# NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions| PDF Download

Here you will find Chapter 13 Direct and Inverse Proportions Class 8 Maths NCERT Solutions that will be helpful in knowing the essential points inside the chapter. These NCERT Solutions for Class 8 Maths are detailed every step so if you find any doubt while solving a question then you can always look from here and move further. Also, you can download PDF of Chapter 13 Direct and Inverse Proportions which is going to help you a lot in the preparation of exams.

You can easily complete your homework by taking help from Chapter 13 Class 8 Maths NCERT Solutions and rectify all those mistakes that can happen in the exams. It will also make you aware of the difficulty of questions. Page No: 208

Exercise 13.1

1. Following are the car parking charges near a railway station upto:
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.

Charges per hour:
C1= 60/4 = Rs.15
C2= 100/8 = Rs.12.50
C3= 140/12 = Rs.11.67
C4= 180/24= Rs.7.50
Here, the charges per hour are not same, i.e., C1 ≠ C2 ≠ C3 ≠ C4
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 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 …… …….. …….. …….

Let the ratio of parts of red pigment and parts of base be a/b
Here a1 = 1, b1 = 8
⇒ a1/b1 = 1/8 = k (say)
When a2 = 4, b2 = ?
k = a2/b2 ⇒ b2 = a2/k = 4/1/8 = 4 × 8 = 32
When a = 7, b = ?
k = a3/b3 ⇒ b3 = a3/k = 7/1/8 = 7 × 8 = 56
When a = 12, b = ?
k = a4/b4 ⇒ b4 = a4/k = 12/1/8 = 12 × 8 = 96
When a5 = 20, b5 =?
k = a5/b5 ⇒ b5 = a5/k = 20/1/8 = 20 × 8 = 160

 Parts of red pigment 1 4 7 12 20 Parts of base 8 32 56 96 160

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?

Let the parts of red pigment mix with 1800 mL base be x.
 Parts of red pigment 1 x Parts of base 75 1800
Since it is in direct proportion.
∴ 1/75 = x/1800
⇒ 75 × x = 1 × 1800
⇒ x = (1 × 1800)/75 = 24 parts
Hence with base 1800 mL, 24 parts red pigment should be mixed.

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

Let the number of bottles filled in five hours be x.
 Hours 6 5 Bottles 840 x
Here ratio of hours and bottles are in direct proportion.
6/840 = 5/x
⇒ 6 × x = 5 × 840
⇒ x = (5 × 840)/6 = 700 bottles
Hence machine will fill 700 bottles in five 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?

Let Actual length of bacteria be 'a'
It is enlarged 50,000 times so 50000 × a = 5 cm
Actual length of bacteria
= 5/50000 = 1/10000 cm = 10-4 cm
Let enlarged length of bacteria be x
 Length 5 x Enlarged length 50,000 20,000
Here length and enlarged length of bacteria are in direct proportion.
∴ 5/50000 = x/20000
⇒ x × 50000 = 5 × 20000
⇒ x = (5 × 20000)/50000 = 2 cm
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?

Let the length of model ship be x.
 Length of actual ship (in m) 12 28 Length of model ship (in cm) 9 x
Here length of mast and actual length of ship are in direct proportion.
∴ 12/9 = 28/x
⇒ x × 12 = 28 ×9
⇒ x = (28 × 9)/12= 21 cm
Hence length of the model ship is 21 cm.

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?

(i) Let sugar crystals be x.
 Weight of sugar (in kg) 2 5 No. of crystals 9 × 106 x
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/(9 × 106) = 5/x
⇒ x × 2 = 5 × 9 × 106
⇒ x = (5 × 9× 106)/2
= 22.5 × 106 = 2.25 × 107
Hence the number of sugar crystals is 2.25 × 107
 Weight of sugar (in kg) 2 1.2 No. of crystals 9× 106 x

(ii) Let sugar crystals be x.
Here weight of sugar and number of crystals are in direct proportion.
∴ 2/(9 × 106) = 1.2/x
⇒ x × 2 = 1.2 ×9 × 106
⇒ x = (1.2 × 9 × 106)/2
= 0.6 × 9 × 106 = 5.4 × 106
Hence the number of sugar crystals is 5.4 × 106

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?

Let distance covered in the map be x.
 Actual distance ( in km) 18 72 Distance covered in map(in cm) 1 x
Here actual distance and distance covered in the map are in direct proportion.
∴ 18/1 = 72/x
⇒ x × 18 = 72 × 1
⇒ x = (72 × 1)/18 = 4 cm
Hence distance covered in 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.

Here height of the pole and length of the shadow are in direct proportion.
And 1 m = 100 cm
5 m 60 cm = 5 × 100 + 60 = 560 cm
3 m 20 cm = 3 × 100 + 20 = 320 cm
10 m 50 cm = 10 × 100 + 50 = 1050 cm
5 m = 5 × 100 = 500 cm

(i) Let the length of the shadow of another pole be x.
 Height of pole ( in cm) 560 1050 Length of shadow (in cm) 320 x
∴ 560/320 = 1050/x
⇒ x ×560 = 1050 × 320
⇒ x = (1050 × 320)/ 560 = 600 cm = 6 m
Hence length of the shadow of another pole is 6 m.

(ii) Let the height of the pole be x.
 Height of pole ( in cm) 560 x Length of shadow (in cm) 320 500
∴ 560/320 = x/500
⇒ x × 320 = 560 × 500
⇒ x = (560 × 500)/320
= 875 cm = 8 m 75 cm
Hence height of the pole is 8 m 75 cm.

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

Let distance covered in 5 hours be x km.
∵ 1 hour = 60 minutes
∴ 5 hours = 5 × 60 = 300 minutes
 Distance (in Km) 14 x Time ( in minutes) 25 300
Here distance covered and time are in direct proportion.
∴ 14/25 = x/300
⇒ x × 25 = 14 × 300
⇒ x = (14 × 300)/25 = 168 km

Page No. 213

Exercise 13.2

1. Which of the following are in inverse proportion:
(i) The number of workers on a job and the time to complete the job.
(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.

(i) The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work.
(ii) Time and distance covered in direct proportion.
(iii) It is a direct proportion because more area of cultivated land will yield more crops.
(iv) Time and speed are inverse proportion because if time is less, speed is more.
(v) It is a inverse proportion. If the population of a country increases, the area of land per person decreases.

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:
 No. of winners 1 2 4 5 8 10 20 Prize for each winner (in’) 1,00,000 50,000 --- --- --- --- ---

Here number of winners and prize money are in inverse proportion because winners are increasing, prize money is decreasing.
When the number of winners are 4, each winner will get = 100000/4 = Rs. 25,000
When the number of winners are 5, each winner will get = 100000/5 = Rs. 20,000
When the number of winners are 8, each winner will get = 100000/8 = Rs. 12,500
When the number of winners are 10, each winner will get = 100000/10 = Rs. 10,000
When the number of winners are 20, each winner will get = 100000/20 = Rs. 5,000

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:
(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°?
 No. of spokes 4 6 8 10 12 Angle between a pair of consecutive spokes 90° 60° --- --- ---

Here the number of spokes are increasing and the angle between a pair of consecutive spokes is decreasing. So, it is a inverse proportion and angle at the centre of a circle is 360°.
When the number of spokes is 8, then angle between a pair of consecutive spokes = 360°/8 = 45°
When the number of spokes is 10, then angle between a pair of consecutive spokes= 360°/10 = 36°
When the number of spokes is 12, then angle between a pair of consecutive spokes= 360°/12 = 30°

 No. 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 a pair of consecutive spokes is in inverse proportion.
(ii) When the number of spokes is 15, then angle between a pair of consecutive spokes= 360°/15 = 24°
(iii) The number of spokes would be needed = 360°/40 = 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?

∵ Each child gets = 5 sweets
∴ 24 children will get 24 × 5 = 120 sweets
Total number of sweets = 120
If the number of children is reduced by 4, then children left = 24 – 4 = 20
Now each child will get sweets = 120/20
= 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?

Let the number of days be x
Total number of animals = 20 + 10 = 30
 Animals 20 30 Days 6 x
Here the number of animals and the number of days are in inverse proportion.
∴ 20/30 = x/6
⇒ 30 × x = 20 × 6
⇒ x = (20 × 6)/30 = 4
Hence the food will last for four 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?

Let time taken to complete the job be x
 Persons 3 4 Days 4 x
Here the number of persons and the number of days are in inverse proportion.
∴ 3/4 = x/4
4 × x = 3 × 4
⇒ x = (3 × 4)/4 = 3 days
Hence they will complete the job in 3 days.

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?

Let the number of boxes be x.
 No. of bottles in each box 12 20 Boxes 25 x
Here the number of bottles and the number of boxes are in inverse proportion.
∴ 12/20 = x/25
⇒ x × 20 = 12 × 25
⇒ x = (12 × 25)/20 = 15
Hence 15 boxes would be filled.

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?

Let the number of machines required be x.
 Days 63 54 Machines 42 x
Here the number of machines and the number of days are in inverse proportion.
∴ 63/54 = x/42
x × 54 = 63 × 42
x = (63 × 42)/54 = 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/hr. How long will it take when the car travels at the speed of 80 km/hr?

Let the number of hours be x.
 Speed (in km/hr) 60 80 Time (in hours) 2 x
Here the speed of car and time are in inverse proportion.
∴ 60/80 = x/2
⇒ x 80 = 60 × 2
⇒ x = (60×2)/80
=3/2
= 1.1/2 hrs.
Hence the car will take 1.1/2 hours to reach its destination.

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

(i) Let the number of days be x
 Persons 2 1 Days 3 x
Here the number of persons and the number of days are in inverse proportion.
∴ 2/1 = x/3
⇒ x × 1 = 2 × 3
⇒ x = (2 × 3)/1 = 6 days

(ii) Let the number of persons be x
 Persons 2 X Days 1 3
Here the number of persons and the number of days are in inverse proportion.
∴ 2/x = 1/3
⇒ x×1 = 2×3
⇒ x = (2×3)/1
= 6 persons

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?

Let the duration of each period be x.
 Period 8 9 Duration of period(in minutes) 45 x
Here the number of periods and the duration of periods are in inverse proportion.
∴ 8/9 = x/45
⇒ x×9 = 8×45
⇒ x = (8×45)/9 = 40 minutes
Hence, duration of each period would be 40 minutes.

## NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions

Chapter 13 NCERT Solutions is best way through which a student can resolve their queries. These NCERT Solutions can be used to clear your doubts and if any occurred while solving exercise. For viewing NCERT Solutions for Class 8 offline, you can download Studyrankers app from Playstore.

• If two quantities x and y vary (change) together in such a manner that the ratio of their corresponding values remains constant, then x and y are said to be in direct proportion. If x and y are in a direct proportion, then (x/y) = constant.

• If two quantities x and y vary (change) in such a manner that an increase in x causes a proportional decrease in y (and vice versa), then x and y are said to be inverse proportion. If x and y are in an inverse variation, then xy = constant.

Below you will get exercisewise Class 8 NCERT Solutions that will help you in increase concentration among students.

You can clear all your doubts through the help of the NCERT Solutions prepared by the subject matter experts of Studyrankers and checking your own answers as well. It will make student confident and help in learning efficiently.

### NCERT Solutions for Class 8 Maths Chapters:

 Chapter 1 Rational Numbers Chapter 2 Linear Equations in Variable Chapter 3 Understanding Quadrilaterals Chapter 4 Practical Geometry Chapter 5 Data Handling Chapter 6 Squares and Square Roots Chapter 7 Cubes and Cube Roots Chapter 8 Comparing Quantities Chapter 9 Algebraic Expressions and Identities Chapter 10 Visualising Solid Shapes Chapter 11 Mensuration Chapter 12 Exponents and Powers Chapter 14 Factorization Chapter 15 Introduction to Graphs Chapter 16 Playing with Numbers

FAQ on Chapter 13 Direct and Inverse Proportions

#### How many exercises in Chapter 13 Direct and Inverse Proportions Class 8 Maths?

There are total 2 exercises in the Chapter 13 of Class 8 Maths that is going to useful in gaining good marks in the examinations. These NCERT Solutions will provide good experience and provide opportunities to learn new things.

#### The circumference (C cm) of a circle is directly proportional to its diameter (d cm). The circumference of a circle of diameter 3.5 cm is 11 cm. What is the circumference of a circle of diameter 4.2 cm?

C is directly proportional to d ⇒ C ∝ d ⇒ C = kd where k is the constant of proportionality.
But C = 11 when d = 3.5
⇒ 11 = k × 3.5
⇒ k = 11/3.5 = 22/7.

#### In a Restaurant, the food provision for 20 persons is for 10 days. How long would the food provision last if there were 5 more persons in that Restaurant?

Number of persons added = 5
∴ Present number of persons = 20 + 5 = 25
Since, for more persons, the food will last less number of days.
∴ It is a case of inverse variation.
∴25 × x = 20 × 10
x = 20 × 10/ 25 = 8
∴ The food will now last for 8 days.

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

Number of bottles filled in 6 hours = 840
Number of bottles filled in 1 hours = 840/6
Number of bottles filled in 5 hours = 840/6 × 5 = 700 bottles.