Ratios, rates and proportional reasoning

A ratio compares quantities in a stated order

A group contains six A-counters and four B-counters. The ratio of A to B is 6:4, read as six to four. Order matters: the ratio of B to A is 4:6. State which quantity comes first and which comes second, rather than leaving a pair of numbers without labels. Both ratios describe the same group, but answer different comparisons.

When the second quantity is nonzero, the numerical ratio a:b can be represented by the quotient a/b. Here A/B = 6/4 = 3/2, so the A-count is one and a half times the B-count. If the second count is zero, that quotient is undefined. A count pair such as 3:0 can record that one category is absent, but it does not make 3/0 a defined division. With 0:0, no unique relative mixture is determined.

A part-to-part ratio is different from a share of the total

The recipe 3 A-counters to 2 B-counters contains five counters per complete bundle. Its part-to-part ratio is 3:2, but the A-share of the whole is 3/(3 + 2) = 3/5 and the B-share is 2/5. The two shares add to one. Writing 3/2 for the A-share would confuse a comparison with B alone with a comparison with the whole group.

For nonnegative counts a and b with a positive total a + b, the part-to-whole fractions are a/(a + b) and b/(a + b). In the group of six A-counters and four B-counters, these are 6/10 = 3/5 and 4/10 = 2/5. The total is a count of counters, not their combined mass or size. If their individual masses differ, the same count ratio need not give the same mass shares. Name the quantity whose parts are being counted or measured.

Preserve a ratio by the same multiplicative change

Two complete 3:2 bundles give 6 A-counters and 4 B-counters; four bundles give 12 and 8. Thus 3:2, 6:4 and 12:8 are equivalent ratios. Multiplying both entries by the same positive factor preserves the relative amounts. In quotient form, a/b = (k × a)/(k × b) when b and k are nonzero. Dividing both whole-number counts by a common positive factor gives a simpler equivalent ratio, as 18:12 becomes 3:2.

Adding the same amount is not a general ratio-preserving rule: adding 1 to both entries of 3:2 gives 4:3, and 3/2 is not 4/3. The difference stayed one, but the multiplicative comparison changed. Scaling the recipe by zero gives an empty selection with counts 0 and 0. The recipe is still known, but the empty selection has no actual part-to-part quotient or part-to-whole shares. For indivisible counters in complete bundles the scale factor is a whole number; continuous measured amounts can allow fractional factors.

Use comparable quantities and a shared unit

The ratio of 2 metres to 50 centimetres is not found by comparing the bare numbers 2 and 50. Convert to one unit: 2 m = 200 cm, so 200 cm:50 cm = 4:1. Their quotient is 4 because the centimetre units cancel. Converting both lengths to metres gives 2 m:0.5 m, with the same numerical ratio. A same-kind measurement ratio should not depend on which shared unit you choose.

A comparison of different kinds of quantities instead retains a compound unit. Ten litres delivered in five minutes give 10 L ÷ 5 min = 2 L/min. This is a rate, expressing volume per time. Keep both units visible; the bare number 2 could otherwise refer to a counter ratio, a length comparison or litres per minute. A ratio between object counts should also retain its labels, as three A-counters for every two B-counters.

A unit rate gives the amount for one unit of the other quantity

A unit rate expresses a comparison per one unit, such as litres per minute, kilometres per hour or euros per kilogram. Divide the first quantity by the second, and retain the resulting units. If 12 litres are delivered in 6 minutes, the average delivery rate over that interval is 2 L/min. If a constant-flow assumption is supplied, this same rate describes every subinterval; the total and elapsed time alone do not prove constant flow.

Convert units before comparing rates. Two litres per minute are 120 litres per hour, because one hour contains sixty minutes: 2 L/min × 60 min/h = 120 L/h. These are equal rates with different numerals, not a change in delivery. For price comparisons, convert package amounts to the same mass unit and compare price per kilogram. A lower unit price addresses that comparison; it does not automatically mean a lower total payment when only whole packages can be bought.

A proportional relationship has one constant multiplier

In a direct proportional relationship, y = k × x, where the multiplier k stays fixed across the allowed values of x. For an initially empty tank receiving water at a constant 2 L/min, volume y in litres equals 2 times elapsed time t in minutes: y = 2 × t. At t = 1, 2 and 3 minutes the volumes are 2, 4 and 6 litres. The multiplier includes its units; it is 2 L/min when these are the chosen volume and time units.

For nonzero x, the quotient y/x equals the constant k. The formula y = k × x also gives y = 0 when x = 0; if zero is in the model’s domain, its graph passes through the origin. The quotient 0/0 is still undefined: the zero case comes from the multiplication rule, not from dividing zero by zero. Doubling x doubles y under this rule. With positive k and nonnegative x, increasing x also increases y. State the model’s range of use, including available capacity and any changes in its conditions.

Keep corresponding quantities aligned

A ratio table records matching values with named columns. For the 3:2 counter recipe, bundle counts 1, 2 and 3 correspond to A-counts 3, 6 and 9 and B-counts 2, 4 and 6. Moving between rows multiplies corresponding quantities by the same factor. A double number line shows the same correspondence: 0, 3, 6, 9 A-counters align with 0, 2, 4, 6 B-counters. The scales differ, but each alignment describes one selection.

For the empty tank, align times 0, 1, 2, 3 minutes with volumes 0, 2, 4, 6 litres. A graph places each corresponding pair at one point; the proportional rule produces a straight line through the origin. Tables, number lines and graphs help expose the structure and catch inconsistent rows. Agreement in a few displayed pairs does not by itself prove that the rule continues everywhere: the relationship must follow from stated conditions or an established mathematical rule.

A proportion states that two ratios are equal

Suppose the 3:2 recipe is used with eight B-counters. Scaling the B-count from 2 to 8 uses factor 4, so the A-count is 3 × 4 = 12. The proportion is 3/2 = A/8. The same answer can be found through a unit ratio: A = (3/2) × 8 = 12. State which entries represent A and B before solving, so that the fractions keep the same order.

For a/b = c/d with nonzero b and d, multiplying both sides by b × d gives a × d = c × b. These cross-products express equality of the ratios. For 3/2 = A/8, this gives 3 × 8 = 2 × A, so A = 12. Verify by substitution: 12/8 = 3/2. Cross-products do not make a zero denominator valid, and they do not establish that a physical relationship is proportional unless the setup warrants that assumption.

A starting amount separates total from added quantity

Compare two tanks with the same constant inflow of 2 L/min. One starts empty; the other already contains 3 litres. Their total volumes are y = 2 × t and y = 3 + 2 × t. In both, each extra minute adds 2 litres, but only the empty tank’s total volume is proportional to elapsed time. The second rule includes a nonzero starting amount; a linear rule with a fixed added constant is called affine.

For the tank starting with 3 litres, total volume divided by elapsed time is 7/2 = 3.5 L/min at 2 minutes, and 11/4 = 2.75 L/min at 4 minutes. These quotients include the initial stock; they are not the actual inflow rate. The inflow is found from changes: (11 − 7) L ÷ (4 − 2) min = 2 L/min. Added volume y − 3 is proportional to time, even though total y is not. Constant increments for equal time steps alone do not show proportionality; the starting amount also matters.

Scale a recipe, then compare two equal inflows

Begin with the counter recipe and factor 2 in the explorer. Count two complete bundles, each containing three A-counters and two B-counters. The selected counts are 6:4 = 3:2, the total is 10, and the shares are 3/5 and 2/5. Change the factor while keeping the recipe fixed. At zero, distinguish the known recipe from the empty selection: the actual counts are 0:0, and its quotient and shares are undefined.

Switch to the tanks and choose the same elapsed time for both. Read the table and graph for total volume, compare their identical added volume, and inspect the quotient total volume/time when time is nonzero. The two lines keep the same constant inflow of 2 L/min, but the tank starting with 3 litres has a varying total-volume/time quotient. At time zero both totals are defined by their formulas, while division by elapsed time is undefined. Explain which quantity is proportional, rather than deciding from a repeated word such as rate.

Distinguish direct scaling, inverse scaling and whole counts

Direct proportionality keeps a quotient fixed: y = k × x. Inverse proportionality instead keeps a nonzero product fixed: y = c/x for nonzero x and a fixed nonzero c. To deliver a fixed 12 litres at a constant positive flow r, the required time is t = 12/r. At 2 L/min it takes 6 minutes; at 4 L/min it takes 3 minutes. Doubling the flow halves the time, and r × t remains 12 litres. This is a different relationship from the growing volume of one tank at a fixed flow.

Check what values the quantities can take. Twelve indivisible counters cannot be split exactly into the 3:2 recipe: the required shares are 36/5 = 7.2 A-counters and 24/5 = 4.8 B-counters. Rounding those counts changes the ratio. For complete counter bundles, totals must be multiples of five. Continuous lengths totaling 12 metres can instead be 7.2 m and 4.8 m in that same ratio. The tank explorer selects whole-minute sample points, but the stated constant-flow model can also describe intermediate times while its physical conditions hold.

Check what stays constant and why

For a ratio, check the order and the shared scaling factor. For a unit rate, check both the numerical value and its units. For a proportional model, check the constant quotient at nonzero inputs and the zero case through the formula. For an inverse model, check the constant product. A straight-line pattern with a starting amount requires the change rate and the initial value, rather than one constant total-to-input ratio.

Return to the conditions after calculating. Is flow constant? Does the tank have room, and is anything leaving it? Are the items divisible or sold only in complete units? Does the chosen comparison concern counts, mass, length or volume? A calculation can be arithmetically correct while describing a different quantity from the one requested. Make the assumed rule explicit and limit predictions to where it applies. The next lesson uses part-to-whole fractions and multiplicative changes to explain percentages and percentage change.

Worked examples

Try it yourself

1. A group has six A-counters and four B-counters. Write the simplified A:B and B:A ratios, then the A- and B-shares of the whole. Does 3/2 describe the A-share?

   Show the answer and reasoning

   A:B = 3:2; B:A = 2:3. The shares are 3/5 and 2/5. The quotient 3/2 compares A with B, not A with the whole.

   Divide both count entries by 2 to simplify each ordered ratio. The total is ten, so the shares are 6/10 and 4/10. Their denominators refer to the total count; the denominator in A/B refers only to the B-count.

2. Starting from 3:2, compare multiplying both entries by 4 with adding 4 to both entries. Then describe a zero-bundle selection: is its actual 0:0 count pair an equivalent numerical ratio or a pair of defined shares?

   Show the answer and reasoning

   Multiplication gives the equivalent ratio 12:8. Addition gives 7:6, which is different. A zero-bundle selection has no defined actual ratio quotient or part-to-whole shares, although its recipe remains known.

   12/8 = 3/2, while 7/6 is not 3/2. The multiplicative comparison is preserved by the same nonzero factor, not generally by the same addition. At zero bundles both entries and their total are zero, so the relevant quotients would be 0/0 and are undefined.

3. Find the ratio of 2 m to 50 cm. Find the rate for 10 litres delivered in five minutes, and express it in litres per hour. Which comparison loses its measurement unit, and which retains compound units?

   Show the answer and reasoning

   The length ratio is 4:1, with dimensionless quotient 4. The delivery rate is 2 L/min = 120 L/h, retaining volume-per-time units.

   Convert 2 m to 200 cm, then divide 200 cm by 50 cm so the equal units cancel. For delivery, divide 10 L by 5 min. Multiplying 2 L/min by 60 min/h converts the time unit and gives the same rate in litres per hour.

4. The ratio A:B is 3:2 and B = 8. Find A by scaling and by the proportion 3/2 = A/8, then check the result. What denominator condition permits the cross-products step?

   Show the answer and reasoning

   A = 12. The check is 12/8 = 3/2. Both denominators must be nonzero.

   Scaling 2 to 8 uses factor 4, so scale 3 to 12. Cross-products give 3 × 8 = 2 × A, hence A = 12. The denominators 2 and 8 are nonzero; a zero denominator would make the fraction undefined before cross-products could be used.

5. For tanks with rules y = 2 × t and y = 3 + 2 × t, find the totals at two and four minutes. Which total is proportional to time? For the second tank, compare y/t with its actual inflow rate.

   Show the answer and reasoning

   The first totals are 4 L and 8 L; the second are 7 L and 11 L. Only the first total is proportional to time. The second has y/t values 3.5 and 2.75 L/min, while its inflow is 2 L/min.

   The first quotient stays 2 at nonzero times, and its starting total is zero. The second quotient includes the initial 3 litres and varies. Its change rate is (11 − 7) L ÷ (4 − 2) min = 2 L/min, showing that a constant inflow does not remove the starting amount.

6. A fixed 12 litres must be delivered at a constant positive flow. Find the time at 2 L/min and 4 L/min; is the time directly or inversely proportional to flow? Separately, can 12 indivisible counters form an exact 3:2 group? Could two lengths totaling 12 m have that ratio?

   Show the answer and reasoning

   Times are 6 minutes and 3 minutes; the relationship is inverse. Twelve indivisible counters cannot form the exact group, but lengths 7.2 m and 4.8 m can have the ratio 3:2.

   Time is 12 divided by flow, and flow × time stays 12 litres. For the group, shares 3/5 and 2/5 require counts 7.2 and 4.8, which are not whole counters; rounding changes the ratio. Continuous lengths can take those fractional amounts, with a positive total and the required quotient 7.2/4.8 = 3/2.

7. A table lists times 1, 2 and 3 minutes with volumes 2, 4 and 6 litres. Does that table alone prove a proportional rule for every future time? State assumptions that justify y = 2 × t for a tank, and predict the volume at four minutes under them.

   Show the answer and reasoning

   The table alone does not prove the rule beyond the displayed pairs. With an empty start, constant inflow 2 L/min, no outflow and sufficient capacity during the interval, the rule gives 8 litres at four minutes.

   The listed quotients agree, which supports checking the proposed rule at those inputs. A different behavior later could still fit all the listed pairs. The stated physical conditions explain why each elapsed minute contributes the same volume and why there is no starting offset; keep predictions within the interval where those conditions hold.

What to carry forward

• A ratio has an order and named quantities. A 3:2 part-to-part ratio gives shares 3/5 and 2/5 of the whole, not 3/2 of the whole.
• Equivalent ratios scale corresponding quantities by the same nonzero factor. An empty 0:0 selection has no unique actual ratio or shares, even when its recipe is known.
• Use shared units for same-kind measurement ratios. Unit rates retain compound units; convert those units before comparing rates.
• Direct proportionality follows y = k × x and has constant y/x when x is nonzero. A nonzero starting amount changes the total rule even if the rate of increase stays constant.
• Cross-products require nonzero denominators; inverse proportionality keeps a product fixed. Check whole-count constraints and the conditions supporting any prediction.