance travelled by the truck and the time taken by the truck are directly proportional to each other.

Therefore, we obtain

$\begin{array}{l}\frac{14}{25}=\frac{\mathrm{x}}{300}\\ ⇒300×14=25\mathrm{x}\\ ⇒\mathrm{x}=\frac{4200}{25}\\ ⇒\mathrm{x}=168\end{array}$

Hence, the distance travelled by the truck is 168 km.

Q.10 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.11 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 x1, x2, x3, x4 and x5.

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

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

Q.12 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 x1, x2 and x3.

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

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

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

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

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

Q.13 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

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

Hence, each student will get 6 sweets.

Q.14 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,

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

Therefore, the food will last for 4 days.

Q.15 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.

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

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

Q.16 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.

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

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

Q.17 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.

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

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

Q.18 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.

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

Q.19 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.20 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.

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

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

Q.21 Following are the car parking charges near a railway station upto

4 hours ₹ 60

8 hours ₹ 100

12 hours ₹ 140

24 hours ₹ 180

Check if the parking charges are in direct proportion to the parking time.

Ans

The given information is represented in a table.

 Number of hours 4 8 12 24 Parking charges (in ₹ ) 60 100 140 180

The ratios of the parking charges to the number of hours are as follows:

$\frac{60}{4}=15,\frac{100}{8}=\frac{25}{2},\frac{140}{12}=\frac{35}{3},\frac{180}{24}=\frac{15}{2}$

Since all the ratios are different from each other, the parking charges are not in a direct proportion to the parking time.

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### 1. How can I make Mathematics my favourite subject?

Mathematics needs a lot of practice. By consistently studying the subject, one can make mathematics their favourite subject. When you consistently study mathematics, an interest in the subject begins to grow.

Interest will motivate you to learn more about the subject and get deeper knowledge. Due to this, you will be able to solve advanced-level questions, thereby making Mathematics your favourite subject and enhancing your self-confidence.

### 2. How will NCERT Solution for Class 8 Mathematics Chapter 13 benefit me in my preparation?

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### 3. What do you understand by direct proportion and indirect proportion?

Direct proportion is the direct relationship between two variables, whereas Indirect proportion is the indirect relationship between two variables.

In direct proportion, if there is an increment in one variable, there is also an increment in other variables. On the other hand, in indirect proportion, if there is an increment in one variable, there is a decrement in other variables.

You can find complete details about this chapter in our NCERT Solution for Class 8 Mathematics Chapter 13, available on the Extramarks’ website.