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

NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.2 focuses on the concept of inverse proportion—where one quantity increases while the other decreases in a fixed manner. This exercise helps students understand how to solve real-life problems like speed and time, number of workers and days, etc., using the idea of inverse relationships.

These NCERT Solutions are prepared according to the official NCERT textbook and provide step-by-step explanations that are easy to follow. With simple examples and clear logic, students can understand how inverse proportion works and how to apply it in different situations. Practicing NCERT Solutions from Exercise 11.2 strengthens students’ skills in solving proportion problems and boosts their overall confidence in maths.

1.0Download NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.2: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.2 PDF with clear, step-by-step answers for easy learning and revision.

2.0Key Concepts in Exercise 11.2 of Class 8 Maths Chapter 11

Exercise 11.2 of Chapter 11 – Direct and Inverse Proportions focuses on the concept of inverse proportion, helping students understand how one quantity increases while the other decreases in a consistent ratio.

• Understanding Inverse Proportion
  Two quantities are said to be in inverse proportion when an increase in one causes a proportional decrease in the other and vice versa.
  For example, if more workers are assigned to a task, the time taken to complete it decreases.
• Mathematical Condition for Inverse Proportion
  If x and y are in inverse proportion, then:
  x×y=constant or $x_1\times y_1=x_2\times y_2$
• Real-Life Examples
  Includes problems like speed and travel time, number of taps and time to fill a tank, number of workers and days to complete a job, etc.
• Table Completion Problems
  Students solve table-based problems by keeping the product of the two variables constant and finding the missing values.

3.0NCERT Class 8 Maths Chapter 11: Other Exercises

4.0NCERT Class 8 Maths Chapter 11 Exercise 11.2: Detailed Solutions

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. Sol. (i) We know that, more is the number of workers to do a job, less is the time taken to finish the job. So, it is a case of inverse proportion. (ii) Clearly, more the time taken, more is the distance covered. So, it is a case of direct variation. (iii) Clearly, more is the area of cultivated land, more is the crop harvested. So, it is a case of direct proportion. (iv) Clearly, more the time taken less is the speed of vehicle. So, it is a case of inverse variation. (v) Clearly, more is the population, less is the area of land per person in a country. So, it is a case of inverse proportion.
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?

Sol. Clearly, more the number of winners, less is the prize for each winner. So, it is a case of inverse proportion. $\therefore 4\times \mathrm{x}=1\times 100000$ $\Rightarrow \mathrm{x}=\frac{100000}{4}=25000$ Thus, for 4 it is ₹ 25000 $5\times y=1\times 100000$ $\Rightarrow \mathrm{y}=\frac{100000}{5}=20000$ Thus, for 5 it is ₹ 20000 . $8\times \mathrm{z}=1\times 100000$ $\Rightarrow \mathrm{z}=\frac{100000}{8}=12500$ Thus, for 8 it is $₹12500$. $10\times t=1\times 100000$ $\Rightarrow \mathrm{t}=\frac{100000}{10}=10000$ Thus, for 10 it is ₹ 10000 . $20\times \mathrm{w}=1\times 100000$ $\Rightarrow \mathrm{w}=\frac{100000}{20}=5000$ Thus, for 20 it is ₹ 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.
   

(i) Are the number of spokes and the angles formed between the pair 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 $40^{\circ }$ ? Sol. A table of the given information is as follows.

From the given table, we obtain $4\times 90^{\circ }=360^{\circ }=6\times 60^{\circ }$ Thus, the number of spokes and the angle between a pair of consecutive spokes are inversely proportional to each other. Therefore, $4\times 90^{\circ }={\mathrm{x}}_1\times 8$ ${\mathrm{x}}_1=\frac{4\times 90^{\circ }}{8}=45^{\circ }$ Similarly, $x_2=\frac{4\times 90^{\circ }}{10}=36^{\circ }$ and ${\mathrm{x}}_3=\frac{4\times 90^{\circ }}{12}=30^{\circ }$ Thus, the following table is obtained.

(i) Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion. (ii) Let the angle between a pair of consecutive spokes on a wheel with 15 spokes be x . Therefore, $4\times 90^{\circ }=15\times \mathrm{x}$ $\mathrm{x}=\frac{4\times 90^{\circ }}{15}=24^{\circ }$ Hence, the angle between a pair of consecutive spokes of a wheel, which has 15 spokes in it, is $24^{\circ }$. (iii) Let the number of spokes in a wheel, which has $40^{\circ }$ angles between a pair of consecutive spokes, be $y$. Therefore, $4\times 90^{\circ }=\mathrm{y}\times 40^{\circ }$ $y=\frac{4\times 90^{\circ }}{40}=9$ 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 would each get, if the number of the children is reduced by 4 ? Sol. Number of remaining children $=24-4$ $=20$ Let the number of sweets which each of the 20 students will get, be $x$. The following table is obtained.

If the number of students is lesser, then each student will get more number of sweets. Since this is a case of inverse proportion, $24\times 5=20\times x$ $\mathrm{x}=\frac{24\times 5}{20}=6$ 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? Sol. Let the food will now last for x days. Then,

Clearly, more is the number of animals, less will be the number of days for food to last. So, it is a case inverse proportion. $\therefore 20\times 6=30\times \mathrm{x}\Rightarrow \mathrm{x}=\frac{20\times 6}{30}=4$ Hence, the food will now last for 4 days.

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? Sol. Let the number of days $=x$ According to question

Since the given problem is in inverse proportion. $\therefore 4\times 3=\mathrm{x}\times 4$ $\Rightarrow \frac{4\times 3}{4}=x\Rightarrow x=3$ $\therefore 4$ persons will take 3 days to complete the job.

7. A batch of bottles were 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?
   
   Sol. Let x boxes be needed when 20 bottles are packed in each box. Then,

Clearly, more is the number of bottles per box, less will be the number of boxes needed for packing. So, it is a case of inverse proportion. $\therefore 25\times 12=\mathrm{x}\times 20\Rightarrow \mathrm{x}=\frac{25\times 12}{20}=15$ Hence, the boxes needed for packing is 15 .

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? Sol. Let the number of machines to produce the articles in 54 days $=x$ According to question

Since number of days are decreasing, number of machines must be increasing. $\therefore$ The problem is in inverse proportion i.e., $x=\frac{63\times 42}{54}\Rightarrow x=49$ Thus, number of machines required to produce the same number of articles in 54 days $=49$.

9. A car takes 2 hours to reach a destination by travelling at the speed of $60\mathrm{km}/\mathrm{h}$. How long will it take when the car travels at the speed of $80\mathrm{km}/\mathrm{h}$ ? Sol. Let the car takes x hours to reach a destination by travelling at the speed of $80\mathrm{km}/\mathrm{h}$. Then,

Clearly, more the speed, less will be the time taken. So, it is a case of inverse proportion. $\therefore 60\times 2=80\times \mathrm{x}\Rightarrow \mathrm{x}=\frac{60\times 2}{80}=\frac{3}{2}$ Hence, the time taken will be $1\frac{1}{2}$ hours.

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? Sol. (i) Let the number of days required by 1 man to fit all the windows be x . The following table is obtained.

Lesser the number of persons, more will be the number of days required to fit all the windows. Hence, this is a case of inverse proportion. Therefore, $2\times 3=1\times x$ $\mathrm{x}=6$ 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 following table is formed.

Lesser the number of days, more will be the number of person required to fit all the windows. Hence, this is a case of inverse proportion. Therefore, $2\times 3=y\times 1$ $y=6$ Hence, 6 person are required to fit all the windows in one day.

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? Sol. Let x minutes be the duration of period when the school has 9 periods a day. Then

Clearly, more the periods, less will be the duration of the period. So, it is a case of inverse proportion. $$\begin{gathered} \therefore 8\times 45=9\times x \\ \Rightarrow x=\frac{8\times 45}{9}=40 \end{gathered}$$ Hence, the duration of period is 40 minutes.

5.0Key Features and Benefits of Class 8 Maths Chapter 11 Exercise 11.2

• Introduction to Inverse Proportion: Exercise 11.2 teaches students how one quantity increases when the other decreases, and vice versa — a key idea in many real-life situations.
• Real-World Applications: This concept is used in examples like speed and time, number of workers and days — making it practical and easy to relate to.
• Clear Problem Sets: The exercise provides well-structured problems that guide students in identifying inverse relationships between quantities.
• Builds Logical Thinking: Solving these questions helps improve reasoning and analytical skills by understanding how two quantities are connected.