Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

ffImage
Last updated date: 13th Jul 2024
Total views: 843.3k
Views today: 18.43k

NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions - FREE PDF Download

NCERT Solutions for Class 8 Maths Chapter Direct and Inverse Proportion Solutions by Vedantu explores the concepts of Direct and Inverse proportions, which are essential for understanding how quantities relate to each other. In direct proportion, as one quantity increases, the other quantity also increases at the same rate. In indirect proportion, as one quantity increases, the other quantity decreases.

toc-symbol
Table of Content
1. NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions - FREE PDF Download
2. Glance on Maths Chapter 11 Class 8 - Direct and Inverse Proportions
3. Access Exercise wise NCERT Solutions for Chapter 11 Maths Class 8
4. Exercises Under NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions
5. Access NCERT Solutions for Class 8 Maths Chapter 11 - Direct and Inverse Proportions
    5.1Exercise 11.1
    5.2Exercise 11.2
6. Class 8 Maths Chapter 11: Exercises Breakdown
7. Other Study Material for CBSE Class 8 Maths Chapter 11
8. Chapter-Specific NCERT Solutions for Class 8 Maths
FAQs


Vedantu’s solutions for this chapter provide clear and detailed explanations to help you understand these concepts. Each solution is presented step-by-step, making it easy to follow and apply the methods to solve problems. By studying these solutions, you will build a solid foundation in Direct and Inverse proportions, which is crucial for higher-level math and real-life applications.


Glance on Maths Chapter 11 Class 8 - Direct and Inverse Proportions

  • In Class 8 Math Direct and Inverse Proportion understand what proportions are and how they relate two quantities.

  • If Two quantities a and b are in direct proportion, the value of a increases or decreases, then the value of b increases or decreases in such a way that the ratio \[\frac{a}{b}\] remains constant is called Direct Proportion.

  • Two quantities a and b are said to be in inverse proportion, if whenever the value of a increases or decreases then the value of b decreases or increases in such a way that \[\frac{a}{b}\] remains constant is called Indirect Proportion.

  • Maths Class 8 Direct and Inverse Proportions article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 11.

  • Class 8 Chapter 11 Maths Direct and Inverse Proportions consists of two exercises (21 fully solved questions) solved by master teachers for 2024-25.


Access Exercise wise NCERT Solutions for Chapter 11 Maths Class 8

Exercises Under NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions

  • Exercise 11.1: This exercise consists of 10 fully solved questions. This exercise focuses on understanding and solving problems related to direct proportion. By using a relationship  $\frac{x}{y}=k$ you can identify situations where two quantities are directly proportional.

  • Exercise 11.2: This exercise consists of 11 fully solved problems. This exercise focuses on solving problems related to indirect (inverse) proportions. By using a relationship  $x\cdot y=k$ you can identify situations where two quantities are indirectly proportional.


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

Exercise 11.1

1. Following are the car parking charges near a railway station up to  $ \text{4}\ \text{hrs}\ \text{Rs}\ \text{60} $ ,  $ \text{8}\ \text{hrs}\ \text{Rs}\ \text{100} $ ,  $ \text{12}\ \text{hrs}\ \text{Rs}\ \text{140} $ , and  $ \text{24}\ \text{hrs}\ \text{Rs}\ \text{180} $ . Check if the parking charges are in direct proportion to the parking time.

Ans: We know the charges will be in direct proportion to the parking time when each of their ratios will be the same.

We can form a table by using the given information as –

Table for Car Parking Charges

Time taken for parking (hrs)

4

8

12

24

Charges (Rs.)

60

100

140

180


Therefore, the ratio of the parking charges to the time taken for parking will be  $\text{=}\dfrac{\text{charge}}{\text{time taken}}$

Hence,

 $ \dfrac{60}{4}=15 $ ,

 $ \Rightarrow \dfrac{100}{8}=\dfrac{25}{2} $ ,

 $ \Rightarrow \dfrac{140}{12}=\dfrac{35}{3} $ ,

 $ \Rightarrow \dfrac{180}{24}=\dfrac{15}{2} $ .

As, we can observe that the ratios are not the same. Hence, we can conclude that the parking charges are not in direct proportion to the parking time.


2. A mixture of paint is prepared by mixing  $ \text{1} $  part of red pigments with  $ \text{8} $  parts of the base. In the following table, find the parts of the base that need to be added.

Parts of red pigment

1

4

7

12

20

Parts of base

8

_

_

_

_


Ans:

Given we have a mixture of paint that is prepared by mixing  $ 1 $  part of red pigments with  $ 8 $  parts of base. We can observe that they are in a direct proportion. Therefore,

Let  $ {{a}_{1}},{{a}_{2}},{{a}_{3}}, $  and  $ {{a}_{4}} $  denote the parts of base of the respective parts of red pigment.

Hence, we have –

 $ \dfrac{8}{1}=\dfrac{{{a}_{1}}}{4} $ 

 $ \Rightarrow {{a}_{1}}=32 $ .

Similarly,  $ \dfrac{8}{1}=\dfrac{{{a}_{2}}}{7} $ 

 $ \Rightarrow {{a}_{2}}=56 $ ,

 $ \dfrac{8}{1}=\dfrac{{{a}_{3}}}{12} $ 

 $ \Rightarrow {{a}_{3}}=96 $ , and

 $ \dfrac{8}{1}=\dfrac{{{a}_{4}}}{20} $ 

 $ \Rightarrow {{a}_{4}}=160 $ .

Therefore, now the given table will be as –

Parts of red pigment

1

4

7

12

20

Parts of base

8

32

56

96

160


3. In Question  $ \text{2} $  above, if  $ \text{1} $  part of a red pigment requires  $ \text{75mL} $  of base, how much red pigment should we mix with  $ \text{1800mL} $  of base?

Ans: Let us assume that  $ a $  denotes the parts of red pigment required. We know that parts of the base are in direct proportion to parts of red pigment.

Therefore,

 $ \dfrac{75}{1}=\dfrac{1800}{a} $ 

 $ \Rightarrow a=\dfrac{1800}{75} $ 

 $ \Rightarrow a=24 $ .

Hence, the table will be as –

Parts of red pigment

1

24

Parts of base

75

1800


4. A machine in a soft drink factory fills  $ \text{840} $  bottles in six hours. How many bottles will it fill in five hours?

Ans: Given we have a soft drink factory that fills  $ 840 $  bottles in six hours. Let us assume that the number of bottles to be filled in five hours will be  $ x $.

Therefore, we can conclude that both the parameters are in direct proportion.

Hence,

 $ \dfrac{840}{6}=\dfrac{x}{5} $ 

 $ \Rightarrow x=\dfrac{840\times 5}{6} $ 

 $ \Rightarrow x=700 $ 

Therefore, the number of bottles to be filled in five hours will be  $ 700 $ .


5. A photograph of a bacteria enlarged  $ \text{50,000} $  times attains a length of  $ \text{5cm} $ . What is the actual length of the bacteria? If the photograph is enlarged  $ \text{20,000} $  times only, what would be its enlarged length?

Ans: Let us assume that the actual length of the bacteria is  $ a\ \text{cm} $  and let us assume that its enlarged length will be  $ b\ \text{cm} $ . The given information implies that the parameters are in direct proportion.

Hence, we can compute a table as –

Length of bacteria (cm)

5

a

b

Number of times bacteria enlarged

50000

1

20000


Therefore, the actual length of the bacteria when the photograph was not enlarged will be –

 $ \dfrac{5}{50000}=\dfrac{a}{1} $ 

 $ \Rightarrow a=\dfrac{1}{10000} $ 

 $ \Rightarrow a={{10}^{-4}}cm $ 

Now, we will calculate the enlarged length of the bacteria when the photograph is enlarged  $ 20,000 $  times.

 $ \dfrac{5}{50000}=\dfrac{b}{20000} $ 

 $ \Rightarrow b=2cm $ .

Therefore, the actual length of the bacteria is  $ 0.0001\ cm $  and the enlarged length is  $ \text{2cm} $ .


6. In a model of a ship, the mast is  $ \text{9}\ \text{cm} $  high, while the mast of the actual ship is  $ \text{12}\ \text{m} $  high. If the length of the ship is  $ \text{28}\ \text{m} $ , how long is the model ship?

Ans:

Let us assume that the length of the model ship is  $ x\ \text{cm} $ . We can observe that the mast and length of the ship are directly proportional to each other. Therefore,

 $ \dfrac{12}{9}=\dfrac{28}{x} $ 

 $ \Rightarrow x=\dfrac{28\times 9}{12} $ 

 $ \Rightarrow x=21\ \text{cm} $ .

Hence, the length of the model ship will be of  $ 21\ \text{cm} $ .


7. Suppose  $ \text{2}\ \text{kg} $  of sugar contains \[\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\] crystals. How many sugar crystals are there in –

Ans:

(i). $ \text{5}\ \text{kg} $  of sugar? 

Let us assume that the  $ 5\ \text{kg} $  of sugar contains  $ x $  number of crystals.

As, weight of sugar is directly proportional to the number of crystals it contains. Therefore,

 $ \dfrac{2}{9\times {{10}^{6}}}=\dfrac{5}{x} $ 

 $ \Rightarrow x=\dfrac{9\times {{10}^{6}}\times 5}{2} $ 

 $ \Rightarrow x=2.25\times {{10}^{7}} $ 

Hence,  $ 5\ \text{kg} $  of sugar contains  $ 2.25\times {{10}^{7}} $  crystals of sugar.

(ii). $ \text{1}\text{.2}\ \text{kg} $  of sugar?

Similarly, let us assume that the  $ 1.2\ \text{kg} $  of sugar contains  $ x $  number of crystals.

As, weight of sugar is directly proportional to the number of crystals it contains. Therefore,

 $ \dfrac{2}{9\times {{10}^{6}}}=\dfrac{1.2}{x} $ 

 $ \Rightarrow x=\dfrac{9\times {{10}^{6}}\times 1.2}{2} $ 

 $ \Rightarrow x=5.4\times {{10}^{6}} $ 

Hence,  $ 1.2\ \text{kg} $  of sugar contains  $ 5.4\times {{10}^{6}} $  crystals of sugar.


8. Rashmi has a road map with a scale of  $ \text{1}\ \text{cm} $  representing \[\text{18}\ \text{km}\]. She drives on a road for \[\text{72}\ \text{km}\]. What would be her distance covered in the map?

Ans:

Given we have a road map with a scale of  $ 1\ cm $  representing \[18\ km\]. Let us assume that the distance covered by her is  $ x\ \text{cm} $ . We can observe that the scale is directly proportional to the distance she covers.

Therefore,

 $ \dfrac{18}{1}=\dfrac{72}{x} $ 

 $ \Rightarrow x=4\ cm $ .

Hence, the distance covered by her represents on the map as  $ 4\ cm $ .


9. A  $ \text{5}\ \text{m}\ \text{60}\ \text{cm} $  high vertical pole casts a shadow  $ \text{3}\ \text{m}\ \text{20}\ \text{cm} $  long. Find at the same time – 

i. The length of the shadow cast by another pole  $ \text{10}\ \text{m}\ \text{50}\ \text{cm} $  high

Ans:

Given we have a vertical pole of height  $ 5\ \text{m}\ \text{60}\ cm $  which casts a shadow of  $ 3\ \text{m}\ 2\text{0}\ cm $ . Let us assume that the length of shadow cast by pole  $ 10\ \text{m}\ 5\text{0}\ cm $  high is  $ x\ \text{m} $ . As we know that the height of the pole is directly proportional to the length of the shadow.

Therefore,

 $ \dfrac{5.60}{3.20}=\dfrac{10.50}{x} $ 

 $ \Rightarrow x=\dfrac{10.50\times 3.20}{5.60} $ 

 $ \Rightarrow x=6 $ 

Hence, the length of the shadow cast by another pole  $ 10\ \text{m}\ 5\text{0}\ cm $  high is  $ 6\ \text{m} $ .

ii. The height of a pole which casts a shadow 5m long.

Ans:

Let us assume that the height of the pole is  $ x\ \text{m} $  which casts a shadow  $ 5\ \text{m} $  long. As we know that the height of the pole is directly proportional to the length of the shadow.

Therefore,

 $ \dfrac{5.60}{3.20}=\dfrac{x}{5} $ 

 $ \Rightarrow x=\dfrac{5\times 5.60}{3.20} $ 

 $ \Rightarrow x=8.75 $ 

Hence, the height of the pole is  $ 8\ \text{m}\ 75\ cm $  which casts a shadow  $ 5\ \text{m} $  long.


10. A loaded truck travels  $ \text{14}\ \text{km} $  in  $ \text{25}\ \text{minutes} $ . If the speed remains the same, how far can it travel in  $ \text{5}\ \text{hours} $ ?

Ans:

Let us assume the distance travelled by truck in  $ 5\ \text{hours} $  is  $ x\ \text{km} $ . As we know that the distance traveled by the loaded truck is directly proportional to the time.

Therefore,

 $ \dfrac{14}{25}=\dfrac{x}{5\times 60} $ 

 $ \Rightarrow x=\dfrac{5\times 60\times 14}{25} $ 

 $ \Rightarrow x=168\ \text{km} $ 

Hence, the distance travelled by the loaded truck in  $ 5\ \text{hours} $  is  $ 168\ \text{km} $ .


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: Since, the number of workers increases, it reduces the time to complete the job. Therefore, we can conclude that this is an inverse proportion.


(ii). The time is taken for a journey and the distance traveled at a uniform speed.

Ans: We know, if we travel at a uniform speed then the distance traveled can be more if we have taken more time. This implies that they are in direct proportion and not in inverse proportion. Hence, we can conclude that this is not in inverse proportion.


(iii). Area of cultivated land and the crop harvested. 

Ans: We know, the more cultivated land we will have, the greater number of crops can be harvested. This implies that they are in direct proportion and not in inverse proportion. Therefore, we can conclude that this is not in inverse proportion.


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

Ans: When we travel with the vehicle at high speed, there is a possibility that we can complete the journey in less time than expected. This implies that this is in inverse proportion.


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

Ans: As the population of a country increases, the area of the land per person will decrease because more people will be requiring the area for themselves. Therefore, this is in inverse proportion.


2. In a Television game show, the prize money of  $ \text{Rs}\ \text{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

100000

50000







Ans:

We know, the more the number of winners will be, the lesser the prize money each winner gets. This implies that these are in inverse proportion.

Now, let us assume the prize for each winner as  $ {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}} $  and  $ {{a}_{5}} $ .

Therefore,

 $ 1\times 100000=4\times {{a}_{1}} $ 

 $ \Rightarrow {{a}_{1}}=\dfrac{100000}{4} $ 

 $ \Rightarrow {{a}_{1}}=Rs\ 25000 $ 

Similarly,

 $ 1\times 100000=5\times {{a}_{2}} $ 

 $ \Rightarrow {{a}_{2}}=\dfrac{100000}{5} $ 

 $ \Rightarrow {{a}_{2}}=Rs\ 20000 $ 

Now, we will calculate  $ {{a}_{3}} $ 

 $ 1\times 100000=8\times {{a}_{3}} $ 

 $ \Rightarrow {{a}_{1}}=\dfrac{100000}{8} $ 

 $ \Rightarrow {{a}_{3}}=Rs\ 12500 $ 

Now, we will calculate  $ {{a}_{4}} $ 

 $ 1\times 100000=10\times {{a}_{4}} $ 

 $ \Rightarrow {{a}_{4}}=\dfrac{100000}{10} $ 

 $ \Rightarrow {{a}_{4}}=Rs\ 10000 $ 

Now, we will calculate  $ {{a}_{5}} $ 

 $ 1\times 100000=20\times {{a}_{5}} $ 

 $ \Rightarrow {{a}_{5}}=\dfrac{100000}{20} $ 

 $ \Rightarrow {{a}_{5}}=Rs\ 5000 $ 

Hence, the amount decreases as the number of winners increases.


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.

Number of spokes

4

6

8

10

12

Angle between 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  $ \text{15} $  spokes. 

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


Ans:

We know, the more the number of spokes will be, the lesser the angle between two spokes will be. This implies that these are in inverse proportion.

Now, let us assume the angle between two consecutive spokes as  $ {{a}_{1}},{{a}_{2}}, $  and  $ {{a}_{3}} $ .

Therefore,

\[4\times 90{}^\circ =8\times {{a}_{1}}\]

 $ \Rightarrow {{a}_{1}}=\dfrac{4\times 90}{8} $ 

 $ \Rightarrow {{a}_{1}}=45{}^\circ  $ 

Similarly,

 $ 4\times 90{}^\circ =10\times {{a}_{2}} $ 

 $ \Rightarrow {{a}_{2}}=\dfrac{4\times 90}{10} $ 

 $ \Rightarrow {{a}_{2}}=36{}^\circ  $ 

Now, we will calculate  $ {{a}_{3}} $ 

 $ 4\times 90{}^\circ =12\times {{a}_{3}} $ 

 $ \Rightarrow {{a}_{3}}=\dfrac{4\times 90}{12} $ 

 $ \Rightarrow {{a}_{3}}=30{}^\circ  $ 

Hence, table will be as –

Number of spokes

4

6

8

10

12

Angle between a pair of consecutive spokes

90o

60o

45o

36o

30o


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

Ans: As the number of spokes increases, we can observe that the angle between the pair of consecutive spokes decreases. Therefore, this implies they are in inverse proportion.


(ii). Calculate the angle between a pair of consecutive spokes on a wheel with  $ \text{15} $  spokes. 

Ans: Let us assume the angle as  $ x{}^\circ  $  of consecutive spokes on a wheel with  $ 15 $  spokes.

Therefore,

\[4\times 90{}^\circ =15\times x\]

 $ \Rightarrow x=\dfrac{4\times 90}{15} $ 

 $ \Rightarrow x=24{}^\circ  $ .

Hence, the angle will be  $ 24{}^\circ  $  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  $ \text{40 }\!\!{}^\circ\!\!\text{ } $ ?

Ans: Let us assume the number of spokes be  $ y $ .

Therefore,

\[4\times 90{}^\circ =y\times 40{}^\circ \]

 $ \Rightarrow y=\dfrac{4\times 90}{40} $ 

 $ \Rightarrow y=9 $ .

Hence,  $ 9 $  spokes would be needed if the angle between a pair of consecutive spokes is  $ 40{}^\circ  $ .


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

Ans: As the number of children is reduced by  $ 4 $ . Hence, the remaining children will be  $ 24-4=20 $  children.

Let us assume that the number of sweets each would get is  $ a $ . We know that the number of sweets each child gets increases whenever the number of children is decreased.

Therefore,

\[24\times 5=20\times a\]

 $ \Rightarrow a=\dfrac{24\times 5}{20} $ 

 $ \Rightarrow a=6 $ .

Hence, each child will get  $ 6 $  sweets.


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

Ans:  Let us assume that the number of days the food will last be  $ x $  days.

He has  $ 10 $  more animals in his cattle. Hence, total animals will be  $ 20+10=30 $ 

We know, the more animals the food will last for less days. This implies these are in inverse proportion.

Therefore,

 $ 20\times 6=30\times x $ 

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

 $ \Rightarrow x=4 $ .

Hence, the food will last for  $ 4 $  days if there were  $ 10 $  more animals in his cattle.


6. A contractor estimates that  $ \text{3} $  persons could rewire Jasminder’s house in  $ \text{4} $  days. If he uses  $ \text{4} $  persons instead of three, how long should they take to complete the job?

Ans: Let us assume the number of days for the persons to complete the job be  $ x $  days.

We know that the number of days and number of persons are in inverse proportion.

Therefore,

 $ 3\times 4=4\times x $ 

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

 $ \Rightarrow x=3 $ .

Hence, the number of days will be  $ 3 $  days if he uses  $ 4 $  persons instead of three to complete the job.


7. A batch of bottles was packed in  $ \text{25} $  boxes with  $ \text{12} $  bottles in each box. If the same batch is packed using  $ \text{20} $  bottles in each box, how many boxes would be filled?

Ans: Let us assume the number of boxes needed to be filled is  $ x $ .

We know that the number of boxes and number of bottles in each box are in inverse proportion.

Therefore,

 $ 25\times 12=x\times 20 $ 

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

 $ \Rightarrow x=15 $ .

Hence, the number of boxes needed to be filled if the same batch is packed using  $ 20 $  bottles in each box will be  $ 15 $  boxes.


8. A factory required  $ \text{42} $  machines to produce a given number of articles in  $ \text{63} $  days. How many machines would be required to produce the same number of articles in  $ \text{54} $  days?

Ans: Let us assume the number of machines required to produce the articles is  $ x $ .

We know that the number of machines and number of days are in inverse proportion.

Therefore,

 $ 42\times 63=x\times 54 $ 

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

 $ \Rightarrow x=49 $ .

Hence, the number of machines required to produce the same number of articles in  $ 54 $  days will be  $ 49 $  machines.


9. A car takes  $ \text{2} $  hours to reach a destination by travelling at the speed of  $ \text{60}\ \text{km/h} $ . how long will it take when the car travels at the speed of  $ \text{80}\ \text{km/h} $ ?

Ans:Let us assume the number of hours the car will take to reach the destination be   $ x\ \text{hrs} $ .

We know that the number of hours and speed of the car are in inverse proportion.

Therefore,

 $ 2\times 60=x\times 80 $ 

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

 $ \Rightarrow x=\dfrac{3}{2} $ .

Hence, the time taken when the car travels at the speed of  $ 80\ \text{km/h} $  will be  $ 1.5\ hrs $ .


10. Two people could fit new windows in the house in  $ \text{3} $  days. 

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

Ans: As, we know that the number of persons and the number of days is in inverse proportion. Let us assume that the number of days for one person to get the job done be  $ x $  days.

Therefore,

 $ 2\times 3=1\times x $ 

 $ \Rightarrow x=\dfrac{2\times 3}{1} $ 

 $ \Rightarrow x=6 $ .

Hence, if one of the persons fell ill before the work started, it will take  $ 6 $  days to fit new windows.

(ii). How many people would be needed to fit the windows in one day?

Ans: As we know that the number of persons and the number of days is in inverse proportion. Let us assume that the number of persons needed to fit the windows is  $ y $ .

Therefore,

 $ 2\times 3=y\times 1 $ 

 $ \Rightarrow y=\dfrac{2\times 3}{1} $ 

 $ \Rightarrow y=6 $ .

Hence,  $ 6 $ person will be needed to fit the windows in one day.


11. A school has  $ \text{8} $  periods a day each of  $ \text{45} $  minutes duration. How long would each period be, if the school has  $ \text{9} $  periods a day, assuming the number of school hours to be the same?

Ans: We know that the time for each period will decrease if we increase the number of periods a day. This implies these are in inverse proportion.

So, let us assume that each period will be of  $ x $  minutes.

Therefore,

 $ 8\times 45=9\times x $ 

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

 $ \Rightarrow x=40 $ .
Hence, each period will be of  $ 40 $  minutes if the school has  $ 9 $  periods a day, assuming the number of school hours to be the same.


Class 8 Maths Chapter 11: Exercises Breakdown

Exercise

Number of Questions

Exercise 11.1

10 Questions and Solutions

Exercise 11.2

11 Questions and Solutions


Conclusion

NCERT Solutions for Class 8 Maths Ch Direct and Inverse Proportion by Vedantu provide a comprehensive understanding of these essential concepts. Direct proportion occurs when two quantities increase or decrease at the same rate, while indirect proportion happens when one quantity increases as the other decreases. It is important to focus on the ability to identify whether a relationship is direct or indirect, and to apply the respective formulas correctly. In previous year exams, around 2-3 questions have been asked from this chapter. Practicing the problems and understanding the solutions provided by Vedantu, students can strengthen their problem-solving skills and perform well in their exams.


Other Study Material for CBSE Class 8 Maths Chapter 11


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.


FAQs on NCERT Solutions for Class 8 Maths Chapter 11 Direct and Inverse Proportions

1. What are the Various Methods to Solve Direct Proportion Problems? Explain any one of them with an Example.

There are two methods that can be used to solve direct proportion problems and they are:

  • Tabular Method

  • Unitary Method - We will understand this method with an example.

Problem - Let us say a worker gets 1000 Rs for working 2 hours. If he gets a salary of 3000 then how much time has he worked in total?


Solution - We know that for two quantities x and y in direct proportion we have x/y = C (constant). Applying this concept here we get:


C = number of hours worked/payment received by the worker for that duration = 2/1000 = 1/500.


So, if x = C * B we get x = 1/500 * 3000 = 12.

So the worker worked for 12 hours to receive a total payment of 3000 Rs.

2. Explain the Concept of Inverse Proportion Between two Quantities with an Example.

Here is a sample problem on the inverse proportion concept:


Problem - If it takes 4 days for 20 laborers to lay the foundation of a building, then how much time will 16 laborers take to do the same job?


Solution - If 20 laborers lay the foundation in 4 days then the duration in which 1 laborer can finish the job = 20 * 4.

To get the duration taken by 16 laborers we have = (20 * 4)/16 = 5 days.

3. What are the key features of using the NCERT Solutions of Chapter 11 of Class 8 Maths?

The key features of using the NCERT Solutions of Chapter 11 Direct and Inverse Proportion of Class 8 Maths are highlighted below:

  • These NCERT Solutions help Class 8 students to understand Chapter 11 Direct and Inverse Proportions clearly. They make all the important concepts of the chapter crystal clear to the students. 

  • The NCERT Solutions enables students to revise significant concepts of the chapter quickly. 

  • They are the best study material to prepare for the Class 8 Maths exam as they are formed after extensive research done by the best Math professionals that tend to provide accurate information. 

4. What do I have to do to download the NCERT Solutions of Chapter 11 of Class 8 Maths?

In order to download the NCERT Solutions of Chapter 11 Direct and Indirect Proportion, you have to go through the given steps:

  • Visit the page NCERT Solutions of Chapter 11 Direct and Indirect Proportion of Class 8 Maths.

  • On that webpage, you will find different study materials other than the NCERT Solutions. It will include revision notes, textbook questions answers, sample papers, mock test papers, etc. 

  • After selecting your desired study material click on the download icon. 

5. What are the topics covered in Chapter 11 of Class 8 Maths?

Chapter 11 Direct and Inverse Proportion covers three topics in total. These are the introduction of the chapter, Direct Proportion, and Inverse Proportion. You can understand these topics thoroughly on Vedantu. This platform helps students to comprehend Maths subjects easily. It also provides various facilities like video lectures, sample papers, and test papers to prepare well for their exams. 

6. What is the actual length of the bacteria if it has enlarged 20,000 times? The initial length of the bacteria is 5cm when it is enlarged 50,000 times. 

Let us consider the actual length of the bacteria to be x cm. If the bacteria attains a length of 5cm after enlarging 50,000 times, then applying the unitary method, the length of the bacteria will be x cm if it enlarges 20,000 times. 

Enlargement                                                                              Length

50,000                                                                                         5 cm

20,000                                                                                         x cm 

After cross multiplying, the original length of the bacteria is 

Original length= 5/50,000 cm = 1/10,000 cm 

Since both the quantities are directly proportional to each other. 

50,000/20,000= 5/x

50,000*x= 5*20,000

x=5*20,000/50,000

x= 2cm

7. What should I do to get good marks in Chapter 11 of Class 8 Maths?

To get good marks in Chapter 11 Direct and Inverse Proportion of Class 8 Maths, you have to attend all the lectures related to the chapter at school. You must pay attention to what your teacher is teaching. Then, invest time in doing a proper revision when you reach home. Make notes of the chapter containing essential concepts and formulas so that you can revise the entire chapter in a short period. Moreover, you are recommended to thoroughly go through the NCERT Solutions of Chapter 11 of Class 8 Maths available free of cost on the Vedantu website and the Vedantu app. 

8. What is the formula for solving direct proportion problems in class 8 math Chapter 11?

The formula for solving direct proportion problems is $\frac{a}{b}=\frac{c}{d}$, where a,b are known quantities and c,d are unknown quantities.

9. What is the formula for solving inverse proportion problems in class 8 math chapter 11?

The formula for solving inverse proportion problems is $a \times b=c \times d$, where a and 𝑏 are the known quantities, and 𝑐 and d are the unknown quantities.