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NCERT Solutions Class 8 Maths Chapter 11 - Direct and Inverse Proportions Exercise 11.2

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NCERT Solutions for Maths Class 8 Chapter 11 Exercise 11.2 - FREE PDF Download

NCERT Class 8 Maths Chapter 11 Exercise 11.2 Solutions, contains complete answers to all questions in Exercise 11.2. Vedantu experts created these answers to help students fully understand the concepts of direct and inverse proportions. Start practising by downloading the FREE Class 8 Maths NCERT Solutions.

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Table of Content
1. NCERT Solutions for Maths Class 8 Chapter 11 Exercise 11.2 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 8 | Vedantu
3. Formulas Used in Class 8 Chapter 11 Exercise 11.2
4. Access NCERT Solutions for Maths Class 8 Chapter 11 - Direct and Inverse Proportions
5. Class 8 Maths Chapter 11: Exercises Breakdown
6. CBSE Class 8 Maths Chapter 11 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 8 Maths
8. Important Related Links for CBSE Class 8 Maths
FAQs


This exercise focuses on solving problems involving direct and inverse proportions, which are necessary for understanding how quantities change to one another. Students can improve their capacity to solve issues and exam preparation by practising these problems. The solutions have been verified with the most recent CBSE Class 8 Maths Syllabus and updated yearly to ensure accuracy and importance.


Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 8 | Vedantu

  • Class 8 Maths Chapter 11 Exercise 11.2 Solutions explains direct and inverse proportions, where direct proportion means two quantities increase or decrease together at the same rate.

  • Inverse proportion in this topic means that when one quantity increases, the other quantity decreases at the same rate. 

  • Direct proportion can be represented as $\frac{x}{y}=k$ where k is a constant. Inverse proportion can be represented as $x\times y=k$. 

  • Solving problems in this exercise involves identifying whether a situation represents direct or inverse proportion and applying the relevant formula. 

  • Understanding these concepts helps in recognizing how quantities relate to each other in various situations.

  • There are 11 fully solved questions in Chapter 11 Exercise 11.2 Direct and Inverse Proportions.


Formulas Used in Class 8 Chapter 11 Exercise 11.2

  • Direct Proportion: $\frac{x}{y}=k$

  • Inverse Proportion: $x\times y=k$

Access NCERT Solutions for Maths Class 8 Chapter 11 - Direct and Inverse Proportions

Exercise  11.2

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: In this situation, observe that the more the workers, the time taken to complete the job will be less. Thus, this is a situation of an inverse proportion.


ii. The time taken for a journey and the distance travelled at a uniform speed.

Ans: If time taken for a journey increases, then the distance travelled at uniform speed remains same. So, it is not the Inverse proportion.


iii. Area of cultivated land and the crop harvested.

Ans: The given situation is not an example of inverse proportion because in more areas, more crops can be harvested.


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

Ans: This is an example of an inverse proportion because with more speed, we can complete a certain distance in less time.


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

Ans: If the population is increasing, then the area of the land per person will be decreasing accordingly. So, they are in inverse proportion.


2. In a Television game show, the prize money of Rs1,00,000 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 for each winner (in Rs.)

100000

50000

-

-

-

-

-


Ans: Assume the missing ratios as ${x_1}$,${x_2}$,${x_3}$,${x_4}$, and ${x_5}$.

Tabulate the data as follows.

Number of winners

1

2

4

5

8

10

20

Prize for each winner (in Rs.)

100000

50000

${x_1}$

${x_2}$

${x_3}$

${x_4}$

${x_5}$


From the table, we get $1 \times 100000 = 100000$ and

$2 \times 50000 = 100000$.

The number of winners and the prize money are inversely proportional to each other.

$1 \times 100000 = 4 \times {x_1}$

Divide both sides by 4 to solve the equation for ${x_1}$.

\[\dfrac{{1 \times 100000}}{4} = {x_1}\]

\[25000 = {x_1}\]

Now, find the value of ${x_2}$.

$1 \times 100000 = 5 \times {x_2}$

Divide both sides by 5 to solve the equation for ${x_2}$.

\[\dfrac{{1 \times 100000}}{5} = {x_2}\]

\[20000 = {x_2}\]

Now, find the value of ${x_3}$.

$1 \times 100000 = 8 \times {x_3}$

Divide both sides by 8 to solve the equation for ${x_3}$.

\[\dfrac{{1 \times 100000}}{8} = {x_3}\]

\[12500 = {x_3}\]

Now, find the value of ${x_4}$.

$1 \times 100000 = 10 \times {x_4}$

Divide both sides by 10 to solve the equation for ${x_4}$.

\[\dfrac{{1 \times 100000}}{{10}} = {x_4}\]

\[10000 = {x_4}\]

Now, find the value of ${x_5}$.

$1 \times 100000 = 20 \times {x_5}$

Divide both sides by 20 to solve the equation for ${x_5}$.

\[\dfrac{{1 \times 100000}}{{20}} = {x_5}\]

\[5000 = {x_5}\]

Thus, the complete table is as follows.

Number of winners

1

2

4

5

8

10

20

Prize for each winner (in Rs.)

100000

50000

25000

20000

12500

10000

5000


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.

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


Number of spokes

4

6

8

10

12

Angle between a pair of consecutive spokes

$90^\circ $

$60^\circ $

-

-

-


Ans: From the given table, we obtain, 

$4 \times 90^\circ  = 360$ and

$6 \times 60^\circ $.

Number of spokes

4

6

8

10

12

Angle between a pair of consecutive spokes

$90^\circ $

$60^\circ $

${x_1}$

${x_2}$

${x_3}$


Thus, we can observe that the number of spokes and the angle between a pair of consecutive spokes are inversely proportional to each other.

$4 \times 90^\circ  = {x_1} \times 8$

Divide both sides by 8.

${x_1} = \dfrac{{4 \times 90^\circ }}{8}$

${x_1} = 45^\circ $

Similarly, find the value of ${x_2}$.

${x_2} = \dfrac{{4 \times 90^\circ }}{{10}}$

${x_2} = 36^\circ $

Now, find the value of ${x_3}$.

${x_3} = \dfrac{{4 \times 90^\circ }}{{12}}$

${x_3} = 30^\circ $

Thus, tabulate the table obtained.

Number of spoked

4

6

8

10

12

Angle between a pair of consecutive spokes

$90^\circ $

$60^\circ $

$45^\circ $

$36^\circ $

$30^\circ $


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

Ans: Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.

ii. Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.

Ans: Let the angle between a pair of consecutive spokes on a wheel with 15 spokes be $x$.

$4 \times 90^\circ  = 15 \times x$

Divide both sides by 15.

$x = \dfrac{{4 \times 90^\circ }}{{15}}$

$x = 24^\circ $

Hence, the angle between the pair of consecutive spokes of a wheel, which has 15 spokes in it, is $24^\circ $.

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

Ans: Let us assume that the number of spokes in a wheel, which has $40^\circ $ angles between a pair of spokes as $y$.

$4 \times 90^\circ  = y \times 40^\circ $

Divide both sides by $40^\circ $.

$y = \dfrac{{4 \times 90^\circ }}{{40^\circ }}$

$y = 9$

Hence, the number of spokes in such a wheel is 9.

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

Ans: It is given that the number of children is reduced by 4 and the number of children is 24. So, the number of children will be, $\left( {24 - 4} \right)$.

Thus, the number of children left are 20.

Now, tabulate the data as follows.

Number of students

24

20

Number of sweets

5

$x$


If the number of students is less, then each student will get a greater number of sweets. Thus, the given problem follows an inverse proportion.

$24 \times 5 = 20 \times x$

Divide both sides by 20.

$\dfrac{{24 \times 5}}{{20}} = x$

$6 = x$

Hence, each student will get 6 sweets.

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

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

Tabulate the data as follows.

Number of animals

20

$20 + 10 = 30$

Number of days

6

$x$


In the given scenario we can see that, more the number of animals, less will be the number of days for which the food will last. Thus, the number of days the food will last and the number of animals is inversely proportional to each other.

$20 \times 6 = 30 \times x$

Divide both sides by 30.

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

$\dfrac{{120}}{{30}} = x$

$4 = x$

Thus, the food will last for 4 days.

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

Ans: Let us assume that the number of days required by 4 persons to complete the job be $x$.Tabulate the data as follows.

Number of days

4

$x$

Number of persons

3

4


We can see that if the number of persons is more, then it will take less time to complete the job. Therefore, the number of days and the persons required are inversely proportional to each other.

$4 \times 3 = x \times 4$

Divide both sides by 4.

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

$\dfrac{{12}}{4} = x$

$3 = x$

Hence, the number of days required to complete the job is 3.

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?


20 bottles in each box


Ans: Assume the number of boxes filles, by using 20 bottles in each box be $x$.

Tabulate the data as follows.


Number of bottles

12

20

Number of boxes

25

$x$


Observe that more the number of bottles, lesser the number of boxes required. Thus, the number of bottles and boxes are inversely proportional to each other.

$12 \times 25 = 20 \times x$

Divide both sides by 20.

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

$15 = x$

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

8. A factory required 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 be $x$.

Tabulate the data as follows.

Number of machines

42

$x$

Number of days

63

54


Observe that more the number of machines, less will be the number of days that it will take to produce the given number of articles.

Thus, the given problem follows an inverse proportion.

$42 \times 63 = 54 \times x$

Divide both sides by 54.

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

$49 = x$

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


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

Ans: Let the time taken by the car to reach the destination by travelling at the speed of 60 km/hr be $x$ hours.

Tabulate the data as follows.

Speed(in km/hr)

60

80

Time taken (In hours)

2

$x$


We can see that if the speed of the car is more then, less time will be taken. Thus, the speed of the car and time taken by the car are inversely proportional to each other.

$60 \times 2 = 80 \times x$

Divide both sides by 80.

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

$\dfrac{{120}}{{80}} = x$

$\dfrac{3}{2} = x$

Thus, the time required by the car to reach the given destination is $\dfrac{3}{2}$ hours.


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

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

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

Tabulate the data as follows.


Number of persons

2

1

Number of days

3

$x$


We can see that a greater number of days will be required if the number of persons are less. Thus, this represents an inverse proportion. Therefore, 

$2 \times 3 = 1 \times x$

Solve the equation.

$6 = x$

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

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

Ans: Let the number of persons required to fit all the windows in one day be $y$.

Tabulate the data as follows.

Number of persons

2

$y$

Number of days

3

1


We can see that a lesser number of days will be required if the number of persons are more. Thus, this represents an inverse proportion. Therefore, 

$2 \times 3 = 1 \times y$

Solve the equation.

$6 = y$

Hence, the number of persons required to fit all the windows in one day is 6.


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 be $x$ minutes.

Tabulate the data as follows.

Duration of each period (in minutes)

45

$x$

Number of periods

8

9


Observe that if a greater number of periods will be held in a day, then the duration of each period will be lesser.

Thus, this is the case of inverse proportion.

$45 \times 8 = x \times 9$

Divide both sides by 9 and solve for $x$.

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

$x = 5 \times 8$

$x = 40$

Hence, the duration of each period will be 40 minutes.


Conclusion

In NCERT Maths Class 8 Chapter 11 Exercise 11.2, students work on understanding direct and inverse proportions. It is essential to understand how quantities connect, whether they increase together or decrease together. By completing these questions, students will improve their ability to identify and use the appropriate formulas for various contexts. This practice is important for developing a strong base with equal ideas, which are frequently applied in real-world situations. To master these necessary concepts, students must practise them carefully.


Class 8 Maths Chapter 11: Exercises Breakdown

Exercise

Number of Questions

Exercise 11.1

10 Questions & Solutions


CBSE Class 8 Maths Chapter 11 Other Study Materials


Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Related Links for CBSE Class 8 Maths

FAQs on NCERT Solutions Class 8 Maths Chapter 11 - Direct and Inverse Proportions Exercise 11.2

1. What is the focus of Class 8 Maths Ch 11 Ex 11.2?

Class 8 Maths Chapter 11 Exercise 11.2 Solutions focuses on the concepts of direct and inverse proportions. It involves understanding how two quantities relate to each other, either increasing together or increasing while the other decreases.

2. Why are direct and inverse proportions important in Class 8 Maths Chapter 11 Exercise 11.2 Solutions?

Direct and inverse proportions are important because they help us understand the relationship between two quantities in various real-life situations, such as speed and travel time, or workers and the time taken to complete a task.

3. How is direct proportion represented mathematically in Class 8 Maths Ch 11 Ex 11.2?

According to Class 8 Maths Ch 11 Ex 11.2, Direct proportion is represented by the formula $\frac{x}{y}=k$, where 𝑥 and 𝑦 are the two quantities, and 𝑘 is a constant. This means that as one quantity increases, the other also increases at a constant rate.

4. How is inverse proportion represented mathematically in Class 8 Maths Ch 11 Ex 11.2?In Class 8 Maths Ch 11 Ex 11.2 Inverse proportion is represented by the formula $x\times y=k$, where 𝑥 and 𝑦 are the two quantities, and 𝑘 is a constant. This means that as one quantity increases, the other decreases at a constant rate.


In Class 8 Maths Ch 11 Ex 11.2 Inverse proportion is represented by the formula $x\times y=k$, where 𝑥 and 𝑦 are the two quantities, and 𝑘 is a constant. This means that as one quantity increases, the other decreases at a constant rate.

5. Can you give an example of a direct proportion problem in Maths Class 8 Chapter 11 Exercise 11.2?

An example of a direct proportion problem is if the cost of 5 apples is Rs. 100, then the cost of 10 apples would be Rs. 200. This is because the cost increases in direct proportion to the number of apples.

6. Can you give an example of an inverse proportion problem in NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2?

An example of an inverse proportion problem is if 4 workers can complete a task in 8 hours, then 8 workers can complete the same task in 4 hours. This is because the time taken decreases as the number of workers increases.

7. Why is it important to understand the constant 𝑘 in proportions in NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2?

Understanding the constant 𝑘 is important because it helps to identify the relationship between the quantities involved. It allows us to predict how one quantity will change when the other changes.

8. How can practising NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 help in exams?

Practising NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 helps students become familiar with the types of problems that can appear in exams. It also enhances their problem-solving skills and understanding of direct and inverse proportions, which are common in many maths problems.

9. Are there any real-life applications of direct and inverse proportions in NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2?

Yes, there are many real-life applications of direct and inverse proportions, such as calculating travel time based on speed, determining the amount of ingredients needed for a recipe, and understanding the relationship between distance and time in physics.

10. What should students focus on when solving NCERT Class 8 Maths Chapter 11 Exercise 11.2 problems?

Students should focus on, correctly identifying whether the problem involves direct or inverse proportion. They should then apply the appropriate formula and carefully solve for the unknown quantity.

11. How do Vedantu's NCERT Solutions help with NCERT Class 8 Maths Chapter 11 Exercise 11.2?

Vedantu's NCERT Solutions provides step-by-step explanations and detailed answers to all the questions in NCERT Class 8 Maths Chapter 11 Exercise 11.2. These solutions help students understand the concepts better and improve their problem-solving skills, making it easier to tackle similar questions in exams.