Percentage Chapter Class 8 CBSE: Notes and Solved Problems

Percentage Chapter Notes for Class 8 CBSE

This post provides comprehensive notes and 26 Solved Examples from NCERT, RS Aggarwal, RD Sharma, and PW Worksheets for the Percentage chapter in Class 8 CBSE Mathematics. These problems cover conversions, percentage calculations, profit and loss, discounts, and complex word problems to help students excel in exams.

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Key Concepts

Definition: Percentage is a fraction expressed as a part of 100, denoted by %. E.g., \( 25\% = \frac{25}{100} = 0.25 \).
Conversions:
- Percent to Fraction: Divide by 100 and simplify, e.g., \( 75\% = \frac{75}{100} = \frac{3}{4} \).
- Percent to Decimal: Divide by 100, e.g., \( 50\% = \frac{50}{100} = 0.5 \).
- Fraction to Percent: Multiply by 100, e.g., \( \frac{3}{5} = \frac{3}{5} \times 100 = 60\% \).
- Ratio to Percent: Convert to fraction, then multiply by 100, e.g., \( 2:5 = \frac{2}{5} = \frac{2}{5} \times 100 = 40\% \).
Applications:
- Finding Percentage: \( \frac{\text{Percent}}{100} \times \text{Quantity} \).
- Increase/Decrease: \( \text{New Value} = \text{Original} \times \left(1 \pm \frac{\text{Percent}}{100}\right) \).
- Profit %: \( \frac{\text{Profit}}{\text{CP}} \times 100 \).
- Simple Interest: \( \text{SI} = \frac{P \times R \times T}{100} \).
Key Formulas:
- If \( x\% \) of a number is \( y \), then Number = \( \frac{y \times 100}{x} \).
- Successive change: \( \text{Final} = \text{Original} \times \left(1 \pm \frac{x}{100}\right)\left(1 \pm \frac{y}{100}\right) \).

26 Solved Examples

Conversions

1. Express 48% as a fraction (RS Aggarwal)

\( 48\% = \frac{48}{100} = \frac{12}{25} \).

2. Convert 220% to a fraction (RS Aggarwal)

\( 220\% = \frac{220}{100} = \frac{11}{5} = 2\frac{1}{5} \).

3. Express 2.5% as a decimal (RS Aggarwal)

\( 2.5\% = \frac{2.5}{100} = 0.025 \).

4. Convert the ratio 4:5 to percentage (RS Aggarwal)

\( 4:5 = \frac{4}{5} = \frac{4}{5} \times 100 = 80\% \).

5. Express \( 1\frac{3}{5} \) as a percentage (NCERT)

\( 1\frac{3}{5} = \frac{8}{5} \). \( \frac{8}{5} \times 100 = 160\% \).

Finding a Percentage of a Quantity

6. Find 16% of 3600 (RS Aggarwal)

\( \frac{16}{100} \times 3600 = 576 \).

7. 20% of x is 25. Find x (NCERT)

\( \frac{20}{100} \times x = 25 \). \( x = 25 \times \frac{100}{20} = 125 \).

8. If 34% of a number is 85, find the number (RS Aggarwal)

Let number = \( x \). \( \frac{34}{100} \times x = 85 \). \( x = \frac{85 \times 100}{34} = 250 \).

Percentage Increase/Decrease

9. A number is increased by 20% and then decreased by 20%. Find the net increase or decrease percent (NCERT)

Let number = 100. Increase: \( 100 \times \left(1 + \frac{20}{100}\right) = 120 \). Decrease: \( 120 \times \left(1 - \frac{20}{100}\right) = 120 \times 0.8 = 96 \). Net decrease = \( 100 - 96 = 4 \). Decrease % = \( \frac{4}{100} \times 100 = 4\% \).

10. The salary of an officer is increased by 25%. By what percent should the new salary be decreased to restore the original salary? (RS Aggarwal)

Let original salary = 100. New salary = \( 100 \times \left(1 + \frac{25}{100}\right) = 125 \). Let decrease % = \( x \). \( 125 \times \left(1 - \frac{x}{100}\right) = 100 \). \( 1 - \frac{x}{100} = \frac{100}{125} = \frac{4}{5} \). \( \frac{x}{100} = 1 - \frac{4}{5} = \frac{1}{5} \). \( x = 20\% \).

11. If A is 20% more than B, by what percent is B less than A? (PW Worksheet)

Let B = 100. A = \( 100 \times \left(1 + \frac{20}{100}\right) = 120 \). Decrease from A to B = \( 120 - 100 = 20 \). % decrease = \( \frac{20}{120} \times 100 = 16.67\% \).

Profit and Loss

12. Vishakha offers a 20% discount on items and still makes a 12% profit. If an article is marked at ₹280, find the cost price (NCERT)

Marked Price = ₹280. Discount = \( \frac{20}{100} \times 280 = 56 \). Selling Price = \( 280 - 56 = 224 \). Let CP = \( x \). Profit = \( \frac{12}{100}x \). SP = \( x + \frac{12}{100}x = \frac{112x}{100} \). \( \frac{112x}{100} = 224 \). \( x = \frac{224 \times 100}{112} = 200 \). CP = ₹200.

13. A CD player was purchased for ₹3200, and ₹560 was spent on repairs. It was sold at a 12.5% gain. How much did the seller receive? (NCERT)

CP = \( 3200 + 560 = 3760 \). Gain = \( 12.5\% = \frac{12.5}{100} = \frac{1}{8} \). SP = \( 3760 \times \left(1 + \frac{12.5}{100}\right) = 3760 \times 1.125 = 4230 \). Seller received ₹4230.

14. Amit’s salary was increased by 20%, and his new salary is ₹30600. What was his original salary? (RS Aggarwal)

Let original salary = \( x \). New salary = \( x \times \left(1 + \frac{20}{100}\right) = 1.2x \). \( 1.2x = 30600 \). \( x = \frac{30600}{1.2} = 25500 \). Original salary = ₹25500.

Real-World Applications

15. A man saves 18% of his monthly income, which is ₹3780. What is his monthly income? (RS Aggarwal)

Let monthly income = \( x \). \( \frac{18}{100} \times x = 3780 \). \( x = \frac{3780 \times 100}{18} = 21000 \). Monthly income = ₹21000.

16. A football team wins 7 games, which is 35% of total games played. How many games were played? (RS Aggarwal)

Let total games = \( x \). \( \frac{35}{100} \times x = 7 \). \( x = \frac{7 \times 100}{35} = 20 \). Total games = 20.

17. Sonal attended school for 204 days, which is 85% of the total school days. Find the total number of school days (RS Aggarwal)

Let total days = \( x \). \( \frac{85}{100} \times x = 204 \). \( x = \frac{204 \times 100}{85} = 240 \). Total school days = 240.

18. A garden has 2000 trees. 12% are mango trees, 18% are lemon trees, and the rest are orange trees. Find the number of orange trees (PW Worksheet)

Mango trees = \( \frac{12}{100} \times 2000 = 240 \). Lemon trees = \( \frac{18}{100} \times 2000 = 360 \). Orange trees = \( 2000 - (240 + 360) = 1400 \).

19. A balanced diet contains 12% proteins, 25% fats, and 63% carbohydrates. If a child needs 2600 calories daily, calculate the calories from proteins (PW Worksheet)

Protein calories = \( \frac{12}{100} \times 2600 = 312 \).

Complex Problems

20. Two-fifths of one-third of three-seventh of a number is 15. What is 40% of that number? (PW Worksheet)

Let number = \( x \). \( \frac{2}{5} \times \frac{1}{3} \times \frac{3}{7} \times x = 15 \). \( \frac{2}{35} \times x = 15 \). \( x = 15 \times \frac{35}{2} = 262.5 \). 40% of \( x \) = \( \frac{40}{100} \times 262.5 = 105 \).

21. In an examination, 65% of candidates passed. If 420 failed, find the total number of candidates (RS Aggarwal)

Pass % = 65%, so fail % = 35%. Let total candidates = \( x \). \( \frac{35}{100} \times x = 420 \). \( x = \frac{420 \times 100}{35} = 1200 \). Total candidates = 1200.

22. A number exceeds 25% of itself by 60. Find the number (RS Aggarwal)

Let number = \( x \). \( x - \frac{25}{100}x = 60 \). \( \frac{75}{100}x = 60 \). \( x = \frac{60 \times 100}{75} = 80 \). Number = 80.

23. The value of a machine depreciates by 10% every year. If its present value is ₹54000, what was its value last year? (RS Aggarwal)

Let last year’s value = \( x \). Present value = \( x \times \left(1 - \frac{10}{100}\right) = 0.9x \). \( 0.9x = 54000 \). \( x = \frac{54000}{0.9} = 60000 \). Value last year = ₹60000.

24. In an examination, 52% failed in English, 42% in Maths, and 17% in both. Find the percentage of those who passed in both subjects (PW Worksheet)

Let total = 100. Failed in English only = \( 52 - 17 = 35\% \). Failed in Maths only = \( 42 - 17 = 25\% \). Failed in both = \( 17\% \). Total failed = \( 35 + 25 + 17 = 77\% \). Passed in both = \( 100 - 77 = 23\% \).

25. A car is marked at ₹300000. The dealer allows successive discounts of 6%, 4%, and 2.5%. Find the selling price (NCERT)

Marked Price = ₹300000. After 6% discount: \( 300000 \times \left(1 - \frac{6}{100}\right) = 300000 \times 0.94 = 282000 \). After 4% discount: \( 282000 \times \left(1 - \frac{4}{100}\right) = 282000 \times 0.96 = 270720 \). After 2.5% discount: \( 270720 \times \left(1 - \frac{2.5}{100}\right) = 270720 \times 0.975 = 263952 \). Selling Price = ₹263952.

26. If 25% of a number is subtracted from a second number, the second number reduces to its five-sixth. Find the ratio of the first number to the second number (PW Worksheet)

Let first number = \( x \), second number = \( y \). \( y - \frac{25}{100}x = \frac{5}{6}y \). \( y - 0.25x = \frac{5}{6}y \). \( y - \frac{5}{6}y = 0.25x \). \( \frac{1}{6}y = 0.25x \). \( y = 1.5x \). Ratio \( \frac{x}{y} = \frac{x}{1.5x} = \frac{2}{3} \). Ratio = \( 2:3 \).