NCERT Solutions Class 8 Maths Chapter 11 Direct and Inverse Proportions

Class 8 Maths Chapter 11, Direct and Inverse Proportions, helps students understand the mathematical relationships between two quantities that vary either directly or inversely. This chapter explains how such relationships can be analyzed using the concepts of direct and inverse proportions. To make these concepts more accessible to students, ALLEN provides detailed NCERT Solutions for Class 8 Maths Chapter 11, helping students grasp the relevant formulas and concepts with ease.

1.0NCERT Solutions for Class 8 Maths Chapter 11 PDF - Free Download

Students can easily download the PDF from the table provided below, ensuring access to the content anytime and anywhere. This convenience allows them to study and revise the material at their own pace, making learning more flexible and efficient.

2.0What Will Students Learn in Chapter 11: Direct and Indirect Proportions?

• Understanding when two quantities increase or decrease in the same ratio and how an increase in one quantity results in a proportional decrease in the other.
• Real-Life Examples: Applying direct and inverse proportions in real-life examples like speed vs time, number of workers vs work completed, and cost vs quantity.
• Recognizing the importance of the constant in maintaining proportional relationships..
• Plotting and interpreting graphs to visualize proportional relationships.

3.0NCERT Questions with Solutions for Class 8 Maths Chapter 11 - Detailed Solutions

EXERCISE : 11.1

• 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.
• 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

• In question 2 , if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base? Sol. 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.

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.

• 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.

• 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 .

• 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.

• 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.

• 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 .

• 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.

• 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 .

EXERCISE : 11.2

• 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.
• 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 .

• 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 between a pair of consecutive spokes	$90^{\circ }$	$60^{\circ }$	---	---	---

  (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.

  Number of spokes	4	6	8	10	12
  Angle between a pair of consecutive spokes	$90^{\circ }$	$60^{\circ }$	$x_1$	$x_2$	$x_3$

  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.

  Number of spokes	$4$	$6$	$8$	$10$	$12$
  Angle between a pair of consecutive spokes	$90^{\circ }$	$60^{\circ }$	$45^{\circ }$	$36^{\circ }$	$30^{\circ }$

  (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 .
  • 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.

  Number of students	$24$	$20$
  Number of sweets	5	x

  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.
  • 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,

  Number of animals	$20$	$30$
  Number of days	6	x

  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.
  • 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

  Number of days	$4$	$\mathbf{x}$
  Number of persons	3	4

  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.
  • 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,

  Number of boxes	$25$	$\mathbf{x}$
  Number of bottles per box	12	20

  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 .
  • 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

  Number of days	$63$	$54$
  Number of machines	42	x

  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 $=49$.
  • 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,

  Speed (in km/hr)	$60$	$80$
  Time (in hours)	2	x

  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.
  • 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.

  Number of persons	$2$	$1$
  Number of days	3	x

  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.

  Number of persons	$2$	$\mathbf{y}$
  Number of days	3	1

  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.
  • 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

  Number of periods	$8$	$9$
  Duration of periods (in minutes)	45	x

  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.