Login / Register
Follow Us
0
0
Register Free!
Register Free!
Online Mock Tests & Exam Preparation | Target Exams
Select Medium / माध्यम चुनें:
Eng (English) Hindi (हिन्दी)
ICSE • Class 8 • Mathematics • Ch 8
Estimated Time: 45 Mins
Study Progress: In Progress

Percentage

In ICSE Class 8 Mathematics, "Percentage" provides an authoritative, mathematically rigorous master study guide investigating the fractional and decimal arithmetic of per-hundred calculations, percentage increase and decrease, and multi-step real-world commercial applications. This comprehensive chapter explores Meaning of Percentage (Per hundred / Parts per hundred, symbol $\%$; Fraction with denominator 100: $x\% = \frac{x}{100}$), Inter-Conversions (Percentage to fraction/decimal, fraction/decimal to percentage), Finding a Percentage of a Given Quantity ($x\% \text{ of } Q = \frac{x}{100} \times Q$), Expressing One Quantity as a Percentage of Another ($\frac{A}{B} \times 100\%$ with identical units of measurement), Percentage Change: Percentage Increase and Percentage Decrease ($\text{Percentage Increase} = \frac{\text{Increase}}{\text{Original Value}} \times 100\%$; $\text{Percentage Decrease} = \frac{\text{Decrease}}{\text{Original Value}} \times 100\%$; Finding original quantity from increased/decreased value), Successive Percentage Changes (Two successive changes of $x\%$ and $y\%$: $\text{Net Change} = \left( x + y + \frac{xy}{100} \right)\%$), Commodity Price-Consumption Elasticity (If price of a commodity increases by $r\%$, reduction in consumption to keep expenditure constant is $\frac{r}{100 + r} \times 100\%$; If price decreases, increase in consumption is $\frac{r}{100 - r} \times 100\%$), Population Growth and Machine Depreciation ($A = P\left(1 \pm \frac{R}{100}\right)^n$), and Examination Marks/Salary Word Problems aligned with the 2026–27 CISCE ICSE curriculum.

Take Chapter CBT Test View All Chapters
Overview Core Concepts Formulas & Identities Summary Self Check Take MCQ Test

Why Does a Store Advertising "Prices Slashed by 20% and Then an Extra 20% Off" Cost You Significantly More Than a 40% Discount?

Walk through a shopping mall during the festival sales season. A neon banner in a clothing store window shouts: "MEGA DOUBLE SAVINGS! Take 20% off all clothes, PLUS an EXTRA 20% off at the checkout register!" Shoppers celebrate, thinking they are getting a $40\%$ discount ($20 + 20 = 40$). But at the billing counter, a shirt priced at $\text{Rs. } 1,000$ does NOT cost $\text{Rs. } 600$! The first $20\%$ cut drops the price to $\text{Rs. } 800$. The second $20\%$ cut is calculated NOT on the original Rs. 1,000, but on the reduced Rs. 800, saving only $\text{Rs. } 160$ and making the final bill Rs. 640! The real net discount was only 36%, not 40%! The store legally kept an extra $4\%$ of your money! Why does a $10\%$ pay raise followed by a $10\%$ pay cut leave you with LESS money than you started with? What is the secret formula to calculate consumption cuts when fuel prices soar? Let's master percentage.

Why This Chapter Matters

Percentage is the universal language of business, banking interest rates, government taxation (GST, income tax), financial inflation metrics, laboratory chemical purity analysis, corporate profit margins, and academic performance grading. Mastering percentage formulas and successive change arithmetic is essential for ICSE Class 8 mathematics.

Before You Begin (Prerequisites)

  • Operations on fractions and decimals from Class 7.
  • Solving simple linear equations in one variable.
  • Units conversion (rupees/paise, meters/centimeters, kilograms/grams).

What You Will Learn (Core Objectives)

  • Convert seamlessly between percentages, fractions, and decimals.
  • Calculate percentage increase and decrease on physical and monetary quantities.
  • Determine the original quantity when given its increased or decreased percentage value.
  • Apply the successive percentage change formula: $\text{Net} = \left( x + y + \frac{xy}{100} \right)\%$.
  • Calculate required reductions in household consumption when commodity prices rise to keep expenditure constant.
  • Solve population growth and machine depreciation problems over multiple years.

Chapter Roadmap & Progression

1 1. Fundamentals & Inter-Conversions
2 2. Percentage Increase & Decrease
3 3. Successive Percentage Changes
4 4. Price-Consumption Elasticity & P...

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamentals & Inter-Conversions

Understand
A. Meaning of Percentage:

The word Percent comes from the Latin *per centum*, meaning "by the hundred" or "out of one hundred". Symbol: $\mathbf{\%}$.

  • To convert a fraction or decimal into a percentage: Multiply by $100\%$. $$\frac{3}{5} = \frac{3}{5} \times 100\% = \mathbf{60\%}; \quad 0.085 = 0.085 \times 100\% = \mathbf{8.5\%}$$
  • To convert a percentage into a fraction or decimal: Divide by $100$ and remove the $\%$ symbol. $$45\% = \frac{45}{100} = \mathbf{\frac{9}{20}} = \mathbf{0.45}$$
B. Expressing One Quantity as a Percentage of Another:
$$\mathbf{\text{Percentage} = \frac{\text{Given Value } (A)}{\text{Base Value } (B)} \times 100\%}$$

Mandatory Rule: Both quantities $A$ and $B$ must be converted into the exact same physical units before calculating!

2. Percentage Increase & Decrease

Percentage Change
A. The Master Change Formulas:
$$\mathbf{\text{Percentage Increase} = \frac{\text{Absolute Increase}}{\text{Original Initial Value}} \times 100\%}$$ $$\mathbf{\text{Percentage Decrease} = \frac{\text{Absolute Decrease}}{\text{Original Initial Value}} \times 100\%}$$

Critical Rule: The denominator is ALWAYS the Original (Initial) Value, never the new value!

B. Direct Calculation of New Values:
  • New Value after $x\%$ Increase: $\mathbf{\text{New Value} = \text{Original} \times \left( 1 + \frac{x}{100} \right)}$
  • New Value after $x\%$ Decrease: $\mathbf{\text{New Value} = \text{Original} \times \left( 1 - \frac{x}{100} \right)}$

3. Successive Percentage Changes

Successive Changes

If a quantity undergoes two successive percentage changes of $x\%$ and $y\%$ (use $+$ for increase, $-$ for decrease):

$$\mathbf{\text{Net Percentage Change} = \left( x + y + \frac{xy}{100} \right)\%}$$
  • The Pay-Cut Paradox: A salary is increased by $10\%$, and then decreased by $10\%$. What is the net change? $$\text{Net Change} = 10 - 10 + \frac{(10)(-10)}{100} = 0 - \frac{100}{100} = \mathbf{-1\% \text{ (A net 1% LOSS!)}}$$

4. Price-Consumption Elasticity & Population Growth

Applications
A. Expenditure Invariance Formula:
  • If the price of a commodity increases by $r\%$, the percentage reduction in consumption required to keep total household expenditure unchanged is: $$\mathbf{\text{Reduction in Consumption} = \left( \frac{r}{100 + r} \times 100 \right)\%}$$
  • If the price decreases by $r\%$, the percentage increase in consumption possible without increasing expenditure is: $$\mathbf{\text{Increase in Consumption} = \left( \frac{r}{100 - r} \times 100 \right)\%}$$
B. Population Growth & Machine Depreciation:
$$\mathbf{A = P \left( 1 \pm \frac{R}{100} \right)^n}$$

($+$ for annual population growth; $-$ for annual vehicle/machinery depreciation; $P$ is initial value, $R$ is annual rate, $n$ is number of years).

Key Formulas, Identities & Theorems

Net Successive Percentage Change
$$\text{Net Change (\%)} = x + y + \frac{xy}{100}$$
Use positive for increase and negative for decrease.
Price-Expenditure Consumption Rule
$$\text{Consumption Cut (\%)} = \frac{r}{100 + r} \times 100$$
Maintains constant expenditure when price rises by r%.

Percentage: Successive Changes & Price-Consumption Dynamics

Percentage: Formulas, Successive Changes & Applications PERCENTAGE CHANGE FORMULAS % Increase = (Increase / Original) × 100% % Decrease = (Decrease / Original) × 100% • Successive Change Formula: Net Change = x + y + (xy / 100) % • The Pay-Cut Trap (Increase 10%, Decrease 10%): 10 - 10 + (10 × -10)/100 = 0 - 1 = -1% (A NET 1% LOSS!) • Always divide by the ORIGINAL INITIAL value! PRICE-EXPENDITURE & GROWTH • Commodity Price Rise (Expenditure Invariant): Consumption Cut = [ r / (100 + r) ] × 100% Example: Sugar price up 25% ⇒ Cut = (25/125)×100 = 20% • Population Growth & Machine Depreciation: A = P (1 ± R/100)n + for annual population growth - for machinery depreciation • Double Discounts: 20% + 20% ≠ 40%! Actual discount = 20 + 20 - 4 = 36% PERCENT = PER HUNDRED • NET CHANGE = x + y + xy/100 • EXPENDITURE CUT = r/(100+r)

Chapter Summary & 10 Key Takeaways

Takeaway 1
Percent means parts per hundred; x% = x / 100.
Takeaway 2
To convert fractions/decimals to percentages, multiply by 100%; to remove %, divide by 100.
Takeaway 3
Percentage increase or decrease is always calculated over the Original Base value.
Takeaway 4
Successive percentage changes x% and y% yield Net Change = (x + y + xy/100)%.
Takeaway 5
An increase of 10% followed by a decrease of 10% results in an overall 1% net reduction.
Takeaway 6
If commodity price rises by r%, consumption must drop by [r / (100 + r)] * 100% to keep spending constant.
Takeaway 7
If commodity price drops by r%, consumption can increase by [r / (100 - r)] * 100%.
Takeaway 8
Population growth after n years: A = P(1 + R/100)^n.
Takeaway 9
Machine depreciation value after n years: A = P(1 - R/100)^n.
Takeaway 10
Two successive discounts of 20% and 20% equal a net single discount of 36% (not 40%).

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
If the price of petrol increases by $25\%$, by what percentage must a motorist reduce his consumption so that his total expenditure on petrol remains unchanged?
Reveal Answer & Explanation
Answer: Let the original price of petrol be $\text{Rs. } 100$ per liter and original consumption be $100\text{ liters}$.
Original Expenditure $= 100 \times 100 = \text{Rs. } 10,000$.
New price of petrol $= 100 + 25 = \text{Rs. } 125$ per liter.
To keep expenditure unchanged at $\text{Rs. } 10,000$, new consumption is:
$$\text{New Consumption} = \frac{10,000}{125} = 80\text{ liters}$$
Reduction in consumption $= 100 - 80 = 20\text{ liters}$.
$$\text{Percentage Reduction} = \frac{20}{100} \times 100\% = \mathbf{20\%}$$

• *Direct Formula Verification:*
$$\text{Reduction} = \left( \frac{r}{100 + r} \times 100 \right)\% = \left( \frac{25}{100 + 25} \times 100 \right)\% = \frac{25}{125} \times 100\% = \mathbf{20\%}$$.
Apply $[r / (100 + r)] \times 100\% = (25 / 125) \times 100\% = 20\%$.
2
An employee's salary was increased by $20\%$, but due to poor company performance, the new salary was subsequently reduced by $20\%$. What is the net percentage change in his salary?
Reveal Answer & Explanation
Answer:

Apply the Successive Percentage Change formula:

$$\text{Net Change} = \left( x + y + \frac{xy}{100} \right)\%$$


Here $x = +20$ (increase) and $y = -20$ (decrease):

$$\text{Net Change} = 20 - 20 + \frac{(20)(-20)}{100}$$


$$= 0 - \frac{400}{100} = \mathbf{-4\%}$$


• The negative sign indicates a reduction.
• The employee suffers a net $4\%$ decrease in his original salary!


Net change $= 20 - 20 + (20 \times -20)/100 = 0 - 4 = -4\%$. (A net 4% decrease).
3
In an examination, $35\%$ marks are required to pass. A student gets $145$ marks and fails by $30$ marks. Find the maximum total marks of the examination.
Reveal Answer & Explanation
Answer:

Step 1: Calculate passing marks required:

$$\text{Passing Marks} = 145 + 30 = 175\text{ marks}$$


Step 2: Let the maximum total marks be $M$. According to the question:

$$35\% \text{ of } M = 175$$


$$\frac{35}{100} \times M = 175$$


$$M = \frac{175 \times 100}{35}$$


Divide $175$ by $35$ ($175 \div 35 = 5$):

$$M = 5 \times 100 = \mathbf{500\text{ marks}}$$

.
The maximum total marks of the examination is $500$.


Pass marks $= 145 + 30 = 175$. Since $35\%$ of Total $= 175$, Total $= (175 \times 100) / 35 = 500$.
4
The population of a town increases at the rate of $5\%$ per annum. If its current population is $185,220$, find its population 3 years ago.
Reveal Answer & Explanation
Answer:

Let the population 3 years ago be $P$.
Annual growth rate $R = 5\%$, Time $n = 3\text{ years}$, Current Population $A = 185,220$.
Apply the Population Growth formula:

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


$$185,220 = P \left( 1 + \frac{5}{100} \right)^3 = P \left( 1 + \frac{1}{20} \right)^3 = P \left( \frac{21}{20} \right)^3$$


$$185,220 = P \times \frac{9261}{8000}$$


Solve for $P$:

$$P = \frac{185,220 \times 8000}{9261}$$


Notice that $185,220 \div 9261 = 20$:

$$P = 20 \times 8000 = \mathbf{160,000}$$

.
The population 3 years ago was $160,000$.


$185,220 = P \times (21/20)^3 = P \times (9261/8000) \implies P = 20 \times 8000 = 160,000$.
5
A scooter was bought for $\text{Rs. } 42,000$. Its value depreciates at the rate of $8\%$ per annum. Find its value after 1 year.
Reveal Answer & Explanation
Answer: Given: Initial cost $P = \text{Rs. } 42,000$, Depreciation rate $R = 8\%$, Time $n = 1\text{ year}$.
Apply Depreciation formula:
$$\text{Value after 1 year} = P \left( 1 - \frac{R}{100} \right)^1$$
$$= 42,000 \times \left( 1 - \frac{8}{100} \right) = 42,000 \times \left( \frac{92}{100} \right)$$
$$= 420 \times 92 = \mathbf{\text{Rs. } 38,640}$$.
Value $= 42,000 \times (1 - 0.08) = 42,000 \times 0.92 = \text{Rs. } 38,640$.
6
What percentage of $2\text{ kilograms}$ is $65\text{ grams}$?
Reveal Answer & Explanation
Answer:

Step 1: Convert both quantities to the same unit ($1\text{ kg} = 1000\text{ g}$):

$$2\text{ kg} = 2 \times 1000\text{ g} = 2000\text{ grams}$$


Step 2: Calculate percentage:

$$\text{Percentage} = \frac{65}{2000} \times 100\% = \frac{65}{20}\% = \frac{13}{4}\% = \mathbf{3.25\%}$$

.


Convert 2 kg to 2000 g: $(65 / 2000) \times 100\% = 3.25\%$.
7
The price of a television set inclusive of $18\%$ GST is $\text{Rs. } 33,040$. Find the original price of the television before GST was added.
Reveal Answer & Explanation
Answer:

Let the original price of the TV be $\text{Rs. } x$.
GST rate $= 18\%$.
Price including GST:

$$x + 18\% \text{ of } x = 33,040$$


$$x \left( 1 + \frac{18}{100} \right) = 33,040$$


$$x \left( \frac{118}{100} \right) = 33,040$$


$$x = \frac{33,040 \times 100}{118}$$


Divide $33,040$ by $118$ ($33,040 \div 118 = 280$):

$$x = 280 \times 100 = \mathbf{\text{Rs. } 28,000}$$

.
The original price of the TV was $\text{Rs. } 28,000$.


Set $1.18x = 33,040 \implies x = 33,040 / 1.18 = \text{Rs. } 28,000$.
8
A number is first increased by $10\%$ and then decreased by $10\%$. Another number is decreased by $20\%$. Compare the results.
Reveal Answer & Explanation
Answer:

• For the first number (let $N = 100$):
Increased by $10\% \to 110$. Then decreased by $10\% \to 110 - 11 = 99$.
Net change is a $1\%$ decrease.
• For the second number: decreased directly by $20\%$ (from 100 to 80).
• The second number suffers a much larger reduction ($20\%$) than the first number ($1\%$).


First number suffers a net 1% decrease ($100 \to 110 \to 99$); second number drops by 20% ($100 \to 80$).
Finished Studying This Chapter?
READY TO PRACTICE?

Timed CBT Practice Tests (Exam Simulator)

Put your concepts to the test with official curriculum-aligned Foundation and Advanced practice tests. Get instant accuracy scores, time metrics, and step-by-step verified explanations.

Foundation Test
15 MCQs • Core Concepts
Advanced Test
15 MCQs • HOTS & Traps

All Class 8 Mathematics Chapters

AI Study Friend

Instant Concept Tutor

Have a doubt in Percentage? Ask our AI study tutor for rapid explanations or customized quizzes.