Study guide
2.2 Proportions & Direct/Inverse Variation
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
2.2 Proportions & Direct/Inverse Variation
Key Takeaways Explain This Topic AI
A proportion states that two ratios are equal (a/b = c/d); it is solved algebraically via the Cross-Multiplication Property (a x d = b x c).
Direct variation follows the linear model y = kx (or y/x = k), representing relationships that always graph as straight lines passing through the origin (0, 0).
Inverse variation follows the hyperbolic model y = k/x (or xy = k), where the product of the two variables remains constant as one variable increases while the other decreases proportionally.
Collaborative work-rate scenarios combine individual reciprocal rates (rate = 1 / time) using the formula 1/t1 + 1/t2 = 1/t_together.
Joint and combined variation models (z = kxy or z = kx/y) are solved by first computing the invariant constant k from baseline conditions.
Last updated: August 2026
Defining and Solving Proportions
A proportion is a mathematical statement establishing that two ratios are strictly equal. It is expressed in the standard fractional form: a b = c d ( b ≠ 0 , d ≠ 0 ) \frac{a}{b} = \frac{c}{d} \quad (b \ne 0, \; d \ne 0) b a = d c ( b = 0 , d = 0 )
In classical terminology, a a a and d d d are called the extremes , while b b b and c c c are called the means .
The Cross-Multiplication Property
The fundamental theorem of proportions states that for any valid proportion, the product of the extremes equals the product of the means: a b = c d ⟺ a ⋅ d = b ⋅ c \frac{a}{b} = \frac{c}{d} \iff a \cdot d = b \cdot c b a = d c ⟺ a ⋅ d = b ⋅ c
This property allows for the immediate conversion of rational equations into linear or polynomial equations.
Solving Linear Proportions with Binomial Expressions
On the ACCUPLACER test, proportions frequently involve algebraic binomials in the numerators or denominators. When applying cross-multiplication, always enclose binomial terms in parentheses to distribute coefficients correctly. 2 x + 5 3 = 5 x − 2 4 \frac{2x + 5}{3} = \frac{5x - 2}{4} 3 2 x + 5 = 4 5 x − 2 4 ( 2 x + 5 ) = 3 ( 5 x − 2 ) 4(2x + 5) = 3(5x - 2) 4 ( 2 x + 5 ) = 3 ( 5 x − 2 ) 8 x + 20 = 15 x − 6 8x + 20 = 15x - 6 8 x + 20 = 15 x − 6 20 + 6 = 15 x − 8 x 20 + 6 = 15x - 8x 20 + 6 = 15 x − 8 x 26 = 7 x ⟹ x = 26 7 26 = 7x \implies x = \frac{26}{7} 26 = 7 x ⟹ x = 7 26
Worked Example 1: Scale Factor & Cartography
On an architectural blueprint for a commercial medical clinic, a scale of 0.75 inches = 12 feet 0.75\text{ inches} = 12\text{ feet} 0.75 inches = 12 feet is used. If the actual length of the surgical recovery ward is 68 feet 68\text{ feet} 68 feet , what is the corresponding length on the blueprint in inches?
Set up a proportion matching blueprint units to real-world units: 0.75 in 12 ft = x in 68 ft \frac{0.75\text{ in}}{12\text{ ft}} = \frac{x\text{ in}}{68\text{ ft}} 12 ft 0.75 in = 68 ft x in
Apply cross-multiplication: 12 ⋅ x = 0.75 ⋅ 68 12 \cdot x = 0.75 \cdot 68 12 ⋅ x = 0.75 ⋅ 68 12 x = 51 12x = 51 12 x = 51
Solve for x x x : x = 51 12 = 4.25 inches ( 4 1 4 inches ) x = \frac{51}{12} = 4.25\text{ inches} \;\left(4\frac{1}{4}\text{ inches}\right) x = 12 51 = 4.25 inches ( 4 4 1 inches )
Direct Variation ( y = k x y = kx y = k x )
Two variables x x x and y y y exhibit direct variation (or are said to be directly proportional ) if a constant ratio exists between them for all non-zero values.
Mathematical Formulation y = k x or y x = k y = kx \quad \text{or} \quad \frac{y}{x} = k y = k x or x y = k
where k k k is a non-zero constant called the constant of proportionality (or constant of variation).
Geometric and Functional Characteristics
Graph : The graph of a direct variation relationship is a straight line passing directly through the origin ( 0 , 0 ) (0, 0) ( 0 , 0 ) with slope equal to k k k .
Behavior : If x x x is multiplied by a factor c c c , y y y is multiplied by the exact same factor c c c . If x x x doubles, y y y doubles; if x x x is halved, y y y is halved.
Proportional Equivalence : For any two paired data points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) : y 1 x 1 = y 2 x 2 \frac{y_1}{x_1} = \frac{y_2}{x_2} x 1 y 1 = x 2 y 2
(Exam Warning: The linear equation y = 3 x + 4 y = 3x + 4 y = 3 x + 4 is NOT direct variation because when x = 0 x = 0 x = 0 , y = 4 ≠ 0 y = 4 \ne 0 y = 4 = 0 . A true direct variation line must have a y y y -intercept of 0 0 0 .)
Worked Example 2: Direct Variation in Physics
Under Hooke's Law, the distance d d d that an elastic spring stretches varies directly with the force F F F applied to it. When a force of 45 Newtons 45\text{ Newtons} 45 Newtons is applied, the spring stretches 7.2 centimeters 7.2\text{ centimeters} 7.2 centimeters . How far will the spring stretch when a force of 75 Newtons 75\text{ Newtons} 75 Newtons is applied?
Find the constant of proportionality k k k : k = d 1 F 1 = 7.2 cm 45 N = 0.16 cm/N k = \frac{d_1}{F_1} = \frac{7.2\text{ cm}}{45\text{ N}} = 0.16\text{ cm/N} k = F 1 d 1 = 45 N 7.2 cm = 0.16 cm/N
Write the specific direct variation equation: d = 0.16 F d = 0.16 F d = 0.16 F
Substitute the new force F = 75 N F = 75\text{ N} F = 75 N : d = 0.16 × 75 = 12.0 centimeters d = 0.16 \times 75 = 12.0\text{ centimeters} d = 0.16 × 75 = 12.0 centimeters
Alternatively, solve directly via proportion: 7.2 45 = d 2 75 ⟹ 45 d 2 = 7.2 × 75 = 540 ⟹ d 2 = 540 45 = 12.0 cm \frac{7.2}{45} = \frac{d_2}{75} \implies 45 d_2 = 7.2 \times 75 = 540 \implies d_2 = \frac{540}{45} = 12.0\text{ cm} 45 7.2 = 75 d 2 ⟹ 45 d 2 = 7.2 × 75 = 540 ⟹ d 2 = 45 540 = 12.0 cm
Inverse Variation ( y = k x y = \frac{k}{x} y = x k or x ⋅ y = k x \cdot y = k x ⋅ y = k )
Two variables x x x and y y y exhibit inverse variation (or are said to be inversely proportional ) if their mathematical product remains constant for all values.
Mathematical Formulation y = k x or x ⋅ y = k y = \frac{k}{x} \quad \text{or} \quad x \cdot y = k y = x k or x ⋅ y = k
where k k k is the non-zero constant of variation.
Geometric and Functional Characteristics
Graph : The graph of an inverse variation relationship is a rectangular hyperbola asymptotic to the x x x -axis and y y y -axis (lying in Quadrants I and III if k > 0 k > 0 k > 0 ).
Reciprocal Scaling : If x x x is multiplied by a constant factor c c c , y y y is divided by c c c (multiplied by 1 c \frac{1}{c} c 1 ). If x x x triples, y y y decreases to 1 3 \frac{1}{3} 3 1 of its original value.
Product Invariance Equation : For any two paired data points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) : x 1 ⋅ y 1 = x 2 ⋅ y 2 x_1 \cdot y_1 = x_2 \cdot y_2 x 1 ⋅ y 1 = x 2 ⋅ y 2
Real-World Applications of Inverse Variation
Speed vs. Travel Time : For a fixed distance D D D , speed r r r and time t t t vary inversely: r ⋅ t = D r \cdot t = D r ⋅ t = D .
Workforce vs. Project Duration : For a fixed total workload, number of workers W W W and completion hours H H H vary inversely: W ⋅ H = Constant Worker-Hours W \cdot H = \text{Constant Worker-Hours} W ⋅ H = Constant Worker-Hours .
Boyle's Law in Chemistry : Pressure P P P and volume V V V of an ideal gas at constant temperature satisfy P ⋅ V = k P \cdot V = k P ⋅ V = k .
Gear and Pulley Mechanics : Gear teeth count N N N and rotational speed S S S satisfy N 1 S 1 = N 2 S 2 N_1 S_1 = N_2 S_2 N 1 S 1 = N 2 S 2 .
Worked Example 3: Workforce Scheduling (Worker-Hours)
A landscaping company determines that a crew of 6 6 6 technicians can clear and sod a corporate park in 14 hours 14\text{ hours} 14 hours . If the client requires the project to be completed in exactly 8 hours 8\text{ hours} 8 hours , how many technicians must be assigned to the crew, assuming all technicians work at the same constant rate?
Identify the constant of variation k k k (total required worker-hours): k = W 1 ⋅ H 1 = 6 technicians × 14 hours = 84 worker-hours k = W_1 \cdot H_1 = 6\text{ technicians} \times 14\text{ hours} = 84\text{ worker-hours} k = W 1 ⋅ H 1 = 6 technicians × 14 hours = 84 worker-hours
Set up the product invariance equation with new time H 2 = 8 hours H_2 = 8\text{ hours} H 2 = 8 hours : W 2 ⋅ H 2 = 84 W_2 \cdot H_2 = 84 W 2 ⋅ H 2 = 84 W 2 ⋅ 8 = 84 W_2 \cdot 8 = 84 W 2 ⋅ 8 = 84
Solve for W 2 W_2 W 2 : W 2 = 84 8 = 10.5 technicians W_2 = \frac{84}{8} = 10.5\text{ technicians} W 2 = 8 84 = 10.5 technicians
Interpret practical constraints:
Since partial technicians cannot be hired, the company must deploy at least 11 11 11 technicians to finish within the 8-hour deadline (or exactly 10.5 10.5 10.5 if fractional labor-hours are modeled).
Comparison of Direct vs. Inverse Variation
- Attribute Direct Variation Inverse Variation
- Core Formula y = k x y = kx y = k x y = k x y = \frac{k}{x} y = x k or x y = k xy = k x y = k
- Constant k k k k = y x k = \frac{y}{x} k = x y k = x ⋅ y k = x \cdot y k = x ⋅ y
Two-Point Relation y 1 x 1 = y 2 x 2 \frac{y_1}{x_1} = \frac{y_2}{x_2} x 1 y 1 = x 2 y 2 x 1 y 1 = x 2 y 2 x_1 y_1 = x_2 y_2 x 1 y 1 = x 2 y 2
- Graphical Form Straight line through ( 0 , 0 ) (0, 0) ( 0 , 0 ) Hyperbola with axes as asymptotes
- Effect of Doubling x x x y y y doubles ( 2 y 2y 2 y ) y y y is halved ( 1 2 y \frac{1}{2}y 2 1 y )
- Classic Example Total cost vs. quantity purchased Vehicle speed vs. travel time
- Joint and Combined Variation
Many physical and economic models involve three or more variables simultaneously.
Definitions and Formulas
Joint Variation ( z z z varies jointly as x x x and y y y ) : z = k ⋅ x ⋅ y z = k \cdot x \cdot y z = k ⋅ x ⋅ y Example : The volume of a cylinder V = π r 2 h V = \pi r^2 h V = π r 2 h varies jointly as h h h and the square of r r r .
Combined Variation ( z z z varies directly as x x x and inversely as y y y ) : z = k ⋅ x y z = \frac{k \cdot x}{y} z = y k ⋅ x Example : Electrical resistance R = ρ L A R = \frac{\rho L}{A} R = A ρ L varies directly with length L L L and inversely with cross-sectional area A A A .
Systematic Two-Step Solving Procedure
Step 1 : Substitute initial baseline data ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) ( x 0 , y 0 , z 0 ) into the model to solve explicitly for the constant k k k .
Step 2 : Re-write the general equation with the calculated numerical value of k k k , then substitute the new given variables to solve for the target unknown.
Collaborative Work-Rate Problems
Work-rate problems represent a high-yield application of rational proportions on the ACCUPLACER test. In collaborative work problems, each entity's rate is expressed as the fraction of the total job completed per unit of time. Work Done = Rate × Time ⟹ Rate = 1 Individual Time t \text{Work Done} = \text{Rate} \times \text{Time} \implies \text{Rate} = \frac{1}{\text{Individual Time } t} Work Done = Rate × Time ⟹ Rate = Individual Time t 1
The Combined Rate Formula
When multiple entities work together simultaneously without interfering with one another, their individual working rates are additive: Rate combined = Rate 1 + Rate 2 + ⋯ + Rate n \text{Rate}_{\text{combined}} = \text{Rate}_1 + \text{Rate}_2 + \dots + \text{Rate}_n Rate combined = Rate 1 + Rate 2 + ⋯ + Rate n 1 t together = 1 t 1 + 1 t 2 \frac{1}{t_{\text{together}}} = \frac{1}{t_1} + \frac{1}{t_2} t together 1 = t 1 1 + t 2 1
Solving for t together t_{\text{together}} t together yields the Product-over-Sum Shortcut for two workers: t together = t 1 ⋅ t 2 t 1 + t 2 t_{\text{together}} = \frac{t_1 \cdot t_2}{t_1 + t_2} t together = t 1 + t 2 t 1 ⋅ t 2
Worked Example 4: Collaborative Drainage Pipes
A commercial storage reservoir has two drainage pipes. Pipe A can completely empty the full reservoir in 6 hours 6\text{ hours} 6 hours when operating alone. Pipe B can empty the reservoir in 9 hours 9\text{ hours} 9 hours when operating alone. If both pipes are opened simultaneously, how long will it take to drain the entire reservoir?
Express individual rates per hour: Rate A = 1 6 of reservoir/hr , Rate B = 1 9 of reservoir/hr \text{Rate}_A = \frac{1}{6}\text{ of reservoir/hr}, \quad \text{Rate}_B = \frac{1}{9}\text{ of reservoir/hr} Rate A = 6 1 of reservoir/hr , Rate B = 9 1 of reservoir/hr
Sum rates to find combined rate: Rate together = 1 6 + 1 9 = 3 18 + 2 18 = 5 18 of reservoir/hr \text{Rate}_{\text{together}} = \frac{1}{6} + \frac{1}{9} = \frac{3}{18} + \frac{2}{18} = \frac{5}{18}\text{ of reservoir/hr} Rate together = 6 1 + 9 1 = 18 3 + 18 2 = 18 5 of reservoir/hr
Invert the combined rate to find total time: t together = 18 5 hours = 3.6 hours = 3 hours and 36 minutes t_{\text{together}} = \frac{18}{5}\text{ hours} = 3.6\text{ hours} = 3\text{ hours and } 36\text{ minutes} t together = 5 18 hours = 3.6 hours = 3 hours and 36 minutes
Common Pitfalls & ACCUPLACER Exam Traps
Applying Direct Variation Logic to Inverse Problems : If 4 4 4 machines build an order in 12 hours 12\text{ hours} 12 hours , 8 8 8 machines will NOT take 24 hours 24\text{ hours} 24 hours . More machines take LESS time: 4 × 12 = 48 ⟹ 8 × t = 48 ⟹ t = 6 hours 4 \times 12 = 48 \implies 8 \times t = 48 \implies t = 6\text{ hours} 4 × 12 = 48 ⟹ 8 × t = 48 ⟹ t = 6 hours .
Averaging Speeds Directly (Harmonic Mean Trap) : If a delivery truck drives to a destination at 40 mph 40\text{ mph} 40 mph and returns along the identical route at 60 mph 60\text{ mph} 60 mph , the average speed for the round trip is NOT 40 + 60 2 = 50 mph \frac{40 + 60}{2} = 50\text{ mph} 2 40 + 60 = 50 mph . Because more time is spent at the slower speed, the true average speed is Total Distance Total Time = 2 D D 40 + D 60 = 2 5 120 = 240 5 = 48 mph \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2D}{\frac{D}{40} + \frac{D}{60}} = \frac{2}{\frac{5}{120}} = \frac{240}{5} = 48\text{ mph} Total Time Total Distance = 40 D + 60 D 2 D = 120 5 2 = 5 240 = 48 mph .
Omitting Parentheses During Cross-Multiplication : In x + 2 4 = x − 1 3 \frac{x+2}{4} = \frac{x-1}{3} 4 x + 2 = 3 x − 1 , forgetting parentheses leads to the common error 3 x + 2 = 4 x − 1 3x + 2 = 4x - 1 3 x + 2 = 4 x − 1 , rather than the correct 3 ( x + 2 ) = 4 ( x − 1 ) ⟹ 3 x + 6 = 4 x − 4 3(x + 2) = 4(x - 1) \implies 3x + 6 = 4x - 4 3 ( x + 2 ) = 4 ( x − 1 ) ⟹ 3 x + 6 = 4 x − 4 .
Linear Equations with Non-Zero Intercepts : Never classify y = m x + b y = mx + b y = m x + b ( b ≠ 0 b \ne 0 b = 0 ) as direct variation. While linear, it fails the proportionality requirement y x = constant \frac{y}{x} = \text{constant} x y = constant .