NCERT Class 8 Maths Chapter 11 Direct and Inverse Proportions Exercise 11.1, 11.2 Solutions
In this page we have provided solutions of the Exercises 11.1, 11.2 of NCERT Class 8 Maths Chapter 11 Direct and Inverse Proportions. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Publishing Organisation 
NCERT 
Class 
8 
Subject 
Mathematics 
Chapter 
11: Direct and Inverse Proportions 
Exercise 
11.1, 11.2 
Exercise – 11.1
(1) Following are the car parking charges near a railway station up to.
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.
Ans: 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 Can be Calculated as
60/4 = 15, 100/8 = 25/2, 140/12 = 35/8, 180/24 = 25/2
As each ratio is not Same, therefore the parking Charges Proportion to the Parking time.
(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 f red Pigment 
1  4  7  12  20 
Parts of base  8  …..  …..  ….. 
…… 
Ans: (i) Here
X_{1} = 1
Y_{1} = 8 and x_{2} = 4
Therefore, x_{1}/y_{1} = x_{2}/y_{2} gives
1/8 = 4/y^{2}
⇒ y_{2} = 8 × 4
⇒ y_{2} = 32
Hence, 32 Parts of the base are needed to be added.
(ii) Here
X_{1} = 1
Y_{1} = 8 and x_{3} = 7
Therefore, x_{1}/y_{1} = x_{3}/y_{3} gives
1/8 = 7/y_{3}
⇒ y_{3} = 8 × 7
⇒ y_{3} = 56
Hence, 56 Parts of the base are needed to be added.
(iii) Here
X_{1} = 1
Y_{1 }= 8
And x_{4 }= 12
Therefore, x_{1}/y_{1 }= x_{4}/y_{4} gives
1/8 = 12/y_{4}
Y_{4} = 12 × 8
Y_{4} = 96
Hence, 96 Parts of the base are needed to be added.
(iv) Here
X_{1} = 1
Y_{1} = 8
and x_{5} = 20
Therefore, x_{1}/y_{1} = x_{5}/y_{5} gives
1/8 = 20/y_{5}
⇒ y_{5}= 8 × 20
⇒ y_{5} = 160
Hence, 160 Parts of the base are needed to be added.
(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?
Ans: 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 a 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 Therefore, We obtain
1/75 = x/1800
⇒ x = 1 × 1800/75
⇒ x = 24
Thus 24 Parts of red Pigments should be mixed with 1800mb safe.
(4) A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Ans: Let the machine fill x bottles in five hours. We put the given information in the form of a table as shown below:
Number of bottles filled 840 × 2
Number of hours 65
More the number of hours, more the number of bottles would be filled. So, the number of bottles filled and the number of hours are directly proportional to each other.
So, x_{1}/x_{2} = y_{1}/y_{2}
⇒ 840/x_{2} = 6/5
⇒ 6_{x2} = 840 × 5
⇒ x_{2} = 840 × 5/6
⇒ y_{2} = 700
Hence, 700 bottles will be filled.
(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?
Ans: Let the actual length of bacteria be x cm and the enlarged length of bacteria by Y cm, if the Photograph is enlarged for 20, 000 times.
The given information in the from 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 of bacteria are indirect Proportional
Therefore, We obtain
5/50,000 = x/1
⇒ x = 1/1000 = 10^{4}
Hence, the actual length bacteria is 10^{4} cm. Let the length of bacteria when the photograph of bacteria is enlarged 20,000 times be y.
5/50,000 = y/20,000
Y = 20,000 × 5/50,000 = 2
Hence, the enlarged length of bacteria 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?
Ans: Let the length of the model Ship be x_{2} cm.
We form a table as shown below:

Actual Ship  Model 
Length of the Ship  28 m 
X_{2} 
Height of the mast 
12 m 
9 cm 
More the length of the Ship, more would be the length of its mast. Hence, this is a case of direct proportion. That is
X_{1}/y_{1} = y_{2}/y_{2}
⇒ 28/12 = x_{2}/9
⇒ 12x_{2} = 28 × 9
⇒ x_{2} = 28 × 9/12
⇒ x_{2} = 21
Hence, the length of the model Ship is 21 m.
(7) Suppose 2 kg of sugar contains 9 × 10^{6} crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?
Ans: (i) Let the number 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 × 10^{6} 
x 
The amount of Sugar of Crystals it Proportional to each other. Therefore, obtain
2/9 × 10^{6} = 5/x
X = 5 × 9 × 10^{6}/2 = 2.25 × 10^{7}
Hence, the table of Sugar Crystals is 2.25 × 10^{7}.
(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 × 10^{6} 
y 
2/9 × 10^{6} = 1.2/y
Y = 1.2 × 9 × 10^{6}/2 = 5.4 × 10^{6}
Hence, the number of sugar crystals is 5.4 × 10^{6}
(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?
Ans: Let the distance covered in the map be x cm. Then
1:18 = x : 72
⇒ 1/18 = x/72
⇒ x = 72/18
⇒ x = 4
Hence, the distance Covered in the map would be 4 cm.
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: The number of workers on jobs and the time to complete the job are in inverse Proportion.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
Ans: The time is taken for a journey and the distance traveled in a uniform Speed are not in inverse Proportion.
(iii) Area of cultivated land and the crop harvested.
Ans: Area of Cultivated land and the crop harvested are not in inverse proportion.
(iv) The time taken for a fixed journey and the speed of the vehicle.
Ans: The time taken for a fixed journey and the speed of the vehicle are in inverse Proportion.
(v) The population of a country and the area of land per person.
Ans: The Popular of a Country and the area of land per person are in 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?
Number of Winners 
1  2  4  5  8  10  20 
Prize for each winner (in ₹) 
1,00,000  50,000  ….  ….  ….  …. 
…. 
Ans: A table of the given information is as follows.
Number of Winners 
1  2  4  5  8  10  20 
Prize for each winner (in ₹)  1,00,000  50,000  X_{1}  X_{2}  X_{3}  X_{4} 
X_{5} 
From the table, we obtain
1 × 100000 = 2 × 50000 = 100000
Thus, the number of winners and the amount given to each Proportional to each other. Therefore,
1 × 100000 = 4 × x_{1}
X_{1} = 100000/4 = 25000
1 × 100000 = 5 × x_{2}
X_{2 }= 100000/5 = 20000
1 × 100000 = 8 × x_{3}
X_{3} = 100000/8 = 12500
1 × 100000 = 10 × x_{4}
X_{4} = 100000/10 = 10000
1 × 100000 = 20 × x_{5}
X_{5} = 100000/20 = 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.
Number of Spokes 
4  6  8  10  12 
Angle between a pair of consecutive spokes  90°  60°  ….  …… 
…… 
Ans: 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°  X_{1}  X_{2} 
X_{3} 
From the given table, We obtain
4 × 90° = 360° = 6 × 60°
Thus, the number of spokes and the angle between a spokes are inversely proportional to other. Therefore.
4 × 90° = x_{1} × 8
X_{1} = 4 × 90°/8 = 45°
Similarly, x_{2} = 4 × 90°/10 = 36° and x_{3} = 4 × 90°/12 = 30°
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) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse Proportion?
Ans: Yes, the number of spokes and the angles formed the pairs of Consecutive spokes are in inverse Proportion.
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
Ans: Let the angle between a pair Consecutive Spokes on a wheel with 15 Spokes be x. Therefore,
4 × 90° = 15 × x
X = 4 × 90°/15 = 24°
Hence, the angle between a pair of Consecutive Spokes of while has 15 spokes in it, is 24°
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Ans: Let the number of spokes in a wheel, which has 40° angles between a pair of consecutive spokes, be y.
Therefore,
4 × 90° = y × 40°
Y = 4 × 90/40 = 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?
Ans: Suppose that each would get y_{2 }sweets.
Thus, we have the following table
Number of Children 
24  24 – 4 = 20 
Number of Sweets  5 
Y_{2} 
Lesson the number of children, more will be the number of sweets each would get. So, this is a case of inverse Proportion.
Hence, 24 × 5 = 20 × y_{2}
⇒ y_{2} = 24 × 5/20
⇒ y_{2} = 6
Hence, each would 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?
Ans: 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 food will last.
Hence, the number of days the food will last and the number are inversely proportional to each other.
Therefore,
20 × 6 = 30^{x}x
X = 20 × 6/30 = 4
Thus, the food will 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?
Ans: Suppose that they take y2 days to complete the job. We have the following table.
Number of Persons 
3  4 
Number of days  4 
Y_{2} 
More the number of persons, lesser will be the number of days. Required to complete the job. So, this is a case of inverse Proportion.
Hence, 3 × 4 = 4 × y_{2}
⇒ y_{2 }= 3 × 4/4
⇒ y_{2} = 3
Hence, they would 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?
Ans: 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 replace these are inversely Proportional to each other.
Therefore,
12 × 25 = 20^{x}x
X = 12 × 25/20 = 15
Hence, the number of boxes required to pack these bottles 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?
Ans: Suppose that x_{2} machines would be required. We have the following table:
Number of machines 
42  X_{2} 
Number of days  63 
54 
Lesser the number of machines, more will be the number of days to produce the same number of articles.
So, this is a case of inverse Proportion
Hence, 42 × 63 = x_{2} × 54
⇒ x_{2} = 42 × 63/54
⇒ x_{2} = 49
Hence, 49 machines would be required.
(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?
Ans: 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 by the destination.
Hence, the speed of the car and the time taken by the proportional to each other. Therefore,
60 × 2 = 80^{x}x
X = 60 × 2/80 = 3/2 = 1 (1/2)
The time required by the car to reach the given destination is 1 (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?
Ans: Let the job would take y_{2} days. We have the following table.
Number of Persons 
2  2 – 1 = 1 
Number of days  3 
Y_{2} 
Clearly, more the number of persons, lesser would be the number of days to do the job. So, the number of persons and number of days very in inverse Proportion.
So, 2 × 3 = 1 × y_{2}
⇒ y_{2} = 6
Thus, the job would now like take 6 days.
(ii) How many persons would be needed to fit the windows in one day?
Ans: Let y_{2} persons be needed. We have the following table:
Number of days 
3  1 
Number of Persons  2 
Y_{3} 
Clearly, more the number of persons, lesser would be the number of days to do the job. So, the number of persons and number of days very in inverse Proportion.
So, 3 × 2 = 1 × y_{3}
⇒ T_{2} = 6
Thus, 6 persons would be needed.
Chapter 10: Exponents and Powers