Ratios, rates, proportions, and units

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Track the quantities before you calculate.

The central idea: preserve meaning while changing form

A ratio compares quantities. A rate is a ratio whose units usually differ. A proportion states that two ratios are equal. These ideas become reliable tools when you label what each number measures, rather than immediately cross-multiplying.

Parts, wholes, and shared quantities

If $A ∶ B = a ∶ b,$ write $A = ak$ and $B = bk$ for a common scale factor k. When A and B are the only parts of a whole, that whole is $(a + b)k.$ Thus

$\frac{A}{B}$ $\frac{A}{A + B}$

These fractions answer different questions. In a $2 ∶ 7$ ratio, the first group is $2/7$ as large as the second group but $2/9$ of their combined total.

For linked ratios, make the shared quantity agree. From $A ∶ B = 2 ∶ 3$ and $B ∶ C = 6 ∶ 5,$ scale the first ratio to $4 ∶ 6.$ Then $A ∶ B ∶ C = 4 ∶ 6 ∶ 5.$ Do not compare A and C until the two representations of B use the same scale.

Direct proportionality versus a general linear relationship

A directly proportional relationship has the form $y = kx,$ where the constant $k = y/x$ is a unit rate for nonzero inputs. Its graph passes through the origin. Multiplying one quantity by a factor multiplies the other by the same factor. A rule $y = mx + b$ with $b \neq 0$ can have constant slope without being proportional: a fixed setup charge destroys a constant cost-per-item ratio.

A supplied statement of proportionality justifies extending a ratio to new values. Several matching pairs in a finite table do not logically establish one unique rule for all inputs unless the problem supplies the model or asks for the best model among choices.

A dependable setup

Name the requested quantity. Write one equality that preserves the labeled ratio, or find the unit rate and multiply. For a recipe, for instance,

$\frac{new servings}{old servings}$

The new-to-old ratio is a multiplier, not an additive increase.

Read the units as instructions

“Dollars per kilogram” means dollars divided by kilograms. Multiplying that rate by kilograms produces dollars. If your expression instead produces kilograms squared per dollar, its arithmetic may be valid but its meaning is wrong. An estimate of the expected size gives a second independent check.

Unit conversions that survive difficult problems

A conversion factor is a ratio equal to 1: its top and bottom describe the same physical quantity. Choose its orientation so the old unit cancels. For example,

$\frac{12 in}{1 ft}$

For a compound rate, convert each component. If converting miles per hour to feet per second, feet belong above miles, while hours belong above seconds. Do not decide operations from whether a unit is “larger” until you have checked its position in the fraction.

Square and cubic units change by powers

If $1 m = 100 cm,$ then

$1 m 2 = (100 cm) 2 = 10,000 cm 2,$ $1 m 3 = 1,000,000 cm 3 .$

The exponent applies to both the number and the unit. But a supplied conversion such as $1 m 3 = 1,000 L$ already compares volume units; do not cube 1,000 again. In a density conversion, mass and volume may move in opposite directions. Converting $g/cm 3$ to $kg/m 3$ changes the numerator by $1/1,000$ and the denominator by $1/1,000,000,$ producing a net factor of 1,000.

Derived units encode relationships

Quantity	Relationship	Useful reading
Speed	$v = d/t$	Distance covered per unit time.
Density	$ρ = m/V$	Mass per unit volume.
Production rate	$r = q/t$	Items completed per unit time.
Energy at constant power	$E = Pt$	Kilowatt-hours are kilowatts times hours.

A word problem may require a chain rather than one formula: operating hours → kilowatt-hours → cost. Write intermediate units so each conversion has a purpose.

Scale drawings

A bare scale such as $1 ∶ 5,000$ compares lengths in matching units. Convert a map length to a real length with the scale, then convert to the requested unit. If the question asks for area or volume and the figures are similar, square or cube the length scale factor. A drawing’s apparent size is not a measurement.

Multistep rates: choose what adds

Average speed is a quotient of totals

For an entire trip,

$\frac{total distance}{total elapsed time}$

Elapsed time includes stated stops. Averaging two speeds directly works only when the time spent at each speed is equal. Equal distances usually take unequal times. In that case, compute each time using $t = d/v,$ add those times, and divide the total distance by their sum. The result must lie between the two speeds when both travel times are positive.

Simultaneous work: add rates, not completion times

If one machine completes a job in a hours at a constant rate, its rate is $1/a$ job per hour. Two machines working independently on the same job at constant rates complete $1/a + 1/b$ job per hour, so their combined time is

$\frac{1}{1/a + 1/b}$

This is not a rule to apply blindly. It assumes their rates can be combined without interference. A draining pipe has a negative contribution to a filling rate. The combined completion time should be shorter than either individual time when both machines contribute positively.

Many factors, one model

When identical machines work at a constant rate, total production is proportional to both machine count and operating time. If only a fraction of the output passes inspection, apply that fraction to the total production, not to a quantity with incompatible units. Record which conditions are held fixed; a changing efficiency invalidates an unchanged unit rate unless the change is included.

Before accepting a rate answer

Check three things: the units are the requested units; the multiplier has the right direction; and the size is reasonable. For a combined-work problem, ask whether the result is faster. For a unit conversion, ask whether you changed the measurement or only its representation. For an average, ask what weights the component rates.

What to practice next

Examples 1.01–1.05 establish ratios and proportionality. Examples 1.06–1.11 build conversion chains. Examples 1.12–1.15 combine rates, changing totals, and multiple scale factors. A calculator helps with arithmetic; the units and the model still have to come from you.

15 worked examples

Try the question before reading the solution. The examples progress from Foundation to Challenge.

1.01. Turn a part‐to‐part ratio into a share of the whole

A box contains only red and blue tiles. The ratio of red tiles to blue tiles is 3 ∶ 5. If there are 64 tiles in the box, how many are red?

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Recognize the structure. The two ratio numbers describe parts, not percentages. The whole contains $3+5 = 8$ equal ratio units.

Work it through. Each ratio unit represents $64/8 = 8$ tiles. Red tiles occupy three of those units, so

$R = 3(8) = 24.$

Equivalently, red tiles make up $3/(3 + 5) = 3/8$ of the box, and $(3/8)(64) = 24.$

Answer: 24 red tiles.

Check. There are $64 - 24 = 40$ blue tiles, and $24 ∶ 40$ simplifies to $3 ∶ 5.$ The two counts add to 64.

Avoid the trap. Using $3/5$ of 64 treats the blue count as the whole. Label the ratio before writing a fraction.

1.02. Find and use a unit price

A store charges $8.40 for 3 kilograms of rice. At the same price per kilogram, how much do 7.5 kilograms cost?

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Recognize the structure. The price per kilogram stays constant, so cost is proportional to mass.

Work it through. First calculate the unit rate, retaining its units:

$\frac{$8.40}{3 kg}$

Then multiply by the desired mass: $($2.80/kg)(7.5 kg) = $21.00.$ The kilograms cancel.

Answer: $21.00.

Check. 7.5 kg is $7.5/3 = 2.5$ times the original mass, and $2.5(8.40) = 21.$ Both quantities scale by the same factor.

Avoid the trap. Dividing 7.5 by 2.80 reverses the meaning of the rate. The units would not produce dollars.

1.03. Scale a recipe without rounding the scale factor

A recipe uses 2 $\frac{1}{2}$ cups of flour to make 6 servings. The ingredients are scaled proportionally. How many cups of flour are needed for 15 servings?

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Recognize the structure. Use new servings divided by original servings as the scale factor.

Work it through. The scale factor is $15/6 = 5/2.$ Write the original flour quantity as $5/2$ cups:

$\frac{5}{2}$

Fractions keep the arithmetic exact and make the cancellation visible.

Answer: 6.25 cups.

Check. The batch is two and a half times as large, so it should use two and a half times as much flour: $2.5(2.5) = 6.25.$

Avoid the trap. The scale factor is not the increase in servings, $15-6 = 9.$ Proportional scaling is multiplicative, not additive.

1.04. Recognize a proportional table

The table gives values of two quantities that are known to be directly proportional.
x
2
5
8
y
7
17.5
28
What is y when x = 12?

x	2	5	8
y	7	17.5	28

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Recognize the structure. Direct proportionality means $y = kx;$ the quotient $y/x$ must be constant.

Work it through. Each given pair has $y/x = 3.5.$ Therefore $y = 3.5x,$ and

$y = 3.5(12) = 42.$

The statement that the quantities are proportional supplies the model. A finite table alone would not force that rule for every possible input.

Answer: 42.

Check. From $x = 8$ to $x = 12,$ the input is multiplied by 1.5. The output is also multiplied by 1.5: $28(1.5) = 42.$

Avoid the trap. A linear rule with a nonzero fixed fee, such as $y = 3.5x + 10,$ is not directly proportional. A constant slope and a constant ratio are different properties.

1.05. Orient a conversion factor

A ribbon is 2.4 meters long. What is its length in centimeters? Use 1 m = 100 cm.

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Recognize the structure. Choose the form of the conversion factor that cancels meters.

Work it through. The ratio $100 cm/1 m$ equals 1, because its numerator and denominator describe the same length. Thus

$\frac{100 cm}{1 m}$

The measurement changes its numerical representation, not its physical size.

Answer: 240 centimeters.

Check. Centimeters are smaller than meters, so the numerical count of units must increase. Dividing by 100 would fail this size check.

Avoid the trap. Memorizing “move the decimal” without tracking units becomes unreliable when a problem combines several conversions.

1.06. Convert both parts of a rate

A vehicle travels at 72 kilometers per hour. What is this speed in meters per second? Use 1 km = 1,000 m and 1 h = 3,600 s.

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Recognize the structure. A rate has both a numerator unit and a denominator unit. Both must become the requested units.

Work it through. Write one chain so every unwanted unit cancels:

$\frac{km}{h}$

Notice that the hour conversion is inverted relative to converting an isolated number of hours into seconds.

Answer: 20 meters per second.

Check. At $20 m/s,$ the vehicle travels $20(3,600) = 72,000 m$ in an hour, or 72 km.

Avoid the trap. Multiplying by 3,600 instead of dividing makes the time unit fail to cancel. Let dimensional cancellation determine the operation.

1.07. Connect a scale drawing to a real distance

On a map with scale 1 ∶ 25,000, a trail measures 3.6 centimeters. What is the trail’s actual length in kilometers? Use 100,000 cm = 1 km.

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Recognize the structure. A numerical scale compares lengths expressed in the same unit.

Work it through. One map centimeter represents 25,000 actual centimeters. The trail therefore has actual length $3.6(25,000) = 90,000 cm = 0.9 km.$

The map-to-reality multiplication comes before, or can be combined with, the centimeter-to-kilometer conversion.

Answer: 0.9 kilometer.

Check. One centimeter on the map represents 0.25 km. Therefore 3.6 map centimeters represent $3.6(0.25) = 0.9 km.$

Avoid the trap. The scale does not mean one centimeter represents 25,000 kilometers. A bare scale ratio uses like units.

1.08. Read a derived unit as an equation

A uniform material has density 2.7 grams per cubic centimeter. A sample has mass 324 grams. What is the sample’s volume, in cubic centimeters?

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Recognize the structure. Density is mass divided by volume, so volume is mass divided by density.

Work it through. Let V be the volume. Then $2.7 = 324/V,$ giving

$\frac{324 g}{2.7 g/ {cm}^{3}}$

Dividing by grams per cubic centimeter leaves cubic centimeters, the requested unit.

Answer: 120 cubic centimeters.

Check. The mass implied by this volume is $(2.7)(120) = 324 g,$ exactly the stated mass.

Avoid the trap. Multiplying mass by density gives units $g 2 /cm 3,$ not volume. Dimensional analysis detects the reversed formula.

1.09. Use a product unit: kilowatt‐hours

A heater uses power at a constant rate of 1.2 kilowatts for 3.5 hours. Electricity costs $0.18 per kilowatt-hour. What is the electricity cost, to the nearest cent?

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Recognize the structure. Energy equals power multiplied by time. A kilowatt-hour is a product unit, not kilowatts per hour.

Work it through. The energy used is $1.2 kW \times 3.5 h = 4.2 kWh.$ Its cost is

$\frac{$0.18}{kWh}$

Round only the final cost: the third decimal digit is 6, so the result is $0.76.

Answer: $0.76.

Check. About 4 kWh at about $0.20 per kWh should cost about $0.80. The answer has the right scale.

Avoid the trap. Dividing power by time changes the physical quantity. Also, rounding intermediate energy values can shift a final cent.

1.10. Square a length conversion for an area

A panel has area 2.5 square meters. What is its area in square centimeters? Use 1 m = 100 cm.

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Recognize the structure. Area contains two factors of length, so both must be converted.

Work it through. A square meter is a 100 cm by 100 cm square. Consequently,

$\frac{100 cm}{1 m}$

The exponent applies to the conversion factor as well as to the unit.

Answer: 25,000 square centimeters.

Check. Two square meters already contain 20,000 square centimeters, so an answer such as 250 is too small.

Avoid the trap. Multiplying by 100 converts only one dimension. For volume, the corresponding factor would be cubed.

1.11. Convert a density across mass and volume units

A material’s density is 1.8 g/cm³. Express the density in kilograms per cubic meter. Use 1 kg = 1,000 g and 1 m = 100 cm.

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Recognize the structure. Convert mass once, but convert the denominator’s length three times.

Work it through. Arrange the chain to cancel grams and cubic centimeters:

$\frac{g}{{cm}^{3}}$

The factor $1003$ is in the numerator because the original cubic centimeters are in the denominator.

Answer: 1,800 kilograms per cubic meter.

Check. A cubic meter contains 1,000,000 cubic centimeters. At 1.8 g each, its mass is 1,800,000 g, or 1,800 kg.

Avoid the trap. A density conversion is not a length conversion. Forgetting the cube, or multiplying by 1,000 when changing grams to kilograms, can produce plausible-looking but incorrect units and magnitudes.

1.12. Add work rates, not completion times

Pump A can fill an empty tank in 4 hours, and pump B can fill the same tank in 6 hours. Each pump works at a constant rate. When both start together and no water leaves the tank, how many hours are required to fill it?

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Recognize the structure. The pumps contribute fractions of one tank per hour. These simultaneous rates add.

Work it through. Pump A fills $1/4$ tank per hour; pump B fills $1/6.$ Together their rate is

$\frac{1}{4}$ tank/hour.

Time equals work divided by rate, so $1 \div (5/12) = 12/5 = 2.4$ hours.

Answer: 2.4 hours.

Check. In 2.4 hours, A fills $2.4/4 = 0.6$ tank and B fills $2.4/6 = 0.4$ tank. Together they fill exactly one tank.

Avoid the trap. Averaging 4 and 6 gives 5 hours, slower than A alone. Two positive simultaneous rates must finish sooner than either alone.

1.13. Compute an average speed from totals

A cyclist travels 60 miles at 30 miles per hour and then another 60 miles at 60 miles per hour. There are no stops. What is the average speed for the entire 120-mile trip?

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Recognize the structure. Average speed is total distance divided by total elapsed time, not usually the arithmetic mean of the speeds.

Work it through. The first leg takes $60/30 = 2$ hours; the second takes $60/60 = 1$ hour. Therefore

$\frac{60 + 60}{2 + 1}$

The cyclist spends twice as much time at the slower speed, so that speed receives more weight.

Answer: 40 miles per hour.

Check. Traveling for 3 hours at 40 mph would also cover 120 miles. The result lies between 30 and 60 and is closer to 30.

Avoid the trap. The two distances are equal, not the two times. An unweighted average of speeds would be justified for equal-duration legs, not equal-distance legs.

1.14. Model a ratio that changes after an addition

A collection initially contains red and blue counters in the ratio 3 ∶ 5. After 12 red counters are added and no blue counters are changed, the numbers of red and blue counters are equal. How many counters were in the original collection?

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Recognize the structure. Represent the original counts with a shared scale factor, then change only the stated count.

Work it through. Write the original counts as $3k$ and $5k.$ The new equality is $3k + 12 = 5k,$ so $k = 6.$ The original total was $3k + 5k = 8k = 48.$

The new total is not the requested quantity.

Answer: 48 counters.

Check. Originally there were 18 red and 30 blue counters. Adding 12 red gives 30 of each.

Avoid the trap. Adding 12 to the ratio number 3 treats a ratio unit as one counter. First determine how many actual counters one ratio unit represents.

Another route. The original blue-red difference is two ratio units. Those two units equal 12 counters, so each unit is 6 counters and all eight units total 48.

1.15. Scale production in several independent dimensions

Four identical machines produce 720 parts in 3 hours. Production is proportional to the number of machines and to operating time. Six machines run for 5 hours at the same rate, and 80% of their parts pass inspection. How many parts pass inspection?

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Recognize the structure. Find output per machine-hour, then scale the machine count, duration, and usable fraction separately.

Work it through. The original production used $4(3) = 12$ machine-hours, giving $720/12 = 60$ parts per machine-hour. The new run uses $6(5) = 30$ machine-hours, so it makes $60(30) = 1,800$ parts. Of these,

$0.80(1,800) = 1,440$

pass inspection.

Answer: 1,440 parts.

Check. The new raw output is multiplied by $(6/4)(5/3) = 2.5;$ $720(2.5) = 1,800.$ Passing output must be less than raw output.

Avoid the trap. Do not scale by machine count while overlooking time. Apply the pass percentage to the new production, not to the original 720.

Section 1: exit questions

Try the five section-exit questions before checking the explanations. Use scratch paper for your reasoning and enter the requested numerical value for a question without choices.

Topic practice