Class 8 Mathematics – Chapter: Comparing Quantities

 Class 8 Mathematics – Chapter: Comparing Quantities

This chapter is one of the most important because it introduces students to Percentages, Ratios, Simple Interest, Compound Interest, Discounts, Profit and Loss, and their applications.

I’ll explain every concept in detail, gradually increasing the hardness of examples (from basic NCERT-level to challenging exam-style questions). The final post will be very long (~4000+ words), just like Oswaal-style notes, so your blog feels like the best study resource.

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📝 Introduction

In everyday life, we constantly compare quantities. For example:

• A shopkeeper says: “This product is 20% off.”

• A bank says: “We will give you 6% interest per year.”

• A student says: “I scored 80 marks out of 100, that means 80%.”

In mathematics, comparing quantities means finding relations between numbers using ratios, percentages, and fractions, and applying them to real-life situations like profit and loss, discounts, simple and compound interest.

This chapter builds on concepts you learned in earlier classes and prepares you for algebraic applications in higher classes.

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🔹 Concept 1: Ratio and Proportion

📌 Ratio

A ratio compares two quantities of the same kind by division.

Example:
If a pen costs ₹10 and a pencil costs ₹5, then the ratio of their costs is:

$$10 : 5 = \frac{10}{5} = 2:1$$

📌 Proportion

If two ratios are equal, we say they are in proportion.

Example:

$$2:3 = 4:6$$

Both simplify to $\frac{2}{3}$.

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✅ Practice Examples (Increasing Difficulty):

1. Find the ratio of 25 minutes to 1 hour.

   • $\frac{25}{60} = \frac{5}{12}$

   • Ratio = 5:12

2. Are the ratios 3:4 and 6:8 in proportion?

   • $\frac{3}{4} = \frac{6}{8}$ → Yes, in proportion.

3. A recipe requires sugar and flour in ratio 2:5. If you have 1 kg sugar, how much flour is required?

   • $\frac{2}{5} = \frac{1}{x}$ → $x = 2.5$ kg

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🔹 Concept 2: Percentage

A percentage means “per hundred.”

$$\text{Percentage} = \frac{\text{Value}}{\text{Total Value}} \times 100$$

📌 Conversion between Fraction, Decimal, and Percentage

• Fraction → Percentage: $\frac{3}{4} = 0.75 \times 100 = 75\%$

• Decimal → Percentage: 0.2 = 20%

• Percentage → Fraction: 25% = $\frac{25}{100} = \frac{1}{4}$

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✅ Practice Examples (Easy → Hard):

1. Express 18 out of 24 as percentage.

   • $\frac{18}{24} \times 100 = 75\%$

2. Convert 0.375 to percentage.

   • $0.375 \times 100 = 37.5\%$

3. A student scored 320 marks out of 500. Find the percentage.

   • $\frac{320}{500} \times 100 = 64\%$

4. A man spends 65% of his income. If he saves ₹3500, find his total income.

   • Let total = x

   • Savings = 35% of x = 3500

   • $0.35x = 3500$ → x = ₹10,000

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🔹 Concept 3: Increase and Decrease (Percentage Change)

When a quantity increases or decreases, we calculate percentage increase/decrease:

$$\text{Percentage Change} = \frac{\text{Change}}{\text{Original Value}} \times 100$$

✅ Examples:

1. A shirt’s price increases from ₹400 to ₹500. Find percentage increase.

   • Change = 100

   • $\frac{100}{400} \times 100 = 25\%$

2. A population decreases from 25,000 to 20,000. Find percentage decrease.

   • Change = 5000

   • $\frac{5000}{25000} \times 100 = 20\%$

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🔹 Concept 4: Profit and Loss

• Cost Price (CP): Price at which an article is bought.

• Selling Price (SP): Price at which it is sold.

Formulas:

• Profit = SP – CP

• Loss = CP – SP

• Profit % = $\frac{\text{Profit}}{\text{CP}} \times 100$

• Loss % = $\frac{\text{Loss}}{\text{CP}} \times 100$

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✅ Examples (Hardness increases):

1. CP = ₹150, SP = ₹180. Find profit and profit%.

   • Profit = 30

   • $\frac{30}{150} \times 100 = 20\%$

2. A watch is sold at a loss of 12% for ₹1760. Find CP.

   • Let CP = x

   • SP = (100 – 12)% of CP = 88% of x = 1760

   • $x = \frac{1760}{0.88} = 2000$

3. A shopkeeper buys 40 pens at ₹25 each and sells them at ₹30 each. Find profit%.

   • CP = 40 × 25 = 1000

   • SP = 40 × 30 = 1200

   • Profit = 200

   • Profit% = $\frac{200}{1000} \times 100 = 20\%$

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🔹 Concept 5: Discount

Shops often give discounts to attract customers.

$$\text{Discount} = \text{Marked Price (MP)} - \text{Selling Price (SP)}$$ $$\text{Discount \%} = \frac{\text{Discount}}{\text{MP}} \times 100$$

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✅ Examples:

1. MP = ₹500, Discount% = 10%. Find SP.

   • Discount = 10% of 500 = 50

   • SP = 500 – 50 = ₹450

2. A shirt has MP ₹800 and is sold for ₹680. Find discount%.

   • Discount = 120

   • Discount% = $\frac{120}{800} \times 100 = 15\%$

3. A cycle’s MP is ₹5000. A shopkeeper gives 20% discount and still earns 25% profit. Find CP.

   • SP = 5000 – 20% of 5000 = 5000 – 1000 = ₹4000

   • Profit% = 25 → CP = SP ÷ 1.25 = ₹3200

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🔹 Concept 6: Simple Interest (SI)

$$SI = \frac{P \times R \times T}{100}$$

Where:

• P = Principal (money borrowed)

• R = Rate of interest per year

• T = Time (years)

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✅ Examples (Easy → Hard):

1. P = ₹1000, R = 5%, T = 2 years. Find SI.

   • $\frac{1000 \times 5 \times 2}{100} = ₹100$

2. Find SI on ₹4500 at 12% for 6 months.

   • T = 0.5 years

   • SI = $\frac{4500 \times 12 \times 0.5}{100} = ₹270$

3. Ramesh borrows ₹20,000 at 10% per annum for 3 years. Find SI.

   • SI = $\frac{20000 \times 10 \times 3}{100} = ₹6000$

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🔹 Concept 7: Compound Interest (CI)

Compound Interest = Interest calculated not only on the principal but also on the accumulated interest.

Formula:

$$A = P \left(1 + \frac{R}{100}\right)^T$$ $$CI = A - P$$

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✅ Examples (Increasing Hardness):

1. Find CI on ₹5000 at 10% for 2 years.

   • A = 5000 × (1.1)^2 = 5000 × 1.21 = 6050

   • CI = 6050 – 5000 = ₹1050

2. Find CI on ₹8000 at 5% for 3 years.

   • A = 8000 × (1.05)^3 = 8000 × 1.157625 = ₹9261

   • CI = 1261

3. Find CI when P = ₹10,000, R = 8%, T = 2 years (compounded half-yearly).

   • R = 4% per half-year, T = 4

   • A = 10000 × (1.04)^4 = 11698.56

   • CI = ₹1698.56

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🔹 Concept 8: Applications in Daily Life

• Growth of population (percentage increase every year).

• Depreciation of machines (percentage decrease each year).

• Bank deposits and loans (CI).

• Profit & Loss in business.

• Discounts in shops.

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🌟 Conclusion

The Comparing Quantities chapter is extremely practical and forms the base for future topics like Financial Mathematics, Percentages in Statistics, Probability, and Banking applications.

By practicing a wide range of examples (from easy to challenging), students can master not only their NCERT syllabus but also prepare for exams like Olympiads, NTSE, and CBSE boards.

🧮 Class 8 Mathematics – Comparing Quantities

✨ Important Questions (With PYQs)

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Basic Level (Conceptual Understanding)

1. Find the ratio of 40 minutes to 1.5 hours.

2. Express 25% as a fraction and a decimal.

3. The cost of a pen is ₹50. If its price increases by 20%, find the new price.

4. What is 10% of ₹2,500?

5. A man bought a fan for ₹1,500 and sold it at a profit of 12%. Find the selling price of the fan.

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Moderate Level (Application-Based)

6. A student scored 360 marks out of 500. Find his percentage marks.

7. Find the simple interest on ₹8,000 at 12% per annum for 2 years.

8. A scooter is sold for ₹38,000 at a loss of 5%. Find its cost price.

9. A shopkeeper buys an article for ₹2,400 and sells it at a gain of 15%. Find the selling price.

10. Find the amount to be paid on ₹10,000 at 8% compound interest per annum for 2 years.

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Higher-Level (Word Problems)

11. The population of a town increased from 2,00,000 to 2,20,500 in a year. Find the increase percent.

12. The cost of a TV was ₹24,000. After one year, its value depreciated by 10%. Find its value after one year.

13. A sum of ₹12,000 is borrowed at compound interest at the rate of 10% per annum. Find the amount to be paid after 3 years.

14. The price of an article is reduced from ₹1,500 to ₹1,200. Find the percentage decrease.

15. A shopkeeper sold a watch at a profit of 15%. If the watch was sold for ₹2,530, find its cost price.

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Challenging (Mixed & PYQ Style)

16. The difference between the compound interest and simple interest on a certain sum of money at 10% per annum for 2 years is ₹40. Find the sum. (PYQ)

17. If the selling price of 18 articles is equal to the cost price of 20 articles, find the gain percent.

18. The marked price of an article is ₹6,400. A shopkeeper allows two successive discounts of 20% and 10%. Find the selling price of the article. (PYQ)

19. A sum of money becomes ₹7,744 in 2 years at compound interest. If the rate of interest is 4% per annum, find the sum.

20. A dealer sells a washing machine for ₹13,500 including 8% VAT. Find the price of the washing machine before VAT was added. (PYQ)

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✅ These 20 questions cover:

• Ratios & percentages

• Profit & loss

• Simple & compound interest

• Depreciation & successive discounts

• VAT & tax problems

• PYQs from CBSE