Class 8 Maths Exercise 11.2 Direct and Inverse Proportions

We will solve Exercise 11.2 for Chapter 11, Direct and Inverse Proportions. Here, we are going to solve questions related to Inverse Proportion. In the Exercise 11.1 we have solved questions on Direct Proportion and this exercise is an extension of that. Let’s have a look at a few basics –

Inverse Proportion

Definition: Inverse proportion occurs when one value increases while the other decreases. For example, if two values x and y are in inverse proportion, then xy = k for some constant k.

Examples:

Suppose you have a certain amount of paint that can cover a wall of 50 square meters. If the wall size doubles to 100 square meters, you’ll need half the amount of paint for each square meter to cover the entire wall.

In this, the amount of paint per square meter and the size of the wall are inversely proportional. If you use 2 liters of paint for 50 square meters, then for 100 square meters, you’ll use 1 liter per 50 square meters. As the area increases, the amount of paint used per unit area decreases, maintaining the relationship xy = k where k is a constant value.

Class 8 Maths Exercise 11.2 Chapter 11 Direct and Inverse Proportions

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 at a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time for a fixed journey and the vehicle’s speed.
(v) The population of a country and the area of land per person

Solution

(i) The number of workers on a job and the time to complete the job are in inverse proportion. As the number of workers increases, the time taken decreases.

(ii) The time taken for a journey and the distance travelled at a uniform speed are in direct proportion, not inverse.

(iii) The cultivated land area and crop harvested are in direct proportion if all other factors are constant. It is not in inverse proportion.

(iv) The time taken for a fixed journey and the vehicle’s speed are inversely proportional. As speed increases, the time it takes decreases.

(v) The population of a country and the area of land per person are in inverse proportion. As the population increases, the area of land available per person decreases.

2. In a Television game show, the prize money of ₹ 1,00,000 will be divided equally amongst the winners. 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	…	…	…	…	…

Solution

The total prize money is fixed at ₹ 1,00,000. As the number of winners increases, the prize for each winner decreases, indicating an inverse proportion.

We will assume the number of winners as a subscript of a and the amount as a subscript of b.

Solving for 4 winners
a₂ = 2, b₂ = 50,000
a₄ = 4, b₄ = x (we do not know the prize money when there are 4 participants; hence we assume it is x)

a₂b₂ = a₄b₄ (inverse proportion)
2 × 50,000 = 4x
x = (2 × 50,000)/4
x = 25,000

For 4 winners, the prize each = ₹ 25,000.

Solving for 5 winners
a₄ = 4, b₄ = 25,000
a₅ = 5, b₅ = x

a₄b₄ = a₅b₅ (inverse proportion)
4 × 25,000 = 5x
x = (4 × 25,000)/5
x = 20,000

For 5 winners, the prize each = ₹ 20,000.

Solving for 8 winners
a₅ = 5, b₅ = 20,000
a₈ = 8, b₈ = x

a₅b₅ = a₈b₈ (inverse proportion)
5 × 20,000 = 8x
x = (5 × 20,000)/8
x = 12500

For 8 winners, the prize each = ₹ 12,500.

Solving for 10 winners
a₈ = 8, b₈ = 12,500
a₁₀ = 10, b₁₀ = x

a₈b₈ = a₁₀b₁₀ (inverse proportion)
8 × 12,500 = 10x
x = (8 × 12,500)/10
x = 10,000

For 10 winners, the prize each = ₹ 10,000.

Solving for 20 winners
a₁₀ = 10, b₁₀ = 10,000
a₂₀ = 10, b₂₀ = x

a₁₀b₁₀ = a₂₀b₂₀ (inverse proportion)
10 × 10,000 = 20x
x = (10 × 10,000)/20
x = 5,000

For 20 winners, the prize each = ₹ 5,000.

Number of winners	1	2	4	5	8	10	20
Prize for each winner (in ₹)	1,00,000	50,000	25,000	20,000	12,500	10,000	5,000

To check if the money paid to winners is directly or inversely proportion –

1 × 1,00,000 = 2 × 50,000 = 4 × 25,000 = 5 × 20,000 = 8 × 12,500 = 10 × 10,000 = 20 × 5,000
1,00,000 = 1,00,000 = 1,00,000 = 1,00,000 = 1,00,000 = 1,00,000 = 1,00,000

a₂b₂ = a₄b₄ = = a₅b₅ = = a₈b₈ = a₁₀b₁₀ = a₂₀b₂₀

The above relations show that the prize money given to winners is inversely proportional to the number of winners.

3. Rehman is making a wheel using spokes. He wants to fix equal spokes so that the angles between any pair of consecutive spokes are equal.

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°?

Solution
From the partial table given, it is clear that the angle decreases with an increase in the number of spokes. The total angle around a point is 360°. The angle between spokes is inversely proportional to the number of spokes.

Like question number 1, we will assume the number of spokes as subscript of a and angle as subscript of b –

Solving for 8 spokes
a₆ = 6, b₆ = 60°
a₈ = 8, b₈ = x

a₆b₆ = a₈b₈ (inverse proportion)
6 × 60 = 8x
x = (6 × 60)/8
x = 45°

For 8 spokes, angle each = 45°

Solving for 10 spokes
a₈ = 8, b₈ = 45°
a₁₀ = 10, b₁₀ = x

a₈b₈ = a₁₀b₁₀ (inverse proportion)
8 × 45 = 10x
x = (8 × 45)/10
x = 36°

For 10 spokes, angle each = 36°

Solving for 10 spokes
a₁₀ = 10, b₁₀ = 36°
a₁₂ = 12, b₁₂ = x

a₁₀b₁₀ = a₁₂b₁₂ (inverse proportion)
10 × 36 = 12x
x = (10 × 36)/12
x = 30°

For 12 spokes, angle each = 30°.

From the above, we know that –
a₆b₆ = a₈b₈ = a₁₀b₁₀ = a₁₂b₁₂ = 360°

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.

Solving for 15 spokes
a₁₂ = 12, b₁₂ = 30°
a₁₅ = 15, b₁₅ = x

a₁₂b₁₂ = a₁₅b₁₅ (inverse proportion)
12 × 30 = 15x
x = (12 × 30)/15
x = 24°

(ii) For 15 spokes, angle each = 24°.

Solving for 40°
a₁₂ = 12, b₁₂ = 30° (we can take any number of spokes. Here, we have taken when the wheel has 12 spokes)
aᵧ = x, bᵧ = 40°

a₁₂b₁₂ = aᵧbᵧ (inverse proportion)
12 × 30 = 30x
x = (12 × 30)/40
x = 9

(iii) For an angle of 40° between spokes, number of spokes = 9 spokes.

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

Solution

To solve this problem, we first determine the total number of sweets and then distribute them among a reduced number of children.

When the number of children decreases, the sweets each child will get are more. This shows that it is a case of inverse proportion.

Initial Scenario:
Number of children = 24
Sweets per child = 5

Revised Scenario:
Number of children reduced by 4 = 24 – 4 = 20
The new distribution of sweets = ?

Since this is a case of inverse proportion –

24 × 5 = 20 × x (Let sweets per child be x)
x = (24 × 5)/20
x = 6

Under the new distribution, each child will get 6 sweets. Therefore, if the number of children is reduced by 4, each child 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?

Solution

In the question, we can see that when the number of animals increases, the food lasts fewer days. Here, we can say that the number of animals and days feed will last are in inverse proportion.

Initial Scenario:
Number of animals = 20
Duration of food availability = 6 days

Revised Scenario:
Increased number of animals = 20 animals + 10 animals = 30 animals
Duration of food availability with an increased number of animals = ? (we can assume it as x)

Since we know that this is a case of inverse proportion –

Number of animals × Duration of food availability = Increased number of animals × Duration of food availability with increased number of animals
20 × 6 = 30 × x
x = (20 × 6)/30
x = 4

Therefore, if there were 10 more animals in his cattle, the food would last 4 days.

6. A contractor estimates 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?

Solution

Initial Scenario:
Number of persons = 3
Time to complete the job = 4 days

Revised Scenario:
New number of persons = 4
Time to complete the job with 4 persons = x (we will assume the time required be x)

Since we know that the case is in inverse proportion –

3 × 4 = 4 × x
x = (3 × 4)/4
x = 3

Therefore, if 4 persons are used instead of 3, they should take 3 days to complete the job.

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?

Solution

Initial Scenario:
Number of boxes = 25
Bottles per box = 12

Revised Scenario:
New number of bottles per box = 20
Number of boxes needed = x (we will assume number of boxes required be x)

Since the case is inverse proportion –

25 × 12 = 20 × x
x = (25 × 12)/20
x = 15

Therefore, if the same batch of bottles is packed using 20 bottles in each box, 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?

Solution

Initial Scenario:
Number of machines = 42
Time to produce articles = 63 days

Revised Scenario:
New time to produce articles = 54 days
Number of machines required = x (we will assume a number of the machines needed to be x)

Since the case is inverse proportion –

42 × 63 = 54 × x
x = (42 × 63)/54
x = 49

Therefore, to produce the same number of articles in 54 days, approximately 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?

Solution

Initial Scenario:
Speed = 60 km/h
Time taken = 2 hours

Revised Scenario:
New speed = 80 km/h
Time taken at new speed = x (we will assume time taken at new speed be x)

Here, we see that this is a case of inverse proportion because when the speed increases, the time taken to reach the destination will decrease.

60 × 2 = 80 × x
x = (60 × 2)/80
x = 1.5

Therefore, when the car travels at 80 km/h, it will take 1.5 hours(1 hour 30 minutes) to reach the 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?

Initial Scenario:
Number of persons = 2
Time taken = 3 days

(i) Revised Scenario with One Person:
New number of persons = 1 (since one person fell ill)
Time taken by 1 person = y days (we will assume time taken by 1 person be y)

Since this is a case of direct proportion:
2 persons × 3 days = 1 person × y days
y = (2 × 3) / 1
y = 6

Therefore, if only one person is available to do the work, it will take 6 days to fit the new windows.

(ii) Scenario to Complete the Job in One Day:
Time taken to complete the job = 1 day
Number of persons required = z (we will assume the number of persons required to be z)

Using the initial scenario as the basis for calculation:
2 persons × 3 days = z persons × 1 day
z = (2 × 3) / 1
z = 6

Therefore, 6 persons would be needed to fit the windows in one day.

11. A school has 8 periods a day, each 45 minutes in 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?

Solution

Initial Scenario:
Number of periods per day = 8
Duration of each period = 45 minutes
Total school hours per day = 8 × 45 minutes

Revised Scenario with 9 Periods:
New number of periods per day = 9
New duration of each period = x minutes (we will assume the new duration to be x)

Since the total number of school hours remains the same:
8 × 45 = 9 × x
x = (8 × 45) / 9
x = 40

Therefore, if the school has 9 periods a day, each period would be 40 minutes long to keep the total number of school hours the same.

Practice Worksheet with Challenging Questions For Class 8 Ex. 11.2 Direct and Inverse Proportions

Questions

1. If 6 workers can complete a task in 8 days, how long will 4 workers take to complete the same task?
2. A car travelling at 70 km/h takes 5 hours to reach its destination. How long will it take at a speed of 100 km/h?
3. 20 machines can manufacture 400 widgets in 5 days. How many widgets can 30 machines manufacture in 10 days?
4. If a pipe can fill a tank in 6 hours, how long will three such pipes fill the same tank?
5. 12 workers can build a wall in 10 days. How many days will 15 workers take to build the same wall?
6. A batch of cookies takes 2 hours to bake in an oven at 180°C. How long would it take at 150°C?
7. If a 200-page book can be printed in 5 hours, how long will it take to print a 300-page book at the same rate?
8. 8 taps can fill a pool in 6 hours. How long will it take for 12 taps to fill the same pool?
9. A farmer can plough a field in 7 hours using a certain tractor. How long will it take with two such tractors?
10. If 5 friends can paint a house in 12 days, how many days will it take 8 friends to paint the same house?
11. 3 pumps can drain a pond in 4 hours. How long will it take for 5 pumps to drain the same pond?
12. 15 buses can transport 600 passengers in one trip. How many passengers can 20 buses transport in one trip?
13. If light travels 300,000 kilometres in 1 second, how far can it travel in 500 seconds?
14. A factory requires 36 workers to complete a task in 18 days. If the factory hires 9 more workers, how many days will the task take?
15. A train covers 120 km in 1.5 hours at a constant speed. How long will it take to cover 200 km at the same speed?

Answers

1. 12 days
2. 3.5 hours
3. 1200 widgets
4. 2 hours
5. 8 days
6. 2.4 hours
7. 7.5 hours
8. 4 hours
9. 3.5 hours
10. 7.5 days
11. 2.4 hours
12. 800 passengers
13. 150,000 kilometers
14. 12 days
15. 2.5 hours