Mathematics NCERT Grade 8, Chapter 13: Direct and inverse proportion- In this chapter, the emphasis is laid on the concept of direct proportions and inverse proportion.
  • It is said that x and y are in direct proportion if x/y = k or x = ky.
  • Two quantities x and y are said to be in direct proportion if they increase or decrease together in such a manner that the ratio of their corresponding values remains constant. 
  • Although variables increasing (or decreasing) together need not always be in direct proportion. 
Ample real life examples like scale of maps, dimensions of different quantities, size of bacteria etc. are given to make the topic more clear.
  • Two quantities are said to be in inverse proportion if an increase in x causes a proportional decrease in y and vice - versa in such a way that the product of their corresponding values remains constant. 
  • Two quantities x and y are said to vary in inverse proportion if there exists a relation of the type xy = k between them, k being a constant.
The chapter consists of two unsolved exercises. Students must go through the solved examples as it will make their task easier to solve the questions given in the unsolved exercises i.e 13.1 and 13.2.
For the purpose of better learning, some questions are also given in the form of 'Try These'. 
Two important points are cited  in the box which is titled 'What Have We Discussed.

Page No 208:

Question 1:

Following are the car parking charges near a railway station up to

4 hours Rs 60

8 hours Rs 100

12 hours Rs 140

24 hours Rs 180

Check if the parking charges are in direct proportion to the parking time.

Answer:

A table of the given information is formed as

Number of hours

4

8

12

24

Parking charges (in Rs)

60

100

140

180

The ratio of parking charges to the respective number of hours (Rs/ hour) can be calculated as

As each ratio is not same, therefore, the parking charges are not in a direct proportion to the parking time.

Page No 208:

Question 2:

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

Parts of red pigment

1

4

7

12

20

parts of base

8

Answer:

The given mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. For more parts of red pigments, the parts of the base will also be more. Therefore, the parts of red pigments and the parts of base are in direct proportion. The given information in the form of a table is as follows.

Parts of red pigment

1

4

7

12

20

Parts of base

8

x1

x2

x3

x4

According to direct proportion,

The table can be drawn as follows.

Parts of red pigment

1

4

7

12

20

Parts of base

8

32

56

96

160

Page No 208:

Question 3:

In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Answer:

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.

Parts of red pigment

1

x

Parts of base (in mL)

75

1800

The parts of red pigment and the parts of base are in direct proportion.

Therefore, we obtain

Thus, 24 parts of red pigments should be mixed with 1800 mL of base.

Page No 208:

Question 4:

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

Answer:

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.

Number of bottles

840

x

Time taken (in hours)

6

5

The number of bottles and the time taken to fill these bottles are in direct proportion. Therefore, we obtain

Thus, 700 bottles will be filled in 5 hours.
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Video Solution for Direct and Inverse Proportions (Page: 208 , Q.No.: 4)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 208 , Question 4

Page No 208:

Question 5:

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

Answer:

Let the actual length of bacteria be x cm and the enlarged length of bacteria be y cm, if the photograph is enlarged for 20,000 times.

The given information in the form of a table is as follows.

Length of bacteria (in cm)

5

x

y

Number of times photograph of Bacteria was enlarged

50000

1

20000

The number of times the photograph of bacteria was enlarged and the length of bacteria are in direct proportion.

Therefore, we obtain

Hence, the actual length of bacteria is 10−4 cm.

Let the length of bacteria when the photograph of bacteria is enlarged 20, 000 times be y.

Hence, the enlarged length of bacteria is 2 cm.
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Video Solution for Direct and Inverse Proportions (Page: 208 , Q.No.: 5)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 208 , Question 5

Page No 208:

Question 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?

Answer:

Let the length of the mast of the model ship be x cm.

The given information in the form of a table is as follows:

-

Height of mast

Length of ship

Model ship

9 cm

x

Actual ship

12 m

28 m

We know that the dimensions of the actual ship and the model ship are directly proportional to each other.

Therefore, we obtain:

Thus, the length of the model ship is 21 cm.
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Video Solution for Direct and Inverse Proportions (Page: 208 , Q.No.: 6)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 208 , Question 6

Page No 208:

Question 7:

Suppose 2 kg of sugar contains 9 × 106 crystals.

How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Answer:

(i) Let the number of sugar crystals in 5 kg of sugar be x.

The given information in the form of a table is as follows.

Amount of sugar (in kg)

2

5

Number of crystals

9 × 106

x

The amount of sugar and the number of crystals it contains are directly proportional to each other. Therefore, we obtain

Hence, the number of sugar crystals is 2.25 × 107.

(ii) Let the number of sugar crystals in 1.2 kg of sugar be y. The given information in the form of a table is as follows.

Amount of sugar (in kg)

2

1.2

Number of crystals

9 × 106

y

Hence, the number of sugar crystals is.

Page No 208:

Question 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?

Answer:

Let the distance represented on the map be x cm.

The given information in the form of a table is as follows.

Distance covered on road in (in km)

18

72

Distance represented on map (in cm)

1

x

The distances covered on road and represented on map are directly proportional to each other. Therefore, we obtain

Hence, the distance represented on the map is 4 cm.
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Video Solution for Direct and Inverse Proportions (Page: 208 , Q.No.: 8)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 208 , Question 8

Page No 208:

Question 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.

Answer:

(i) Let the length of the shadow of the other pole be x m.

1 m = 100 cm

The given information in the form of a table is as follows.

Height of pole (in m)

5.60

10.50

Length of shadow (in m)

3.20

x

More the height of an object, more will be the length of its shadow.

Thus, the height of an object and length of its shadow are directly proportional to each other. Therefore, we obtain

Hence, the length of the shadow will be 6 m.

(ii) Let the height of the pole be y m.

The given information in the form of a table is as follows.

Height of pole (in m)

5.60

y

Length of shadow (in m)

3.20

5

The height of the pole and the length of the shadow are directly proportional to each other. Therefore,

Thus, the height of the pole is 8.75 m or 8 m 75 cm.

Page No 208:

Question 10:

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

Answer:

Let the distance travelled by the truck in 5 hours be x km.

We know, 1 hour = 60 minutes

∴ 5 hours = (5 × 60) minutes = 300 minutes

The given information in the form of a table is as follows.

Distance travelled (in km)

14

x

Time (in min)

25

300

The distance travelled by the truck and the time taken by the truck are directly proportional to each other. Therefore,

Hence, the distance travelled by the truck is 168 km.



Page No 213:

Question 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.

Answer:

(i) These are in inverse proportion because if there are more workers, then it will take lesser time to complete that job.

(ii) No, these are not in inverse proportion because in more time, we may cover more distance with a uniform speed.

(iii) No, these are not in inverse proportion because in more area, more quantity of crop may be harvested.

(iv) These are in inverse proportion because with more speed, we may complete a certain distance in a lesser time.

(v) These are in inverse proportion because if the population is increasing, then the area of the land per person will be decreasing accordingly.

Page No 213:

Question 2:

In a Television game show, the prize money of Rs 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 for each winner (in Rs)

100000

50000

Answer:

A table of the given information is as follows.

Number of winners

1

2

4

5

8

10

20

Prize for each winner (in Rs)

100000

50000

x1

x2

x3

x4

x5

From the table, we obtain

1 × 100000 = 2 × 50000 = 100000

Thus, the number of winners and the amount given to each winner are inversely proportional to each other. Therefore,

1 × 100000 = 4 × x1

1 × 100000 = 5 × x2

1 × 100000 = 8 × x3

1 × 100000 = 10 × x4

1 × 100000 = 20 × x5

Page No 213:

Question 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

90°

60°

(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 15 spokes.

(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?

Answer:

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°

60°

x1

x2

x3

From the given table, we obtain

4 × 90° = 360° = 6 × 60°

Thus, the number of spokes and the angle between a pair of consecutive spokes are inversely proportional to each other. Therefore,

4 × 90° = x1 × 8

Similarly, and

Thus, the following table is obtained.

Number of spokes

4

6

8

10

12

Angle between a pair of consecutive spokes

90°

60°

45°

36°

30°

(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 × 90° = 15 × x

Hence, the angle between a pair of consecutive spokes of a wheel, which has 15 spokes in it, is 24°.

(iii) Let the number of spokes in a wheel, which has 40º angles between a pair of consecutive spokes, be y.

Therefore,

4 × 90° = y × 40°

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



Page No 214:

Question 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?

Answer:

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 × 5 = 20 × x

Hence, each student will get 6 sweets.
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Video Solution for Direct and Inverse Proportions (Page: 214 , Q.No.: 4)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 214 , Question 4

Page No 214:

Question 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?

Answer:

Let the number of days that the food will last if there were 10 more animals in the cattle be x. The following table is obtained.

Number of animals

20

20 + 10 = 30

Number of days

6

x

More the number of animals, lesser will be the number of days for which the food will last.

Hence, the number of days the food will last and the number of animals are inversely proportional to each other.

Therefore,

20 × 6 = 30 × x

Thus, the food will last for 4 days.
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Video Solution for Direct and Inverse Proportions (Page: 214 , Q.No.: 5)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 214 , Question 5

Page No 214:

Question 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?

Answer:

Let the number of days required by 4 persons to complete the job be x.

The following table is obtained.

Number of days

4

x

Number of persons

3

4

If the number of persons is more, then it will take lesser time to complete the job.

Hence, the number of days and the number of persons required to complete the job are inversely proportional to each other.

Therefore,

4 × 3 = x × 4

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

Page No 214:

Question 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?

Answer:

Let the number of boxes filled, by using 20 bottles in each box, be x.

The following table is obtained.

Number of bottles

12

20

Number of boxes

25

x

More the number of bottles, lesser will be the number of boxes.

Hence, the number of bottles and the number of boxes required to pack these are inversely proportional to each other.

Therefore,

12 × 25 = 20× x

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

Page No 214:

Question 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?

Answer:

Let the number of machines required to produce articles in 54 days be x. The following table is obtained.

Number of machines

42

x

Number of days

63

54

More the number of machines, lesser will be the number of days that it will take to produce the given number of articles. Thus, this is a case of inverse proportion. Therefore,

42 × 63 = 54× x

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

Page No 214:

Question 9:

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

Answer:

Let the time taken by the car to reach the destination, while travelling with a speed of 80 km/hr, be x hours.

The following table is obtained.

Speed (in km/hr)

60

80

Time taken (in hours)

2

x

More the speed of the car, lesser will be the time taken by it to reach the destination.

Hence, the speed of the car and the time taken by the car are inversely proportional to each other. Therefore,

60 × 2 = 80× x

The time required by the car to reach the given destination is hours.
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Video Solution for Direct and Inverse Proportions (Page: 214 , Q.No.: 9)

NCERT Solution for Class 8 math - Direct and Inverse Proportions 214 , Question 9

Page No 214:

Question 10:

Two persons could fit new windows in 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?

Answer:

(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 × 3 = 1× x

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

Y

Number of days

3

1

Lesser the number of days, more will be the number of persons required to fit all the windows. Hence, this is a case of inverse proportion. Therefore,

2 × 3 = y × 1

y = 6

Hence, 6 persons are required to fit all the windows in one day.

Page No 214:

Question 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?

Answer:

Let the duration of each period, when there are 9 periods a day in the school, be x minutes. The following table is obtained.

Duration of each period (in minutes)

45

x

Number of periods

8

9

If there is more number of periods a day in the school, then the duration of each period will be lesser. Hence, this is a case of inverse proportion. Therefore

45 × 8 = x× 9

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



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