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1.2 Fractions, Decimals, & Mixed Numbers Operations
ACCUPLACER Quantitative Reasoning, Algebra, and Statistics Exam
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1 Chapter 1: Rational Numbers & Operations
1.1 Rational Number Arithmetic & Signed Numbers 1.2 Fractions, Decimals, & Mixed Numbers Operations 1.3 Absolute Value, Number Line, & Estimation
2 Chapter 2: Ratios, Proportions & Rates
2.1 Ratios, Unit Rates, & Dimensional Analysis 2.2 Proportions & Direct/Inverse Variation 2.3 Percent Applications, Markup, Discount, & Simple Interest
3 Chapter 3: Exponents & Scientific Notation
3.1 Integer Exponents & Laws of Exponents 3.2 Scientific Notation & Operations
4 Chapter 4: Algebraic Expressions
4.1 Evaluating & Translating Algebraic Expressions 4.2 Simplifying, Combining Like Terms, & Factoring
5 Chapter 5: Linear Equations & Inequalities
5.1 Solving Single-Variable Linear Equations 5.2 Solving & Graphing Linear Inequalities 5.3 Systems of Linear Equations
6 Chapter 6: Linear Applications & Graphing
6.1 Slope & Rate of Change 6.2 Forms of Linear Equations & Graphing 6.3 Linear Modeling & Inequality Graphs
7 Chapter 7: Probability & Set Theory
7.1 Fundamental Probability Concepts 7.2 Set Theory, Operations, & Venn Diagrams
8 Chapter 8: Descriptive Statistics & Data Displays
8.1 Measures of Center: Mean, Median, Mode, & Weighted Average 8.2 Measures of Spread: Range, IQR, Variance, & Standard Deviation 8.3 Data Displays: Histograms, Box Plots, & Scatter Plots
9 Chapter 9: Geometry Concepts & Measurement
9.1 2D Geometry: Perimeter, Area, & Circles 9.2 3D Geometry: Surface Area & Volume of Solids 9.3 Triangles, Angles, & the Pythagorean Theorem
1.2 Fractions, Decimals, & Mixed Numbers Operations
Key Takeaways Explain This Topic AI
To convert a mixed number W n/d to an improper fraction, compute (W · d + n)/d; to convert back, divide the numerator by denominator to determine the whole quotient and remainder.
A fraction in simplest form converts to a terminating decimal if and only if the prime factorization of its denominator contains no prime factors other than 2 and 5.
Repeating decimals can be converted into exact rational fractions by setting x equal to the decimal, multiplying by powers of 10 to align the repeating blocks, and subtracting the equations to eliminate the repeat.
When adding or subtracting fractions with unlike denominators, find the Least Common Denominator (LCD) using prime factorization, rename numerators, and simplify the resulting fraction.
Division of fractions is performed by multiplying by the reciprocal of the divisor: (a/b) ÷ (c/d) = (a/b) · (d/c) = (ad)/(bc) ('Keep-Change-Flip').
Last updated: August 2026
Anatomy and Classification of Fractions
A fraction a b \frac{a}{b} b a represents the division of an integer numerator a a a by a non-zero integer denominator b b b . The denominator specifies the total number of equal parts into which a unit is partitioned, while the numerator specifies the number of parts selected.
Fractions fall into three distinct structural categories:
Proper Fractions : Fractions where the absolute value of the numerator is strictly less than the absolute value of the denominator ( ∣ a ∣ < ∣ b ∣ |a| < |b| ∣ a ∣ < ∣ b ∣ ). Proper fractions represent values strictly between − 1 -1 − 1 and + 1 +1 + 1 . Examples: 3 8 , 5 12 , − 7 10 \text{Examples: } \frac{3}{8}, \quad \frac{5}{12}, \quad -\frac{7}{10} Examples: 8 3 , 12 5 , − 10 7
Improper Fractions : Fractions where the absolute value of the numerator is greater than or equal to the absolute value of the denominator ( ∣ a ∣ ≥ ∣ b ∣ |a| \ge |b| ∣ a ∣ ≥ ∣ b ∣ ). Improper fractions represent values with magnitude greater than or equal to 1 1 1 . Examples: 11 4 , 7 7 = 1 , − 29 6 \text{Examples: } \frac{11}{4}, \quad \frac{7}{7} = 1, \quad -\frac{29}{6} Examples: 4 11 , 7 7 = 1 , − 6 29
Mixed Numbers : An alternative representation of an improper fraction consisting of a non-zero whole integer combined with a proper fraction ( W n d W \frac{n}{d} W d n ). Examples: 2 3 4 , 5 1 6 , − 4 2 5 \text{Examples: } 2\frac{3}{4}, \quad 5\frac{1}{6}, \quad -4\frac{2}{5} Examples: 2 4 3 , 5 6 1 , − 4 5 2
Converting Between Mixed Numbers and Improper Fractions
Mixed Number to Improper Fraction :
Multiply the whole number part W W W by the denominator d d d , add the numerator n n n , and place the result over the original denominator d d d : W n d = W ⋅ d + n d W\frac{n}{d} = \frac{W \cdot d + n}{d} W d n = d W ⋅ d + n Example: 4 5 8 = 4 ⋅ 8 + 5 8 = 32 + 5 8 = 37 8 \text{Example: } 4\frac{5}{8} = \frac{4 \cdot 8 + 5}{8} = \frac{32 + 5}{8} = \frac{37}{8} Example: 4 8 5 = 8 4 ⋅ 8 + 5 = 8 32 + 5 = 8 37
Note for Negative Mixed Numbers : Retain the negative sign on the entire fraction: − 3 2 5 = − ( 3 ⋅ 5 + 2 5 ) = − 17 5 -3\frac{2}{5} = -\left(\frac{3 \cdot 5 + 2}{5}\right) = -\frac{17}{5} − 3 5 2 = − ( 5 3 ⋅ 5 + 2 ) = − 5 17 .
Improper Fraction to Mixed Number :
Divide the numerator by the denominator using integer long division. The quotient becomes the whole number, the remainder becomes the new numerator, and the denominator remains unchanged: 43 6 ⟹ 43 ÷ 6 = 7 with remainder 1 ⟹ 7 1 6 \frac{43}{6} \implies 43 \div 6 = 7 \text{ with remainder } 1 \implies 7\frac{1}{6} 6 43 ⟹ 43 ÷ 6 = 7 with remainder 1 ⟹ 7 6 1
- Decimals, Fractions, and Benchmark Equivalencies
- 1. Terminating versus Repeating Decimals
A rational fraction a b \frac{a}{b} b a in reduced form produces a terminating decimal if and only if the prime factorization of its denominator contains only 2s, only 5s, or both ( b = 2 m ⋅ 5 n b = 2^m \cdot 5^n b = 2 m ⋅ 5 n ). If the reduced denominator contains any other prime factor (such as 3 , 7 , 11 , 13 3, 7, 11, 13 3 , 7 , 11 , 13 ), the decimal will repeat indefinitely .
7 40 \frac{7}{40} 40 7 : Denominator 40 = 2 3 ⋅ 5 1 ⟹ 40 = 2^3 \cdot 5^1 \implies 40 = 2 3 ⋅ 5 1 ⟹ Terminates ( 0.175 0.175 0.175 ).
13 60 \frac{13}{60} 60 13 : Denominator 60 = 2 2 ⋅ 3 1 ⋅ 5 1 ⟹ 60 = 2^2 \cdot 3^1 \cdot 5^1 \implies 60 = 2 2 ⋅ 3 1 ⋅ 5 1 ⟹ Contains prime factor 3 ⟹ 3 \implies 3 ⟹ Repeats ( 0.21666 ⋯ = 0.21 6 ‾ 0.21666\dots = 0.21\overline{6} 0.21666 ⋯ = 0.21 6 ).
2. Converting Repeating Decimals to Fractions (The Algebraic Method)
To convert a repeating decimal into an exact rational fraction, use algebraic subtraction to eliminate the infinite repeating tail:
- Demonstration: Convert 0. 27 ‾ = 0.272727 … 0.\overline{27} = 0.272727\dots 0. 27 = 0.272727 … to a fraction
- Let x = 0.272727 … x = 0.272727\dots x = 0.272727 …
Because the repeating cycle has 2 2 2 digits, multiply by 10 2 = 100 10^2 = 100 1 0 2 = 100 : 100 x = 27.272727 … 100x = 27.272727\dots 100 x = 27.272727 …
Subtract the original equation from this new equation: 100 x = 27.272727 … − x = 0 0.272727 … 99 x = 27 \begin{aligned} 100x &= 27.272727\dots \\ -\quad x &= \phantom{0}0.272727\dots \\ \hline 99x &= 27 \end{aligned} 100 x − x 99 x = 27.272727 … = 0 0.272727 … = 27
Solve for x x x and reduce to simplest form by dividing by the greatest common divisor ( 9 9 9 ): x = 27 99 = 3 11 x = \frac{27}{99} = \frac{3}{11} x = 99 27 = 11 3
- Demonstration: Convert 0.4 6 ‾ = 0.46666 … 0.4\overline{6} = 0.46666\dots 0.4 6 = 0.46666 … to a fraction
- Let x = 0.46666 … x = 0.46666\dots x = 0.46666 …
Multiply by 10 10 10 to bring the non-repeating part in front of the decimal: 10 x = 4.6666 … 10x = 4.6666\dots 10 x = 4.6666 …
Multiply by 100 100 100 to bring one full repeating cycle in front: 100 x = 46.6666 … 100x = 46.6666\dots 100 x = 46.6666 …
Subtract the two equations: 100 x − 10 x = 46. 6 ‾ − 4. 6 ‾ ⟹ 90 x = 42 100x - 10x = 46.\overline{6} - 4.\overline{6} \implies 90x = 42 100 x − 10 x = 46. 6 − 4. 6 ⟹ 90 x = 42
Solve for x x x and reduce by dividing by 6 6 6 : x = 42 90 = 7 15 x = \frac{42}{90} = \frac{7}{15} x = 90 42 = 15 7
3. Essential Benchmark Equivalencies Table
Instant recall of these benchmark conversions accelerates test speed on the ACCUPLACER:
- Fraction Decimal Percentage Fraction Decimal Percentage
- 1 2 \frac{1}{2} 2 1 0.5 0.5 0.5 50 % 50\% 50% 1 8 \frac{1}{8} 8 1 0.125 0.125 0.125 12.5 % 12.5\% 12.5%
1 3 \frac{1}{3} 3 1 0. 3 ‾ 0.\overline{3} 0. 3 33. 3 ‾ % 33.\overline{3}\% 33. 3 % 3 8 \frac{3}{8} 8 3 0.375 0.375 0.375 37.5 % 37.5\% 37.5%
2 3 \frac{2}{3} 3 2 0. 6 ‾ 0.\overline{6} 0. 6 66. 6 ‾ % 66.\overline{6}\% 66. 6 % 5 8 \frac{5}{8} 8 5 0.625 0.625 0.625 62.5 % 62.5\% 62.5%
1 4 \frac{1}{4} 4 1 0.25 0.25 0.25 25 % 25\% 25% 7 8 \frac{7}{8} 8 7 0.875 0.875 0.875 87.5 % 87.5\% 87.5%
3 4 \frac{3}{4} 4 3 0.75 0.75 0.75 75 % 75\% 75% 1 6 \frac{1}{6} 6 1 0.1 6 ‾ 0.1\overline{6} 0.1 6 16. 6 ‾ % 16.\overline{6}\% 16. 6 %
1 5 \frac{1}{5} 5 1 0.2 0.2 0.2 20 % 20\% 20% 5 6 \frac{5}{6} 6 5 0.8 3 ‾ 0.8\overline{3} 0.8 3 83. 3 ‾ % 83.\overline{3}\% 83. 3 %
- 2 5 \frac{2}{5} 5 2 0.4 0.4 0.4 40 % 40\% 40% 1 10 \frac{1}{10} 10 1 0.1 0.1 0.1 10 % 10\% 10%
- 3 5 \frac{3}{5} 5 3 0.6 0.6 0.6 60 % 60\% 60% 1 20 \frac{1}{20} 20 1 0.05 0.05 0.05 5 % 5\% 5%
- 4 5 \frac{4}{5} 5 4 0.8 0.8 0.8 80 % 80\% 80% 1 50 \frac{1}{50} 50 1 0.02 0.02 0.02 2 % 2\% 2%
- Lowest Common Denominator (LCD) & Equivalent Fractions
To add or subtract fractions with different denominators, you must convert them to equivalent fractions sharing the Least Common Denominator (LCD) , which is the Least Common Multiple (LCM) of the denominators.
Finding the LCD via Prime Factorization
Find the prime factorization of each denominator.
For every distinct prime factor present, take the highest power that appears in any single factorization.
Multiply these highest powers together.
- Example: Find the LCD of 5 24 \frac{5}{24} 24 5 and 7 36 \frac{7}{36} 36 7
- 24 = 2 3 ⋅ 3 1 24 = 2^3 \cdot 3^1 24 = 2 3 ⋅ 3 1
- 36 = 2 2 ⋅ 3 2 36 = 2^2 \cdot 3^2 36 = 2 2 ⋅ 3 2
- Highest power of 2 2 2 : 2 3 = 8 2^3 = 8 2 3 = 8
- Highest power of 3 3 3 : 3 2 = 9 3^2 = 9 3 2 = 9
- LCD = 2 3 ⋅ 3 2 = 8 ⋅ 9 = 72 \text{LCD} = 2^3 \cdot 3^2 = 8 \cdot 9 = 72 LCD = 2 3 ⋅ 3 2 = 8 ⋅ 9 = 72
Convert to equivalent fractions: 5 24 = 5 ⋅ 3 24 ⋅ 3 = 15 72 , 7 36 = 7 ⋅ 2 36 ⋅ 2 = 14 72 \frac{5}{24} = \frac{5 \cdot 3}{24 \cdot 3} = \frac{15}{72}, \qquad \frac{7}{36} = \frac{7 \cdot 2}{36 \cdot 2} = \frac{14}{72} 24 5 = 24 ⋅ 3 5 ⋅ 3 = 72 15 , 36 7 = 36 ⋅ 2 7 ⋅ 2 = 72 14
The Four Fundamental Operations with Fractions
1. Addition and Subtraction a d + b d = a + b d , a d − b d = a − b d \frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}, \qquad \frac{a}{d} - \frac{b}{d} = \frac{a - b}{d} d a + d b = d a + b , d a − d b = d a − b
Regrouping (Borrowing) in Mixed Number Subtraction :
When subtracting mixed numbers where the fraction of the subtrahend is larger than that of the minuend, borrow 1 1 1 from the whole number part: 5 1 6 − 2 5 6 = ( 4 + 1 1 6 ) − 2 5 6 = 4 7 6 − 2 5 6 = ( 4 − 2 ) + 7 − 5 6 = 2 2 6 = 2 1 3 5\frac{1}{6} - 2\frac{5}{6} = \left(4 + 1\frac{1}{6}\right) - 2\frac{5}{6} = 4\frac{7}{6} - 2\frac{5}{6} = (4 - 2) + \frac{7 - 5}{6} = 2\frac{2}{6} = 2\frac{1}{3} 5 6 1 − 2 6 5 = ( 4 + 1 6 1 ) − 2 6 5 = 4 6 7 − 2 6 5 = ( 4 − 2 ) + 6 7 − 5 = 2 6 2 = 2 3 1
2. Multiplication & Cross-Simplification
Multiply numerators together and denominators together: a b ⋅ c d = a ⋅ c b ⋅ d \frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d} b a ⋅ d c = b ⋅ d a ⋅ c
Cross-Simplification : Always cancel common factors between any numerator and any denominator before multiplying. This avoids massive numbers and simplifies arithmetic. 14 25 ⋅ 15 28 = 14 1 25 5 ⋅ 15 3 28 2 = 1 ⋅ 3 5 ⋅ 2 = 3 10 \frac{14}{25} \cdot \frac{15}{28} = \frac{\cancel{14}^1}{\cancel{25}_5} \cdot \frac{\cancel{15}^3}{\cancel{28}_2} = \frac{1 \cdot 3}{5 \cdot 2} = \frac{3}{10} 25 14 ⋅ 28 15 = 25 5 14 1 ⋅ 28 2 15 3 = 5 ⋅ 2 1 ⋅ 3 = 10 3
Rule for Mixed Numbers : Never multiply whole numbers and fractions separately. Always convert mixed numbers to improper fractions prior to multiplication: 2 2 3 ⋅ 1 7 8 = 8 3 ⋅ 15 8 = 8 1 3 1 ⋅ 15 5 8 1 = 5 1 = 5 2\frac{2}{3} \cdot 1\frac{7}{8} = \frac{8}{3} \cdot \frac{15}{8} = \frac{\cancel{8}^1}{\cancel{3}_1} \cdot \frac{\cancel{15}^5}{\cancel{8}_1} = \frac{5}{1} = 5 2 3 2 ⋅ 1 8 7 = 3 8 ⋅ 8 15 = 3 1 8 1 ⋅ 8 1 15 5 = 1 5 = 5
3. Division & The Reciprocal Rule ("Keep-Change-Flip")
Dividing by a fraction is algebraically equivalent to multiplying by its reciprocal: a b ÷ c d = a b ⋅ d c = a d b c ( c , d ≠ 0 ) \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc} \quad (c, d \neq 0) b a ÷ d c = b a ⋅ c d = b c a d ( c , d = 0 )
Keep the first fraction exactly as written.
Change the division symbol ( ÷ \div ÷ ) to multiplication ( ⋅ \cdot ⋅ ).
Flip the second fraction to its reciprocal ( c d → d c \frac{c}{d} \to \frac{d}{c} d c → c d ).
Complex Fractions
A complex fraction has a fraction in its numerator, denominator, or both: 3 4 9 10 = 3 4 ÷ 9 10 = 3 1 4 2 ⋅ 10 5 9 3 = 5 6 \frac{\frac{3}{4}}{\frac{9}{10}} = \frac{3}{4} \div \frac{9}{10} = \frac{\cancel{3}^1}{\cancel{4}_2} \cdot \frac{\cancel{10}^5}{\cancel{9}_3} = \frac{5}{6} 10 9 4 3 = 4 3 ÷ 10 9 = 4 2 3 1 ⋅ 9 3 10 5 = 6 5
- Step-by-Step Worked Examples
- Worked Example 1: Multi-Step Fraction Evaluation
Evaluate the following expression and write the answer as a simplified mixed number: 2 1 3 − 1 5 6 3 4 ⋅ 8 9 + 1 2 \frac{2\frac{1}{3} - 1\frac{5}{6}}{\frac{3}{4} \cdot \frac{8}{9} + \frac{1}{2}} 4 3 ⋅ 9 8 + 2 1 2 3 1 − 1 6 5
Step 1: Simplify the Numerator
Convert mixed numbers to improper fractions: 2 1 3 = 7 3 , 1 5 6 = 11 6 2\frac{1}{3} = \frac{7}{3}, \qquad 1\frac{5}{6} = \frac{11}{6} 2 3 1 = 3 7 , 1 6 5 = 6 11
Find common denominator (LCD = 6 = 6 = 6 ): 7 3 = 14 6 \frac{7}{3} = \frac{14}{6} 3 7 = 6 14
Subtract: Numerator = 14 6 − 11 6 = 3 6 = 1 2 \text{Numerator} = \frac{14}{6} - \frac{11}{6} = \frac{3}{6} = \frac{1}{2} Numerator = 6 14 − 6 11 = 6 3 = 2 1
Step 2: Simplify the Denominator
Evaluate multiplication first: 3 4 ⋅ 8 9 = 3 1 4 1 ⋅ 8 2 9 3 = 2 3 \frac{3}{4} \cdot \frac{8}{9} = \frac{\cancel{3}^1}{\cancel{4}_1} \cdot \frac{\cancel{8}^2}{\cancel{9}_3} = \frac{2}{3} 4 3 ⋅ 9 8 = 4 1 3 1 ⋅ 9 3 8 2 = 3 2
Add 1 2 \frac{1}{2} 2 1 (LCD of 3 3 3 and 2 2 2 is 6 6 6 ): 2 3 + 1 2 = 4 6 + 3 6 = 7 6 \frac{2}{3} + \frac{1}{2} = \frac{4}{6} + \frac{3}{6} = \frac{7}{6} 3 2 + 2 1 = 6 4 + 6 3 = 6 7 Denominator = 7 6 \text{Denominator} = \frac{7}{6} Denominator = 6 7
Step 3: Perform the Division Numerator Denominator = 1 2 7 6 = 1 2 ÷ 7 6 = 1 2 1 ⋅ 6 3 7 = 3 7 \frac{\text{Numerator}}{\text{Denominator}} = \frac{\frac{1}{2}}{\frac{7}{6}} = \frac{1}{2} \div \frac{7}{6} = \frac{1}{\cancel{2}_1} \cdot \frac{\cancel{6}^3}{7} = \frac{3}{7} Denominator Numerator = 6 7 2 1 = 2 1 ÷ 6 7 = 2 1 1 ⋅ 7 6 3 = 7 3
The simplified result is 3 7 \frac{3}{7} 7 3 .
Worked Example 2: Carpentry & Construction Measurement with Kerf Waste
A custom furniture maker has a single solid oak board measuring 8 feet 8\text{ feet} 8 feet in length ( 96 inches 96\text{ inches} 96 inches ). The builder needs to cut 5 5 5 identical shelving pieces, each measuring 14 3 8 inches 14\frac{3}{8}\text{ inches} 14 8 3 inches long. The table saw blade creates a "kerf" (material turned into sawdust) that removes exactly 1 8 inch \frac{1}{8}\text{ inch} 8 1 inch of length for each cut made between pieces. If all 5 5 5 shelves are cut consecutively from the single board using 4 4 4 cuts, how many inches of the oak board remain as leftover scrap?
Step 1: Calculate the total wood length required for the 5 shelves Total Shelving Length = 5 × 14 3 8 = 5 × 115 8 = 575 8 inches \text{Total Shelving Length} = 5 \times 14\frac{3}{8} = 5 \times \frac{115}{8} = \frac{575}{8}\text{ inches} Total Shelving Length = 5 × 14 8 3 = 5 × 8 115 = 8 575 inches
Convert to a mixed number: 575 8 = 71 7 8 inches \frac{575}{8} = 71\frac{7}{8}\text{ inches} 8 575 = 71 8 7 inches
Step 2: Calculate the total material lost to saw blade kerf Cutting 5 5 5 pieces from a continuous board requires 4 4 4 interior cuts: Total Kerf Waste = 4 × 1 8 = 4 8 = 1 2 inch \text{Total Kerf Waste} = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2}\text{ inch} Total Kerf Waste = 4 × 8 1 = 8 4 = 2 1 inch
Step 3: Calculate the total oak wood consumed Total Wood Consumed = 71 7 8 + 4 8 = 71 11 8 = 72 3 8 inches \text{Total Wood Consumed} = 71\frac{7}{8} + \frac{4}{8} = 71\frac{11}{8} = 72\frac{3}{8}\text{ inches} Total Wood Consumed = 71 8 7 + 8 4 = 71 8 11 = 72 8 3 inches
Step 4: Subtract total consumed wood from the initial 96-inch board 96 − 72 3 8 = 95 8 8 − 72 3 8 = ( 95 − 72 ) + ( 8 − 3 8 ) = 23 5 8 inches 96 - 72\frac{3}{8} = 95\frac{8}{8} - 72\frac{3}{8} = (95 - 72) + \left(\frac{8 - 3}{8}\right) = 23\frac{5}{8}\text{ inches} 96 − 72 8 3 = 95 8 8 − 72 8 3 = ( 95 − 72 ) + ( 8 8 − 3 ) = 23 8 5 inches
The leftover length of the oak board is 23 5 8 inches 23\frac{5}{8}\text{ inches} 23 8 5 inches .
Common Pitfalls & ACCUPLACER Exam Traps
Adding Denominators Directly : Writing 2 5 + 1 3 = 2 + 1 5 + 3 = 3 8 \frac{2}{5} + \frac{1}{3} = \frac{2+1}{5+3} = \frac{3}{8} 5 2 + 3 1 = 5 + 3 2 + 1 = 8 3 is completely false. You must find an LCD: 6 15 + 5 15 = 11 15 \frac{6}{15} + \frac{5}{15} = \frac{11}{15} 15 6 + 15 5 = 15 11 .
Forgetting to Invert the Divisor in Fraction Division : Students sometimes invert the dividend (first fraction) rather than the divisor (second fraction). Remember: a b ÷ c d = a b ⋅ d c \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} b a ÷ d c = b a ⋅ c d .
Multiplying Mixed Numbers Without Converting to Improper Fractions : Multiplying 3 1 2 × 2 1 4 3\frac{1}{2} \times 2\frac{1}{4} 3 2 1 × 2 4 1 as ( 3 × 2 ) + ( 1 2 × 1 4 ) = 6 1 8 (3 \times 2) + (\frac{1}{2} \times \frac{1}{4}) = 6\frac{1}{8} ( 3 × 2 ) + ( 2 1 × 4 1 ) = 6 8 1 ignores the cross-terms of the distributive property. The correct procedure is 7 2 × 9 4 = 63 8 = 7 7 8 \frac{7}{2} \times \frac{9}{4} = \frac{63}{8} = 7\frac{7}{8} 2 7 × 4 9 = 8 63 = 7 8 7 .
Kerf and Post Counts in Measurement Problems : In partitioning problems, N N N pieces require N − 1 N - 1 N − 1 cuts. Multiplying the kerf waste by 5 5 5 instead of 4 4 4 cuts leads to an off-by-one error.