Ratios comparing like quantities and rates comparing unlike ones, unit rates and why they make comparison possible, averaging a rate by totalling both quantities, unit analysis as a way of converting units and checking a setup, and using a rate to compute a total.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Ratios and Rates
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-182 — the lesson these objectives are drawn from
Warm-up
Lesson 1.1 attached units to answers. This lesson lets the units do some of the work.
Discussion prompt
A car travels 150 miles on 6 gallons. Without deciding anything, write down the two quantities with their units and ask what you would have to do to get miles per gallon.
Hint: Read the phrase miles per gallon as an instruction.
Answer:
\[ \tfrac{150 \text{ miles}}{6 \text{ gallons}} = 25 \text{ miles per gallon} \]
The word per is a division sign, so miles per gallon means miles divided by gallons. The phrase names the operation, and once you notice that, most rate problems tell you what to do with their own wording.
Concept
The ratio of a to b is a over b. If a and b are measured in different units, the quotient is called the rate of a per b. Whether the units cancel is exactly what distinguishes a ratio from a rate.
rate — A comparison of two quantities measured in different units, such as 60 miles per gallon. A comparison of two quantities in the same unit is a ratio, and it has no units.
A rate expressed per one unit is called a unit rate.
Figure (svg): Two columns contrasting a ratio, which compares like quantities, with a rate, which compares unlike ones
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-178
Section
Section 1
Concept
A ratio compares two quantities measured in the same unit. Because the units cancel, the ratio itself has no units, and it is simplified like any other fraction.
A ratio of five thirds may be written five to three or five colon three, and all three notations mean the same thing.
Figure (svg): Sixteen matches split into ten wins and six losses, with the ratio simplified
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177 — Example 1, Find a Ratio, and the Writing Algebra note
Picture it
Both counts are numbers of matches, so the unit cancels.
Figure (svg): Sixteen matches split into ten wins and six losses, with the ratio simplified
Ten over six simplifies to five over three, read five to three. The answer carries no unit, because matches divided by matches leaves nothing behind.
Worked example
This is Example 1 from the textbook. The team won 10 of its 16 matches.
\[ \text{Find the ratio of wins to losses.} \]
Work out both quantities being compared
Why: Ten wins, and sixteen minus ten, which is six losses.
Write them in the order named
Why: Wins to losses, so wins on top.
\[ \frac{10}{6} \]
Simplify
Why: Two divides both, giving five over three.
\[ \frac{5}{3} \]
State it in words
Why: Five to three.
\[ 5\text{ to } 3 \]
Figure (svg): Sixteen matches split into ten wins and six losses, with the ratio simplified
\[ \tfrac{10}{6} = \tfrac{5}{3}, \text{ read five to three} \]
Verify: check that the parts account for the whole
Why: Five plus three is eight, and ten plus six is sixteen, which is twice eight — so the simplified ratio really does describe the same split. And the answer has no unit, since matches divided by matches cancels.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177
Sorting
Ask whether the two quantities share a unit.
Sort into buckets
Sort each comparison by whether it is a ratio or a rate.
The test never involves the numbers. It is entirely about whether the two quantities are measured in the same thing.
Worked example
Guided Practice 1. Eight wins out of fifteen games, with no ties.
\[ \text{Find the team's ratio of wins to losses.} \]
Find the number of losses
Why: Fifteen games minus eight wins is seven losses.
\[ 7\text{ losses} \]
Write the ratio in the named order
Why: Wins to losses puts eight on top.
\[ \frac{8}{7} \]
Check whether it simplifies
Why: Eight and seven share no factor other than one, so it is already simplest.
State it in words
Why: Eight to seven.
\[ 8\text{ to } 7 \]
Figure (svg): The solution to Worked example a ratio from a total shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{8}{7}, \text{ read eight to seven} \]
Verify: check against the total
Why: Eight and seven total fifteen, which is the number of games played, so no game has been lost or double-counted. That check works because there were no ties — with ties the wins and losses would not account for the whole.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177
Trap
The team won 10 of its 16 matches, so the win-loss ratio is 10 to 16.
Use the two numbers the question gives
Why: Ten and sixteen are both stated, so they look like the pair being asked for.
Sixteen is the total, not the losses. Ten to sixteen is the ratio of wins to matches played, which is a different comparison.
\[ \text{losses} = 16 - 10 = 6 \;\Longrightarrow\; \tfrac{10}{6} = \tfrac{5}{3} \]
Work out both quantities named in the question before writing anything
Why: Wins to losses needs the losses, which have to be computed from the total.
Read the question's wording carefully: to losses, to matches and to games are three different denominators, and only one of them was asked for.
Elimination
A team won 10 of its 16 matches, with no draws.
Eliminate the wrong options
What is the ratio of wins to losses?
Survives elimination: A
Why: The losses are sixteen minus ten, which is six, and ten to six simplifies to five to three. Options B and D are the same comparison at different levels of simplification, and both answer a question about matches played rather than about losses.
Faded example
Compute the second quantity, then simplify.
Fill in the blanks
\text7 \;\rightarrow\; \text7 = ___ \;\rightarrow\; \text___ = \tfrac______}
Why: Fifteen games minus eight wins leaves seven losses, so the ratio of wins to losses is eight to seven. It happens to be already in simplest form, since eight and seven share no common factor.
Socratic
Every other quotient in this book carries a unit.
Discussion prompt
Explain why the ratio of ten matches to six matches has no unit, using the cancellation idea from Lesson 1.1. Then say what would change if the two quantities were measured in different units.
Hint: Treat the word matches as a factor that can cancel.
Answer:
Ten matches divided by six matches has matches on the top and matches on the bottom, and they cancel exactly as a common numerical factor would. What is left is the pure number five thirds, with nothing attached — which is why a ratio can be read as five to three without naming any quantity.
If the units differed, nothing would cancel and the quotient would carry a compound unit such as kilometres per minute. That is precisely the definition of a rate, so the difference between a ratio and a rate is entirely a question of whether the cancellation happens.
Section
Section 2
Concept
A rate compares two quantities in different units, so its value carries a compound unit. Expressing it per one unit gives a unit rate, which is what makes two rates comparable.
\[ \tfrac{10 \text{ km}}{50 \text{ min}} = 0.2 \text{ km per minute} \]
A unit rate is a rate per one given unit, such as sixty miles per one gallon.
Figure (svg): A rate of 10 kilometres in 50 minutes reduced to a unit rate per one minute
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177 — the definition of unit rate and Example 2
Picture it
Dividing makes the denominator one.
Figure (svg): A rate of 10 kilometres in 50 minutes reduced to a unit rate per one minute
Ten kilometres in fifty minutes is 0.2 kilometres in one minute. Both describe the same running, and only the second can be compared directly with another runner's pace.
Worked example
This is Example 2 from the textbook. A 10 kilometre race in 50 minutes.
\[ \text{Find the average speed in kilometres per minute.} \]
Read the required unit from the question
Why: Kilometres per minute means kilometres divided by minutes.
\[ \text{km} / \min \]
Write the rate as a fraction with those units
Why: Ten kilometres over fifty minutes.
\[ 10 \text{km} / 50 \min \]
Divide to make the denominator one
Why: Ten over fifty is one fifth.
\[ 0.2 \]
Attach the unit
Why: Nought point two kilometres per minute.
\[ 0.2 \text{km} / \min \]
Figure (svg): A rate of 10 kilometres in 50 minutes reduced to a unit rate per one minute
\[ \tfrac{10 \text{ km}}{50 \text{ min}} = 0.2 \text{ km/min} \]
Verify: scale the unit rate back up
Why: Nought point two kilometres a minute for fifty minutes is ten kilometres, which is the race distance. A unit rate multiplied by the original denominator must return the original numerator.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177
Translation
The word per names the operation and the order.
Match the pairs
Why: In every case the unit named before per goes on top and the one named after it goes underneath. The first and last are the same two quantities in opposite orders, giving 0.2 and 5 respectively for the race — both correct answers to different questions.
Worked example
Guided Practice 2 and 3. Read the required unit from the wording each time.
\[ \text{A plane flies } 1200 \text{ miles in } 4 \text{ hours. You earn } 45 \text{ dollars for mowing } 3 \text{ lawns.} \]
Set up the first as a fraction
Why: Twelve hundred miles over four hours.
\[ 1200 \text{mi} / 4 h \]
Divide and attach the unit
Why: Three hundred miles per hour.
\[ 300 \text{mi} / h \]
Set up the second
Why: Forty-five dollars over three lawns.
\[ 45\text{ dollars } / 3\text{ lawns} \]
Divide and attach the unit
Why: Fifteen dollars per lawn.
Figure (svg): The solution to Worked example two unit rates from guided practice shown as a ladder of expressions, one row per algebraic move
\[ 300 \text{ mi/h} \qquad 15 \text{ dollars per lawn} \]
Verify: scale each back up
Why: Three hundred miles an hour for four hours is 1200 miles; fifteen dollars a lawn for three lawns is forty-five dollars. Both return the figures given, which confirms the divisions were set up the right way round.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-177
Error analysis
The student computed three unit rates. Two are wrong.
Annotate
On: \( \tfrac{10 \text{ km}}{50 \text{ min}} = 5 \text{ km/min} \qquad \tfrac{1200 \text{ mi}}{4 \text{ h}} = 300 \text{ mi/h} \qquad \tfrac{45}{3 \text{ lawns}} = 0.067 \text{ dollars/lawn} \)
Both errors are the same one, and both are caught by asking whether the answer is a plausible size for the quantity described. The required unit tells you which quantity goes on top.
Estimation
A size check catches a reversed division immediately.
Predict first
A runner covers 10 kilometres in 50 minutes. Which answer is plausible for their speed in kilometres per minute?
Correct: 0.2.
\[ \tfrac{10}{50} = 0.2 \text{ km/min} = 12 \text{ km/h} \]
Why: A runner covers a fraction of a kilometre each minute, so a value well below one is expected. Five kilometres a minute would be three hundred kilometres an hour, and five hundred would be faster than sound. Knowing roughly what size a familiar rate should be is often quicker than checking the units.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Runner | Rate as given | Unit rate |
|---|---|---|
| A | 10 km in 50 min | 0.2 km per minute |
| B | 6 km in 24 min | 0.25 km per minute |
| Faster | B | visible only once both are per minute |
The rates as given cannot be compared directly, since neither the distances nor the times match. Converting both to a per-minute basis is what makes the comparison possible at all.
Socratic
The rate as given already describes the situation completely.
Discussion prompt
Explain why a unit rate makes two rates comparable when the original forms do not, using two grocery prices as your example. Then say what a supermarket's price-per-unit labels are doing.
Hint: Think about two packets of different sizes at different prices.
Answer:
Two packets at 3.20 for 400 grams and 4.50 for 600 grams cannot be compared directly, because both the prices and the sizes differ. Reducing each to a price per hundred grams — eighty pence and seventy-five pence — puts both on the same denominator, and only then does one of them visibly win.
A supermarket's unit-price labels do exactly this calculation for every product on the shelf, precisely because shoppers cannot do it reliably in their heads. The labels exist because comparing rates with different denominators is genuinely hard, and converting to a common one is the only reliable method.
Section
Section 3
Concept
To find an average rate over several trips, add all the numerators and all the denominators and divide once. Averaging the individual rates gives a different and usually wrong answer.
\[ \text{average mileage} = \tfrac{\text{total miles}}{\text{total gallons}} \]
The average rate is the single rate that would have produced the same totals.
Figure (svg): Five trips totalled and divided to give an average mileage in miles per gallon
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 178-178 — Example 3, Find a Rate
Picture it
Total the miles, total the gallons, divide once.
Figure (svg): Five trips totalled and divided to give an average mileage in miles per gallon
Eleven hundred and forty-nine miles on 47.4 gallons gives about 24.2 miles per gallon. Averaging the five individual mileages would give a slightly different number, and it would answer a different question.
Worked example
This is Example 3 from the textbook. Five trips with their miles and gallons recorded.
\[ \text{Find the average mileage in miles per gallon, to the nearest tenth.} \]
Total the miles
Why: 290 plus 242 plus 196 plus 237 plus 184.
\[ 1149\text{ miles} \]
Total the gallons
Why: 12.1 plus 9.8 plus 8.2 plus 9.5 plus 7.8.
\[ 47.4\text{ gallons} \]
Divide once
Why: Eleven hundred and forty-nine over 47.4.
\[ 24.24... \]
Round and attach the unit
Why: About 24.2 miles per gallon.
\[ 24.2 \text{mi} / \text{gal} \]
Figure (svg): Five trips totalled and divided to give an average mileage in miles per gallon
\[ \tfrac{1149 \text{ mi}}{47.4 \text{ gal}} \approx 24.2 \text{ mi/gal} \]
Verify: check the answer against the individual trips
Why: The five individual mileages range from about 23.6 to 25.1 miles per gallon, and 24.2 sits inside that range as an average must. An average outside the range of its inputs would be impossible.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 178-178
Elimination
Five trips are recorded with their miles and gallons.
Eliminate the wrong options
How do you find the average miles per gallon?
Survives elimination: A
Why: The average rate is the single rate that would have produced the same totals, so both quantities are totalled and divided once. Option B is the tempting one and is genuinely different: it treats each trip equally rather than weighting by distance.
Worked example
The two methods give different answers, and only one answers the question.
\[ \text{A car does } 60 \text{ miles at } 30 \text{ mi/h and } 60 \text{ miles at } 60 \text{ mi/h. Find its average speed.} \]
Compute the time for each leg
Why: Sixty miles at thirty is two hours; sixty at sixty is one hour.
\[ 2 h\text{ and } 1 h \]
Total the distance and the time
Why: One hundred and twenty miles in three hours.
\[ 120 \text{mi}, 3 h \]
Divide once
Why: One hundred and twenty over three.
\[ 40 \text{mi} / h \]
Compare with averaging the rates
Why: Averaging thirty and sixty would give forty-five, which is wrong.
\[ \text{not } 45 \]
Figure (svg): The solution to Worked example why averaging the rates differs shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{120 \text{ mi}}{3 \text{ h}} = 40 \text{ mi/h} \]
Verify: check by asking which method matches the definition
Why: Average speed means total distance over total time, and the journey really did take three hours for a hundred and twenty miles. Averaging the two speeds would only be right if equal times had been spent at each, and here the slow leg took twice as long.
Trap
Five trips gave mileages of about 24.0, 24.7, 23.9, 24.9 and 23.6.
Add the five mileages and divide by five
Why: Averaging is adding and dividing by how many, and there are five numbers.
That answers the question what was the average of my five mileage figures, which is not the same as what was my mileage over the whole period.
\[ \tfrac{1149 \text{ total miles}}{47.4 \text{ total gallons}} \approx 24.2 \text{ mi/gal} \]
Total both quantities and divide once, which weights each trip by its size
Why: A long trip should count for more than a short one, and totalling does that automatically.
The two answers are close here because the trips were similar in length. When the parts differ greatly in size, the two methods diverge sharply — as the sixty-miles-each example shows.
Prediction
A car drives 60 miles at 30 mph and 60 miles at 60 mph.
Predict first
Comparing total-distance-over-total-time with averaging the two speeds, which gives the larger answer?
Correct: Averaging the speeds, which gives 45 against 40.
\[ \tfrac{120}{3} = 40 \quad \text{against} \quad \tfrac{30 + 60}{2} = 45 \]
Why: The slow leg takes twice as long as the fast one, so it should carry twice the weight. Totalling does that automatically and gives forty; averaging the speeds treats both legs equally and gives forty-five. The correct answer is always the lower one when equal distances are driven at different speeds, because more time is spent going slowly.
Faded example
Total both quantities, then divide.
Fill in the blanks
\text1149 290 + 242 + 196 + 237 + 184 = 24.2, \quad \text___ 47.4 \;\rightarrow\; \text___ \approx ___
Why: The total distance is 1149 miles and the total fuel is 47.4 gallons, so the average mileage is about 24.2 miles per gallon. Totalling first and dividing once is what weights each trip by its own length.
Socratic
They gave nearly the same answer for the trips and very different ones for the journey.
Discussion prompt
Describe the condition under which averaging the individual rates gives the same answer as totalling. Then say which of the two examples in this lesson came closer to satisfying it.
Hint: Think about the sizes of the denominators.
Answer:
They agree exactly when all the denominators are equal — the same number of gallons in each trip, or the same time spent on each leg. Then every rate carries the same weight and averaging them is the same as totalling.
The five truck trips came close, since their fuel amounts ranged only from 7.8 to 12.1 gallons, so the two methods differ by a fraction of a mile per gallon. The two-leg journey did not come close at all: one leg took two hours and the other one, so the weights differed by a factor of two and the answers differed by five miles per hour.
Section
Section 4
Concept
Writing the units alongside the quantities is called unit analysis. Units multiply and divide like numbers, so a conversion is a multiplication by a fraction whose top and bottom are equal quantities in different units.
\[ 3 \text{ hours} \cdot \tfrac{60 \text{ minutes}}{1 \text{ hour}} = 180 \text{ minutes} \]
Figure (svg): A conversion carried out by multiplying by a fraction equal to one, with the units cancelling
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 178-178 — the Unit Analysis paragraph and Example 4
Picture it
Hours on the bottom cancel hours on the top.
Figure (svg): A conversion carried out by multiplying by a fraction equal to one, with the units cancelling
Because the fraction equals one, multiplying by it cannot change the quantity — only how it is expressed. The surviving unit tells you what you have.
Worked example
This is Example 4 from the textbook. Hours to minutes, and inches to feet.
\[ \text{Convert } 3 \text{ hours to minutes, and } 72 \text{ inches to feet.} \]
Write the relating fact as a fraction
Why: Sixty minutes equals one hour, so sixty minutes over one hour equals one.
\[ 60 \min / 1 h = 1 \]
Orient it so hours cancel and multiply
Why: Hours are on top in the quantity, so hours go on the bottom of the fraction.
\[ 3 h \cdot(60 \min / 1 h) = 180 \min \]
Set up the second conversion
Why: One foot equals twelve inches, so one foot over twelve inches equals one.
\[ 1 \text{ft} / 12\text{ in } = 1 \]
Orient and multiply
Why: Inches must cancel, so inches go on the bottom.
Figure (svg): A conversion carried out by multiplying by a fraction equal to one, with the units cancelling
\[ 3 \text{ h} = 180 \text{ min} \qquad 72 \text{ in} = 6 \text{ ft} \]
Verify: check the direction of each answer
Why: Minutes are smaller than hours, so the number of minutes must be larger — and 180 is larger than 3. Feet are larger than inches, so the number of feet must be smaller — and 6 is smaller than 72. Both directions are right.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 178-178
Matching
The unit you have must go on the bottom.
Match the pairs
Why: In each case the starting unit appears on the bottom of the fraction so that it cancels, leaving the target unit on top. Two of these convert to a smaller unit and make the number bigger; two convert to a larger unit and make it smaller.
Worked example
Guided Practice 4 and 5. Choose the orientation each time.
\[ \text{Convert } 8 \text{ pounds to ounces, and } 84 \text{ days to weeks.} \]
Set up the first
Why: One pound is sixteen ounces, so sixteen ounces over one pound equals one.
\[ 16 \text{oz} / 1 \text{lb} \]
Multiply and cancel
Why: Pounds cancel and ounces survive.
\[ 8 \text{lb} \cdot(16 \text{oz} / 1 \text{lb}) = 128 \text{oz} \]
Set up the second
Why: One week is seven days, so one week over seven days equals one.
\[ 1\text{ week } / 7\text{ days} \]
Multiply and cancel
Why: Days cancel and weeks survive.
Figure (svg): The solution to Worked example two from guided practice shown as a ladder of expressions, one row per algebraic move
\[ 8 \text{ lb} = 128 \text{ oz} \qquad 84 \text{ days} = 12 \text{ weeks} \]
Verify: check each direction
Why: Ounces are smaller than pounds, so the count grows from 8 to 128. Weeks are larger than days, so the count shrinks from 84 to 12. Both behave as changing to a smaller or larger unit requires.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 178-178
Trap
\[ 3 \text{ hours} \cdot \tfrac{1 \text{ hour}}{60 \text{ minutes}} \]
Write the conversion fraction whichever way it comes to mind
Why: Both orientations equal one, so both look equally valid.
\[ = 0.05 \tfrac{\text{hours}^2}{\text{minutes}} \]
Nothing cancelled, and the surviving unit is hours squared per minute — which describes nothing. The arithmetic was fine and the setup was not.
\[ 3 \text{ hours} \cdot \tfrac{60 \text{ minutes}}{1 \text{ hour}} = 180 \text{ minutes} \]
Put the unit you already have on the bottom of the fraction so that it cancels
Why: The orientation is decided by what needs to disappear, not by which number looks bigger.
Writing the units out is what makes the choice automatic. Without them both orientations look identical and the choice becomes a guess.
Prediction
The direction is decided by which unit is bigger.
Predict first
Converting 5 kilometres to metres, will the number grow or shrink?
Correct: Grow, because a metre is smaller than a kilometre.
\[ 5 \text{ km} \cdot \tfrac{1000 \text{ m}}{1 \text{ km}} = 5000 \text{ m} \]
Why: The same distance measured in smaller units needs more of them, so five kilometres becomes five thousand metres. The quantity is unchanged but the number describing it grows. Predicting the direction before converting catches an upside-down fraction immediately.
Faded example
The unit you have goes on the bottom.
Fill in the blanks
72 inches × (1 foot / 12 inches) = 6 feet
Why: Inches must cancel, so inches go underneath and feet on top. The fraction one foot over twelve inches equals one, so multiplying by it changes the units without changing the length.
Socratic
Multiplying by it changes the number, and it is still a multiplication by one.
Discussion prompt
Explain why sixty minutes over one hour equals one, and why that means multiplying by it does not change the quantity. Then say what it does change.
Hint: Ask whether the top and bottom describe the same amount of time.
Answer:
Sixty minutes and one hour are the same duration written two ways, so the fraction is a quantity divided by itself, which is one. Multiplying by one leaves any quantity unchanged, by the identity property from Lesson 2.5.
What changes is the unit it is expressed in, and therefore the number attached. Three hours and 180 minutes are the same duration, so nothing about the quantity moved — only the scale on which it is being measured. That is exactly why unit conversion is safe: every conversion is a multiplication by one.
Section
Section 5
Concept
When a rate problem does not make the operation obvious, write the units and see which arrangement produces the unit you want. The units decide the setup before any arithmetic happens.
Figure (svg): A tank of 18 gallons multiplied by a mileage rate to give a distance
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 180-180 — Example 5, estimating the distance on a full tank
Picture it
Gallons times miles per gallon leaves miles.
Figure (svg): A tank of 18 gallons multiplied by a mileage rate to give a distance
Eighteen gallons at 24.2 miles per gallon gives about 436 miles. The units confirmed the multiplication before a single digit was computed.
Worked example
Example 5 from the textbook. The truck averages 24.2 miles per gallon and the tank holds 18 gallons.
\[ \text{Estimate how far the truck can travel on } 18 \text{ gallons.} \]
State the unit the answer must have
Why: The question asks how far, so the answer is in miles.
Arrange the quantities so gallons cancel
Why: Gallons times miles per gallon has gallons on top and bottom.
\[ 18 \text{gal} \cdot 24.2 \text{mi} / \text{gal} \]
Confirm the surviving unit
Why: Gallons cancel and miles survive.
Compute
Why: Eighteen times 24.2 is about 436.
\[ \text{about } 436\text{ miles} \]
Figure (svg): A tank of 18 gallons multiplied by a mileage rate to give a distance
\[ 18 \text{ gal} \cdot 24.2 \tfrac{\text{mi}}{\text{gal}} \approx 436 \text{ miles} \]
Verify: check the direction and the size
Why: More gallons should give more miles, and 436 is much larger than either input — which is right, since each gallon contributes over twenty miles. Dividing instead would have given about 0.74, which is not a distance any tank produces.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 180-180
Sorting
Decide from the units what the answer needs.
Sort into buckets
Sort each problem by the operation the units call for.
The three pairs use the same rates in both directions. Which operation you need depends entirely on which of the rate's two units you already have.
Worked example
The same two quantities can be combined two ways, and only one gives a sensible unit.
\[ \text{Given } 24.2 \text{ mi/gal and } 18 \text{ gal, decide whether to multiply or divide.} \]
Try multiplying
Why: Gallons times miles per gallon leaves miles, which is a distance.
Try dividing the rate by the gallons
Why: Miles per gallon over gallons leaves miles per gallon squared.
\[ \text{mi} / \text{gal} ^{2} \]
Try dividing the gallons by the rate
Why: Gallons over miles per gallon leaves gallons squared per mile.
\[ \text{gal} ^{2} / \text{mi} \]
Choose the arrangement giving miles
Why: Only the multiplication produces a unit anybody measures.
Figure (svg): The solution to Worked example the units choose the operation shown as a ladder of expressions, one row per algebraic move
\[ 18 \text{ gal} \cdot 24.2 \tfrac{\text{mi}}{\text{gal}} = 436 \text{ mi} \]
Verify: confirm the two rejected units describe nothing
Why: Miles per gallon squared and gallons squared per mile are not quantities anybody measures or has a name for. When only one arrangement produces a meaningful unit, the units have decided the setup on their own.
Trap
Given 24.2 miles per gallon and 18 gallons, divide because the answer should be smaller.
\[ \tfrac{24.2}{18} \approx 1.34 \]
Choose the operation by which answer looks reasonable
Why: Dividing often makes answers smaller, and small numbers feel safer.
One point three four what? The unit is miles per gallon squared, which measures nothing, and the answer is not a distance at all.
\[ 18 \text{ gal} \cdot 24.2 \tfrac{\text{mi}}{\text{gal}} = 436 \text{ miles} \]
Write the units first and choose the arrangement that leaves the unit you want
Why: The operation follows from the units rather than from a guess about the size of the answer.
This is the same discipline as Lesson 1.6's labels stage, now doing real work: the units are not decoration, they are the instructions.
Elimination
You have 18 gallons and a rate of 24.2 miles per gallon.
Eliminate the wrong options
Which arrangement produces an answer in miles?
Survives elimination: A
Why: Only the multiplication has gallons on both the top and the bottom, so only it cancels them and leaves miles. Three of the four options fail a units check before any arithmetic, which is what makes unit analysis a genuine tool rather than a formality.
Estimation
A rough answer confirms the setup as well as the arithmetic.
Predict first
Roughly how far can a truck go on 18 gallons at about 24 miles per gallon?
Correct: About 430 miles.
\[ 18 \cdot 24 = 432 \quad \text{so about } 430 \text{ miles} \]
Why: Eighteen times twenty-four is a little over four hundred. The other options correspond to dividing, to dividing the other way, and to a decimal-point slip, and an estimate rules out all three at once. Estimating first is often faster than a units check and catches the same class of error.
Socratic
Treating a unit as a factor is not merely a mnemonic.
Discussion prompt
Explain why cancelling gallons in the expression gallons times miles per gallon is legitimate, treating the unit as an algebraic factor. Then say what this means about how a rate should be written down.
Hint: Write the rate as an explicit fraction and look at what appears twice.
Answer:
\[ 18 \text{ gal} \cdot \tfrac{24.2 \text{ mi}}{1 \text{ gal}} = \tfrac{18 \cdot 24.2 \text{ gal} \cdot \text{mi}}{1 \text{ gal}} \]
The word gallons appears once on the top and once on the bottom, and a common factor cancels whether it is a number or a unit. That is why the cancellation is legitimate: units obey the same multiplication and division rules that numbers do.
It means a rate should always be written as an explicit fraction with both units shown, rather than as a bare number with the unit remembered. Written as a fraction the cancellation is visible; written as 24.2 with the unit in your head it is a guess.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Ratio | Rate | Unit rate | |
|---|---|---|---|
| Units of the two quantities | the same | different | different |
| Unit of the answer | none | a compound unit | a compound unit |
| Denominator | anything | anything | one |
A unit rate is just a rate with its denominator reduced to one, which is what makes two rates comparable at a glance.
Pattern
Whether the question asks for a ratio, a rate, a conversion or a total, the same five moves cover it.
Step three is where the units do the work. If two arrangements are possible, only one of them leaves a unit anybody measures.
OpenStax Elementary Algebra 2e, §8.7 Solve Proportion and Similar Figure Applications §8.7
Check
A ratio. Compute both quantities first.
Check your understanding
A team plays 20 games and wins 12. What is the ratio of wins to losses?
Answer: A
Why: The losses are twenty minus twelve, which is eight, so the ratio is twelve to eight. Dividing both by four gives three to two. The answer carries no unit, since games divided by games cancels.
Check
A unit rate. Read the required unit from the wording.
Check your understanding
A plane flies 1750 miles in 5 hours. What is its speed in miles per hour?
Answer: A
Why: Miles per hour means miles divided by hours, so 1750 over 5 is 350. Scaling back up confirms it: 350 miles an hour for five hours is 1750 miles.
Check
Unit analysis. Choose the orientation that cancels.
Check your understanding
Convert 5 pounds to ounces, given that 1 pound is 16 ounces.
Answer: A
Why: Multiplying five pounds by sixteen ounces per pound cancels the pounds and leaves eighty ounces. Ounces are smaller than pounds, so the number must grow, which it does from five to eighty.
Real world
Two petrol stations. One sells at 1.42 dollars per litre, the other at 5.20 dollars per US gallon. One US gallon is about 3.785 litres.
Discussion prompt
Convert the second price to dollars per litre using unit analysis, showing which way up your conversion fraction goes and why. Then say which station is cheaper and by how much per litre.
Hint: You want dollars per litre, so litres must end up on the bottom.
Answer:
\[ 5.20 \tfrac{\text{dollars}}{\text{gallon}} \cdot \tfrac{1 \text{ gallon}}{3.785 \text{ litres}} = 1.374 \tfrac{\text{dollars}}{\text{litre}} \]
The gallon unit had to cancel, so gallons went on the top of the conversion fraction to meet the gallons on the bottom of the rate. The surviving units are dollars over litres, which is what was wanted.
The second station is cheaper at about 1.37 dollars per litre against 1.42, a saving of about five cents a litre. Without the conversion the two prices cannot be compared at all, since 5.20 and 1.42 describe different amounts of fuel.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A car drives 60 miles at 30 mph and 60 miles at 60 mph. What is its average speed?
Correct: 40 miles per hour, total distance over total time.
\[ \tfrac{60}{30} = 2 \text{ h}, \quad \tfrac{60}{60} = 1 \text{ h}, \quad \tfrac{120}{3} = 40 \text{ mi/h} \]
Why: The slow leg takes two hours and the fast one takes one, so the journey covers 120 miles in three hours, giving forty. Averaging the two speeds would be right only if equal times had been spent at each, and here twice as long was spent going slowly. Averaging rates rather than totalling both quantities is one of the most common errors with rates, and it always overstates the answer when equal distances are driven at different speeds.
Explain it
They can divide confidently and have never used units as part of the working.
Discussion prompt
In no more than four sentences, explain how the units tell you whether to multiply or divide in a rate problem. Use gallons and miles per gallon as your example, and give them the one thing to write down that makes it work.
Hint: The thing to write down is easy to leave out.
Answer:
A usable answer: write the units next to every number, treating them like letters that can cancel. If you have gallons and a rate in miles per gallon, multiplying puts gallons on the top and the bottom so they cancel and leave miles. Dividing would leave a unit like gallons squared per mile, which means nothing, so multiplying must be right.
The thing to write down is the rate as a proper fraction, with both units shown — miles over gallons rather than just 24.2. If the units are only in your head, nothing can cancel on the page and the choice becomes a guess.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Ratio-or-rate is fixed by asking whether the two quantities share a unit. Unit-rate direction is fixed by reading the required unit from the wording, since per names the order. Averaging is fixed by totalling both quantities and dividing once rather than averaging the rates. Conversion orientation is fixed by putting the unit you already have on the bottom so it cancels. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write one ratio and one rate from your own life, with both quantities and their units shown, and mark which one has units that cancel. Underneath, take your rate and reduce it to a unit rate, showing the division and the unit of the answer. In the middle, work one unit conversion by multiplying by a conversion fraction, drawing a line through each unit that cancels and circling the one that survives. Near the bottom, write a rate problem where you must decide between multiplying and dividing, and show both arrangements with their resulting units so that the wrong one is visibly meaningless. Finally, in the margin, write the rule for finding an average rate over several trips.
In your conversion, exactly one unit should survive and it should be the one you wanted. If two units survive or the surviving one is squared, the fraction went in upside down.
Recap
Five things, and the last one turns the units from decoration into instructions.
| If the question says | Your first move is |
|---|---|
| Find the ratio of wins to losses | Compute the losses from the total |
| Find the speed in km per minute | Put kilometres on top and minutes underneath |
| Find the average mileage | Total the miles and total the gallons |
| Convert 3 hours to minutes | Put hours on the bottom of the fraction |
| How far on 18 gallons | Arrange so gallons cancel and miles survive |
Lesson 3.9 finishes the chapter with percents, which are ratios with a fixed denominator of one hundred — and with a verbal model that turns all three kinds of percent question into the same equation.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.8 Ratios and Rates §3.8, pp. 177-182 — everything on these slides traces back here
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