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5.2 Fractions, Mixed Numbers & Operations

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ACCUPLACER Suite Study Guide Exam

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1 Chapter 1: ACCUPLACER Suite Overview & Testing Mechanics

1.1 The ACCUPLACER Testing System & Computer-Adaptive Mechanics 1.2 Placement Policies, Score Scales & Institutional Cut Scores 1.3 Test-Taking Strategies, Tool Mechanics & Pacing

2 Chapter 2: Next-Generation Reading

2.1 The Reading Test Blueprint: Set-Based and Discrete Items 2.2 Information and Ideas: Close Reading & Central Themes 2.3 Summarizing and Understanding Relationships 2.4 Rhetoric, Text Structure & Author's Purpose 2.5 Analyzing Point of View and Authorial Purpose 2.6 Analyzing Arguments: Claims, Counterclaims, and Evidence 2.7 Synthesis & Paired Text Comparative Analysis 2.8 Vocabulary in Context & Figurative Language 2.9 Passage Genre, Content Area, Mode, and Text Complexity 2.10 Answer-Choice Traps and Evidence Discipline

3 Chapter 3: Next-Generation Writing: Expression of Ideas

3.1 The Writing Test Blueprint: Passage Sets and Content Distribution 3.2 Development: Proposition and Supporting Evidence 3.3 Focus: Adding, Retaining, and Deleting for Relevance 3.4 Organization: Logical Sequence and Paragraph Order 3.5 Introductions, Conclusions, and Transitional Cohesion 3.6 Effective Language Use: Precision and Word Choice 3.7 Concision: Eliminating Wordiness and Redundancy 3.8 Style, Tone, and Voice Consistency 3.9 Syntax: Sentence Combining for Rhetorical Purpose

4 Chapter 4: Next-Generation Writing: Standard English Conventions

4.1 Sentence Boundaries: Fragments, Run-ons, and Comma Splices 4.2 Subordination and Coordination 4.3 Parallel Structure 4.4 Modifier Placement: Misplaced and Dangling Modifiers 4.5 Inappropriate Shifts in Verb Tense 4.6 Inappropriate Shifts in Verb Voice and Mood 4.7 Inappropriate Shifts in Pronoun Person and Number 4.8 Subject-Verb Agreement, Pronoun Case, and Verb Forms 4.9 Possessive Determiners, Contractions, and Confusable Adverbs 4.10 Pronoun Clarity and Pronoun-Antecedent Agreement 4.11 Frequently Confused Words, Noun Agreement, and Logical Comparison 4.12 Within-Sentence Punctuation: Semicolons, Colons, Dashes, and Apostrophes 4.13 Items in a Series and Nonrestrictive Elements 4.14 End-of-Sentence Punctuation, Hyphenation, and Unnecessary Punctuation

5 Chapter 5: Next-Generation Arithmetic

5.1 Whole Number Operations, Estimation & Multi-Step Applications 5.2 Fractions, Mixed Numbers & Operations 5.3 Decimal Operations & Place-Value Conversions 5.4 Percent Calculations, Increase, Decrease, and Applications 5.5 Number Comparisons and Equivalents Across Formats

6 Chapter 6: Next-Generation Quantitative Reasoning, Algebra, and Statistics (QAS)

6.1 Rational Numbers, Absolute Value, and Order of Operations 6.2 Exponents, Radicals, and Scientific Notation 6.3 Ratio, Proportional Relationships, and Unit Conversions 6.4 Creating and Evaluating Algebraic Expressions 6.5 Linear Equations, Inequalities, and Systems of Two Equations 6.6 Linear Applications, Functions, and Graphs 6.7 Probability and Sets 6.8 Descriptive Statistics: Center, Spread, and Graphical Displays 6.9 Geometry for Pre-Algebra and Algebra 1

7 Chapter 7: Next-Generation Advanced Algebra, Quadratics & Polynomials

7.1 Advanced Linear Equations, Systems, and Coordinate Geometry 7.2 Factoring Methods for Quadratics, Cubics, and Polynomials 7.3 Quadratic Equations, Vertex Form & Parabolic Functions 7.4 Polynomial Equations, Roots, and Graphs 7.5 Radical and Rational Equations and Functions 7.6 Geometry for Algebra 1 and Algebra 2

8 Chapter 8: Functions, Exponentials, Logarithms & Trigonometry

8.1 Function Notation, Transformations, Compositions & Inverses 8.2 Domain, Range, and Interpreting Functions in Context 8.3 Exponential & Logarithmic Functions, Laws & Modeling 8.4 Right-Triangle Trigonometry, the Unit Circle, and Radian Measure 8.5 Law of Sines, Law of Cosines, and Trigonometric Equations 8.6 Choosing Your Math Test: Arithmetic, QAS, or AAF

9 Chapter 9: WritePlacer & WritePlacer ESL Essays

9.1 WritePlacer Prompt Format, Planning, and the Writing Process 9.2 The WritePlacer Scoring Rubric & 6 Core Dimensions 9.3 Reading the WritePlacer Score Scale: What Each Band Demonstrates 9.4 Essay Architecture: Thesis, Evidence Integration & Structure 9.5 WritePlacer ESL Scoring Standards & Synthesis

10 Chapter 10: ACCUPLACER ESL Placement Suite

10.1 The ESL Placement Suite: Structure, Scoring, and Score Interpretation 10.2 ESL Reading Skills & Passage Comprehension 10.3 ESL Sentence Meaning & Context Clues 10.4 ESL Language Use & Grammatical Mechanics 10.5 ESL Listening Comprehension & Spoken Contexts

5.2 Fractions, Mixed Numbers & Operations

Key Takeaways Explain This Topic AI

A fraction represents part-to-whole division a/b, classified as proper (a < b), improper (a >= b), or mixed numbers (W a/b), where conversions require exact modular arithmetic.

Finding the Least Common Multiple (LCM) of denominators establishes the Least Common Denominator (LCD), which is mathematically mandatory before executing addition or subtraction.

The Greatest Common Factor (GCF) determined via prime factorization allows fractions to be reduced to simplest terms by dividing numerator and denominator by their greatest common divisor.

Fraction multiplication multiplies numerators directly and denominators directly, made highly efficient by cross-canceling common factors prior to multiplication.

Fraction division operates by multiplying the dividend by the reciprocal (multiplicative inverse) of the divisor: (a/b) ÷ (c/d) = (a/b) × (d/c).

Last updated: August 2026

5.2 Fractions, Mixed Numbers & Operations

Core Concept: Fractions represent rational numbers expressed as the quotient of two integers a b \frac{a}{b} b a ​ (where b ≠ 0 b \neq 0 b  = 0 ). Success on the ACCUPLACER requires total fluency with equivalent fraction transformations, common denominator conversions, mixed number regrouping, cross-canceling, and dividing by reciprocals without a calculator.

1. Anatomy of Fractions, Classifications & Structural Conversions

A fraction a b \frac{a}{b} b a ​ consists of a numerator a a a (representing the number of equal parts considered) and a denominator b b b (representing the total number of equal parts comprising one whole unit). +-------------------------------------------------------------------------------------------------------+

  • | FRACTION CLASSIFICATION MATRIX |
  • +-------------------+---------------------------+-----------------------------------+-------------------+
  • | Classification | Mathematical Condition | Defining Characteristic | Examples |
  • +-------------------+---------------------------+-----------------------------------+-------------------+
  • | Proper Fraction | Numerator < Denominator | Value strictly between 0 and 1 | 3/8, 5/12, 7/9 |
  • | Improper Fraction | Numerator >= Denominator | Value greater than or equal to 1 | 11/4, 9/9, 43/6 |
  • | Mixed Number | Integer + Proper Fraction | Explicit whole number plus part | 2 3/4, 7 1/6 |
  • +-------------------+---------------------------+-----------------------------------+-------------------+
  • Bidirectional Conversion Algorithms

Mixed Number to Improper Fraction: Multiply the whole integer W W W by the denominator d d d , add the numerator n n n , and place the resulting sum over the original denominator d d d : W n d = ( W × d ) + n d W \frac{n}{d} = \frac{(W \times d) + n}{d} W d n ​ = d ( W × d ) + n ​

Example: Convert 5 3 8 5 \frac{3}{8} 5 8 3 ​ to an improper fraction: 5 3 8 = ( 5 × 8 ) + 3 8 = 40 + 3 8 = 43 8 5 \frac{3}{8} = \frac{(5 \times 8) + 3}{8} = \frac{40 + 3}{8} = \frac{43}{8} 5 8 3 ​ = 8 ( 5 × 8 ) + 3 ​ = 8 40 + 3 ​ = 8 43 ​

Improper Fraction to Mixed Number: Divide the numerator n n n by the denominator d d d using integer division. The quotient q q q becomes the whole number, and the remainder r r r becomes the new numerator over d d d : n d = q r d , where n = ( q × d ) + r \frac{n}{d} = q \frac{r}{d}, \quad \text{where } n = (q \times d) + r d n ​ = q d r ​ , where n = ( q × d ) + r

Example: Convert 79 6 \frac{79}{6} 6 79 ​ to a mixed number: 79 ÷ 6 = 13 with a remainder of 1 ⟹ 13 1 6 79 \div 6 = 13 \text{ with a remainder of } 1 \implies \mathbf{13 \frac{1}{6}} 79 ÷ 6 = 13 with a remainder of 1 ⟹ 13 6 1 ​

2. Prime Factorization, Greatest Common Factor (GCF) & Least Common Multiple (LCM)

Simplifying fractions and creating common denominators both rely on the prime factorization of integers.

Simplifying Fractions via Greatest Common Factor (GCF)

A fraction is in simplest form (lowest terms) when its numerator and denominator share no common factors other than 1 1 1 (i.e., gcd ⁡ ( a , b ) = 1 \gcd(a, b) = 1 g cd ( a , b ) = 1 ). To reduce a fraction in a single step, divide both terms by their GCF.

Example: Reduce 84 140 \frac{84}{140} 140 84 ​ to simplest terms:

Find prime factorizations: 84 = 2 2 × 3 × 7 84 = 2^2 \times 3 \times 7 84 = 2 2 × 3 × 7 140 = 2 2 × 5 × 7 140 = 2^2 \times 5 \times 7 140 = 2 2 × 5 × 7

Extract common prime factors with lowest exponents: GCF ( 84 , 140 ) = 2 2 × 7 = 28 \text{GCF}(84, 140) = 2^2 \times 7 = 28 GCF ( 84 , 140 ) = 2 2 × 7 = 28

Divide numerator and denominator by 28 28 28 : 84 ÷ 28 140 ÷ 28 = 3 5 \frac{84 \div 28}{140 \div 28} = \mathbf{\frac{3}{5}} 140 ÷ 28 84 ÷ 28 ​ = 5 3 ​

Finding the Least Common Denominator (LCD) via LCM

To add or subtract fractions with unlike denominators, you must convert them to equivalent fractions sharing the Least Common Denominator (LCD) , which is the Least Common Multiple (LCM) of the denominators.

Example: Find the LCD for fractions with denominators 12 12 12 , 18 18 18 , and 30 30 30 :

Prime factorizations: 12 = 2 2 × 3 1 12 = 2^2 \times 3^1 12 = 2 2 × 3 1 18 = 2 1 × 3 2 18 = 2^1 \times 3^2 18 = 2 1 × 3 2 30 = 2 1 × 3 1 × 5 1 30 = 2^1 \times 3^1 \times 5^1 30 = 2 1 × 3 1 × 5 1

Take the highest power of every unique prime factor: LCM ( 12 , 18 , 30 ) = 2 2 × 3 2 × 5 1 = 4 × 9 × 5 = 180 \text{LCM}(12, 18, 30) = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = \mathbf{180} LCM ( 12 , 18 , 30 ) = 2 2 × 3 2 × 5 1 = 4 × 9 × 5 = 180

The LCD is 180 180 180 .

  • 3. Addition and Subtraction with Like and Unlike Denominators
  • Fractions with Unlike Denominators

Unlike denominators represent unequal partition sizes; they cannot be combined directly. You must first scale each fraction by a form of 1 (such as k k \frac{k}{k} k k ​ ) to match the LCD. a b ± c d = a ⋅ ( LCD / b ) LCD ± c ⋅ ( LCD / d ) LCD \frac{a}{b} \pm \frac{c}{d} = \frac{a \cdot (\text{LCD}/b)}{\text{LCD}} \pm \frac{c \cdot (\text{LCD}/d)}{\text{LCD}} b a ​ ± d c ​ = LCD a ⋅ ( LCD / b ) ​ ± LCD c ⋅ ( LCD / d ) ​

Worked Example: Compute 7 12 + 5 18 − 1 8 \frac{7}{12} + \frac{5}{18} - \frac{1}{8} 12 7 ​ + 18 5 ​ − 8 1 ​ .

Find LCD: Denominators are 12 = 2 2 × 3 12 = 2^2 \times 3 12 = 2 2 × 3 , 18 = 2 × 3 2 18 = 2 \times 3^2 18 = 2 × 3 2 , 8 = 2 3 8 = 2^3 8 = 2 3 . LCD = 2 3 × 3 2 = 8 × 9 = 72 \text{LCD} = 2^3 \times 3^2 = 8 \times 9 = 72 LCD = 2 3 × 3 2 = 8 × 9 = 72

Convert each fraction: 7 12 = 7 × 6 12 × 6 = 42 72 \frac{7}{12} = \frac{7 \times 6}{12 \times 6} = \frac{42}{72} 12 7 ​ = 12 × 6 7 × 6 ​ = 72 42 ​ 5 18 = 5 × 4 18 × 4 = 20 72 \frac{5}{18} = \frac{5 \times 4}{18 \times 4} = \frac{20}{72} 18 5 ​ = 18 × 4 5 × 4 ​ = 72 20 ​ 1 8 = 1 × 9 8 × 9 = 9 72 \frac{1}{8} = \frac{1 \times 9}{8 \times 9} = \frac{9}{72} 8 1 ​ = 8 × 9 1 × 9 ​ = 72 9 ​

Combine Numerators: 42 + 20 − 9 72 = 53 72 \frac{42 + 20 - 9}{72} = \mathbf{\frac{53}{72}} 72 42 + 20 − 9 ​ = 72 53 ​

Mixed Number Addition and Subtraction with Regrouping (Borrowing)

When subtracting mixed numbers, if the fractional part of the minuend is smaller than that of the subtrahend, you must borrow 1 1 1 whole ( 1 = d d 1 = \frac{d}{d} 1 = d d ​ ) from the whole-number integer.

Worked Example: Evaluate 8 1 6 − 3 3 4 8 \frac{1}{6} - 3 \frac{3}{4} 8 6 1 ​ − 3 4 3 ​ .

Establish Common Denominator ( 12 12 12 ): 8 1 6 = 8 2 12 , 3 3 4 = 3 9 12 8 \frac{1}{6} = 8 \frac{2}{12}, \quad 3 \frac{3}{4} = 3 \frac{9}{12} 8 6 1 ​ = 8 12 2 ​ , 3 4 3 ​ = 3 12 9 ​

Regroup (Borrow 1 1 1 whole): Because 2 12 < 9 12 \frac{2}{12} < \frac{9}{12} 12 2 ​ < 12 9 ​ , borrow 1 1 1 from 8 8 8 : 8 2 12 = 7 + 12 12 + 2 12 = 7 14 12 8 \frac{2}{12} = 7 + \frac{12}{12} + \frac{2}{12} = 7 \frac{14}{12} 8 12 2 ​ = 7 + 12 12 ​ + 12 2 ​ = 7 12 14 ​

Subtract Whole Numbers and Fractions Separately: ( 7 − 3 ) + ( 14 12 − 9 12 ) = 4 + 5 12 = 4 5 12 (7 - 3) + \left(\frac{14}{12} - \frac{9}{12}\right) = 4 + \frac{5}{12} = \mathbf{4 \frac{5}{12}} ( 7 − 3 ) + ( 12 14 ​ − 12 9 ​ ) = 4 + 12 5 ​ = 4 12 5 ​

4. Multiplication, Cross-Cancellation & Division Algorithms +-------------------------------------------------------------------------------------------------------+

  • | FRACTION MULTIPLICATION & DIVISION RULES |
  • +----------------+--------------------------------------+-----------------------------------------------+
  • | Operation | Algorithmic Formula | Critical Step |
  • +----------------+--------------------------------------+-----------------------------------------------+
  • | Multiplication | (a/b) × (c/d) = (a × c) / (b × d) | Cross-cancel common factors BEFORE multiplying|
  • | Division | (a/b) ÷ (c/d) = (a/b) × (d/c) | Invert divisor (reciprocal) and multiply |
  • | Mixed Numbers | Convert all to improper fractions | Never multiply mixed whole numbers and |
  • | | before computing | fractions separately |
  • +----------------+--------------------------------------+-----------------------------------------------+
  • Cross-Cancellation in Multiplication

Never multiply large raw numerators and denominators without checking for shared diagonal factors first. Cross-canceling eliminates common prime factors early and keeps arithmetic manageable.

Worked Example: Calculate 14 25 × 15 21 × 5 8 \frac{14}{25} \times \frac{15}{21} \times \frac{5}{8} 25 14 ​ × 21 15 ​ × 8 5 ​ .

Pair 14 21 \frac{14}{21} 21 14 ​ : Divide both by 7 ⟹ 14 → 2 7 \implies 14 \to 2 7 ⟹ 14 → 2 and 21 → 3 21 \to 3 21 → 3 .

Pair 15 25 \frac{15}{25} 25 15 ​ : Divide both by 5 ⟹ 15 → 3 5 \implies 15 \to 3 5 ⟹ 15 → 3 and 25 → 5 25 \to 5 25 → 5 .

The expression becomes: 2 5 × 3 3 × 5 8 = 2 5 × 1 × 5 8 \frac{2}{5} \times \frac{3}{3} \times \frac{5}{8} = \frac{2}{5} \times 1 \times \frac{5}{8} 5 2 ​ × 3 3 ​ × 8 5 ​ = 5 2 ​ × 1 × 8 5 ​ .

Pair the 5 5 5 in the denominator with the 5 5 5 in the numerator: 2 1 × 1 8 = 2 8 \frac{2}{1} \times \frac{1}{8} = \frac{2}{8} 1 2 ​ × 8 1 ​ = 8 2 ​ .

Reduce 2 8 = 1 4 \frac{2}{8} = \mathbf{\frac{1}{4}} 8 2 ​ = 4 1 ​ .

Division by Multiplying by the Reciprocal

Division by a non-zero fraction is mathematically identical to multiplication by its multiplicative inverse (reciprocal) : a b ÷ c d = a b × d c \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} b a ​ ÷ d c ​ = b a ​ × c d ​

Worked Example: Evaluate 4 2 3 ÷ 2 4 5 4 \frac{2}{3} \div 2 \frac{4}{5} 4 3 2 ​ ÷ 2 5 4 ​ .

Convert mixed numbers to improper fractions: 4 2 3 = 14 3 , 2 4 5 = 14 5 4 \frac{2}{3} = \frac{14}{3}, \quad 2 \frac{4}{5} = \frac{14}{5} 4 3 2 ​ = 3 14 ​ , 2 5 4 ​ = 5 14 ​

Invert the divisor and change ÷ \div ÷ to × \times × : 14 3 ÷ 14 5 = 14 3 × 5 14 \frac{14}{3} \div \frac{14}{5} = \frac{14}{3} \times \frac{5}{14} 3 14 ​ ÷ 5 14 ​ = 3 14 ​ × 14 5 ​

Cross-cancel the 14 14 14 in numerator and denominator: 1 3 × 5 1 = 5 3 = 1 2 3 \frac{1}{3} \times \frac{5}{1} = \mathbf{\frac{5}{3} = 1 \frac{2}{3}} 3 1 ​ × 1 5 ​ = 3 5 ​ = 1 3 2 ​

  • 5. Applied Fraction Word Problems
  • Real-World Carpentry & Measurement Problem

Problem: A carpenter has a raw wooden timber measuring 14 1 2 14 \frac{1}{2} 14 2 1 ​ feet in length. The carpenter cuts 4 4 4 identical structural braces from this timber, each measuring 2 3 4 2 \frac{3}{4} 2 4 3 ​ feet long. Each saw cut destroys 1 8 \frac{1}{8} 8 1 ​ foot of wood as sawdust kerf waste (requiring a total of 4 4 4 cuts). What is the exact length of the remaining wood piece?

Step-by-Step Solution:

Calculate Length of 4 Braces: 4 × 2 3 4 = 4 × 11 4 = 11 feet 4 \times 2 \frac{3}{4} = 4 \times \frac{11}{4} = 11 \text{ feet} 4 × 2 4 3 ​ = 4 × 4 11 ​ = 11 feet

Calculate Total Kerf Sawdust Waste: 4 × 1 8 = 4 8 = 1 2 foot 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \text{ foot} 4 × 8 1 ​ = 8 4 ​ = 2 1 ​ foot

Sum Total Lumber Consumed: 11 + 1 2 = 11 1 2 feet 11 + \frac{1}{2} = 11 \frac{1}{2} \text{ feet} 11 + 2 1 ​ = 11 2 1 ​ feet

Subtract from Original Timber: 14 1 2 − 11 1 2 = 3 feet 14 \frac{1}{2} - 11 \frac{1}{2} = \mathbf{3 \text{ feet}} 14 2 1 ​ − 11 2 1 ​ = 3 feet

Conclusion: Exactly 3 3 3 feet of lumber remains.