Study guide
8.2 Fractions, Ratios, Proportions & Percentage Mastery
Chhattisgarh State Eligibility Test CG SET
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1 Chapter 1: Exam Structure, Blueprint & Preparation Strategy
1.1 Exam Blueprint, Marking Scheme & Qualifying Cutoffs 1.2 Eligibility Requirements, Master's Degree Criteria & Domicile Policy 1.3 Paper I Strategic Study Plan, Time Allocation & OMR Tactics
2 Chapter 2: Teaching Aptitude — Concepts, Objectives & Learner Characteristics
2.1 Concepts, Objectives & Levels of Teaching 2.2 Learner Characteristics & Individual Differences 2.3 Factors Influencing Teaching Effectiveness & Instructional Facilities
3 Chapter 3: Teaching Aptitude — Methods, Support Systems & Evaluation
3.1 Teacher-Centered vs. Learner-Centered Methods & Modern Pedagogy 3.2 Teaching Support Systems & Online Digital Platforms (SWAYAM, MOOCs) 3.3 Evaluation Systems, CBCS & Assessment Innovations
4 Chapter 4: Research Aptitude — Paradigms, Types & Methodologies
4.1 Meaning, Objectives, Characteristics & Types of Research 4.2 Epistemological Approaches: Positivism vs. Post-Positivist Philosophies 4.3 Core Research Methods: Descriptive, Experimental & Historical
5 Chapter 5: Research Aptitude — Process, Ethics & Scholarly Writing
5.1 Systematic Steps of Research & Hypothesis Testing 5.2 Thesis, Dissertation & Scholarly Article Writing 5.3 Application of ICT in Research & Research Ethics
6 Chapter 6: Reading Comprehension — Critical Analysis & Interpretation
6.1 Critical Reading Strategies: Structure, Main Ideas & Skimming vs. Scanning 6.2 Inferential Analysis, Contextual Vocabulary & Author's Tone 6.3 Question-Solving Tactics & Eliminating Distractors in Academic Passages
7 Chapter 7: Communication in Education & Mass Media
7.1 Meaning, Types & Characteristics of Communication 7.2 Effective Classroom Communication: Verbal/Non-Verbal Dynamics & Immediacy 7.3 Communication Barriers & The Role of Mass Media in Society
8 Chapter 8: Mathematical Reasoning & Quantitative Aptitude
8.1 Number Series, Letter Series, Coding-Decoding & Relationships 8.2 Fractions, Ratios, Proportions & Percentage Mastery 8.3 Profit & Loss, Interest, Discounting, Speed-Distance & Averages
9 Chapter 9: Logical Reasoning — Western Deductive & Inductive Logic
9.1 Structure of Arguments & Categorical Propositions 9.2 Formal & Informal Fallacies & Connotation vs. Denotation 9.3 Classical Square of Opposition & Venn Diagrams for Validity
10 Chapter 10: Logical Reasoning — Indian Logic & Nyaya Epistemology
10.1 Indian Logic: Six Pramanas (Means of Valid Knowledge) 10.2 Structure and Types of Anumana (Inference), Vyapti & Hetu 10.3 Hetvabhasa: Fallacies of Inference in Nyaya Logic
11 Chapter 11: Data Interpretation & Quantitative Representation
11.1 Sources, Acquisition & Classification of Quantitative and Qualitative Data 11.2 Graphical Representation: Bar Charts, Histograms, Pie Charts & Line Graphs 11.3 Table Charts, Cross-Tabulation & Data Interpretation for Governance 11.4 Mapping of Data & Data in Governance
12 Chapter 12: Information & Communication Technology (ICT)
12.1 Foundational ICT Terminology, Acronyms & Computer Architecture 12.2 Internet Architecture, Intranets, Email & Video Conferencing Mechanics 12.3 National Digital Initiatives in Higher Education & ICT Governance
13 Chapter 13: People, Development & Environment — Goals & Pollution
13.1 Development and Environment: Millennium & Sustainable Development Goals 13.2 Anthropogenic Impacts & Environmental Pollution Types 13.3 Health Impacts of Pollutants, Heavy Metals & Climate Dynamics
14 Chapter 14: People, Development & Environment — Energy, Hazards & Treaties
14.1 Renewable & Non-Renewable Energy Resources and India's Energy Transition 14.2 Natural Hazards, Disasters & Disaster Management Architecture 14.3 Environmental Legislation, NAPCC & International Conventions
15 Chapter 15: Higher Education System — Historical Evolution & Modalities
15.1 Higher Learning & Education in Ancient India 15.2 Evolution of Higher Learning in Post-Independence India & NEP 2020 15.3 Oriental, Conventional, Non-Conventional & Vocational Education
16 Chapter 16: Higher Education System — Governance, Ethics & Policy
16.1 Value Education & Environmental Education in Academic Curricula 16.2 Regulatory Landscape: UGC, Professional Councils, NAAC & NIRF 16.3 Higher Education Governance & Administration in Chhattisgarh State
8.2 Fractions, Ratios, Proportions & Percentage Mastery
Key Takeaways Explain This Topic AI
Fractions represent division of quantities; comparing fractions is rapidly executed via cross-multiplication for two terms or LCM of denominators for multi-term sets.
Ratios quantify relative magnitudes through division; properties of proportion ( a : b = c : d ⟺ a d = b c a:b = c:d \iff ad = bc a : b = c : d ⟺ a d = b c ) establish foundational equations for third, fourth, and mean proportionals ( a b \sqrt{ab} ab ).
Partnership problems equate profit shares directly to the product of capital invested and duration of investment ( Profit Ratio = C 1 T 1 : C 2 T 2 : C 3 T 3 \text{Profit Ratio} = C_1 T_1 : C_2 T_2 : C_3 T_3 Profit Ratio = C 1 T 1 : C 2 T 2 : C 3 T 3 ).
Memorizing standard benchmark fraction-to-percentage conversions ( 1 / 2 = 50 % 1/2 = 50\% 1/2 = 50% down to 1 / 16 = 6.25 % 1/16 = 6.25\% 1/16 = 6.25% ) accelerates arithmetic operations and eliminates long division under timed conditions.
Successive percentage changes follow the net formula a + b + a b 100 % a + b + \frac{ab}{100}\% a + b + 100 ab % , while price-consumption inverse relations dictate that a price rise of r % r\% r % requires a consumption reduction of r 100 + r × 100 % \frac{r}{100+r} \times 100\% 100 + r r × 100% .
Last updated: October 2026
8.2 Fractions, Ratios, Proportions & Percentage Mastery
Core Principle: Fractions, ratios, and percentages are not three separate mathematical subjects; they are three isomorphic notations for expressing the relative comparison of quantities. In competitive aptitude, fluency in instantly translating between fractions, ratios, and percentages is the single most powerful accelerator of calculation speed and conceptual precision.
1. Fractions and Decimals: Taxonomy, Comparison & Recurring Forms
A fraction represents a part of a whole or the quotient of two quantities expressed in the form a b \frac{a}{b} b a , where a a a is the numerator and b b b is the non-zero denominator ( b ≠ 0 b \neq 0 b = 0 ).
1.1 Structural Classification of Fractions
Proper Fraction: Numerator is strictly less than the denominator ( a < b a < b a < b ). The numerical value is strictly less than one (e.g., 3 7 , 5 12 \frac{3}{7}, \frac{5}{12} 7 3 , 12 5 ). Proper fractions represent genuine fractional parts.
Improper Fraction: Numerator is greater than or equal to the denominator ( a ≥ b a \ge b a ≥ b ). The numerical value is greater than or equal to one (e.g., 9 4 , 13 5 \frac{9}{4}, \frac{13}{5} 4 9 , 5 13 ).
Mixed Fraction: An improper fraction expressed as the combination of an integer quotient and a proper fraction remainder: a b = Q R b = Q + R b ( e.g., 17 5 = 3 2 5 ) \frac{a}{b} = Q \frac{R}{b} = Q + \frac{R}{b} \quad \left(\text{e.g., } \frac{17}{5} = 3 \frac{2}{5}\right) b a = Q b R = Q + b R ( e.g., 5 17 = 3 5 2 )
Equivalent Fractions: Fractions whose numerical values remain identical when both numerator and denominator are multiplied or divided by the same non-zero scalar k k k : a b = a × k b × k = a ÷ k b ÷ k \frac{a}{b} = \frac{a \times k}{b \times k} = \frac{a \div k}{b \div k} b a = b × k a × k = b ÷ k a ÷ k
1.2 High-Speed Fraction Comparison Techniques
In CG SET Paper I, questions frequently ask candidates to identify the largest, smallest, or ascending/descending order of fractions. Three standard methods exist:
Method 1: The Cross-Multiplication Method (Optimal for 2 Fractions)
To determine whether a b > c d \frac{a}{b} > \frac{c}{d} b a > d c (where b , d > 0 b, d > 0 b , d > 0 ):
Multiply a × d a \times d a × d (left product) and b × c b \times c b × c (right product).
If a ⋅ d > b ⋅ c a \cdot d > b \cdot c a ⋅ d > b ⋅ c , then a b > c d \frac{a}{b} > \frac{c}{d} b a > d c .
If a ⋅ d < b ⋅ c a \cdot d < b \cdot c a ⋅ d < b ⋅ c , then a b < c d \frac{a}{b} < \frac{c}{d} b a < d c .
Demonstration: Compare 7 11 \frac{7}{11} 11 7 and 9 14 \frac{9}{14} 14 9 . 7 × 14 = 98 vs. 11 × 9 = 99 7 \times 14 = 98 \quad \text{vs.} \quad 11 \times 9 = 99 7 × 14 = 98 vs. 11 × 9 = 99 Since 98 < 99 98 < 99 98 < 99 , it follows immediately that 7 11 < 9 14 \frac{7}{11} < \frac{9}{14} 11 7 < 14 9 .
Method 2: The Lowest Common Multiple (LCM) Method (Optimal for 3 to 4 Fractions)
Convert all denominators to their LCM:
Compare 2 3 , 3 4 , 5 6 , 7 12 \frac{2}{3}, \frac{3}{4}, \frac{5}{6}, \frac{7}{12} 3 2 , 4 3 , 6 5 , 12 7 .
LCM ( 3 , 4 , 6 , 12 ) = 12 \text{LCM}(3, 4, 6, 12) = 12 LCM ( 3 , 4 , 6 , 12 ) = 12 .
Equivalent numerators: 8 12 , 9 12 , 10 12 , 7 12 \frac{8}{12}, \frac{9}{12}, \frac{10}{12}, \frac{7}{12} 12 8 , 12 9 , 12 10 , 12 7 .
Ascending order: 7 12 < 2 3 < 3 4 < 5 6 \frac{7}{12} < \frac{2}{3} < \frac{3}{4} < \frac{5}{6} 12 7 < 3 2 < 4 3 < 6 5 .
Method 3: The Characteristic Difference Theorem
For positive proper fractions where the difference between denominator and numerator is identical ( ( D − N ) = k (D - N) = k ( D − N ) = k ):
Rule for Proper Fractions ( N < D N < D N < D ): As the magnitude of the terms increases, the value of the fraction increases . If 3 4 , 5 6 , 7 8 , 11 12 ( D − N = 1 ) ⟹ 3 4 < 5 6 < 7 8 < 11 12 \text{If } \frac{3}{4}, \frac{5}{6}, \frac{7}{8}, \frac{11}{12} \quad (D - N = 1) \implies \frac{3}{4} < \frac{5}{6} < \frac{7}{8} < \frac{11}{12} If 4 3 , 6 5 , 8 7 , 12 11 ( D − N = 1 ) ⟹ 4 3 < 6 5 < 8 7 < 12 11
Rule for Adding a Constant: If x > 0 x > 0 x > 0 is added to both numerator and denominator of a proper fraction ( N D < 1 \frac{N}{D} < 1 D N < 1 ): N + x D + x > N D \frac{N + x}{D + x} > \frac{N}{D} D + x N + x > D N (e.g., 2 3 = 0.667 ⟹ 2 + 5 3 + 5 = 7 8 = 0.875 > 0.667 \frac{2}{3} = 0.667 \implies \frac{2+5}{3+5} = \frac{7}{8} = 0.875 > 0.667 3 2 = 0.667 ⟹ 3 + 5 2 + 5 = 8 7 = 0.875 > 0.667 ) .
1.3 Terminating vs. Recurring Decimals
A rational fraction p q \frac{p}{q} q p in irreducible form generates a terminating decimal if and only if the prime factorization of the denominator q q q contains no prime factors other than 2 or 5 ( q = 2 m × 5 n q = 2^m \times 5^n q = 2 m × 5 n ). If q q q contains any other prime factor (such as 3, 7, 11, 13), the decimal expansion is non-terminating and recurring .
Pure Recurring Decimals to Vulgar Fractions
In pure recurring decimals, repetition begins immediately after the decimal point. The equivalent fraction is the repeating block divided by as many 9s as there are repeating digits:
0. a ‾ = a 9 , 0. a b ‾ = a b 99 , 0. a b c ‾ = a b c 999 0.\overline{a} = \frac{a}{9}, \quad 0.\overline{ab} = \frac{ab}{99}, \quad 0.\overline{abc} = \frac{abc}{999} 0. a = 9 a , 0. ab = 99 ab , 0. ab c = 999 ab c
Examples: 0. 7 ‾ = 7 9 0.\overline{7} = \frac{7}{9} 0. 7 = 9 7 ; 0. 45 ‾ = 45 99 = 5 11 0.\overline{45} = \frac{45}{99} = \frac{5}{11} 0. 45 = 99 45 = 11 5 ; 0. 123 ‾ = 123 999 = 41 333 0.\overline{123} = \frac{123}{999} = \frac{41}{333} 0. 123 = 999 123 = 333 41 .
Mixed Recurring Decimals to Vulgar Fractions
In mixed recurring decimals, non-repeating digits appear between the decimal point and the repeating cycle. To convert to a vulgar fraction:
Fraction = Full Number (Non-repeating and repeating digits) − Non-repeating digits Denominator formed by as many 9s as repeating digits, followed by as many 0s as non-repeating digits \text{Fraction} = \frac{\text{Full Number (Non-repeating and repeating digits)} - \text{Non-repeating digits}}{\text{Denominator formed by as many 9s as repeating digits, followed by as many 0s as non-repeating digits}} Fraction = Denominator formed by as many 9s as repeating digits, followed by as many 0s as non-repeating digits Full Number (Non-repeating and repeating digits) − Non-repeating digits
Algebraic Formula: 0. a b ‾ = a b − a 90 , 0. a b c d ‾ = a b c d − a b 9900 , 0. a b c d ‾ = a b c d − a 9990 \text{Algebraic Formula: } 0.a\overline{b} = \frac{ab - a}{90}, \quad 0.ab\overline{cd} = \frac{abcd - ab}{9900}, \quad 0.a\overline{bcd} = \frac{abcd - a}{9990} Algebraic Formula: 0. a b = 90 ab − a , 0. ab c d = 9900 ab c d − ab , 0. a b c d = 9990 ab c d − a
Worked Demonstration: Convert 0.4 16 ‾ 0.4\overline{16} 0.4 16 to a vulgar fraction: Fraction = 416 − 4 990 = 412 990 = 206 495 \text{Fraction} = \frac{416 - 4}{990} = \frac{412}{990} = \frac{206}{495} Fraction = 990 416 − 4 = 990 412 = 495 206
- 2. Ratios, Proportions & Partnership Mathematics
- 2.1 Foundational Concept and Properties of Ratios
A ratio is a mathematical comparison of two quantities of the same kind and measured in the same units, expressed as a : b a : b a : b or a b \frac{a}{b} b a . In the ratio a : b a : b a : b , a a a is called the antecedent and b b b is called the consequent .
Unit Independence: Because a ratio represents relative comparison through division, it is a dimensionless pure number ( a : b a : b a : b has no physical units).
Scalar Invariance: Multiplying or dividing both antecedent and consequent by a non-zero constant k k k leaves the ratio unchanged ( a : b = k a : k b a : b = ka : kb a : b = ka : kb ).
- Derived Ratios
- Duplicate Ratio: a 2 : b 2 a^2 : b^2 a 2 : b 2
- Sub-duplicate Ratio: a : b \sqrt{a} : \sqrt{b} a : b
- Triplicate Ratio: a 3 : b 3 a^3 : b^3 a 3 : b 3
- Sub-triplicate Ratio: a 3 : b 3 \sqrt[3]{a} : \sqrt[3]{b} 3 a : 3 b
Compound Ratio: The compound ratio of ( a : b ) (a : b) ( a : b ) and ( c : d ) (c : d) ( c : d ) is ( a × c : b × d ) = a c : b d (a \times c : b \times d) = ac : bd ( a × c : b × d ) = a c : b d .
Inverse (Reciprocal) Ratio: The inverse of a : b a : b a : b is 1 a : 1 b = b : a \frac{1}{a} : \frac{1}{b} = b : a a 1 : b 1 = b : a .
2.2 Proportions and Algebraic Theorems
An equality of two ratios constitutes a proportion . The expression a : b : : c : d a : b :: c : d a : b :: c : d (or a b = c d \frac{a}{b} = \frac{c}{d} b a = d c ) asserts that the ratio a : b a : b a : b equals c : d c : d c : d . Here, a a a and d d d are called the extremes , while b b b and c c c are called the means .
Fundamental Law of Proportion: a ⋅ d = b ⋅ c ⟺ Product of Extremes = Product of Means \text{Fundamental Law of Proportion: } a \cdot d = b \cdot c \iff \text{Product of Extremes} = \text{Product of Means} Fundamental Law of Proportion: a ⋅ d = b ⋅ c ⟺ Product of Extremes = Product of Means
Continued Proportion & Proportional Quantities
Three quantities a , b , c a, b, c a , b , c are in continued proportion if a b = b c \frac{a}{b} = \frac{b}{c} b a = c b .
Mean Proportional (Geometric Mean): a x = x b ⟹ x 2 = a b ⟹ x = a b \frac{a}{x} = \frac{x}{b} \implies x^2 = ab \implies x = \sqrt{ab} x a = b x ⟹ x 2 = ab ⟹ x = ab
Third Proportional: The third proportional to a a a and b b b is the value x x x such that: a : b : : b : x ⟹ x = b 2 a a : b :: b : x \implies x = \frac{b^2}{a} a : b :: b : x ⟹ x = a b 2
Fourth Proportional: The fourth proportional to three quantities a , b , c a, b, c a , b , c is the value x x x such that: a : b : : c : x ⟹ x = b ⋅ c a a : b :: c : x \implies x = \frac{b \cdot c}{a} a : b :: c : x ⟹ x = a b ⋅ c
Proportional Transformation Theorems
Invertendo: If a b = c d \frac{a}{b} = \frac{c}{d} b a = d c , then b a = d c \frac{b}{a} = \frac{d}{c} a b = c d .
Alternando: If a b = c d \frac{a}{b} = \frac{c}{d} b a = d c , then a c = b d \frac{a}{c} = \frac{b}{d} c a = d b .
Componendo: If a b = c d \frac{a}{b} = \frac{c}{d} b a = d c , then a + b b = c + d d \frac{a + b}{b} = \frac{c + d}{d} b a + b = d c + d .
Dividendo: If a b = c d \frac{a}{b} = \frac{c}{d} b a = d c , then a − b b = c − d d \frac{a - b}{b} = \frac{c - d}{d} b a − b = d c − d .
Componendo & Dividendo (C&D Rule): If a b = c d \frac{a}{b} = \frac{c}{d} b a = d c , then: a + b a − b = c + d c − d \frac{a + b}{a - b} = \frac{c + d}{c - d} a − b a + b = c − d c + d
2.3 Partitioning Quantities in Given Ratios
When a total quantity Q Q Q is partitioned into parts according to the ratio a : b : c a : b : c a : b : c :
Total Proportional Parts = a + b + c \text{Total Proportional Parts} = a + b + c Total Proportional Parts = a + b + c Part 1 = ( a a + b + c ) × Q , Part 2 = ( b a + b + c ) × Q , Part 3 = ( c a + b + c ) × Q \text{Part } 1 = \left(\frac{a}{a + b + c}\right) \times Q, \quad \text{Part } 2 = \left(\frac{b}{a + b + c}\right) \times Q, \quad \text{Part } 3 = \left(\frac{c}{a + b + c}\right) \times Q Part 1 = ( a + b + c a ) × Q , Part 2 = ( a + b + c b ) × Q , Part 3 = ( a + b + c c ) × Q
2.4 Partnership Mathematics
In business partnerships, profit is distributed based on both financial capital and time duration of the investment:
Profit ( P ) ∝ Capital ( C ) × Time ( T ) \text{Profit } (P) \propto \text{Capital } (C) \times \text{Time } (T) Profit ( P ) ∝ Capital ( C ) × Time ( T )
Simple Partnership: All partners invest capital for identical durations ( T 1 = T 2 = T 3 T_1 = T_2 = T_3 T 1 = T 2 = T 3 ): P 1 : P 2 : P 3 = C 1 : C 2 : C 3 P_1 : P_2 : P_3 = C_1 : C_2 : C_3 P 1 : P 2 : P 3 = C 1 : C 2 : C 3
Compound Partnership: Partners invest varying capitals for varying periods ( T 1 , T 2 , T 3 T_1, T_2, T_3 T 1 , T 2 , T 3 ): P 1 : P 2 : P 3 = ( C 1 × T 1 ) : ( C 2 × T 2 ) : ( C 3 × T 3 ) P_1 : P_2 : P_3 = (C_1 \times T_1) : (C_2 \times T_2) : (C_3 \times T_3) P 1 : P 2 : P 3 = ( C 1 × T 1 ) : ( C 2 × T 2 ) : ( C 3 × T 3 )
Active vs. Sleeping Partners: An active partner manages the enterprise and receives a specified salary or percentage of gross profit before the remaining profit is partitioned according to the investment-time ratio.
3. Percentages: The Universal Quantitative Bridge
The word percentage originates from the Latin per centum ("by the hundred"). A percentage is simply a fraction with a denominator of 100:
- x % = x 100 = x × 0.01 x\% = \frac{x}{100} = x \times 0.01 x % = 100 x = x × 0.01
- 3.1 Benchmark Fraction-to-Percentage Equivalence Table
Memorizing benchmark equivalents is essential for rapid paper solving in CG SET Paper I:
- Fraction Percentage Exact Mixed Fraction Decimal Multiplier
- 1 / 1 1/1 1/1 100 % 100\% 100% 100 % 100\% 100% 1.00 1.00 1.00
- 1 / 2 1/2 1/2 50 % 50\% 50% 50 % 50\% 50% 0.50 0.50 0.50
- 1 / 3 1/3 1/3 33.33 % 33.33\% 33.33% 33 1 3 % 33 \frac{1}{3}\% 33 3 1 % 0.333 0.333 0.333
- 1 / 4 1/4 1/4 25 % 25\% 25% 25 % 25\% 25% 0.25 0.25 0.25
- 1 / 5 1/5 1/5 20 % 20\% 20% 20 % 20\% 20% 0.20 0.20 0.20
- 1 / 6 1/6 1/6 16.67 % 16.67\% 16.67% 16 2 3 % 16 \frac{2}{3}\% 16 3 2 % 0.1667 0.1667 0.1667
- 1 / 7 1/7 1/7 14.28 % 14.28\% 14.28% 14 2 7 % 14 \frac{2}{7}\% 14 7 2 % 0.1428 0.1428 0.1428
- 1 / 8 1/8 1/8 12.50 % 12.50\% 12.50% 12 1 2 % 12 \frac{1}{2}\% 12 2 1 % 0.125 0.125 0.125
- 1 / 9 1/9 1/9 11.11 % 11.11\% 11.11% 11 1 9 % 11 \frac{1}{9}\% 11 9 1 % 0.111 0.111 0.111
- 1 / 10 1/10 1/10 10 % 10\% 10% 10 % 10\% 10% 0.10 0.10 0.10
- 1 / 11 1/11 1/11 9.09 % 9.09\% 9.09% 9 1 11 % 9 \frac{1}{11}\% 9 11 1 % 0.0909 0.0909 0.0909
- 1 / 12 1/12 1/12 8.33 % 8.33\% 8.33% 8 1 3 % 8 \frac{1}{3}\% 8 3 1 % 0.0833 0.0833 0.0833
- 1 / 14 1/14 1/14 7.14 % 7.14\% 7.14% 7 1 7 % 7 \frac{1}{7}\% 7 7 1 % 0.0714 0.0714 0.0714
- 1 / 15 1/15 1/15 6.67 % 6.67\% 6.67% 6 2 3 % 6 \frac{2}{3}\% 6 3 2 % 0.0667 0.0667 0.0667
- 1 / 16 1/16 1/16 6.25 % 6.25\% 6.25% 6 1 4 % 6 \frac{1}{4}\% 6 4 1 % 0.0625 0.0625 0.0625
- 1 / 20 1/20 1/20 5 % 5\% 5% 5 % 5\% 5% 0.05 0.05 0.05
- 1 / 25 1/25 1/25 4 % 4\% 4% 4 % 4\% 4% 0.04 0.04 0.04
- 3.2 Percentage Change vs. Percentage Points
Percentage Change: The relative change compared against the original base value: Percentage Change = ( Final Value − Initial Value Initial Value ) × 100 % \text{Percentage Change} = \left(\frac{\text{Final Value} - \text{Initial Value}}{\text{Initial Value}}\right) \times 100\% Percentage Change = ( Initial Value Final Value − Initial Value ) × 100%
Percentage Points: The arithmetic difference between two percentage values: Δ points = P 2 % − P 1 % \Delta_{\text{points}} = P_2\% - P_1\% Δ points = P 2 % − P 1 %
Example: If an institution's pass percentage rises from 40 % 40\% 40% to 50 % 50\% 50% , the absolute increase is 10 percentage points , but the relative percentage increase is 50 − 40 40 × 100 % = 25 % \frac{50 - 40}{40} \times 100\% = 25\% 40 50 − 40 × 100% = 25% . Confusing these two concepts is a frequent exam pitfall.
- 4. Advanced Percentage Formulations & Economic Applications
- 4.1 Successive Percentage Changes
When a quantity undergoes two successive percentage changes of a % a\% a % and b % b\% b % :
Net Percentage Change = ( a + b + a ⋅ b 100 ) % \text{Net Percentage Change} = \left(a + b + \frac{a \cdot b}{100}\right)\% Net Percentage Change = ( a + b + 100 a ⋅ b ) %
(Where positive values denote increases and negative values denote decreases) .
Two Successive Increases ( + a % , + b % +a\%, +b\% + a % , + b % ): Net increase = a + b + a b 100 = a + b + \frac{ab}{100} = a + b + 100 ab .
Demonstration: Two successive salary hikes of 20 % 20\% 20% and 10 % 10\% 10% : Net Hike = 20 + 10 + 20 × 10 100 = 30 + 2 = 32 % \text{Net Hike} = 20 + 10 + \frac{20 \times 10}{100} = 30 + 2 = 32\% Net Hike = 20 + 10 + 100 20 × 10 = 30 + 2 = 32%
Equal Successive Increase and Decrease ( + x % , − x % +x\%, -x\% + x % , − x % ): Net Change = + x − x + ( + x ) ( − x ) 100 = − x 2 100 % \text{Net Change} = +x - x + \frac{(+x)(-x)}{100} = -\frac{x^2}{100}\% Net Change = + x − x + 100 ( + x ) ( − x ) = − 100 x 2 % Key Invariant: When a quantity is increased by x % x\% x % and then decreased by x % x\% x % , there is always an overall net decrease of x 2 100 % \frac{x^2}{100}\% 100 x 2 % .
4.2 Expenditure, Price, and Consumption Elasticity
Total expenditure ( E E E ) is the product of price per unit ( P P P ) and total consumption quantity ( C C C ):
Expenditure ( E ) = Price ( P ) × Consumption ( C ) \text{Expenditure } (E) = \text{Price } (P) \times \text{Consumption } (C) Expenditure ( E ) = Price ( P ) × Consumption ( C )
If total expenditure must remain strictly constant when the price changes:
Price Increases by r % r\% r % : Required Reduction in Consumption = ( r 100 + r ) × 100 % \text{Required Reduction in Consumption} = \left(\frac{r}{100 + r}\right) \times 100\% Required Reduction in Consumption = ( 100 + r r ) × 100%
Price Decreases by r % r\% r % : Possible Increase in Consumption = ( r 100 − r ) × 100 % \text{Possible Increase in Consumption} = \left(\frac{r}{100 - r}\right) \times 100\% Possible Increase in Consumption = ( 100 − r r ) × 100%
Demonstration: If the price of petrol rises by 25 % 25\% 25% ( r = 25 r = 25 r = 25 ), to keep expenditure constant: Reduction in Consumption = ( 25 100 + 25 ) × 100 % = 25 125 × 100 % = 1 5 × 100 % = 20 % \text{Reduction in Consumption} = \left(\frac{25}{100 + 25}\right) \times 100\% = \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\% Reduction in Consumption = ( 100 + 25 25 ) × 100% = 125 25 × 100% = 5 1 × 100% = 20%
4.3 Demographic & Depreciation Exponential Models
Population Growth: If initial population P 0 P_0 P 0 grows at annual rate r % r\% r % for n n n years: P n = P 0 ( 1 + r 100 ) n P_n = P_0 \left(1 + \frac{r}{100}\right)^n P n = P 0 ( 1 + 100 r ) n
Asset Depreciation: If machinery value V 0 V_0 V 0 depreciates at annual rate r % r\% r % for n n n years: V n = V 0 ( 1 − r 100 ) n V_n = V_0 \left(1 - \frac{r}{100}\right)^n V n = V 0 ( 1 − 100 r ) n
- 5. Worked Exemplars with Step-by-Step Algebraic Breakdowns
- Exemplar 1: Continued Proportion & Ratios
Problem: If A : B = 3 : 4 A : B = 3 : 4 A : B = 3 : 4 and B : C = 8 : 9 B : C = 8 : 9 B : C = 8 : 9 , find A : C A : C A : C and the unified compound ratio A : B : C A : B : C A : B : C .
Solution Step 1 (Equalize Intermediate Term B): B B B corresponds to 4 in the first ratio and 8 in the second. Multiply the first ratio by 2: A : B = ( 3 × 2 ) : ( 4 × 2 ) = 6 : 8 A : B = (3 \times 2) : (4 \times 2) = 6 : 8 A : B = ( 3 × 2 ) : ( 4 × 2 ) = 6 : 8
Solution Step 2 (Unify Ratios): Since B = 8 B = 8 B = 8 in both representations, A : B : C = 6 : 8 : 9 A : B : C = 6 : 8 : 9 A : B : C = 6 : 8 : 9 .
Solution Step 3 (Extract A : C A : C A : C ): A : C = 6 : 9 = 2 : 3 A : C = 6 : 9 = 2 : 3 A : C = 6 : 9 = 2 : 3
Exemplar 2: Mixture and Ratio Replacement
Problem: A 60-liter vessel contains a mixture of milk and water in the ratio 2 : 1 2 : 1 2 : 1 . How many liters of water must be added to make the ratio 1 : 2 1 : 2 1 : 2 ?
Solution Step 1 (Calculate Initial Quantities): Milk = 2 3 × 60 = 40 liters , Water = 1 3 × 60 = 20 liters \text{Milk} = \frac{2}{3} \times 60 = 40\text{ liters}, \quad \text{Water} = \frac{1}{3} \times 60 = 20\text{ liters} Milk = 3 2 × 60 = 40 liters , Water = 3 1 × 60 = 20 liters
Solution Step 2 (Formulate Equation for Added Water w w w ): Milk volume remains invariant at 40 liters. Water becomes 20 + w 20 + w 20 + w : 40 20 + w = 1 2 ⟹ 40 × 2 = 20 + w ⟹ 80 = 20 + w ⟹ w = 60 liters \frac{40}{20 + w} = \frac{1}{2} \implies 40 \times 2 = 20 + w \implies 80 = 20 + w \implies w = 60\text{ liters} 20 + w 40 = 2 1 ⟹ 40 × 2 = 20 + w ⟹ 80 = 20 + w ⟹ w = 60 liters Answer: Exactly 60 liters of water must be added.
6. Exam Traps & Common Calculation Blunders
Trap 1: The Base Inversion Trap in Percentages: Calculating percentage change against the final value rather than the initial base value. Remember: Percentage Change = Difference INITIAL Value × 100 % \text{Percentage Change} = \frac{\text{Difference}}{\text{INITIAL Value}} \times 100\% Percentage Change = INITIAL Value Difference × 100% .
Trap 2: Linear Summation of Successive Changes: Assuming a + 20 % +20\% + 20% increase followed by a + 10 % +10\% + 10% increase equals + 30 % +30\% + 30% . The true value is 32 % 32\% 32% because the second percentage acts upon the enlarged base.
Trap 3: Unit Mismatch in Ratio Formulation: Comparing quantities measured in different units without normalization (e.g., forming a ratio between 45 minutes and 2 hours as 45 : 2 45 : 2 45 : 2 ). Always convert to uniform units ( 45 : 120 = 3 : 8 45 : 120 = 3 : 8 45 : 120 = 3 : 8 ).
Trap 4: Forgetting the Constant Term in C&D: Applying Componendo and Dividendo to only one side of an equation. The transformation must be applied simultaneously to both sides: a + b a − b = c + d c − d \frac{a+b}{a-b} = \frac{c+d}{c-d} a − b a + b = c − d c + d .