Constant Rate Problems — I can solve problems involving constant speed or constant density using unit rates. - Pre Algebra

SECTION 1

Historical Context & Motivation

Have you ever wondered how people figured out how fast a horse could run or how heavy a bag of gold was compared to its size? People have been working with rates — comparisons between two different measurements — for thousands of years. Understanding rates helped ancient traders, builders, and explorers solve everyday problems.

~250 BC • Archimedes and Density

The Greek mathematician Archimedes discovered how to measure density (mass per unit of volume). He used this idea to figure out if a king's crown was made of pure gold.

~1600 • Galileo Studies Speed

Galileo Galilei rolled balls down ramps and carefully measured the distance they traveled over time. He was one of the first scientists to study constant speed using math.

1687 • Newton's Laws of Motion

Isaac Newton published his famous laws of motion. These laws describe how objects move at constant or changing speeds. His work made the idea of a unit rate (like miles per hour) essential for science.

Today • Rates Are Everywhere

We use constant rates every day — tracking speed on a GPS, comparing prices at the grocery store, or measuring how much a material weighs per cubic inch.

The big question these thinkers were all trying to answer is: If something stays the same — the speed doesn't change, the material is uniform — how can we use one simple number to predict and solve problems? That simple number is a unit rate, and that's exactly what you'll learn in this lesson.

SECTION 2

Core Principles & Definitions

Before we dive into solving problems, let's make sure we understand the key ideas. A rate is a ratio that compares two quantities with different units. For example, "60 miles in 2 hours" is a rate. A unit rate is a special rate where the second quantity is exactly 1. So "30 miles per 1 hour" (or 30 mph) is a unit rate.

Rate

A comparison of two quantities with different units. Example: 120 miles in 3 hours, or 15 grams per 5 cubic centimeters.

Unit Rate

A rate simplified so the second quantity equals one. Example: 40 miles per hour, or 3 grams per cubic centimeter. You find it by dividing.

Constant Rate

When a rate never changes. A car going exactly 50 mph for the whole trip has a constant speed. A block of iron has the same density throughout.

Constant Speed

A specific type of constant rate: the same distance per unit of time, every single time period. Speed = distance ÷ time.

Constant Density

A specific type of constant rate: the same mass per unit of volume, throughout the whole object. Density = mass ÷ volume.

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KEY TAKEAWAY

Think of a unit rate like a recipe for one serving. If you know a smoothie recipe makes one serving, you can easily multiply to make two, five, or ten servings. A unit rate works the same way. Once you know the amount for "one" (one hour, one cubic centimeter), you can multiply to find any amount you need.

SECTION 3

Visual Explanation — Seeing Constant Rates

One of the best ways to understand a constant rate is to see it on a graph. When a rate is constant, the graph of the two quantities always makes a perfectly straight line that passes through the origin (the point 0, 0). The steeper the line, the bigger the unit rate.

This graph shows two cars traveling at different constant speeds. Car A (solid line) travels at 40 mph — notice how it covers more distance each hour. Car B (dashed line) travels at 25 mph — a less steep line. Both lines are straight because the speed is constant.

Notice something important: both lines start at the origin (0, 0). That makes sense because at time 0, neither car has traveled any distance yet. Also, the steeper line belongs to the faster car. The steepness of the line is actually the unit rate! This connection between graphs and rates will be useful throughout your math journey.

SECTION 4

Mathematical Framework

Constant rate problems always follow the same basic pattern. Two quantities are related by multiplication. Let's look at the two main formulas you'll use.

CONSTANT SPEED FORMULA

distance = speed × time or d = r × t

d = distance traveled (miles, kilometers, etc.) • r = rate of speed (the unit rate, like mph) • t = time (hours, seconds, etc.)

You can rearrange this formula to solve for any of the three quantities. If you know distance and time, divide to find speed: r = d ÷ t. If you know distance and speed, divide to find time: t = d ÷ r.

CONSTANT DENSITY FORMULA

mass = density × volume or m = D × V

m = mass (grams, kilograms, etc.) • D = density (the unit rate, like grams per cm³) • V = volume (cm³, liters, etc.)

Just like the speed formula, you can rearrange this one too. To find density: D = m ÷ V. To find volume: V = m ÷ D.

GENERAL CONSTANT RATE PATTERN

total amount = unit rate × number of units

This pattern works for any constant rate problem — cost per item, calories per serving, words per minute, and more.

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KEY TAKEAWAY

All constant rate formulas follow the exact same shape: total = rate × quantity. It's like filling bags with candy — if each bag holds the same amount (the rate), the total candy equals the number of bags times the amount per bag. Speed, density, and price problems all follow this one simple pattern.

SECTION 5

Comparing Speed and Density Problems

Speed and density problems look different on the surface, but they share the same structure underneath. Let's compare them side-by-side so you can see the pattern.

Side-by-side comparison of a speed problem and a density problem. Both follow the same steps: find the unit rate by dividing, then use multiplication to predict.

Feature	Speed	Density
"Total" quantity	Distance	Mass
"Per" quantity	Time	Volume
Unit rate	Speed (e.g., 60 mph)	Density (e.g., 8 g/cm³)
Formula	d = r × t	m = D × V

Speed vs. Density — Same structure, different labels

SECTION 6

Worked Example

Example 1: Constant Speed

A train travels at a constant speed. It covers 180 miles in 3 hours. How far will it travel in 7 hours?

Solving a Constant Speed Problem

1
Step 1 — Identify What You Know

The train travels 180 miles in 3 hours at a constant speed. We want to find the distance in 7 hours.
2
Step 2 — Find the Unit Rate

Divide the distance by the time to find the speed (the unit rate). speed = distance ÷ time = 180 ÷ 3 = 60

The unit rate is 60 miles per hour.
3
Step 3 — Use the Unit Rate to Solve

Now multiply the unit rate by the new time. distance = speed × time = 60 × 7 = 420

The train will travel 420 miles in 7 hours.

Example 2: Constant Density

A block of aluminum has a constant density. A piece with a volume of 4 cm³ has a mass of 10.8 grams. What is the mass of a piece with a volume of 15 cm³?

Solving a Constant Density Problem

1
Step 1 — Identify What You Know

Mass = 10.8 grams. Volume = 4 cm³. We want to find the mass when volume = 15 cm³.
2
Step 2 — Find the Unit Rate (Density)

Divide mass by volume to find the density. density = mass ÷ volume = 10.8 ÷ 4 = 2.7

The density is 2.7 grams per cm³.
3
Step 3 — Use the Unit Rate to Solve

Multiply the density by the new volume. mass = density × volume = 2.7 × 15 = 40.5

The piece of aluminum has a mass of 40.5 grams.

SECTION 7

Common Strengths & Pitfalls

Now that you've seen how these problems work, let's talk about what makes them easy — and where students often get tripped up.

Strengths of Unit Rate Approach	Common Pitfalls to Avoid
Once you find the unit rate, you can solve for any value — big or small.	Dividing in the wrong order: always divide the "total" by the "per" quantity (e.g., miles ÷ hours, not hours ÷ miles).
The same pattern works for speed, density, price, and many other topics.	Forgetting to check that the rate is constant. If a car speeds up and slows down, you can't use one unit rate for the whole trip.
You can check your answer easily: multiply your unit rate by the original denominator — you should get the original total back.	Mixing up units. Always label your answer (miles, grams, hours, cm³) so you know what your number means.
It connects naturally to graphs — the unit rate is the slope of the line.	Rounding too early. When a unit rate is a decimal, keep the full number until the very end.

Strengths vs. Pitfalls when using unit rates

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AVOID THIS COMMON MISTAKE

Always ask yourself: "What am I measuring per ONE of what?" For speed, it's distance per one hour (or one second). For density, it's mass per one cm³ (or one liter). If you set up the division with "per one" on the bottom, you'll always get the right unit rate.

SECTION 8

Connection to Advanced Topics

The ideas you're learning now are the foundation for lots of exciting math coming your way. Here's how constant rate thinking connects to bigger topics.

What You Know Now	What's Coming Next
Unit rates as a single number (like 40 mph)	In algebra, this becomes the slope of a line in the equation y = mx.
Constant rate graphs are straight lines through the origin	These are called proportional relationships. You'll study them deeply in 7th and 8th grade.
d = r × t (constant speed)	In physics, you'll add acceleration and study changing speed with the equation d = ½at².
m = D × V (constant density)	In chemistry, density helps you identify unknown substances and understand buoyancy (why things float or sink).

From constant rates to advanced math and science

The key idea is that mastering unit rates now gives you a head start. When you see y = mx in algebra, you'll already understand what the m (the slope) really means — it's just a unit rate! You're building a skill that keeps growing with you.

SECTION 9

Practice Problems

Try these five problems on your own. They start simple and get more challenging. Remember: find the unit rate first, then use it!

Question 1: CONCEPTUAL

A car drives at a constant speed. In your own words, explain what "constant speed" means. If the car travels 50 miles in the first hour, how far does it go in the second hour? Why?

Show answer

"Constant speed" means the car covers the same distance every hour — it doesn't speed up or slow down. If it travels 50 miles in the first hour, it also travels 50 miles in the second hour, because the rate never changes. The unit rate is 50 miles per hour, and it stays the same for every hour of the trip.

Question 2: BASIC CALCULATION

A swimmer moves at a constant speed and covers 200 meters in 4 minutes. What is her speed in meters per minute? How far will she swim in 10 minutes?

Show answer

First, find the unit rate: speed = 200 ÷ 4 = 50 meters per minute. Next, multiply the unit rate by the new time: distance = 50 × 10 = 500 meters. The swimmer will cover 500 meters in 10 minutes. You can check: 50 meters per minute × 4 minutes = 200 meters, which matches the original information.

Question 3: INTERMEDIATE

A block of steel has a constant density. A piece with a volume of 6 cm³ has a mass of 46.8 grams. What is the density? Another piece of the same steel has a mass of 117 grams. What is its volume?

Show answer

First, find the density (unit rate): D = 46.8 ÷ 6 = 7.8 grams per cm³. For the second part, you know the mass and need the volume, so rearrange: V = m ÷ D = 117 ÷ 7.8 = 15 cm³. The density is 7.8 g/cm³, and the volume of the second piece is 15 cm³. This problem requires two steps: finding the unit rate, then dividing to find the missing quantity.

Question 4: APPLIED

Two friends are biking on a trail. Mia bikes at a constant speed of 12 miles per hour. Leo bikes at a constant speed of 9 miles per hour. They both start at the same time. After 3 hours, how much farther ahead is Mia than Leo?

Show answer

Calculate each person's distance separately. Mia: d = 12 × 3 = 36 miles. Leo: d = 9 × 3 = 27 miles. Subtract to find the difference: 36 − 27 = 9 miles. After 3 hours, Mia is 9 miles ahead of Leo. Notice that you can also think of the difference in their speeds (12 − 9 = 3 mph) and multiply by the time (3 × 3 = 9 miles) to get the same answer.

Question 5: CRITICAL THINKING

A scientist has two metal samples. Sample A has a mass of 54 grams and a volume of 20 cm³. Sample B has a mass of 42 grams and a volume of 15 cm³. The scientist says they are the same metal because their densities are close. Calculate each density. Are they exactly the same? What might explain a small difference?

Show answer

Sample A: D = 54 ÷ 20 = 2.7 g/cm³. Sample B: D = 42 ÷ 15 = 2.8 g/cm³. The densities are close but not exactly equal. A small difference could be caused by measurement errors — maybe the volume or mass wasn't measured perfectly. It could also mean one sample has tiny air bubbles or impurities mixed in. In real science, small differences like this are normal, and scientists decide whether the values are "close enough" to call them the same material. Both values are near the density of aluminum (2.7 g/cm³).

SUMMARY

Lesson Summary

Summary

A constant rate means a quantity stays the same for every unit — like a car driving the same number of miles every hour, or a material having the same mass for every cubic centimeter. A unit rate is the amount per one unit, found by dividing. For constant speed, the formula is d = r × t (distance equals speed times time). For constant density, the formula is m = D × V (mass equals density times volume).

Both follow the same master pattern: total = unit rate × number of units. To solve any constant rate problem, first find the unit rate by dividing, then multiply to find your answer. On a graph, constant rates appear as straight lines through the origin, and the steepness of the line shows the size of the unit rate. This skill is the foundation for slope and proportional relationships in algebra.