The Percentage Formula is one of the most fundamental mathematical tools used to express a number as a fraction of 100, and it appears across NCERT Mathematics from Class 6 right through to Class 12. Written as Percentage = (Part / Whole) × 100, this formula helps students solve problems in profit and loss, data interpretation, and statistics. It is equally important for CBSE board exams, JEE Main, and NEET quantitative sections. This article covers the complete percentage formula, its derivation, a full formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Percentage Formulas at a Glance
Quick reference for the most important percentage formulas used in CBSE and competitive exams.
- Basic percentage: \( P\% = \dfrac{\text{Part}}{\text{Whole}} \times 100 \)
- Finding the part: \( \text{Part} = \dfrac{P\%}{100} \times \text{Whole} \)
- Finding the whole: \( \text{Whole} = \dfrac{\text{Part} \times 100}{P\%} \)
- Percentage change: \( \%\,\text{Change} = \dfrac{\text{New Value} – \text{Old Value}}{\text{Old Value}} \times 100 \)
- Percentage increase: \( \%\,\text{Increase} = \dfrac{\text{Increase}}{\text{Original Value}} \times 100 \)
- Percentage decrease: \( \%\,\text{Decrease} = \dfrac{\text{Decrease}}{\text{Original Value}} \times 100 \)
- Converting fraction to percent: \( \dfrac{a}{b} \times 100\% \)
What is the Percentage Formula?
The Percentage Formula expresses a number as a part of 100. The word “per cent” comes from the Latin phrase per centum, meaning “by the hundred”. In everyday mathematics, a percentage tells us how much of a whole quantity a given part represents, always scaled to a base of 100.
In NCERT Mathematics, the concept of percentage is introduced in Class 6 (Chapter 8 — Decimals and Fractions) and is revisited in Class 7 (Chapter 8 — Comparing Quantities) and Class 8 (Chapter 8 — Comparing Quantities), where it extends to profit, loss, discount, and simple interest. By Class 11 and 12, percentage calculations underpin statistics, probability, and data interpretation.
At its core, the Percentage Formula states: if a part P exists within a whole W, then the percentage value is found by dividing the part by the whole and multiplying by 100. This single relationship generates a family of derived formulas that students use across dozens of problem types. Understanding the base formula thoroughly makes every derived application straightforward.
Percentage Formula — Expression and Variables
The standard Percentage Formula is written as:
\[ P\% = \frac{\text{Part}}{\text{Whole}} \times 100 \]
This can be rearranged into three equivalent forms depending on which quantity is unknown:
\[ \text{Part} = \frac{P\%}{100} \times \text{Whole} \]
\[ \text{Whole} = \frac{\text{Part} \times 100}{P\%} \]
| Symbol | Quantity | SI Unit / Remark |
|---|---|---|
| \( P\% \) | Percentage value | Dimensionless (expressed with % sign) |
| Part | The portion being considered | Same unit as Whole |
| Whole | The total or base quantity | Any consistent unit (marks, rupees, kg, etc.) |
| 100 | Scaling constant | Dimensionless |
Derivation of the Percentage Formula
The derivation begins with the idea of a ratio. Any ratio of Part to Whole can be written as a fraction: \( \dfrac{\text{Part}}{\text{Whole}} \). To convert this fraction into a “per hundred” form, we scale it by multiplying by \( \dfrac{100}{100} \), which equals 1 and does not change the value. This gives \( \dfrac{\text{Part}}{\text{Whole}} \times 100 \) as the numerator, which is the percentage. The denominator becomes 100, confirming the “per hundred” interpretation. Therefore:
\[ P\% = \frac{\text{Part}}{\text{Whole}} \times 100 \]
This derivation shows that percentage is simply a standardised ratio, making comparisons between different wholes easy and consistent.
Complete Percentage Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Percentage | \( P\% = \dfrac{\text{Part}}{\text{Whole}} \times 100 \) | Part = given value; Whole = total value | Dimensionless (%) | Class 7, Ch 8 |
| Finding the Part | \( \text{Part} = \dfrac{P}{100} \times \text{Whole} \) | P = percentage; Whole = base | Same as Whole | Class 7, Ch 8 |
| Finding the Whole | \( \text{Whole} = \dfrac{\text{Part} \times 100}{P} \) | Part = known value; P = percentage | Same as Part | Class 7, Ch 8 |
| Percentage Change | \( \%\,\text{Change} = \dfrac{\text{New} – \text{Old}}{\text{Old}} \times 100 \) | New = new value; Old = original value | Dimensionless (%) | Class 8, Ch 8 |
| Percentage Increase | \( \%\,\text{Increase} = \dfrac{\text{Increase}}{\text{Original}} \times 100 \) | Increase = New − Old (positive) | Dimensionless (%) | Class 8, Ch 8 |
| Percentage Decrease | \( \%\,\text{Decrease} = \dfrac{\text{Decrease}}{\text{Original}} \times 100 \) | Decrease = Old − New (positive) | Dimensionless (%) | Class 8, Ch 8 |
| Profit Percentage | \( \%\,\text{Profit} = \dfrac{\text{Profit}}{\text{CP}} \times 100 \) | CP = cost price; Profit = SP − CP | Dimensionless (%) | Class 8, Ch 8 |
| Loss Percentage | \( \%\,\text{Loss} = \dfrac{\text{Loss}}{\text{CP}} \times 100 \) | CP = cost price; Loss = CP − SP | Dimensionless (%) | Class 8, Ch 8 |
| Discount Percentage | \( \%\,\text{Discount} = \dfrac{\text{Discount}}{\text{MP}} \times 100 \) | MP = marked price; Discount = MP − SP | Dimensionless (%) | Class 8, Ch 8 |
| Fraction to Percent | \( \dfrac{a}{b} \times 100\% \) | a = numerator; b = denominator | Dimensionless (%) | Class 6, Ch 8 |
| Successive Percentage Change | \( \text{Net}\% = x + y + \dfrac{xy}{100} \) | x, y = successive percentage changes | Dimensionless (%) | Class 8, Ch 8 |
Percentage Formula — Solved Examples
Example 1 (Class 7-8 Level — Basic Percentage Calculation)
Problem: Riya scored 78 marks out of 120 in her half-yearly examination. What is her percentage score?
Given: Part (marks obtained) = 78, Whole (total marks) = 120
Step 1: Write the basic Percentage Formula: \( P\% = \dfrac{\text{Part}}{\text{Whole}} \times 100 \)
Step 2: Substitute the given values: \( P\% = \dfrac{78}{120} \times 100 \)
Step 3: Simplify the fraction first: \( \dfrac{78}{120} = 0.65 \)
Step 4: Multiply by 100: \( P\% = 0.65 \times 100 = 65\% \)
Answer
Riya’s percentage score = 65%
Example 2 (Class 9-10 Level — Percentage Change)
Problem: The price of a laptop was ₹45,000 last year. This year, the price increased to ₹52,200. Find the percentage increase in price.
Given: Old price = ₹45,000; New price = ₹52,200
Step 1: Find the actual increase: \( \text{Increase} = 52200 – 45000 = ₹7200 \)
Step 2: Apply the Percentage Increase Formula: \( \%\,\text{Increase} = \dfrac{\text{Increase}}{\text{Original Value}} \times 100 \)
Step 3: Substitute values: \( \%\,\text{Increase} = \dfrac{7200}{45000} \times 100 \)
Step 4: Simplify: \( \dfrac{7200}{45000} = 0.16 \)
Step 5: Multiply by 100: \( \%\,\text{Increase} = 0.16 \times 100 = 16\% \)
Answer
The percentage increase in the laptop price = 16%
Example 3 (JEE/NEET Level — Successive Percentage Change)
Problem: The population of a town increases by 20% in the first year and then decreases by 10% in the second year. Find the net percentage change in the population after two years.
Given: First change \( x = +20\% \); Second change \( y = -10\% \)
Step 1: Use the successive percentage change formula: \( \text{Net}\% = x + y + \dfrac{xy}{100} \)
Step 2: Substitute values: \( \text{Net}\% = 20 + (-10) + \dfrac{(20)(-10)}{100} \)
Step 3: Simplify each term: \( \text{Net}\% = 20 – 10 + \dfrac{-200}{100} \)
Step 4: Calculate: \( \text{Net}\% = 10 – 2 = 8\% \)
Step 5: Verify using multiplier method: Final population \( = P \times 1.20 \times 0.90 = P \times 1.08 \), confirming a net increase of 8%.
Answer
Net percentage change in population = +8% increase
CBSE Exam Tips 2025-26
- Memorise all three rearrangements of the basic formula. CBSE questions often give you the percentage and the whole, asking for the part — or vice versa. Knowing all three forms saves time.
- Always identify the base (Whole) carefully. In percentage change problems, the base is always the original (old) value, not the new one. We recommend underlining the base quantity before solving.
- Use the multiplier method for multi-step problems. Converting percentages to decimal multipliers (e.g., 15% increase = × 1.15) reduces calculation errors in 2025-26 board paper Section C questions.
- Check units consistency. The Part and the Whole must be in the same unit before applying the formula. Convert units first if they differ.
- Practice data interpretation problems. CBSE Class 10 and 12 boards frequently include bar chart or pie chart questions that require the Percentage Formula to extract values.
- Revise profit/loss and discount formulas as derived percentage applications. These appear in 2-3 mark questions in Class 8 and Class 10 CBSE papers every year.
Common Mistakes to Avoid
- Using the wrong base: Many students divide by the new value instead of the original value when calculating percentage change. Always use the original (old) value as the denominator in percentage increase or decrease problems.
- Forgetting to multiply by 100: The ratio Part/Whole gives a decimal, not a percentage. Failing to multiply by 100 is the most common arithmetic error. The result is only a percentage after the multiplication step.
- Adding percentages directly: Two successive percentage changes cannot be simply added. A 10% increase followed by a 10% decrease does NOT give 0% net change. Use the successive percentage formula \( x + y + \dfrac{xy}{100} \) instead.
- Confusing percentage with percentage points: If a rate rises from 20% to 25%, the increase is 5 percentage points but the percentage increase is \( \dfrac{5}{20} \times 100 = 25\% \). These are different quantities.
- Incorrect unit conversion before applying the formula: If marks are given as a fraction (e.g., 3/4 of 80), students sometimes apply the percentage formula to the fraction directly without first finding the actual part. Always compute the part numerically first.
JEE/NEET Application of Percentage Formula
In our experience, JEE aspirants encounter the Percentage Formula primarily in the quantitative aptitude and data interpretation sections of JEE Main Paper 2, as well as in Chemistry and Biology numerical problems in NEET. Here are the three most important application patterns:
Pattern 1: Successive Percentage Change (JEE Main)
JEE Main frequently tests successive discounts, successive population changes, and compound interest approximations. The key formula is \( \text{Net}\% = x + y + \dfrac{xy}{100} \). For three successive changes, apply the formula twice. This pattern appears in approximately 1-2 questions per JEE Main session.
Pattern 2: Percentage Composition in Chemistry (NEET)
NEET Chemistry uses the Percentage Formula to calculate the percentage composition of elements in a compound. The formula becomes: \( \%\,\text{Element} = \dfrac{\text{Molar mass of element in formula}}{\text{Molar mass of compound}} \times 100 \). For example, the percentage of carbon in CO₂ (molar mass 44 g/mol) is \( \dfrac{12}{44} \times 100 = 27.27\% \). This is a direct application of the basic Percentage Formula.
Pattern 3: Percentage Error in Physics (JEE Advanced)
JEE Advanced tests percentage error in measurements. If a quantity \( Z = \dfrac{A^m B^n}{C^p} \), then the maximum percentage error is: \( \dfrac{\Delta Z}{Z} \times 100 = \left( m\,\dfrac{\Delta A}{A} + n\,\dfrac{\Delta B}{B} + p\,\dfrac{\Delta C}{C} \right) \times 100 \). This is a critical application of the Percentage Formula in experimental physics. Our experts suggest practising at least 10 such problems before the JEE Advanced exam.
FAQs on Percentage Formula
For more related algebra formulas, explore our Complete Algebra Formulas hub. You may also find these articles helpful: Quadratic Function Formula for solving quadratic equations with percentage-based coefficients, Confidence Interval Formula for statistical percentage applications, and our comprehensive All Maths Formulas reference page. For the official NCERT syllabus, visit ncert.nic.in.