NCERTCh 11Free

Chapter 11

🎓 Class 8📖 Mathematics📖 2 notes🧠 15 Q&A⏱️ ~5 min
Chapter 10Chapter 11 of 13Factorisation

Chapter 11Study Notes

NCERT-aligned · 2 notes · 3 shown free

11.1 Introduction

Explanation

11.1 Introduction

The chapter 'Direct and Inverse Proportions' begins with practical examples that students encounter in daily life, illustrating how changes in one quantity affect another. The introductory section presents two scenarios: Mohan preparing tea for himself and his sister using fixed quantities of ingredients, and students arranging chairs for an assembly. These examples highlight the concept of variation — how one quantity changes in response to another. For instance, if Mohan needs to prepare tea for five persons instead of two, how much of each ingredient is required? Similarly, if two students take 20 minutes to arrange chairs, how long will five students take? These questions prompt the study of proportional relationships. The section further provides everyday examples where quantities vary together: (i) Increase in the number of articles purchased leads to an increase in total cost. (ii) More money deposited in a bank results in more interest earned. (iii) Increase in vehicle speed decreases the time taken to cover a fixed distance. (iv) More workers reduce the time needed to complete a job. Such observations emphasize that changes in one quantity often cause changes in another. To analyze and quantify these relationships, the chapter introduces the concepts of direct and inverse proportion. Understanding these concepts enables students to solve problems involving proportional relationships efficiently.

  • Variation in one quantity often causes variation in another.
  • Direct and inverse proportions help quantify these relationships.
  • Real-life examples include cost vs. quantity, speed vs. time, workers vs. time.
  • The introductory examples set the stage for studying proportionality.
  • Understanding proportions aids in solving practical problems.
  • Observation of ratios and products helps identify types of proportion.
  • 📌 Variation: Change in one quantity causing change in another.
  • 📌 Direct Proportion: When two quantities increase or decrease together maintaining a constant ratio.
  • 📌 Inverse Proportion: When one quantity increases while the other decreases such that their product remains constant.

11.2 Direct Proportion

Concept

11.2 Direct Proportion

Direct proportion describes a relationship between two quantities where both increase or decrease together such that their ratio remains constant. This means if one quantity doubles, the other also doubles; if one triples, the other triples, and so forth. The section begins with an example: If 1 kg of sugar costs ₹36, then 3 kg costs ₹108. This proportionality is tabulated to show that as the weight of sugar increases, the cost increases maintaining a constant ratio of cost to weight. Another example involves petrol consumption and distance travelled by a car. If 4 litres of petrol allow the car to travel 60 km, then 12 litres (three times) will allow it to travel 180 km (also three times). This shows that petrol consumed (x) and distance travelled (y) satisfy the relation x/y = k, a constant. Mathematically, two quantities x and y are in direct proportion if x/y = k or x = ky, where k is a positive constant. For two pairs of values (x1, y1) and (x2, y2), the relation x1/y1 = x2/y2 holds. The section also includes activities like measuring the angle turned by a clock's minute hand over time, showing that angle turned and time are directly proportional. Another activity compares the ages of a friend and their mother over time, illustrating that not all variables that increase together are in direct proportion. Students are encouraged to think of more examples and verify direct proportionality by checking if the ratio remains constant. The section concludes with solved examples involving cloth cost, shadows, weights, and uniform speed, applying the direct proportion formula to find unknown quantities. Understanding direct proportion helps in solving practical problems involving scaling, cost estimation, speed-distance-time relations, and map reading using scales.

  • Direct proportion means two quantities increase or decrease together maintaining a constant ratio.
  • Mathematically, x and y are directly proportional if x/y = k or x = ky.
  • Examples include cost and quantity, petrol consumption and distance, angle turned and time.
  • Activities help verify direct proportion by checking constant ratios.
  • Not all quantities increasing together are directly proportional.
  • Direct proportion formula: (x1/y1) = (x2/y2) is used to solve problems.
  • 📌 Direct Proportion: Relation where ratio of two quantities is constant.
  • 📌 Constant of Proportionality (k): The constant ratio x/y in direct proportion.

Practice QuestionsChapter 11

Includes NCERT exercise questions with answers

Q1.1. 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.

Answer:

To check if parking charges are in direct proportion to parking time, we check if the ratio (charge/time) is constant. For 4 hours: 60/4 = 15 For 8 hours: 100/8 = 12.5 For 12 hours: 140/12 ≈ 11.67 For 24 hours: 180/24 = 7.5 Since the ratios are not equal, the parking charges are not in direct proportion to the parking time.

Explanation:

Direct proportion means the ratio of charges to time should be constant. Here, the ratio decreases as time increases, so charges are not directly proportional to time.

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Q2.2. 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, ..., ..., ..., ...

Answer:

Given ratio of red pigment to base is 1:8. For 4 parts red pigment, base = 4 × 8 = 32 For 7 parts red pigment, base = 7 × 8 = 56 For 12 parts red pigment, base = 12 × 8 = 96 For 20 parts red pigment, base = 20 × 8 = 160 So, the completed table is: | Parts of red pigment | 1 | 4 | 7 | 12 | 20 | |---------------------|---|---|---|----|----| | Parts of base | 8 | 32| 56| 96 |160 |

Explanation:

Since the parts of base are directly proportional to parts of red pigment with ratio 1:8, multiply red pigment parts by 8 to get base parts.

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Q3.3. 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?

Answer:

Given: 1 part red pigment requires 75 mL base. Let the parts of red pigment be x. Then, base required = 75 × x mL. Given base = 1800 mL. So, 75 × x = 1800 => x = 1800 / 75 = 24 Therefore, 24 parts of red pigment should be mixed with 1800 mL of base.

Explanation:

Using direct proportion, base = 75 × (parts of red pigment). Equate and solve for parts of red pigment.

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Q4.4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Answer:

Bottles filled in 6 hours = 840 Bottles filled in 1 hour = 840 / 6 = 140 Bottles filled in 5 hours = 140 × 5 = 700 Therefore, the machine will fill 700 bottles in 5 hours.

Explanation:

Number of bottles filled is directly proportional to time. Find bottles per hour and multiply by 5.

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Q5.5. 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?

Answer:

Given: Enlargement factor = 50,000 Length of enlarged bacteria = 5 cm Actual length = Enlarged length / Enlargement factor = 5 cm / 50,000 = 0.0001 cm = 1 × 10⁻⁴ cm If enlarged 20,000 times: Enlarged length = Actual length × 20,000 = 1 × 10⁻⁴ × 20,000 = 2 cm Therefore, actual length is 0.0001 cm and enlarged length at 20,000 times is 2 cm.

Explanation:

Actual length = Enlarged length divided by enlargement factor. For new enlargement, multiply actual length by new factor.

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Q6.6. 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?

Answer:

Given: Height of model mast = 9 cm = 0.09 m Height of actual mast = 12 m Length of actual ship = 28 m Scale factor = model mast / actual mast = 0.09 / 12 = 0.0075 Length of model ship = scale factor × actual length = 0.0075 × 28 = 0.21 m = 21 cm Therefore, the model ship is 21 cm long.

Explanation:

Use ratio of corresponding heights to find scale factor, then apply to length.

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Q7.7. 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?

Answer:

(i) Number of crystals in 2 kg = 9 × 10⁶ Number of crystals in 1 kg = (9 × 10⁶) / 2 = 4.5 × 10⁶ Number of crystals in 5 kg = 4.5 × 10⁶ × 5 = 22.5 × 10⁶ = 2.25 × 10⁷ (ii) Number of crystals in 1.2 kg = 4.5 × 10⁶ × 1.2 = 5.4 × 10⁶ Therefore, (i) 2.25 × 10⁷ crystals, (ii) 5.4 × 10⁶ crystals.

Explanation:

Number of crystals is directly proportional to mass. Find crystals per kg and multiply by given mass.

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Q8.8. 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?

Answer:

Scale: 1 cm = 18 km Distance driven = 72 km Distance on map = 72 / 18 = 4 cm Therefore, Rashmi's distance covered on the map is 4 cm.

Explanation:

Use scale to convert actual distance to map distance by dividing actual distance by scale factor.

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