Class 8 Maths - Direct and Inverse Proportions NCERT Solutions

Chapter 13: Direct and Inverse Proportions (NCERT Solutions)

Exercise 13.1

Q1. Following are the car parking charges near a railway station upto:
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.

Let's find the ratio of parking charges to the parking time:
C₁ / T₁ = 60 / 4 = 15
C₂ / T₂ = 100 / 8 = 12.5
C₃ / T₃ = 140 / 12 = 11.66
C₄ / T₄ = 180 / 24 = 7.5

Since the ratios are not equal (15 ≠ 12.5 ≠ 11.66 ≠ 7.5), the parking charges are not in direct proportion to the parking time.

Q2. 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 parts of base needed for 4, 7, 12, 20 parts of red pigments be x₁, x₂, x₃, and x₄ respectively.
This is a case of direct proportion.
1 / 8 = 4 / x₁ ⇒ x₁ = 8 × 4 = 32
1 / 8 = 7 / x₂ ⇒ x₂ = 8 × 7 = 56
1 / 8 = 12 / x₃ ⇒ x₃ = 8 × 12 = 96
1 / 8 = 20 / x₄ ⇒ x₄ = 8 × 20 = 160

Q3. 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 be x.
This is a case of direct proportion.
1 / 75 = x / 1800
75x = 1800
x = 1800 / 75 = 24 parts.

Q4. 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 5 hours be x.
Since the number of bottles filled and the time taken are in direct proportion:
840 / 6 = x / 5
140 = x / 5
x = 140 × 5 = 700 bottles.

Q5. 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?

Actual length = (Enlarged Length) / (Enlargement ratio)
Actual length = 5 / 50000 = 1 / 10000 = 10^-4 cm.

Let the new enlarged length be x cm.
Since enlargement and enlarged length are in direct proportion:
5 / 50000 = x / 20000
x = (5 × 20000) / 50000 = 10 / 5 = 2 cm.

Q6. 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 the model ship be x cm.
Since the measurements of the model and the actual ship are proportional:
Height of model / Height of actual ship = Length of model / Length of actual ship
9 / 12 = x / 28
12x = 9 × 28
x = (9 × 28) / 12 = 3 × 7 = 21 cm.

Q7. 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 the number of crystals in 5 kg be x.
Direct proportion.
2 / (9 × 10⁶) = 5 / x
2x = 5 × 9 × 10⁶
x = (45 / 2) × 10⁶ = 22.5 × 10⁶ = 2.25 × 10⁷ crystals.

(ii) Let the number of crystals in 1.2 kg be y.
Direct proportion.
2 / (9 × 10⁶) = 1.2 / y
2y = 1.2 × 9 × 10⁶
y = (10.8 / 2) × 10⁶ = 5.4 × 10⁶ crystals.

Q8. 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 the distance on the map be x cm.
Direct proportion.
1 / 18 = x / 72
18x = 72
x = 72 / 18 = 4 cm.

Q9. 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 5m long.

Convert all units to cm: 5 m 60 cm = 560 cm. 3 m 20 cm = 320 cm. 10 m 50 cm = 1050 cm. 5 m = 500 cm.

(i) Let the length of the shadow be x cm.
Direct proportion (Height/Shadow).
560 / 320 = 1050 / x
7 / 4 = 1050 / x
7x = 4200
x = 600 cm = 6 m.

(ii) Let the height of the pole be y cm.
Direct proportion.
560 / 320 = y / 500
7 / 4 = y / 500
4y = 3500
y = 3500 / 4 = 875 cm = 8 m 75 cm.

Q10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Convert hours to minutes: 5 hours = 5 × 60 = 300 minutes.
Let the distance traveled be x km.
Direct proportion.
14 / 25 = x / 300
25x = 14 × 300
x = (14 × 300) / 25 = 14 × 12 = 168 km.

Exercise 13.2

Q1. 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) Yes (Inverse proportion). More workers will take less time to complete the job.

(ii) No (Direct proportion). More time is taken to cover a larger distance at uniform speed.

(iii) No (Direct proportion). More cultivated land yields a larger harvest.

(iv) Yes (Inverse proportion). Faster speed means lesser time taken for the journey.

(v) Yes (Inverse proportion). Higher population means lesser land area per person.

Q2. 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 ₹): 100000, 50000, ...

Let the number of winners be x and the prize money be y.
Here, x × y = 1 × 100000 = 100000 (constant). So, they are inversely proportional.

For x = 4, 4 × y = 100000 ⇒ y = 25000
For x = 5, 5 × y = 100000 ⇒ y = 20000
For x = 8, 8 × y = 100000 ⇒ y = 12500
For x = 10, 10 × y = 100000 ⇒ y = 10000
For x = 20, 20 × y = 100000 ⇒ y = 5000

Q3. 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 angle 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°?

Number of spokes (x) and Angle (y). Total angle = 360°.
x × y = 4 × 90 = 360 (constant). So they are in inverse proportion.

For x = 8, 8y = 360 ⇒ y = 45°
For x = 10, 10y = 360 ⇒ y = 36°
For x = 12, 12y = 360 ⇒ y = 30°

(i) Yes, they are in inverse proportion.

(ii) For 15 spokes (x=15), 15y = 360 ⇒ y = 360 / 15 = 24°.

(iii) For angle 40° (y=40), 40x = 360 ⇒ x = 360 / 40 = 9 spokes.

Q4. 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?

Reduced number of children = 24 - 4 = 20.
Let the number of sweets each gets be x.
Inverse proportion (fewer children → more sweets each).
24 × 5 = 20 × x
120 = 20x
x = 120 / 20 = 6 sweets.

Q5. 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?

New number of animals = 20 + 10 = 30.
Let the food last for x days.
Inverse proportion (more animals → food lasts for fewer days).
20 × 6 = 30 × x
120 = 30x
x = 120 / 30 = 4 days.

Q6. A contractor estimates that 3 persons could rewire Jasminder's house in 4 days. If he uses 4 persons instead of 3, how long should they take to complete the job?

Let the number of days taken by 4 persons be x.
Inverse proportion.
3 × 4 = 4 × x
12 = 4x
x = 12 / 4 = 3 days.

Q7. 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.
Inverse proportion.
25 × 12 = x × 20
300 = 20x
x = 300 / 20 = 15 boxes.

Q8. 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.
Inverse proportion (Fewer days need more machines).
42 × 63 = x × 54
x = (42 × 63) / 54
x = (42 × 7) / 6 = 7 × 7 = 49 machines.

Q9. 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?

Let the time taken be x hours.
Inverse proportion (More speed means less time).
60 × 2 = 80 × x
120 = 80x
x = 120 / 80 = 1.5 hours = 1 hour 30 minutes.

Q10. 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) Only 1 person is working.
Let the days required be x. Inverse proportion.
2 × 3 = 1 × x
x = 6 days.

(ii) Let the number of persons needed be y.
Inverse proportion.
2 × 3 = y × 1
y = 6 persons.

Q11. 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 minutes.
Inverse proportion (More periods mean lesser duration per period).
8 × 45 = 9 × x
360 = 9x
x = 360 / 9 = 40 minutes.

Class 8 Maths - Direct and Inverse Proportions Practice Questions

Chapter 13: Direct and Inverse Proportions (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. If the cost of 12 pens is ₹ 138, find the cost of 14 such pens.

Let the cost of 14 pens be ₹ x.
This is a case of direct proportion.
12 / 138 = 14 / x
12x = 138 × 14
x = (138 × 14) / 12 = ₹ 161.

Q2. A worker is paid ₹ 1110 for 6 days. If his total wages during a month are ₹ 4625, for how many days did he work?

Let the number of days be x.
This is a case of direct proportion.
1110 / 6 = 4625 / x
1110 × x = 4625 × 6
x = (4625 × 6) / 1110 = 25 days.

Q3. A car can cover a distance of 522 km on 36 litres of petrol. How far can it travel on 14 litres of petrol?

Let the distance covered be x km.
Direct proportion: distance and petrol consumption.
522 / 36 = x / 14
36x = 522 × 14
x = (522 × 14) / 36 = 203 km.

Q4. 8 men can complete a piece of work in 20 days. In how many days can 20 men complete the same work?

Let the days required be x.
More men will take fewer days. This is an inverse proportion.
8 × 20 = 20 × x
160 = 20x
x = 160 / 20 = 8 days.

Q5. If 36 animals can eat a certain quantity of food in 15 days, for how many days will the same food last if there are 45 animals?

Let the number of days be x.
More animals will finish the food in fewer days. Inverse proportion.
36 × 15 = 45 × x
x = (36 × 15) / 45 = 12 days.

Q6. A sum of money is divided among 25 persons such that each gets ₹ 95. If the same sum is divided among 19 persons, how much will each get?

Let the amount each gets be ₹ x.
Fewer persons means each gets more money. Inverse proportion.
25 × 95 = 19 × x
x = (25 × 95) / 19 = 25 × 5 = ₹ 125.

Q7. 120 men had food provision for 200 days. After 5 days, 30 men died due to an epidemic. How long will the remaining food last?

After 5 days, food left is sufficient for 120 men for (200 - 5) = 195 days.
Remaining men = 120 - 30 = 90 men.
Let the food last for x days for 90 men. This is inverse proportion.
120 × 195 = 90 × x
x = (120 × 195) / 90 = 4 × 195 / 3 = 4 × 65 = 260 days.

Q8. A map is given with a scale of 2 cm = 1000 km. What is the actual distance between the two places in kms, if the distance in the map is 2.5 cm?

Let the actual distance be x km.
Direct proportion.
2 / 1000 = 2.5 / x
2x = 2500
x = 2500 / 2 = 1250 km.

Q9. 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used?

1 hr 20 mins = 80 minutes.
Let the time taken by 5 pipes be x minutes.
Fewer pipes will take more time. Inverse proportion.
6 × 80 = 5 × x
x = 480 / 5 = 96 minutes = 1 hour 36 minutes.

Q10. At a speed of 60 km/h, a journey takes 4.5 hours. What should be the speed if the journey is to be completed in 3 hours?

Let the new speed be x km/h.
Less time means higher speed. Inverse proportion.
60 × 4.5 = x × 3
270 = 3x
x = 270 / 3 = 90 km/h.

Q11. The mass of an aluminium rod varies directly with its length. If a 16 cm long rod has a mass of 192 g, find the length of the rod whose mass is 105 g.

Let the length of the rod be x cm.
Direct proportion: length and mass.
16 / 192 = x / 105
x = (16 × 105) / 192
x = 105 / 12 = 8.75 cm.

Q12. A tree 24 m high casts a shadow of 15 m. At the same time, another tree casts a shadow of 6 m. Find the height of the second tree.

Let the height of the second tree be x m.
Direct proportion: Height and shadow length.
24 / 15 = x / 6
x = (24 × 6) / 15 = 144 / 15 = 9.6 m.

Q13. In a camp, there is enough food for 500 soldiers for 30 days. If 100 soldiers leave the camp, how many days will the food last?

Remaining soldiers = 500 - 100 = 400.
Let the food last for x days.
Inverse proportion.
500 × 30 = 400 × x
x = (500 × 30) / 400 = 15000 / 400 = 37.5 days.

Q14. An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.

Let the height of the tree be x m.
Direct proportion.
14 / 10 = x / 15
10x = 210
x = 210 / 10 = 21 m.

Q15. A contractor estimates that 4 persons could rewire Jasminder's house in 4 days. If he uses 8 persons instead, how long should they take?

Let the days taken be x.
More persons will take fewer days. Inverse proportion.
4 × 4 = 8 × x
16 = 8x
x = 16 / 8 = 2 days.

Q16. The cost of 5 metres of a particular quality of cloth is ₹ 210. Find the cost of 13 metres.

Let the cost be x.
Direct proportion.
5 / 210 = 13 / x
5x = 210 × 13
x = (210 × 13) / 5 = 42 × 13 = ₹ 546.

Q17. 45 cows can graze a field in 13 days. How many cows will graze the same field in 9 days?

Let the number of cows be x.
To finish grazing in fewer days, more cows are needed. Inverse proportion.
45 × 13 = x × 9
x = (45 × 13) / 9 = 5 × 13 = 65 cows.

Q18. If x and y vary inversely and x = 18 when y = 8. Find x when y = 16.

For inverse proportion, x × y = constant.
18 × 8 = x × 16
144 = 16x
x = 144 / 16 = 9.

Q19. A train is moving at a uniform speed of 75 km/hour. (i) How far will it travel in 20 minutes? (ii) Find the time required to cover a distance of 250 km.

(i) Distance and time are in direct proportion.
Speed: 75 km in 60 minutes.
Let distance be x km in 20 mins.
75 / 60 = x / 20
x = (75 × 20) / 60 = 75 / 3 = 25 km.

(ii) Let time be t minutes for 250 km.
75 / 60 = 250 / t
75t = 15000
t = 15000 / 75 = 200 minutes = 3 hours 20 minutes.

Q20. Reema types 540 words during half an hour. How many words would she type in 8 minutes?

Half an hour = 30 minutes.
Let the words typed be x.
Direct proportion: Words typed and time.
540 / 30 = x / 8
18 = x / 8
x = 18 × 8 = 144 words.

Class 8 Maths - Direct and Inverse Proportions Summary

Chapter 13: Direct and Inverse Proportions (Concepts & Formulas)

1. Introduction

Variations describe how two quantities relate to each other. If the value of one quantity changes, causing a corresponding change in the other quantity, they are said to be in variation or proportion.

Examples: If the number of articles purchased increases, the total cost increases.
If the speed of a car increases, the time taken to cover a fixed distance decreases.

2. Direct Proportion

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.

If x and y are in direct proportion, then x/y = k, where k is a positive number (constant).

Alternatively, if y₁, y₂ are the values of y corresponding to the values x₁, x₂ of x respectively, then:

x₁ / y₁ = x₂ / y₂

Example: Cost of articles and the number of articles.

3. Inverse Proportion

Two quantities x and y are said to be in inverse proportion if an increase in x causes a proportional decrease in y (and vice versa) in such a manner that the product of their corresponding values remains constant.

If x and y are in inverse proportion, then x × y = k, where k is a positive number (constant).

Alternatively, if y₁, y₂ are the values of y corresponding to the values x₁, x₂ of x respectively, then:

x₁ × y₁ = x₂ × y₂

Example: Speed of a vehicle and the time required to cover a fixed distance. Or, the number of workers and the days required to complete a piece of work.

4. Quick Problem Solving Strategy

Identify the two quantities involved in the problem.
Determine the relationship between them:
- If an increase in one leads to an increase in the other → Direct Proportion.
- If an increase in one leads to a decrease in the other → Inverse Proportion.
Create a table with rows for the two variables (e.g., x and y).
Fill in the known values (x₁, y₁, x₂). Let the unknown be y₂ (or x).
Apply the correct formula and solve the equation.

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