A quantity doubles. What should happen to the other quantity? Sometimes it doubles too. Sometimes it halves. Sometimes it increases by a fixed amount. Sometimes the information is not enough to decide. Ratio, proportion, rate and percentage become manageable when a student stops guessing from words such as more or less and identifies the relationship that stays true.
This Secondary 2 Mathematics Learning Guide connects multiplicative comparison, direct and inverse proportion, unit rates, map scales and percentage change. Each calculation begins with a reference quantity, a unit and a stated assumption. The purpose is not to memorise a different trick for every story. It is to recognise which mathematical relationship the story actually supports.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 1, Guide 2. Companion guides cover algebraic factorisation, linear graphs, and geometry and similarity.
Course boundary. The MOE G2 and G3 Mathematics syllabuses place direct and inverse proportion and map scales in Secondary Two. This guide also revisits the ratio, rate and percentage foundations those topics use. Match the depth to your current school course; this is not a complete syllabus checklist. All prices, rates and scenarios below are hypothetical teaching examples, not current quotations or financial advice.
Navigate: Ratio and reference · Direct and inverse proportion · Rates and units · Percentage change · Map scales · Practice and answers · Teaching and transfer.
1. A comparison needs a named reference
Suppose a box contains 12 blue counters and 18 red counters. Blue to red is 12:18 = 2:3. Blue to total is 12:30 = 2:5. The blue fraction of all counters is 2/5, and the blue percentage is 40%. All four statements describe the same collection, but they compare different quantities. Confusing part-to-part with part-to-whole changes the answer even when the arithmetic is flawless.
Write the labels before simplifying. B:R = 2:3 is not interchangeable with R:B = 2:3. The first says there are two blue parts for every three red parts. The second reverses the quantities. A ratio without its labels is an unfinished mathematical description because the order carries meaning.
A proportion states that two ratios are equal. The conventional form a/b = c/d requires non-zero denominators. OpenStax’s definitions of proportion and percentage change provide a concise reference for this language. Our examples below focus on identifying the reference correctly before manipulating the equality.
Equivalent ratios preserve multiplication, not addition
The ratios 2:3, 4:6 and 10:15 are equivalent because both parts have been multiplied by the same positive factor. Adding the same number to both parts does not generally preserve the ratio. For example, 2:3 becomes 3:4 after adding 1 to each part, but 2/3 is not equal to 3/4.
Students sometimes treat every change as an additive step because that works for differences. Here it is the multiplicative relationship that must stay unchanged. Compare 20 and 30: their difference is 10 and their ratio is 2:3. Doubling them preserves the ratio but doubles the difference. Adding 10 to both preserves the difference but changes the ratio. Ask which property the question says is fixed.
Worked example 1: share a total in a ratio
Share 420 counters between A and B in the ratio 3:4. The total contains 3 + 4 = 7 equal ratio parts. One part is 420 ÷ 7 = 60 counters. A receives 180 and B receives 240. Check both conditions: 180 + 240 = 420, and 180:240 simplifies to 3:4.
Dividing 420 by 3 and by 4 separately would not share one total into seven equal parts. It would create two unrelated calculations. The ratio numbers tell you the relative number of equal units; they are not independent divisors of the whole. Naming one part makes that relationship visible.
Worked example 2: the difference is known instead of the total
A and B have counters in the ratio 3:5. B has 24 more than A. How many does each have? The difference represents 5 − 3 = 2 parts. Therefore one part is 12, giving A = 36 and B = 60. Their total is 96, but that total was not the supplied information.
The surface resembles the previous question, yet the known quantity has changed role. In the sharing problem, the given number represented the sum of the parts. Here it represents their difference. Mark the meaning of the given number before dividing; otherwise a familiar method may be applied to the wrong whole.
Worked example 3: connect two ratios through the shared quantity
A:B = 2:3 and B:C = 4:5. Find A:B:C. The two representations of B must match before the ratios can be joined. Multiply the first ratio by 4 to obtain 8:12. Multiply the second by 3 to obtain 12:15. Therefore A:B:C = 8:12:15.
Writing 2:3:5 would combine incompatible sizes of ratio unit. The 3 in the first ratio and the 4 in the second both describe B, but in different scales. Matching the common quantity is the bridge. Check the result by reducing 8:12 to 2:3 and 12:15 to 4:5.
When quantities change, identify what is conserved
A and B have 24 and 40 counters. If B transfers 8 to A, the new amounts are 32 and 32. The total remains 64, but the ratio changes from 3:5 to 1:1. A transfer within a closed pair conserves the total. It does not conserve the difference or the original ratio.
If both people instead receive the same extra amount from outside, the difference stays unchanged while the total grows. These are different before-and-after structures. Do not reuse the same ratio unit across two time states unless the question establishes that it remains the same. The labels before and after are part of the mathematics, not merely organisational headings.
Worked example 4: equal additions create a new ratio
Two amounts are initially in the ratio 3:5. After 10 is added to each, the ratio becomes 2:3. Find the original amounts. Let them be 3k and 5k. Then (3k + 10)/(5k + 10) = 2/3. Cross-multiplying gives 9k + 30 = 10k + 20, so k = 10.
The original amounts are 30 and 50. Afterwards they are 40 and 60, whose ratio is 2:3. Their difference stays 20. Notice why an equation helps: the original ratio unit is k, but the new ratio does not automatically use that same unit. The equality connects the states without pretending their scales are identical.
2. Direct proportion: the quotient stays constant
In a direct proportion y = kx, one quantity is a constant multiple of the other. For non-zero x, y/x = k. When x is multiplied by a factor, y is multiplied by the same factor. The constant k carries the conversion between them. This is stronger than saying that both quantities increase together.
Imagine that each identical booklet uses 12 sheets. The sheet count is S = 12n for n booklets, provided the production rule stays unchanged. Doubling the booklet count doubles the sheets required. The quotient S/n remains 12 sheets per booklet for n greater than zero. The equation also gives S = 0 when n = 0, under this simplified model.
Worked example 5: find the constant, then use the model
y is directly proportional to x. When x = 6, y = 42. Find y when x = 11. Write y = kx. The given pair gives 42 = 6k, so k = 7. Hence y = 7x and the required value is 77.
A scaling route also works: x changes by the factor 11/6, so y changes by the same factor, giving 42 × 11/6 = 77. These routes express the same relationship. The equation is convenient when several values are needed; direct scaling is efficient when only one comparison is required. Neither route should be used without the direct-proportion condition.
Both increasing is not enough evidence
Suppose a fictional delivery cost is C = 4 + 3n dollars, where n is the number of boxes. Both n and C increase, but C is not directly proportional to n because of the fixed 4-dollar charge. One box costs 7 dollars and two boxes cost 10 dollars; doubling the box count does not double the total price.
The differences are constant: each additional box adds 3 dollars. The quotient C/n is not constant. This is a linear relationship with a non-zero intercept, not direct proportion. It connects directly to the linear graphs guide: a straight line need not pass through the origin.
A small table may support a proposed model without proving it for every future case. Before extending the pattern, ask whether the problem states a constant rate or a direct proportion. Real arrangements may contain minimum charges, capacity limits or changing rates. In a mathematical exercise, use the conditions supplied; in a model, state the conditions you are assuming.
Inverse proportion: the product stays constant
For positive quantities in an inverse proportion, y = k/x and xy = k. Multiplying x by a factor divides y by that factor. A fixed amount of work completed by identical workers at a constant rate is a familiar model, but the assumptions matter: equal productivity, work that can be shared, and no added coordination delay.
More workers do not automatically imply exact inverse proportion in every real task. One person may have to finish before another can begin; equipment may be limited; some time may be fixed. The mathematics is valid when the chosen model’s conditions hold. Identifying those conditions is part of the answer, especially in unfamiliar contexts.
Worked example 6: fixed work, different number of workers
Six identical workers complete a task in 15 days. Under a constant-productivity, fully shareable-work model, how long would 10 workers take? The task requires 6 × 15 = 90 worker-days. With 10 workers, the time is 90 ÷ 10 = 9 days.
The direction check is useful before calculation. More workers should mean fewer days in this model, so 25 days is immediately suspicious. That incorrect value would arise from multiplying 15 by 10/6 as though time were directly proportional to the worker count. The preserved quantity is total work, not the time-per-worker quotient.
Worked example 7: inverse proportion from a data pair
y is inversely proportional to x. When x = 4, y = 15. Find y when x = 10. The constant product is 4 × 15 = 60, so y = 60/x. At x = 10, y = 6. Check that 10 × 6 returns the same product 60.
Do not confuse an inverse proportion with an inverse operation. Subtracting 4 undoes adding 4, but y = 20 − x is not an inverse proportion merely because one quantity decreases as the other increases. Its sum is fixed at 20. In y = 60/x, the product is fixed. State the invariant rather than relying only on direction.
Combining two proportional changes
Three identical printers produce 240 pages in two hours. At the same constant rate, how many pages would five printers produce in four hours? The original setup uses 6 printer-hours, so the rate is 40 pages per printer-hour. The new setup uses 20 printer-hours, giving 800 pages.
Alternatively, multiply 240 by 5/3 for the change in printer count and by 4/2 for the change in time. The result is again 800. This assumes independent printers, sufficient supplies and unchanged output rates. The phrase same constant rate is doing mathematical work; without it, the two pieces of information alone would not determine a unique answer.
3. Rates connect quantities through units
A rate compares quantities with different units: kilometres per hour, litres per minute, dollars per kilogram or pages per printer-hour. The word per indicates division. Keeping the units visible helps determine whether to multiply or divide. A speed of 60 km/h multiplied by 2 h gives 120 km because the time unit cancels.
The familiar relationship distance = speed × time assumes that speed is constant over the interval when a single constant speed is used. Average speed is defined using total distance divided by total elapsed time. For a journey with several speeds, do not choose a formula by appearance; decide which totals the question asks you to compare.
Worked example 8: change the time unit before calculating
A vehicle travels at a constant 72 km/h for 25 minutes. Find the distance. Convert the time to 25/60 hours. Then distance = 72 × 25/60 = 30 km. Multiplying 72 by 25 would incorrectly combine kilometres per hour with a time expressed in minutes.
You could instead convert 72 km/h to 1.2 km/min and multiply by 25 min. Both routes preserve the unit relationship. Choose the route that makes the arithmetic easier, but never change a number without changing its unit consistently. A unit conversion is an equivalence, not an extra adjustment made after an answer has been obtained.
Why metres per second and kilometres per hour differ by 3.6
There are 3,600 seconds in an hour and 1,000 metres in a kilometre. A rate of 1 m/s corresponds to 3,600 m/h = 3.6 km/h. Therefore 18 m/s = 64.8 km/h. In the other direction, 54 km/h = 54 ÷ 3.6 = 15 m/s.
Rather than memorising multiply in one direction and divide in the other without meaning, reconstruct one example. The same motion covers many metres over an hour, then those metres are regrouped into kilometres. A magnitude check follows: the numerical value in km/h is larger than the numerical value in m/s for the same positive speed.
Worked example 9: average speed is not usually the average of two speeds
A journey consists of 120 km at 60 km/h and another 120 km at 40 km/h. Find the average speed. The times are 2 hours and 3 hours. Total distance is 240 km and total time is 5 hours. Average speed is therefore 48 km/h, not 50 km/h.
The slower part takes longer, so it contributes more time to the total. A simple arithmetic mean gives each speed equal weight, which would be appropriate for equal time intervals, not equal distances. For one hour at 60 km/h and one hour at 40 km/h, the average really is 100/2 = 50 km/h. The same two speeds can give different averages because the weighting differs.
If the journey includes a stop and the question asks for average speed over the whole journey, include the stopped time in total elapsed time. If it explicitly asks for average speed while moving, use only moving time. Read the interval definition before calculating. A mathematically correct quotient can still answer the wrong version of the question.
Unit prices require comparable quantities
Suppose one fictional pack costs 4.80 dollars for 600 g and another costs 6.30 dollars for 900 g. The first costs 0.80 dollars per 100 g; the second costs 0.70 dollars per 100 g. The second has the lower unit price, although the total price is higher.
This answers a particular mathematical question: price per equal quantity. It does not by itself decide which purchase a person should make. The required quantity, waste, quality and other conditions may matter. In a school problem, state the requested comparison precisely rather than turning one calculated rate into a broader claim that the question did not justify.
4. Percentage: always ask, percentage of what?
Percent means per hundred. A percentage represents a comparison with a chosen reference: part/reference × 100%. The reference might be an original price, a total number of students, a target quantity or a current value. The words increased by, decreased by and is what percentage of do not all choose the same denominator.
For definitions and basic conversions, see OpenStax: Understand Percent. For translating the relationship into an equation, see OpenStax: General Applications of Percent. The examples here emphasise reference changes and reverse reasoning.
Percentage increase and percentage decrease use the starting value
A quantity rises from 80 to 100. The increase is 20, compared with the original 80, so the percentage increase is 20/80 × 100% = 25%. A fall from 100 to 80 is a decrease of 20 compared with the original 100, so the percentage decrease is 20%.
The same absolute difference produces different percentages because the reference changes. This is why reversing a 25% increase does not require a 25% decrease. In the first direction the base is 80; in the reverse direction it is 100. Write the base in words if the arithmetic is easy but the interpretation is not.
Multipliers keep the whole calculation visible
An increase of p% multiplies the original by 1 + p/100. A decrease of p% multiplies it by 1 − p/100. Thus increasing 250 by 12% gives 250 × 1.12 = 280. Decreasing 250 by 12% gives 250 × 0.88 = 220.
The multiplier includes the original whole as well as the change. Multiplying only by 0.12 gives the size of the change, not the final amount. Before writing the answer, check which output the question requires. The words find the increase and find the increased amount refer to different quantities.
Worked example 10: reverse a percentage decrease
After a 30% reduction, a hypothetical price is 84 dollars. Find the original price. The final price represents 70% of the original. Let the original be P. Then 0.70P = 84, giving P = 120 dollars.
Adding 30% of 84 would give 109.20 dollars, which is not the original. That calculation uses the reduced amount as its reference, whereas the stated reduction was based on the original amount. Verify forwards: 30% of 120 is 36, and 120 − 36 = 84. Reverse percentage problems are equations about an unknown base.
Worked example 11: successive changes multiply
A quantity increases by 20%, then decreases by 20% of its new value. What is the overall change? The combined multiplier is 1.20 × 0.80 = 0.96. The final amount is 96% of the original, an overall 4% decrease.
Starting with 100 makes the reference shift visible: 100 becomes 120, then the decrease is 24, leaving 96. The equal percentage labels do not cancel because they act on different bases. More generally, increasing by a proportion r and then decreasing by that same proportion gives the multiplier (1 + r)(1 − r) = 1 − r².
For two percentage-only multipliers applied to the whole running amount, the multiplication order does not change the final product. However, fixed amounts, minimum charges, caps and rounding rules can change the situation. A 10-dollar reduction followed by 20% off is not generally equivalent to 20% off followed by a 10-dollar reduction. Read the actual operation sequence rather than extending a rule beyond its conditions.
Worked example 12: two discounts are not added
A hypothetical price is reduced by 20%, then by a further 10% of the reduced price. The multiplier is 0.80 × 0.90 = 0.72. The final amount is 72% of the original, so the overall reduction is 28%, not 30%.
At an original price of 100 dollars, the first reduction leaves 80 dollars. The second removes 8 dollars, not 10, leaving 72 dollars. Multipliers prevent the two denominators from being silently treated as the same. A direct written equation is usually clearer than an unlabelled chain of percentage calculations.
Percentage points and relative percentage change are different
A participation rate rises from 30% to 36%. The increase is 6 percentage points. Relative to the original rate, it is also a 20% increase, because 6/30 × 100% = 20%. Both descriptions can be correct, but they answer different comparisons.
Writing increased by 6% without clarification is ambiguous here. In a numerical exercise, state whether you are subtracting two percentages or measuring a relative change in the rate. This distinction is especially useful when reading charts, survey summaries or claims built around large-looking percentage changes.
Worked example 13: combine counts before combining percentages
In one group, 12 of 20 students participate. In another, 18 of 60 participate. What percentage of the combined group participates? Add the participants and totals: 30 of 80. The combined rate is 37.5%.
The individual rates are 60% and 30%. Their simple average, 45%, incorrectly weights the groups equally even though one contains three times as many students. The combined reference is the total number of students, not the number of groups. This is the same weighting issue encountered in average speed: the correct denominator determines how the pieces combine.
5. Map scales: distance scales once, area scales twice
A scale of 1:25,000 means one unit on the map represents 25,000 of the same unit in reality. It might mean 1 cm represents 25,000 cm, or 1 mm represents 25,000 mm. You must establish a common unit before simplifying or applying a scale ratio.
At this scale, 1 cm represents 250 m, or 0.25 km. The conversion is useful because map distances are commonly measured in centimetres while real distances may be expressed in metres or kilometres. Keep the geometric scale and the unit conversion as distinct steps, even when they are eventually combined into one calculation.
Worked example 14: convert in both directions
A map has scale 1:25,000. Two points are 6 cm apart on the map. Find their represented straight-line distance. Since each centimetre represents 0.25 km, the distance is 6 × 0.25 = 1.5 km. This does not automatically give the length of a winding walking route between the points.
In reverse, a straight-line distance of 2 km appears as 2 ÷ 0.25 = 8 cm on the map. Multiplying by 0.25 again would move in the wrong direction. Name the known and required units before choosing the operation; the same conversion relationship can be read forwards or backwards.
Worked example 15: area is not a distance
At the same scale, a region has map area 4 cm². Find its represented area in km². One map centimetre represents 0.25 km in each perpendicular direction. Therefore one square centimetre represents 0.25 × 0.25 = 0.0625 km². Four square centimetres represent 0.25 km².
Multiplying 4 by 0.25 only once would apply a length scale to an area. The squared factor is not an extra rule to memorise in isolation; area combines two lengths. This reasoning is developed further in CIMT’s similarity material and in our geometry and similarity guide.
6. Practice: name the invariant before calculating
For each question, first write the quantity or relationship that remains fixed: total, difference, ratio, quotient, product, unit rate or percentage reference. Then calculate. The method note can be one sentence. Its purpose is to distinguish a reasoned route from a number operation chosen only because it resembles the preceding example.
Questions 1–6. 1. A box contains 15 green and 25 yellow counters. Find green:yellow and the percentage that is green. 2. Share 560 in the ratio 3:5. 3. Two quantities are in the ratio 4:7 and differ by 36. Find them. 4. A:B = 3:4 and B:C = 2:5. Find A:B:C. 5. y is directly proportional to x; y = 24 when x = 8. Find y when x = 13. 6. Is C = 5 + 2n directly proportional to n? Explain.
Questions 7–12. 7. y is inversely proportional to x; y = 18 when x = 5. Find y when x = 12. 8. Eight identical workers take 15 days for fixed, fully shareable work. Find the time for 12 workers under the same rate assumption. 9. Convert 15 m/s to km/h. 10. Find the distance travelled at 90 km/h for 20 minutes. 11. A journey covers 60 km at 30 km/h and 60 km at 60 km/h. Find average speed over the journey. 12. Increase 350 by 8%.
Questions 13–18. 13. After a 25% reduction, an amount is 90. Find the original. 14. An amount increases by 10%, then decreases by 10% of its new value. Find the overall percentage change. 15. A rate changes from 40% to 50%. State the percentage-point change and relative percentage increase. 16. One group has 9 participants out of 15, another 21 out of 35. Find the combined participation percentage. 17. At scale 1:50,000, what real distance does 7 cm represent? 18. At that scale, what real area does 3 cm² represent in km²?
Explained answers: questions 1–6
1. Green:yellow = 3:5; green percentage = 15/40 × 100% = 37.5%. The ratio denominator is not the total. 2. There are 8 parts, each worth 70, so the shares are 210 and 350. 3. Three parts represent 36, so one part is 12 and the quantities are 48 and 84.
4. A:B:C = 3:4:10. Multiply the second ratio by 2 to match B. 5. The constant quotient is 24/8 = 3, so y = 39. 6. No. The fixed 5 means doubling n does not double C. The relationship is linear with a non-zero intercept.
Explained answers: questions 7–12
7. xy = 90, so y = 90/12 = 7.5. 8. Total work is 120 worker-days, so 12 workers take 10 days. 9. 15 × 3.6 = 54 km/h. 10. 20 minutes is 1/3 hour, so the distance is 30 km.
11. The times are 2 hours and 1 hour. Average speed is 120/3 = 40 km/h, not 45 km/h. 12. The final amount is 350 × 1.08 = 378. The increase itself is 28; the question asks for the increased amount.
Explained answers: questions 13–18
13. 0.75P = 90 gives P = 120. 14. 1.10 × 0.90 = 0.99, an overall 1% decrease. 15. The rise is 10 percentage points and 10/40 × 100% = 25% relative to the original rate.
16. There are 30 participants among 50 people, so the combined rate is 60%. Here the separate percentages are also both 60%, but the count method remains valid even when they differ. 17. One centimetre represents 0.5 km, so 7 cm represents 3.5 km. 18. One cm² represents 0.5² = 0.25 km²; therefore 3 cm² represents 0.75 km².
7. A complete case: planning a small exhibition
A fictional student exhibition needs 180 information cards. Blue and white cards are required in the ratio 2:3. A printer produces cards at a constant 30 cards per minute. The hypothetical printing charge is a fixed 6 dollars plus 12 cents per card. A 10% reduction applies only to the per-card charge, not to the fixed charge. Find the card counts, printing time and total charge.
The ratio contains five parts, each 36 cards, so the counts are 72 blue and 108 white. The printing time is 180/30 = 6 minutes under the stated constant-rate model. The variable charge is 180 × 0.12 = 21.60 dollars. Reducing that part by 10% gives 19.44 dollars, then adding the unchanged fixed charge gives 25.44 dollars.
Three relationships operate in the same situation. The total determines the ratio unit. Output divided by rate determines time. The percentage reduction has a restricted reference: only the variable charge. Applying 10% off the whole undiscounted total would produce 24.84 dollars, which contradicts the condition that the fixed charge is unchanged.
Now suppose the printing rate rises by 20% while the card count stays 180. The new rate is 36 cards per minute, so the time becomes 5 minutes. A 20% increase in rate does not mean a 20% decrease in time. The fixed output creates an inverse relationship. Time changes by the factor 1/1.2 = 5/6, a decrease of 16⅔%.
This case is a useful transfer test because the topic labels have disappeared. The learner has to select the relationship for each part rather than apply one formula to every number in the paragraph. Ask for the units and the preserved condition at each step. A final numerical answer alone does not reveal whether those decisions were controlled.
Diagnose the error by the reference that went missing
If a student uses 2/3 as the blue fraction, the missing reference is the total of five parts. If the student multiplies by 30 to find time, the missing relationship is cards divided by cards per minute. If the student discounts the fixed charge, the missing condition is the stated percentage base. If the student reduces time by 20%, the missing distinction is between direct and inverse change.
These are not interchangeable careless mistakes. A useful correction names the specific comparison and supplies a fresh example that tests it. The next task should resemble the underlying decision, not necessarily the exhibition story. For example, use a recipe for ratio, a filling tank for rate, or a fixed-distance journey for inverse proportion between speed and time.
A suggested teaching sequence
Start with two quantities and ask for their difference, their ratio and each one’s fraction of the total. Then change the quantities by a common multiplier and by a common addition. Ask which comparisons remain unchanged. This separates additive and multiplicative reasoning before formal proportion equations are introduced.
Next compare three models: y = 3x, y = 3x + 5 and y = 24/x for positive x. Build a short table for each. In the first, the quotient stays constant. In the second, equal input steps produce equal output steps but the quotient changes. In the third, the product stays constant. Let the learner explain these patterns before naming the methods.
Finally mix a ratio task, a rate task, a reverse-percentage problem and an area-scale problem. Require the reference and unit before the first calculation. This is a proposed practice structure; adjust it to the student’s actual errors and school sequence rather than treating it as a universal timetable or a guaranteed route to marks.
How to check without repeating the same assumption
For a ratio share, check the sum and the simplified ratio. For a direct proportion, compare the quotient. For an inverse proportion, compare the product. For a rate, cancel the units. For reverse percentage, run the original change forwards. For a map area, reconstruct a one-unit square. Each check asks whether the answer returns to the relationship that created the problem.
Magnitude checks add another layer. A 30% reduction should leave less than the original but more than half of it. More identical workers should take less time in a fixed-work inverse model. A region’s area should increase by more than its length factor when every length is enlarged by a factor greater than one. These checks do not replace exact reasoning, but they can reveal an impossible direction before a wrong answer is accepted.
What independent progress looks like in this guide
The learner can name the two compared quantities, identify the denominator, state what remains fixed, choose consistent units and explain why the selected model fits. The learner also knows when a claim is not justified: both increasing does not prove direct proportion; both decreasing and increasing in opposite directions does not prove inverse proportion; two group percentages cannot always be averaged equally.
Parents can ask one useful question before the answer is shown: what is your calculation comparing? A student who can answer precisely has made the mathematics easier to inspect. A student who cannot may need a clearer representation, not a larger collection of formulas.
8. Connect the relationship to the rest of Mathematics
Algebra provides the language y = kx, xy = k and final = multiplier × original. Graphs make constant rate and intercept visible. Geometry supplies scale factors and the difference between length, area and volume. These are not isolated chapters. They are different representations of how quantities relate while conditions are preserved.
Use Algebraic Factorisation and Structural Control when the equation is understood but the manipulation is unstable. Use Linear Graphs, Coordinates and Relationships to compare direct proportion with other straight-line relationships. Use Geometry, Similarity and Mathematical Constraints when a scale factor acts on a shape.
Before asking whether to multiply or divide, ask what the quantities mean, which reference is being used, and what relationship must remain true.
