NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1

NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1 helps students understand the concept of direct proportion—a situation where two quantities increase or decrease together at the same rate. This exercise includes real-life examples such as cost calculation, distance-time problems, and more to make the concept clear and easy to understand.

These NCERT Solutions are prepared as per the official NCERT guidelines and explained in simple steps to help students learn effectively. Each question is solved in a step-by-step format so students can understand the logic and method easily. Practicing NCERT Solutions from Exercise 11.1 builds a strong foundation in solving proportion problems and improves overall mathematical thinking

1.0Download NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1 PDF with simple, step-by-step solutions for better understanding and practice.

2.0Key Concepts in Exercise 11.1 of Class 8 Maths Chapter 11

Exercise 11.1 of Chapter 11 – Direct and Inverse Proportions introduces the concept of direct proportion, which helps students understand how two quantities increase or decrease in the same ratio.

• Understanding Direct Proportion
  Two quantities are said to be in direct proportion when they increase or decrease together in the same ratio.
  That means, if one quantity doubles, the other also doubles.
• Mathematical Condition for Direct Proportion
  If two quantities x and y are in direct proportion, then:
  $\frac{x_1}{y_1}=\frac{x_2}{y_2}orx\alpha y$
• Real-Life Applications
  Examples include speed and distance, cost and quantity, number of workers and total wages (if time is constant), etc.
• Table-Based Questions
  Students solve problems by completing missing values in a table using cross-multiplication or direct proportion logic.
• Problem-Solving Practice
  Students apply logical reasoning to identify if the quantities given in word problems are in direct proportion.

3.0NCERT Class 8 Maths Chapter 11: Other Exercises

4.0NCERT Class 8 Maths Chapter 11 Exercise 11.1: Detailed Solutions

1. Followings 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. Sol. Since $\frac{4}{60}\ne \frac{8}{100}\ne \frac{12}{140}\ne \frac{24}{180}$ $\therefore$ The parking charges are not in direct proportion to the parking time.
2. A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. In the following table, find the parts of base that need to be added.

Sol. It is given that parts of red pigment, say $x$ and parts of base, say y are in direct proportion. Therefore, the ratio of the corresponding values of x and y remains constant. We have, $\frac{1}{8}=\frac{1}{8}$. So, x and y are in direct variation with the constant of variation equals to $\frac{1}{8}$. This means that x is $\frac{1}{8}$ of y and y is eight times of $x$. Thus, the required entries are $\frac{4}{32},\frac{7}{56},\frac{12}{96},\frac{20}{160}$. Thus, table becomes

The parts of red pigment and the parts of base are in direct proportion. Therefore, we obtain $\frac{1}{75}=\frac{\mathrm{x}}{1800}$ $\Rightarrow \mathrm{x}=\frac{1\times 1800}{75}\Rightarrow \mathrm{x}=24$ Thus, 24 parts of red pigment should be mixed with 1800 mL of base.

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

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain $\frac{840}{6}=\frac{x}{5}$ $\mathrm{x}=\frac{840\times 5}{6}=700$ Thus, 700 bottles will be filled in 5 hours.

5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
   
   Sol. Actual length of the bacteria $=\frac{5}{50000}\mathrm{cm}$ $=\frac{1}{10000}\mathrm{cm}=\frac{1}{10^4}\mathrm{cm}=10^{-4}\mathrm{cm}$ Suppose x be the enlarged length of the bacteria when its photograph is enlarged 20000 times. Then the information can be put in the following tabular form :

Clearly, it is a case of direct variation. $$\begin{gathered} \therefore \frac{5}{50000}=\frac{x}{20000} \\ \Rightarrow x=\frac{20000\times 5}{50000}=2\mathrm{cm} \end{gathered}$$ Hence, its enlarged length is 2 cm .

6. 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?
   
   Sol. Let the length of the mast of the model ship be xcm . The given information in the form of a table is as follows :

We know that the dimensions of the actual ship and the model ship are directly proportional to each other. Therefore, we obtain : $\frac{12}{9}=\frac{28}{x}$ $\mathrm{x}=\frac{28\times 9}{12}=21$ Thus, the length of the model ship is 21 cm.

7. Suppose 2 kg of sugar contains $9\times 10^6$ crystals. How many sugar crystals are there in (i) 5 kg of sugar (ii) 1.2 kg of sugar Sol. Let x and y crystals are in 5 kg of sugar and 1.2 kg of sugar. Then, the given information can be exhibited in the following tabular form :

Clearly, it is a case of direct variation. (i) $\frac{9\times 10^6}{2}=\frac{x}{5}\Rightarrow \mathrm{x}=\frac{5\times 9\times 10^6}{2}$ $\Rightarrow \mathrm{x}=\frac{9\times 10\times 10^6}{4}=\frac{9}{4}\times 10^7=2.25\times 10^7$ Hence, 5 kg of sugar contains $2.25\times 10^7$ crystals. (ii) $\frac{9\times 10^6}{2}=\frac{y}{1.2}\Rightarrow y=\frac{1.2\times 9\times 10^6}{2}$ $\Rightarrow \mathrm{y}=0.6\times 9\times 10^6=5.4\times 10^6$ Hence 1.2 kg of sugar contains $5.4\times 10^6$ crystals.

8. 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? Sol. Let the distance represented on the map be xcm . The given information in the form of a table is as follows.

The distances covered on road and represented on map are directly proportional to each other. Therefore, we obtain $\frac{18}{1}=\frac{72}{x}\Rightarrow x=\frac{72}{18}=4$ Hence, the distance represented on the map is 4 cm .

9. 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 5 m long. Sol. Let x m be the length of the pole whose shadow is of length 10 m 50 cm . Let y m be the length of the pole whose shadow is 5 m long. Then, the given information can be exhibited in the following tabular form :

Clearly, it is a case of direct variation. (i) $\frac{5.60}{3.20}=\frac{10.50}{x}$ $\Rightarrow \mathrm{x}=10.50\times \frac{3.20}{5.60}=6$ Hence, the length of the shadow is 6 m . (ii) $\frac{5.60}{3.20}=\frac{y}{5}\Rightarrow y=5\times \frac{5.60}{3.20}=8.75$ Hence, the length of the pole is 8.75 metres.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours? Sol. Let the distance travelled by the truck in 5 hours be xkm. We know, 1 hour $=60$ minutes $\therefore 5$ hours $=(5\times 60)$ minutes $=300$ minutes The given information in the form of a table is as follows.

The distance travelled by the truck and the time taken by the truck are directly proportional to each other. Therefore, $\frac{14}{25}=\frac{\mathrm{x}}{300}$ $\mathrm{x}=\frac{14\times 300}{25}=168$ Hence, the distance travelled by the truck is 168 km .

5.0Key Features and Benefits of Class 8 Maths Chapter 11 Exercise 11.1

• Introduction to Direct Proportion: Exercise 11.1 helps students understand how two quantities increase or decrease together in the same ratio.
• Simple Real-Life Examples: Concepts are explained using easy day-to-day examples like distance and time or cost and quantity, making learning relatable.
• Step-by-Step Practice: The questions help students apply the formula of direct proportion in a clear and structured way.
• Improves Problem-Solving Skills: Regular practice of this exercise builds accuracy and speed in solving ratio-based problems.
• NCERT-Aligned Questions: All problems are taken from the official NCERT textbook, which ensures students are well-prepared for tests and exams.