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Algebra 1 · N.Q.A.1 · N.Q.A.2

Units, quantities and a sensible level of accuracy

A problem hands you numbers with units attached. How do the units tell you what to multiply, what to divide, whether the answer can be right, and how many of its digits to believe?

Where you are 12 in Measurement

When you reach for it

Most problems about the real world arrive as a handful of quantities, each a number with a unit: 14 turns per minute, 6.4 liters per hundred kilometers, 86{,}000 people on 412 square kilometers. The question asks for something in a different unit: turns in a year, kilometers on one liter, people per square kilometer of dry land. Three decisions stand between the givens and the answer: which quantities matter, how to combine them (mostly whether to multiply or divide, and which way up to write each fraction), and how many digits the answer deserves.

Units help with all three: they show which way up a fraction must go, they show at once when a set-up does not produce the quantity asked for, and they separate exact numbers from measured ones. It builds on conversion factors and on significant figures, and it is the working method behind every rate in rates from measurements.

The procedure

Units multiply, divide and cancel like letters

A quantity such as 80\text{ km/h} is a number times a unit, and the unit “kilometers per hour” is a quotient, \displaystyle \frac{\text{km}}{\text{h}}. Units obey the same rules as letters in algebra: they multiply, divide, and cancel when the same unit appears above and below a fraction bar. Driving 3 hours at 80\text{ km/h}:

3\text{ h} \times \frac{80\text{ km}}{1\text{ h}} = 240\text{ km}

The hours cancel, leaving kilometers, which is what a distance is measured in. Dividing instead, \displaystyle 3\text{ h} \div \frac{80\text{ km}}{1\text{ h}}, would give the unit \displaystyle \frac{\text{h}^2}{\text{km}}, which measures nothing anybody wants. The arithmetic gives no warning; the unit does.

Two quantities can be added or subtracted only when they have the same unit: 2\text{ km} + 300\text{ m} is 2.3\text{ km} or 2300\text{ m}, never 302 of anything.

A conversion factor is a fraction equal to one

Since 1\text{ h} = 3600\text{ s} exactly, the fraction \displaystyle \frac{3600\text{ s}}{1\text{ h}} has the same quantity on top and bottom, so it equals 1, and so does its reciprocal \displaystyle \frac{1\text{ h}}{3600\text{ s}}. Multiplying by 1 never changes an amount; it changes only the unit it is written in. That is the whole idea of a conversion factor.

Which way up to write it is decided by the unit you want to get rid of: it goes on the opposite side of the bar from where it sits now, so the two cancel. To turn hours into seconds, hours must cancel, so hours go underneath:

2.5\text{ h} \times \frac{3600\text{ s}}{1\text{ h}} = 9000\text{ s}

A rate has a unit on top and a unit underneath, and each is converted by its own factor. Factors can be chained in one line, and the unit left at the end is the unit of the answer. Write the chain before doing any arithmetic: if the units left over are not the units you want, the set-up does not answer the question.

The facts this lesson uses are exact by definition, apart from the mile:

FactStatus
1\text{ km} = 1000\text{ m}, 1\text{ kg} = 1000\text{ g}exact
1\text{ L} = 1000\text{ mL}, 1\text{ m}^3 = 1000\text{ L}exact
1\text{ h} = 3600\text{ s}, 1\text{ day} = 24\text{ h}exact
1\text{ km}^2 = 100\text{ ha}exact
1\text{ inch} = 2.54\text{ cm}exact
1\text{ mile} \approx 1.609\text{ km}four figures

Square and cubic units need their length factor used two or three times, which is converting square and cubic units.

Choosing the quantities

Before any arithmetic, decide what the answer depends on, and give each quantity a unit as well as a name: “the trip in kilometers”, not just “the trip”. A model always leaves things out; a fuel estimate uses distance and fuel use and ignores the hills. Leaving a quantity out is fine when its effect is smaller than the uncertainty already in the answer. When one name could mean two quantities (an island’s area with or without its lake), say which one you used.

Reporting to a sensible accuracy

An answer can be no more accurate than the quantities it was built from. The rules of thumb are in significant figures, and only their use is repeated here:

  • A product or quotient keeps about as many significant figures as the measured input with the fewest.
  • Exact numbers never limit the answer: unit definitions such as 3600\text{ s} in an hour, counts, and the 100 inside “liters per 100 km”, which is part of the unit’s name.
  • Carry every digit through the chain and round once, at the end.
  • The situation can demand something coarser: milk comes in whole cartons, and a model that assumes an average rate all year earns only “about” a round number.

Worked examples

Example 1

Kilometers per hour and meters per second

(a) A high-speed train runs at 288\text{ km/h}. Write its speed in meters per second. (b) A gust of wind reaches 25\text{ m/s}. Write it in kilometers per hour. (c) A road sign gives a limit of 60 miles per hour. About how many kilometers per hour is that?

Solution

(a) Kilometers are on top and must become meters, so the factor with kilometers underneath is used. Hours are underneath and must become seconds, so the factor with hours on top is used:

\begin{aligned} &\frac{288\text{ km}}{1\text{ h}} \times \frac{1000\text{ m}}{1\text{ km}} \\ &\quad \times \frac{1\text{ h}}{3600\text{ s}} \\ &= \frac{288 \times 1000}{3600}\ \frac{\text{m}}{\text{s}} = 80\text{ m/s} \end{aligned}

Kilometers cancel against kilometers and hours against hours, leaving meters over seconds.

(b) Now every factor is turned over, because the units to cancel are meters (on top) and seconds (underneath):

\begin{aligned} &\frac{25\text{ m}}{1\text{ s}} \times \frac{3600\text{ s}}{1\text{ h}} \\ &\quad \times \frac{1\text{ km}}{1000\text{ m}} \\ &= \frac{25 \times 3600}{1000}\ \frac{\text{km}}{\text{h}} = 90\text{ km/h} \end{aligned}

The two factors together multiply by \displaystyle \frac{3600}{1000} = 3.6, so 1\text{ m/s} = 3.6\text{ km/h} exactly. Check (a) with it: 80 \times 3.6 = 288.

(c) Only the top unit changes:

\frac{60\text{ mi}}{1\text{ h}} \times \frac{1.609\text{ km}}{1\text{ mi}} = 96.54\text{ km/h}

The factor 1.609 is itself rounded, and nobody reads a speedometer to a hundredth, so this is about 97\text{ km/h}.

Example 2

A recipe scaled, and a price per kilogram

Ginevra’s blueberry crumble serves 10 and uses 240\text{ g} of flour, 150\text{ g} of blueberries and 350\text{ mL} of milk for the custard. She wants to serve 24. Blueberries cost 16.50 dollars per kilogram and milk comes in 500\text{ mL} cartons. How much of each does she need, how much do the blueberries cost, and how many cartons of milk does she buy?

Solution

The scale factor compares the servings wanted with the servings the recipe makes:

\frac{24\text{ servings}}{10\text{ servings}} = 2.4

The units cancel, so the factor is a pure number, and multiplying by it keeps each ingredient in its own unit: 240 \times 2.4 = 576\text{ g} of flour, 150 \times 2.4 = 360\text{ g} of blueberries, and 350 \times 2.4 = 840\text{ mL} of milk.

The price is per kilogram, so the grams must cancel before the price can act:

\begin{aligned} &360\text{ g} \times \frac{1\text{ kg}}{1000\text{ g}} \times \frac{16.50\text{ dollars}}{1\text{ kg}} \\ &= 5.94\text{ dollars} \end{aligned}

Without the middle factor the product is 360 \times 16.50 = 5940, in the unit \displaystyle \frac{\text{g} \times \text{dollars}}{\text{kg}}. Grams and kilograms do not cancel, which is how the units show that a conversion belongs in the chain.

The milk is 840 \div 500 = 1.68 cartons. Part of a carton cannot be bought, so she buys 2 cartons. Here the situation, not the arithmetic, decided the rounding, and it rounded up, since 1 carton would be too little.

Example 3

Turns of a wind turbine in a year

The data logger on a wind turbine shows that its blades averaged 14 turns per minute over a whole day. Assuming that rate holds, about how many turns do the blades make in a year?

Solution

The rate is 14 turns per minute, and the time is 1 year, taken as 365 days. Minutes must cancel, so the chain runs from minutes up through hours and days to years:

\begin{aligned} &\frac{14\text{ turns}}{1\text{ min}} \times \frac{60\text{ min}}{1\text{ h}} \\ &\quad \times \frac{24\text{ h}}{1\text{ day}} \times \frac{365\text{ days}}{1\text{ year}} \\ &= 7{,}358{,}400\ \frac{\text{turns}}{\text{year}} \end{aligned}

Every unit but turns per year canceled, so the set-up gives the quantity asked for. Every conversion factor is exact, so the only measured input is the 14, which has two significant figures. The answer keeps two:

\text{about } 7.4 \text{ million turns a year}

Ignoring leap years is safe: a quarter of a day more adds 14 \times 60 \times 24 \times 0.25, about 5000 turns, less than a tenth of a percent. The larger doubt is the model: calm days, storms and repair stops all differ from the one day measured.

Example 4

Liters per 100 km and kilometers per liter

Jolyon’s car uses 6.4 liters of fuel per 100 kilometers. (a) How far does it go on one liter? (b) How much fuel does a 340\text{ km} trip need? (c) His friend’s car does 20 kilometers per liter. Which car uses less fuel?

Solution

(a) The given rate is \displaystyle \frac{6.4\text{ L}}{100\text{ km}}. Kilometers per liter is the same relationship written the other way up:

\frac{100\text{ km}}{6.4\text{ L}} = 15.625\ \frac{\text{km}}{\text{L}}

The 100 is part of the unit, so it is exact; the 6.4 is measured to two significant figures. The car goes about 16\text{ km} on a liter.

(b) The rate can be written either way up, and the units decide which one the trip needs. With kilometers on top, the product has the unit \displaystyle \text{km} \times \frac{\text{km}}{\text{L}} = \frac{\text{km}^2}{\text{L}}, which is not an amount of fuel. With kilometers underneath they cancel and leave liters:

340\text{ km} \times \frac{6.4\text{ L}}{100\text{ km}} = 21.76\text{ L}

which is about 22\text{ L}. The unrounded answer from (a) gives the same: 340 \div 15.625 = 21.76. The rounded 16\text{ km/L} gives 340 \div 16 = 21.25\text{ L}, half a liter adrift: round once, at the end.

(c) Put both cars in the same unit. Turning 20 kilometers per liter over gives \displaystyle \frac{1}{20} of a liter per kilometer, and a hundred kilometers takes a hundred times as much:

\frac{1\text{ L}}{20\text{ km}} \times 100\text{ km} = 5\text{ L}

so the friend’s car uses 5 liters per hundred kilometers, less than Jolyon’s 6.4. A smaller number of liters per 100\text{ km} and a larger number of kilometers per liter both mean a thriftier car.

Example 5

People per square kilometer

A made-up island has about 86{,}000 people, a figure given to the nearest thousand. Its area is 412\text{ km}^2, of which a lake covers 62\text{ km}^2. (a) Find the population density of the whole island. (b) Find the density on the land alone, per square kilometer and per hectare.

Solution

(a) Density is people divided by area, so its unit is people per square kilometer:

\frac{86{,}000\text{ people}}{412\text{ km}^2} = 208.7\ldots\ \frac{\text{people}}{\text{km}^2}

The population has two significant figures (the thousands were rounded), so this is about 210 people per square kilometer.

(b) Nobody lives on the lake. The land is 412 - 62 = 350\text{ km}^2, and

\frac{86{,}000\text{ people}}{350\text{ km}^2} = 245.7\ldots\ \frac{\text{people}}{\text{km}^2}

about 250 people per square kilometer. A square kilometer is 100 hectares, and the area unit sits underneath, so the factor puts square kilometers on top to cancel:

\begin{aligned} &245.7\ \frac{\text{people}}{\text{km}^2} \times \frac{1\text{ km}^2}{100\text{ ha}} \\ &= 2.457\ \frac{\text{people}}{\text{ha}} \end{aligned}

which is about 2.5 people per hectare. Which density is “right” depends on the question: to compare with other places, use their definition; to ask how crowded the land is, leave the lake out. Either way, say which area was used.

Example 6

Water from a dripping tap

Fearghal’s kitchen tap drips. He leaves a measuring jug under it for 10 minutes and it collects 45\text{ mL}; the jug is marked every 5\text{ mL}. About how many liters does the tap waste in a year?

Solution

The model: the tap drips at a steady rate all year, and a year is 365 days. Minutes must cancel on the way to years, then milliliters on the way to liters:

\begin{aligned} &\frac{45\text{ mL}}{10\text{ min}} \times \frac{60\text{ min}}{1\text{ h}} \times \frac{24\text{ h}}{1\text{ day}} \\ &\quad \times \frac{365\text{ days}}{1\text{ year}} \times \frac{1\text{ L}}{1000\text{ mL}} \\ &= 2365.2\ \frac{\text{L}}{\text{year}} \end{aligned}

How accurate is that? The jug was read to the nearest 5\text{ mL}, so the true amount was between about 42.5 and 47.5\text{ mL}. The chain multiplies the milliliters by \displaystyle \frac{60 \times 24 \times 365}{10 \times 1000} = 52.56, so the yearly waste lies between

\begin{aligned} 42.5 \times 52.56 &= 2233.8\text{ L} \\ 47.5 \times 52.56 &= 2496.6\text{ L} \end{aligned}

Even the hundreds digit is uncertain, so two figures is the most the answer can claim: about 2400 liters a year.

Boundaries

Only like units add. A sum or difference needs every term in the same unit, and quantities of different kinds (a length and an area, a mass and a volume) cannot be added at all. Products and quotients can mix units freely; that is how rates are made.

A conversion factor must be a true equality. \displaystyle \frac{3600\text{ s}}{1\text{ h}} is exactly 1. \displaystyle \frac{1.609\text{ km}}{1\text{ mi}} is 1 only to four figures, which matters only in an answer that needs more than four.

Not every unit change is a multiplication. Celsius and Fahrenheit have different zeros, so no single factor converts them: 0^\circ\text{C} is not 0^\circ\text{F}. That conversion multiplies and then adds, which is converting between Celsius and Fahrenheit. Factors equal to one work only for units that share a zero.

Squared and cubed units take the factor more than once. A square kilometer is a million square meters, not a thousand.

Right units do not prove a right answer. The unit check catches a fraction written the wrong way up or a missing conversion. It cannot catch a wrong number inside a correct unit: 7 days per month has the units of a correct factor and the wrong value, and a pure number such as a scale factor carries no unit at all.

A rate is an average. Liters per 100 kilometers, turns per minute and people per square kilometer each describe a whole stretch as if it were even. Using one elsewhere assumes the rate holds there too, and that assumption can limit the answer more than any measurement does.

Where this goes next

Every rate built from two measurements, speed, density or flow rate, is worked out and checked by carrying its units this way; rates from measurements collects them. Rates with fractions inside them are unit rates with fractions. Why metric units convert by powers of ten while miles and inches do not is the story of systems of units. And every slope in algebra is a rate with units, so reading a slope’s unit is how a graph’s rate of change gets its meaning.

Exercises

Converting rates

  1. Write 126\text{ km/h} in meters per second.
  2. Write 12\text{ m/s} in kilometers per hour.
  3. Hywel runs a mile in 6 minutes. Taking 1\text{ mile} \approx 1.609\text{ km}, find his speed in kilometers per hour and in meters per second.
  4. A printer prints 300 dots per inch. Using 1\text{ inch} = 2.54\text{ cm}, how many dots is that per centimeter, to the nearest whole dot?

Prices and recipes

  1. A 250\text{ g} punnet of mushrooms costs 2.15 dollars. Find the price per kilogram.
  2. Rice costs 4.20 dollars for a 1.5\text{ kg} bag, or 1.20 dollars for a 400\text{ g} bag. Which is cheaper per kilogram, and by how much?
  3. A recipe for 8 scones uses 180\text{ g} of butter. How much butter do 20 scones need, and what does it cost at 9.60 dollars per kilogram?
  4. A dessert for 6 people uses 400\text{ mL} of cream. Kittie makes it for 15 people, and cream is sold in 300\text{ mL} tubs. How many tubs does she buy?

Fuel, turns, people and water

  1. A car uses 7.5 liters per 100 kilometers. About how far does it go on a liter, and about how much fuel does a 250\text{ km} trip take?
  2. A car does 18 kilometers per liter. Write its fuel use in liters per 100 kilometers.
  3. A metronome ticks steadily at 75 beats per minute. About how many days does it take to tick a million times?
  4. A made-up town has 5{,}400 people on 3.6\text{ km}^2. Find its population density per square kilometer and per hectare.
  5. A region has 120 people per square kilometer. Taking 1\text{ mile} \approx 1.609\text{ km}, about how many people is that per square mile?
  6. A tap drips once every 2 seconds, and each drop is about 0.05\text{ mL}. About how many liters does it waste in a 365-day year?

Checking a set-up by its units

  1. A car uses 6 liters per 100 kilometers on a 150\text{ km} trip. Find the unit of \displaystyle 150\text{ km} \times \frac{100\text{ km}}{6\text{ L}} and the unit of \displaystyle 150\text{ km} \times \frac{6\text{ L}}{100\text{ km}}. Which product is the fuel for the trip, and how much fuel is it?
  2. Write the conversion of 15\text{ m/s} to kilometers per hour as a chain of factors equal to one. Does the chain multiply 15 by 3.6 or divide it by 3.6? Give the speed in kilometers per hour.
  3. Fearghal fills his car’s tank, drives 538\text{ km}, and fills it again with 41.3 liters. Find the car’s fuel use in liters per 100 kilometers and in kilometers per liter, each to a sensible accuracy.

Reasoning

  1. Explain why multiplying a length by \displaystyle \frac{1000\text{ m}}{1\text{ km}} changes the number but not the length, and how you decide which way up to write the fraction.
  2. A model of a trip’s fuel uses only the distance and the car’s liters per 100 kilometers. Name two things it leaves out that change how much fuel is really used, and say when leaving them out is reasonable.
  3. In the wind turbine example, explain why the factors 60, 24 and 365 do not limit how many significant figures the answer keeps, while the 14 does.
Answers
  1. 35\text{ m/s}. 126 \times 1000 \div 3600 = 35.
  2. 43.2\text{ km/h}. 12 \times 3600 \div 1000 = 43.2.
  3. About 16.1\text{ km/h} and about 4.47\text{ m/s}. 6 minutes is a tenth of an hour, so 1.609 \times 10 = 16.09; and 1609\text{ m} \div 360\text{ s} = 4.469\ldots
  4. 118 dots per centimeter. 300 \div 2.54 = 118.1\ldots
  5. 8.60 dollars per kilogram. 2.15 \div 0.25 = 8.60.
  6. The 1.5\text{ kg} bag, by 0.20 dollars per kilogram. 4.20 \div 1.5 = 2.80 against 1.20 \div 0.4 = 3.00.
  7. 450\text{ g}, costing 4.32 dollars. \displaystyle 180 \times \frac{20}{8} = 450, and 0.45 \times 9.60 = 4.32.
  8. 4 tubs. 400 \times 15 \div 6 = 1000\text{ mL}, and 1000 \div 300 = 3.33\ldots; three are too few.
  9. About 13\text{ km} per liter, and about 19\text{ L}. 100 \div 7.5 = 13.33\ldots, and 250 \times 7.5 \div 100 = 18.75; the 7.5 has two significant figures.
  10. About 5.6\text{ L} per 100\text{ km}. 100 \div 18 = 5.55\ldots
  11. About 9.3 days. A day holds 75 \times 60 \times 24 = 108{,}000 ticks, and 1{,}000{,}000 \div 108{,}000 = 9.259\ldots
  12. 1500 people per square kilometer, which is 15 people per hectare. 5400 \div 3.6 = 1500, and a square kilometer is 100 hectares.
  13. About 310 people per square mile. A square mile is about 1.609^2 \approx 2.589\text{ km}^2, and 120 \times 2.589 = 310.6\ldots
  14. About 800\text{ L}. 0.025\text{ mL} per second for 31{,}536{,}000 seconds is 788{,}400\text{ mL} = 788.4\text{ L}; the drop size has one significant figure.
  15. The first gives \displaystyle \text{km} \times \frac{\text{km}}{\text{L}} = \frac{\text{km}^2}{\text{L}}, not an amount of fuel. The second leaves liters: \displaystyle 150 \times \frac{6}{100} = 9\text{ L}.
  16. 54\text{ km/h}, by multiplying by 3.6. The chain is 15\text{ m/s} \times (3600\text{ s}/1\text{ h}) \times (1\text{ km}/1000\text{ m}); the two factors together multiply by 3600/1000 = 3.6, so the speed is 15 \times 3.6 = 54\text{ km/h}.
  17. About 7.68\text{ L} per 100\text{ km} and about 13.0\text{ km} per liter. 41.3 \div 538 \times 100 = 7.676\ldots and 538 \div 41.3 = 13.02\ldots; both measurements have three significant figures.
  18. The fraction equals 1, because 1000\text{ m} and 1\text{ km} are the same length, and multiplying by 1 leaves a quantity unchanged; the number grows to make up for the smaller unit. The unit to be removed goes on the opposite side of the bar, so kilometers on top of the length means kilometers underneath in the factor.
  19. Any two of: speed, hills, traffic, the load carried, the weather, air conditioning. Leaving them out is reasonable when the fuel figure is an average over similar journeys, or when their effect is smaller than the accuracy the answer is reported to.
  20. 60 and 24 come from definitions, and 365 from the choice of a 365-day year, so they carry no measurement error. The 14 was measured, good to about two figures, and the answer can be no better than that.
kg1128 · updated 2026-10-10 · CC-BY-SA-4.0