NCERT Solutions for Class 8 Maths Ex. 11.1 Direct and Inverse Proportions

Direct Proportions: When two quantities increase or decrease in the same ratio, they are said to be in direct proportion. For example, if the number of items bought increases, the total cost also increases. The simultaneous increase maintains a constant ratio.

Inverse Proportions: When 1 quantity increases and the other decreases in such a way that their product remains constant, they are called as in inverse proportion. For example, if the speed of a vehicle increases, the time taken to cover the same distance decreases. This keeps the product of speed and time constant.

Examples and Solutions

Example 1: Direct Proportion (Cost and Quantity)

Question: If 5 pens cost ₹10, how much would 8 pens cost?
Solution:

1. Let the cost of 8 pens be ₹X.
2. Since the cost is directly proportional to the number of pens, we set up a proportion: 5 pens / ₹10 = 8 pens / ₹X.
3. To find X, cross-multiply and solve: 5X = 8 * 10, so X = (8 * 10) / 5 = ₹16.
4. Therefore, 8 pens would cost ₹16.

Example 2: Inverse Proportion (Speed and Time)

Question: If a car travels 60 km in 2 hours, how long will it take to travel 120 km at the same speed?
Solution:

1. Let the time to travel 120 km be T hours.
2. Since speed and time are inversely proportional (speed up, time down), the product of distance and time is constant: 60 km * 2 hours = 120 km * T hours.
3. Solve for T: 120 * T = 120, so T = 120 / 120 = 1 hour.
4. It will take 1 hour to travel 120 km at the same speed.

Example 3: Direct Proportion (Map Scale)

Question: On a map with a scale of 1:100000, a distance of 3 cm represents an actual distance of 3 km. What actual distance does 5 cm on the map represent?
Solution:

1. Let the actual distance for 5 cm be D km.
2. Using direct proportion, set up the equation: 3 cm / 3 km = 5 cm / D km.
3. Cross-multiply and solve for D: 3D = 5 * 3, so D = 15 / 3 = 5 km.
4. Therefore, 5 cm on the map represents an actual distance of 5 km.

NCERT Solutions for Class 8 Maths Exercise 11.1 Chapter 11 Direct and Inverse Proportions

1. 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.

Solution:
Parking time (hours) = t
Parking charge (₹) = C

For direct proportion, C/t should be constant.

For 4 hours: C/t = 60/4 = 15
For 8 hours: C/t = 100/8 ≈ 12.5
For 12 hours: C/t = 140/12 ≈ 11.67
For 24 hours: C/t = 180/24 = 7.5

Since C/t is not constant, parking charges are not in direct proportion to 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.

Solution:
Parts of red pigment = x
Parts of base = y
Proportion = 1 part red pigment : 8 parts base

For 1 part red pigment: y = 1 * 8 = 8 parts base
For 4 parts red pigment: y = 4 * 8 = 32 parts base
For 7 parts red pigment: y = 7 * 8 = 56 parts base
For 12 parts red pigment: y = 12 * 8 = 96 parts base
For 20 parts red pigment: y = 20 * 8 = 160 parts base

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?

Solution:
1 part red pigment requires 75 mL base.
Let the required parts of red pigment = z

1 part pigment : 75 mL base = z parts pigment : 1800 mL base

We know that x₁/y₁ = x₂/y₂
Here z parts pigment = x₂
Here x₁ = 1, y₁ = 75, x₂ = z, and y₂ = 1800
therefore, 1/75 = x₂/1800
Cross multiplying, x₂ = (1800 / 75)
x₂ = (1800 × 1)/ 75 = 24 parts

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

Solution:
Bottles filled in 6 hours = 840
Let bottles filled in 5 hours = B

6 hours : 840 bottles = 5 hours : B bottles
B = (840 * 5) / 6 = 700 bottles

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?

Solution:
Enlarged length at 50,000 times = 5 cm
Let actual length = A cm

50,000 times : 5 cm = 20,000 times : Enlarged length
Enlarged length at 20,000 times = (5 * 20,000) / 50,000 = 2 cm
Actual length = 5 cm / 50,000 = 0.0001 cm

Actual length of bacteria = 0.0001 cm
Enlarged length at 20,000 times = 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?

Solution:
Mast height of model ship = 9 cm
Mast height of actual ship = 12 m (1200 cm)
Length of actual ship = 28 m (2800 cm)
Let the length of the model ship = L cm

1200 cm : 2800 cm = 9 cm : L cm
L = (9 * 2800) / 1200 = 21 cm

Length of the model ship = 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?

Solution:

2 kg sugar = 9 × 10⁶ crystals

(i) For 5 kg of sugar:
Let the number of crystals = N₁
2 kg : 9 × 10⁶ = 5 kg : N₁
N₁ = (5 * 9 × 10⁶) / 2 = 22.5 × 10⁶ crystals

(ii) For 1.2 kg of sugar:
Let the number of crystals = N₂
2 kg : 9 × 10⁶ = 1.2 kg : N₂
N₂ = (1.2 * 9 × 10⁶) / 2 = 5.4 × 10⁶ crystals

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?

Solution:
Map scale = 1 cm : 18 km
Actual distance driven = 72 km
Let map distance = M cm

18 km : 1 cm = 72 km : M cm
M = 72 / 18 = 4 cm

Map distance = 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 5m long.

Solution:

Vertical pole height = 5 m 60 cm (560 cm)
Shadow length = 3 m 20 cm (320 cm)

(i) For another pole 10 m 50 cm (1050 cm) high:
Let shadow length = S₁ cm
560 cm : 320 cm = 1050 cm : S₁ cm
S₁ = (1050 * 320) / 560 = 600 cm (6 m)

(ii) For a pole with 5 m (500 cm) long shadow:
Let pole height = H cm
560 cm : 320 cm = H cm : 500 cm
H = (560 * 500) / 320 ≈ 875 cm (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?

Solution:
14 km in 25 minutes
Let distance covered in 5 hours (300 minutes) = D km

25 minutes : 14 km = 300 minutes : D km
D = (14 * 300) / 25 = 168 km

Distance traveled in 5 hours = 168 km

Direct and Inverse Exercise 11.1 Proportions Worksheet

Questions

1. If 3 apples cost $1.50, how much will 10 apples cost?
2. A car travels 200 km in 4 hours. How long will it take to travel 350 km at the same speed?
3. If 8 workers can build a house in 240 days, how many days will 6 workers take to build the same house?
4. A recipe needs 4 cups of flour for 2 cups of sugar. How much flour is needed for 3 cups of sugar?
5. If a 7 cm tall toy is a scale model of a 21 m tall monument, what is the scale of the model?
6. 12 machines take 18 hours to complete a job. How many hours will 16 machines take to do the same job?
7. A photograph is enlarged to twice its length and breadth. How does the area change?
8. If 5 liters of paint cover an area of 75 square meters, how much area will 15 liters of paint cover?
9. A train covers 60 km in 40 minutes. How long will it take to cover 150 km at the same speed?
10. A map scale is 1 cm to 50 km. How far apart are two cities that are 3.5 cm apart on the map?
11. It takes 5 hours to fill a swimming pool with two pipes. How long will it take if only one pipe is used?
12. A rubber band stretched to a length of 8 cm has a width of 2 cm. If it is stretched to 10 cm, what is the new width?
13. If 6 batteries last 30 hours in a flashlight, how long will 4 batteries last?
14. 20 kg of rice is enough for 40 people. How many people can be fed with 35 kg of rice?
15. A fan running at high speed makes 900 revolutions per minute. How many revolutions will it make in 10 seconds?

Answers

1. $5.00 for 10 apples.
2. 7 hours to travel 350 km.
3. 320 days for 6 workers.
4. 6 cups of flour for 3 cups of sugar.
5. Scale of 1 cm to 3 m.
6. 13.5 hours for 16 machines.
7. Area becomes 4 times larger.
8. 225 square meters covered by 15 liters of paint.
9. 1 hour and 40 minutes for 150 km.
10. 175 km apart.
11. 10 hours with one pipe.
12. New width is 1.6 cm.
13. 20 hours for 4 batteries.
14. 70 people can be fed with 35 kg of rice.
15. 150 revolutions in 10 seconds.