SAT Math Proportional Relationships

Direct, inverse, and constant-rate word problems

Proportional relationships are an overlay on other topics — proportionality shows up in linear functions, percentages, and geometry. But the SAT also tests the concept itself: distinguishing direct ( $y = k x$ ) from inverse ( $y = \frac{k}{x}$ ) and applying it to rate, work, or mixture problems.

Problems here are usually word problems — "a car going 60 km/h", "4 workers finish a job in 6 days", "two solutions mixed in a 2:3 ratio". The key is recognizing the relationship type and picking the right formula.

What the SAT actually tests

Direct proportionality: $y = k x$ with constant $k$
Inverse proportionality: $y = \frac{k}{x}$ — as $x$ grows, $y$ shrinks
Rate problems: speed, work, density, unit cost
Mixtures: two solutions of different concentrations
Scale (maps, models): proportionally scaled objects

Key concepts

Direct proportionality

$y = k x$ . When $x$ doubles, $y$ doubles. Graph: a line through the origin.

Inverse proportionality

$y = \frac{k}{x}$ . When $x$ doubles, $y$ halves. Classic: work time is inversely proportional to number of workers.

Rate

Speed = distance / time. Work rate = work / time. All built on direct proportionality.

Cross-multiplication

When $\frac{a}{b} = \frac{c}{d}$ , then $a \cdot d = b \cdot c$ . The fastest method for proportion equations.

Worked examples

Example 1

A car travels 240 km in 4 hours. How far will it travel in 7 hours at the same speed?

Solution

Speed: $240/4 = 60$ km/h. Over 7 hours: $60 \cdot 7 = 420$ km. (Or cross-multiply: $\frac{240}{4} = \frac{x}{7}$ , so $x = 420$ .)

💡 Compute the per-unit rate first (km/h), then multiply. Universal technique.

Example 2

4 workers finish a project in 12 days. How long will it take 6 workers at the same pace?

Solution

This is INVERSE proportionality — more workers = less time. Total work: $4 \cdot 12 = 48$ worker-days. For 6 workers: $48/6 = 8$ days.

💡 For team-work problems, compute "worker-days" as the constant, then divide.

Common pitfalls

Confusing direct with inverse. "More workers = less time" is inverse, NOT direct.
Using direct proportion on team-work problems. "4 workers → 12 days, so 6 workers → $\frac{6}{4} \cdot 12 = 18$ days" — wrong.
Dropping units. A result of "420 m/s" in a car problem should raise a flag.
Mixture problems — weighting volume instead of concentration.

Exam strategy

Identify the relationship type first: when one quantity grows, does the other grow or shrink? Grows together → direct ( $y = k x$ ). Grows oppositely → inverse ( $y = \frac{k}{x}$ ). For rate problems, compute the per-unit value (km/h, L/min) and multiply by target units. For worker problems, compute "worker-days" as a constant and divide by the new worker count.

Frequently asked questions

Direct vs inverse proportion — what's the difference?

Direct: $y = k x$ . Double $x$ , double $y$ . Inverse: $y = k / x$ . Double $x$ , halve $y$ . The product $x \cdot y$ is constant in inverse.

How do I solve rate problems?

Use $distance = rate \cdot time$ . Convert each piece to the same units. Example: 60 km/h × 0.5 h = 30 km.

How do I solve worker problems?

Compute "work" as workers × days — a constant for the job. If 5 workers finish in 8 days, work = 40 worker-days. 10 workers finish in 40/10 = 4 days.

How do I recognize proportion from a graph?

Direct: straight line through $(0, 0)$ . Inverse: hyperbola — curve in quadrants I and III, asymptotic to the axes.

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