6.1 Ratios, Proportions & Dimensional Analysis

6.1 Ratios, Proportions & Dimensional Analysis

Problem-Solving and Data Analysis questions make up approximately 15% (5 to 7 questions) of the Digital SAT Math section. At the core of this domain is proportional reasoning—the ability to model, convert, and scale quantities across diverse scientific, economic, and geometric contexts.

Mastery of this domain requires fluency in ratio algebra, multi-step dimensional analysis (converting rates and higher-dimensional units), and volumetric and population density calculations.

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1. Ratio Fundamentals: Part-to-Part vs. Part-to-Whole

A ratio is a mathematical comparison of two or more numerical quantities. Ratios can be expressed in three equivalent notations: $a:b$, $\frac{a}{b}$, or "$a$ to $b$".

+-----------------------------------------------------------------------------+
|                     PART-TO-PART VS. PART-TO-WHOLE RATIOS                   |
|                                                                             |
|   Example: A chemistry lab mixture contains 3 parts acid and 5 parts water. |
|                                                                             |
|   [ Part-to-Part Ratio ]           [ Part-to-Whole Ratio ]                  |
|   Acid : Water = 3 : 5             Acid : Total Mixture = 3 : (3 + 5) = 3:8 |
|   Water : Acid = 5 : 3             Water : Total Mixture = 5 : 8            |
|                                                                             |
|   Fraction of acid = 3/5 of water  Fraction of mixture that is acid = 3/8   |
+-----------------------------------------------------------------------------+

The Common Multiplier Method ($x$-Method)

When a ratio is given along with a total quantity or a difference between components, assign a common variable multiplier $x$ to each part of the ratio.

$$\text{Part } A = a x, \quad \text{Part } B = b x, \quad \text{Total} = (a + b)x$$

Worked Example 1: Three-Part Ratio Mixture

Problem: An alloy is created by combining copper, zinc, and tin in the ratio $5 : 3 : 2$ by weight. If a metallurgist needs to produce $140\text{ kilograms}$ of this alloy, how many kilograms of zinc are required?

Step-by-Step Solution:

1. Define the parts using multiplier $x$: $$\text{Copper} = 5x, \quad \text{Zinc} = 3x, \quad \text{Tin} = 2x$$
2. Set up the total weight equation: $$5x + 3x + 2x = 140 \implies 10x = 140 \implies x = 14$$
3. Calculate the required mass of zinc: $$\text{Zinc} = 3x = 3(14) = 42\text{ kg}$$

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2. Proportions and Cross-Multiplication Mechanics

A proportion is an equation stating that two rational ratios are equal:

$$\frac{a}{b} = \frac{c}{d} \iff a d = b c \quad (b, d \ne 0)$$

Direct vs. Inverse Proportionality

• Direct Proportion ($y = kx$): As $x$ increases, $y$ increases at a constant rate $k = \frac{y}{x}$.
• Inverse Proportion ($y = \frac{k}{x}$ or $xy = k$): As $x$ increases, $y$ decreases such that their product remains constant.

Worked Example 2: Inverse Proportionality

Problem: A construction crew of 6 workers can complete a roofing installation in 14 hours. Assuming all workers work at the same constant rate, how many total hours would it take a crew of 8 workers to complete the same installation?

Step-by-Step Solution:

1. Identify the relationship: The number of workers $w$ and time $t$ are inversely proportional. The total labor in worker-hours is constant: $$k = w_1 \cdot t_1 = 6 \times 14 = 84\text{ worker-hours}$$
2. Solve for the new time $t_2$ with $w_2 = 8$: $$w_2 \cdot t_2 = k \implies 8 \cdot t_2 = 84 \implies t_2 = \frac{84}{8} = 10.5\text{ hours}$$

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3. Unit Rates and Linear Slopes

A unit rate expresses how many units of the numerator correspond to exactly 1 unit of the denominator. Common unit rates include:

• Speed: $\frac{\text{miles}}{\text{hour}}$ or $\frac{\text{meters}}{\text{second}}$
• Unit Pricing: $\frac{\text{dollars}}{\text{pound}}$ or $\frac{\text{cents}}{\text{fluid ounce}}$
• Flow Rate: $\frac{\text{gallons}}{\text{minute}}$ or $\frac{\text{liters}}{\text{second}}$

In algebraic models of the form $y = mx + b$, the slope $m$ is the unit rate of change:

$$m = \frac{\Delta y}{\Delta x} = \frac{\text{Change in Output}}{\text{Change in Input}}$$

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4. Dimensional Analysis (The Unit-Fraction Method)

Dimensional analysis is a systematic method of unit conversion that treats measurement units algebraically, allowing them to cancel out when positioned in opposing numerators and denominators.

+-----------------------------------------------------------------------------+
|                     DIMENSIONAL ANALYSIS UNIT CANCELLATION                  |
|                                                                             |
|   Given Quantity       Conversion Factor 1       Conversion Factor 2        |
|                                                                             |
|   [ Unit A ]        [   Unit B   ]            [   Unit C   ]                |
|   ----------   x    --------------       x    --------------  =  [ Unit C ] |
|       1             [   Unit A   ]            [   Unit B   ]                |
|                           ^                         ^                       |
|               (Unit A cancels out)      (Unit B cancels out)                |
+-----------------------------------------------------------------------------+

Step-by-Step Multi-Unit Conversion Workflow

1. Identify the Given Rate: Write the initial rate as a fraction with explicit units.
2. Construct Unit Fractions: Form conversion fractions equal to $1$ (e.g., $\frac{1\text{ mi}}{5{,}280\text{ ft}} = 1$) such that unwanted units appear on the opposite side of the fraction bar.
3. Cancel Units and Multiply: Cross out matching units and compute the numerical product.

Worked Example 3: Converting Velocity (ft/s to mph)

Problem: An experimental drone cruises at a constant speed of $88\text{ feet per second}$. Given that $1\text{ mile} = 5{,}280\text{ feet}$ and $1\text{ hour} = 3{,}600\text{ seconds}$, what is the drone's speed in miles per hour?

Step-by-Step Solution:

1. Set up the dimensional analysis expression: $$\frac{88\text{ ft}}{1\text{ s}} \times \frac{1\text{ mile}}{5{,}280\text{ ft}} \times \frac{3{,}600\text{ s}}{1\text{ hour}}$$
2. Cancel units algebraically: $$\frac{88\ \cancel{\text{ft}}}{1\ \cancel{\text{s}}} \times \frac{1\text{ mile}}{5{,}280\ \cancel{\text{ft}}} \times \frac{3{,}600\ \cancel{\text{s}}}{1\text{ hour}} = \frac{88 \times 3{,}600}{5{,}280}\ \frac{\text{miles}}{\text{hour}}$$
3. Calculate the numerical value: $$\frac{316{,}800}{5{,}280} = 60\text{ miles per hour}$$

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5. Multi-Dimensional Conversions (Area & Volume Rules)

A common source of error on the SAT occurs when converting units of area (squared units) or volume (cubed units). Conversion factors must be raised to the corresponding power.

+-----------------------------------------------------------------------------+
|                 SQUARE AND CUBE UNIT CONVERSION EXPONENTS                   |
|                                                                             |
|   Linear:     1 yd  = 3 ft                                                  |
|                                                                             |
|   Area:       1 yd² = (3 ft)²  = 3² ft²  =  9 ft²   (NOT 3 ft²!)            |
|                                                                             |
|   Volume:     1 yd³ = (3 ft)³  = 3³ ft³  = 27 ft³   (NOT 3 ft³!)            |
|                                                                             |
|   Metric:     1 m   = 100 cm                                                |
|               1 m²  = (100 cm)² = 10,000 cm²                                |
|               1 m³  = (100 cm)³ = 1,000,000 cm³                             |
+-----------------------------------------------------------------------------+

Worked Example 4: Area Conversion Trap

Problem: A solar panel array covers an area of $12\text{ square meters}$. How many square centimeters ($\text{cm}^2$) does this array cover?

Step-by-Step Solution:

1. Linear equivalence: $1\text{ m} = 100\text{ cm}$.
2. Square both sides: $$1\text{ m}^2 = (100\text{ cm})^2 = 10{,}000\text{ cm}^2$$
3. Multiply by the area: $$12\text{ m}^2 \times \frac{10{,}000\text{ cm}^2}{1\text{ m}^2} = 120{,}000\text{ cm}^2$$

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6. Volumetric and Population Density Models

Density is a ratio describing how tightly packed a quantity is within a spatial boundary.

1. Volumetric Density (Mass per Unit Volume)

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Mass} = \text{Density} \times \text{Volume}$$ $$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$

2. Population Density (Count per Unit Area)

$$\text{Population Density} = \frac{\text{Population}}{\text{Land Area}} \implies \text{Population} = \text{Density} \times \text{Area}$$

Worked Example 5: Multi-Step Density and Geometry

Problem: A solid cylindrical bronze rod has a radius of $2\text{ cm}$, a length (height) of $15\text{ cm}$, and a mass of $1{,}601.4\text{ grams}$. Using $\pi \approx 3.14$, what is the density of the bronze in grams per cubic centimeter ($\text{g/cm}^3$)?

Step-by-Step Solution:

1. Calculate the volume of the cylinder: $$V = \pi r^2 h = 3.14 \times (2)^2 \times 15 = 3.14 \times 4 \times 15 = 3.14 \times 60 = 188.4\text{ cm}^3$$
2. Apply the volumetric density formula: $$\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{1{,}601.4\text{ g}}{188.4\text{ cm}^3} = 8.5\text{ g/cm}^3$$