MathematicsClass 811.1 Introduction

11.1 Introduction | Class 8 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

11.1 Introduction – this guide gives you a concise, exam-ready overview of 11.1 Introduction from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

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.

📊 Diagram: See figure_3: Observe that as weight of sugar increases, cost also increases in such a manner that their ratio remains constant.; See figure_4:

🧪 Activity: Activity: Fix the minute hand of a clock at 12 and record the angle turned and time passed at intervals; observe the ratio of time to angle to verify direct proportion.

🔗 Connection: This section lays the foundation for understanding direct proportion, which is contrasted with inverse proportion in the next section.

Frequently asked questions

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.

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.

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, ..., ..., ..., ...

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:

Petrol in litres (x)	4	8	12	15	20	25
Distance in km (y)	60	...	180	...	...

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?

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.

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

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.

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11.1 Introduction | Class 8 Mathematics Notes

11.2 Direct Proportion

Frequently asked questions

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