SECONDARY 1 MATHEMATICS LEARNING GUIDE · GUIDE 6
A ratio describes a multiplicative comparison. Proportion asks whether that comparison stays consistent as quantities change. The central skill is not cross-multiplication. It is knowing what is being compared, what one equal part means, and which relationship must remain unchanged.
Suppose two quantities are in the ratio 2:3. They could be 2 and 3, 10 and 15, or 80 and 120. The ratio tells us their relative size, not their individual amounts. One extra condition—a total, a difference or the value of one quantity—can fix the scale. Without that condition, there may be many possible pairs.
This is why a learner can simplify ratios successfully but still become stuck in a word problem. Simplification changes the form of an already known comparison. A word problem asks the learner to construct the comparison, connect it to a constraint and decide which method is justified.
This guide belongs to the Secondary Mathematics Hub and S1–S4 Capability Map. Use Numbers, Number Lines, Approximation and Estimation for fraction and numerical foundations. The companion guides on Percentages and Rates and Speed develop related but distinct comparisons.
Using this guide: begin with labelled ratios and equal-part reasoning. Move to algebraic changes and inverse proportion only when the prerequisite work is secure. Extension sections are not a claim that every Secondary 1 class studies the same material at the same time. For the official subject-level context, consult the MOE secondary mathematics syllabus directory.
Navigate: meaning and labels · equal parts · linked ratios · direct proportion · changing ratios · practice · answers · teaching and repair.
1. A ratio needs labels and an order
The statement red:blue = 2:3 means that the red quantity corresponds to two equal parts and the blue quantity to three of those same-sized parts. The comparison blue:red is therefore 3:2, not 2:3. The order of the labels is part of the information.
Write the labels before the numbers when a question is unfamiliar. This prevents a common silent reversal: the arithmetic is correct for blue compared with red, but the question asked for red compared with blue. A final bare ratio may hide that the order has changed.
For positive quantities, A:B = a:b can be expressed as A/B = a/b. That fraction compares A with B. It does not automatically describe A as a fraction of the total. To find A’s fraction of the combined amount, use a/(a + b).
Worked example: part-to-part versus part-to-whole
A box contains 12 red counters and 18 blue counters, with no other counters. Red:blue = 12:18 = 2:3. Red as a fraction of all the counters is 12/30 = 2/5. Red as a fraction of blue is 12/18 = 2/3.
All three statements can be true. They have different reference quantities. The denominator answers a question: compared with blue, or compared with the whole box? Much of ratio accuracy comes from keeping that reference stable.
For an additional introductory treatment of ratio notation and comparisons, see OpenStax, Ratios and Rate. This guide uses colon notation extensively because it keeps the ordered quantities visible.
2. A ratio does not tell us the size of the group
If A:B = 3:5, then A = 3k and B = 5k for some common positive scale k. The ratio identifies the relationship but leaves k unknown. The pair 6 and 10 uses k = 2; the pair 21 and 35 uses k = 7.
This is a useful bridge into algebra. The variable k does not stand for “whatever number I happen to be finding”. It has a precise role: the amount represented by one ratio part. Every part in the same comparison must have the same size.
When A and B are counts of indivisible objects and 3:5 is in simplest whole-number form, possible counts come in multiples of three and five. When A and B are lengths or masses, k may be a fractional or decimal measurement. The context controls which values are allowed.
Missing-information checkpoint
“Two ribbons have lengths in the ratio 3:5. Find their lengths.” This is incomplete as a numerical question. We need another condition, such as their combined length or one known length. Do not choose 3 cm and 5 cm merely because those are the displayed ratio numbers.
A correct response can identify what is known and what remains unknown. Recognising insufficient information is a mathematical success, not a failure to calculate quickly enough.
3. Convert comparable measurements to a common unit
Before simplifying a ratio of lengths, masses or times, express both quantities in the same unit. The numbers alone do not reveal whether they are comparable. A length of 1.2 m compared with 80 cm is not 1.2:80 when the units are silently removed.
Convert 1.2 m to 120 cm. The ratio becomes 120:80 = 3:2. Alternatively, convert 80 cm to 0.8 m and simplify 1.2:0.8. Both routes preserve the same comparison.
Worked example: time units
Find the ratio of 45 minutes to 1.5 hours. Since 1.5 hours is 90 minutes, the ratio is 45:90 = 1:2. The conversion comes before simplification.
Do not convert 1.5 hours to 150 minutes. Decimal hours divide an hour into tenths; clock minutes divide it into sixtieths. The companion guide on Rate, Speed and Unit Conversion gives a fuller treatment.
Ratio and rate are connected, but units still matter
A same-kind comparison such as 120 cm to 80 cm can become a dimensionless ratio. A comparison such as 120 km in 2 hours produces a rate with a unit, 60 km/h. Do not remove unlike units merely because the numerical fraction can be simplified.
The dividing line is practical: what meaning must the answer retain? A ratio of ingredients should preserve their relative quantities; a speed must preserve distance per unit of time.
4. Equivalent ratios are made by common scaling
Multiplying or dividing every term by the same positive factor preserves a positive ratio. Thus 4:6, 8:12 and 20:30 all describe 2:3. Simplest whole-number form removes common factors without changing the relationship.
Adding the same amount to both terms generally does not preserve a ratio. Starting with 2:3 and adding 4 gives 6:7, not another form of 2:3. Multiplicative comparison and additive difference are different structures.
Why the additive shortcut fails
In 2:3, the first quantity is two-thirds of the second. In 6:7, it is six-sevenths. The difference between the displayed terms remains 1, but their relative size changes. Holding a difference constant does not hold a ratio constant.
A useful contrast is 2:3 → 4:6 versus 2:3 → 4:5. In the first, both quantities are doubled. In the second, both gain 2. Ask the learner which relationship each transformation preserves.
Three-term ratios
The same rule applies to 6:9:15. Divide every term by 3 to obtain 2:3:5. Dividing only the first two terms would not preserve the comparison among all three quantities.
Before simplifying, check the number of labelled categories. A three-term ratio belongs to three quantities, not to one fraction plus an unrelated extra number.
5. Ratios containing decimals and fractions
A ratio containing decimals can be scaled to whole numbers. For 0.75:1.2, multiply both terms by 100 to obtain 75:120, then divide by 15 to obtain 5:8. Multiplying both by 20 would also work, giving 15:24 before simplification.
The scale need not be a power of ten; it only needs to preserve both terms and make the work clearer. What matters is applying the same factor to every term.
Worked example: fractional terms
Simplify 2/3 : 5/6. Multiply both terms by 6, a common denominator. The result is 4:5, already in simplest whole-number form. Equivalently, dividing 2/3 by 5/6 gives 4/5, which confirms the first quantity is four-fifths of the second.
Do not simply compare the numerators 2 and 5 while ignoring the denominators. Two-thirds and five-sixths count different unit fractions. Put the quantities on a common scale first.
Mixed-number example
Simplify 1 1/2 : 2 1/4. Convert to improper fractions: 3/2 : 9/4. Multiplying both by 4 gives 6:9, so the ratio is 2:3.
Read the final result back against the original values. One and a half is two-thirds of two and a quarter. This quick interpretation is more informative than checking only that the arithmetic was tidy.
6. Divide a known total by the sum of the parts
If two positive quantities are in the ratio 3:4, their combined amount corresponds to seven equal parts. A known total therefore determines one part by division by 7, not by 3 or by 4.
This is often called the unitary method: find the value of one unit or part, then scale to the required amount. It is not a collection of separate tricks for each wording. The same relationship works for money, length, mass and counts, provided the context permits the resulting values.
Worked example: share 84 stickers
Asha and Ben share 84 stickers in the ratio 3:4. The total is seven parts. One part is 84 ÷ 7 = 12 stickers. Asha receives 3 × 12 = 36 stickers and Ben receives 4 × 12 = 48 stickers.
Use two checks. First, 36 + 48 = 84. Second, 36:48 simplifies to 3:4. A total check alone is insufficient because many incorrect pairs could still add to 84.
Three-way sharing
Three groups share 180 cards in the ratio 2:3:5. There are ten parts, so one part is 18 cards. The shares are 36, 54 and 90 cards. They add to 180 and simplify to the requested three-term ratio.
A clear diagram can show ten equal parts grouped as two, three and five. The diagram and the arithmetic express the same structure. Use whichever representation makes that structure easier to inspect.
7. A known difference uses the difference of the parts
A ratio problem does not always provide a total. If A:B = 5:8 and B exceeds A by 27, the difference of 27 corresponds to three parts, not thirteen.
One part is 27 ÷ (8 − 5) = 9. Therefore A = 5 × 9 = 45 and B = 8 × 9 = 72. The total happens to be 117, but it was not the quantity given at the start.
Why the wrong divisor can look convincing
A student who automatically divides 27 by 13 may have memorised “add the ratio numbers” without knowing what the supplied amount measures. The right question is not “What do I always do with a ratio?” but “Which collection of equal parts corresponds to this amount?”
If the amount is the total, use the sum. If it is a difference, use the difference. If it is one named quantity, use that quantity’s number of parts. The same equal-part model explains all three cases.
A known component
If flour:sugar = 5:2 by mass and the sugar mass is 140 g, two parts correspond to 140 g. One part is 70 g, so the flour mass is 350 g. Adding the ratio numbers would answer a different question.
After calculating, label the result. If the question asks for flour but your intermediate value is one part, that intermediate number is not yet the requested answer.
8. Ratios connect to fractions of the whole
If A:B = 3:7 and those are the only two categories, A is 3/10 of the total and B is 7/10. The first quantity is also 3/7 of the second. These comparisons should be translated deliberately, not memorised as interchangeable formulas.
Conversely, if 3/8 of a collection is red and the rest is blue, red:blue = 3:5. The remaining part is 5/8, not 8/8. The ratio 3:8 would compare red with the whole collection.
Worked example: a fraction statement becomes a ratio
Two-fifths of the books in a small collection are fiction, and all the remaining books are non-fiction. Fiction:non-fiction = 2:3. If there are 45 non-fiction books, three parts correspond to 45, so there are 30 fiction books and 75 books altogether.
The phrase “all the remaining books” matters. Without it, there might be another category and the two-part ratio would not follow. Mathematical reading protects the assumptions behind the comparison.
Why this matters for percentages
A ratio of 2:3 does not mean the first quantity is 2/3 of the total. It is 2/5 of the total, or 40%, when there are only two categories. The first is 66 2/3% of the second, a different reference. The Percentages and Reverse Percentages guide develops this denominator discipline in detail.
9. Combine ratios by matching the shared quantity
Suppose A:B = 2:3 and B:C = 4:5. The symbol B refers to the same quantity in both statements. Yet the displayed ratio part counts for B are 3 and 4. We must put them on a common scale before combining the comparisons.
Multiply A:B by 4 to obtain 8:12. Multiply B:C by 3 to obtain 12:15. Now the shared B has the same part count, so A:B:C = 8:12:15.
Do not write 2:3:5 by taking the first, middle and last visible numbers. That would generally fail the second ratio. The labels may look aligned, but the units represented by the ratio parts are not yet aligned.
Check both source relationships
From 8:12:15, A:B = 8:12 = 2:3, while B:C = 12:15 = 4:5. Both original statements survive. This is an example of changing representation while preserving the defining relationships.
When the ratios cannot simply be joined
If A:B describes one classroom and B:C describes another classroom with a different B population, the shared letter does not guarantee the same quantity. Similarly, ratios referring to different times cannot automatically be joined unless the connecting quantity is unchanged or another relation is supplied.
Before using a common multiple, ask: “Is this genuinely the same B?” Algebraic convenience must not erase the context. The operation is valid because a quantity is shared, not because a letter happens to be repeated.
10. Direct proportion means a constant multiplier
Two positive quantities are directly proportional when one is a constant multiple of the other. In the form y = kx, the constant k stays fixed throughout the relationship. Doubling x doubles y; multiplying x by a factor multiplies y by the same factor.
Suppose identical exercise books cost $2.40 each in a constructed price model with no delivery charge, discount or minimum purchase. The total cost C for n books is C = 2.40n. The cost per book is constant, so cost is directly proportional to the number of books in this model.
A table makes the invariant visible
| Number of books n | Total cost C | C ÷ n |
|---|---|---|
| 2 | $4.80 | $2.40 per book |
| 5 | $12.00 | $2.40 per book |
| 8 | $19.20 | $2.40 per book |
The stable quotient is the evidence within the given model. The statement “both quantities increase” is not enough. Many non-proportional relationships also increase.
Graph interpretation
For a directly proportional continuous relationship, the graph of y = kx is a straight line through the origin. For a count such as a number of books, only whole-number inputs may be physically relevant. The same formula still describes those permitted points, but half a book may not be a valid purchase.
The origin condition is important. A straight line with a non-zero starting charge is not direct proportion, even though equal increases in x produce equal increases in y.
11. A fixed charge breaks direct proportion
Consider another invented model: an order costs $5 for preparation plus $2 for each item. The total is C = 5 + 2n. For two items, the cost is $9. For four items, it is $13, not $18. Doubling the number of items does not double the total charge.
The per-item variable charge remains $2, but the average total cost per item changes because the fixed $5 is spread across a different number of items. Students need to distinguish a constant rate for one component from direct proportion of the entire total.
The test before using proportion
Ask whether zero of the input would produce zero of the output in the model. Ask whether the ratio output/input is constant for non-zero inputs. Ask whether a fixed amount, threshold, discount or changing condition has been introduced.
These tests are not a demand to overcomplicate every simple question. If the question explicitly states a constant unit price with no other conditions, use the given model. The important habit is not to infer direct proportion solely because two numbers appear in a word problem.
What “same conditions” is protecting
A machine producing 50 items in five minutes does not automatically prove it will produce 600 in an hour. That prediction needs a constant-rate assumption over the hour. There might be a setup stage, a pause or a changing rate. In an elementary exercise, the constant-rate model is often stated; keep it attached to your conclusion.
12. Solving a proportion without blind cross-multiplication
Under the same constant-price model, six identical folders cost $15. What do ten cost? The unitary route gives $15 ÷ 6 = $2.50 per folder, then 10 × $2.50 = $25.
A scale-factor route gives the same result: ten is 10/6 times six, so the cost is 15 × 10/6 = $25. A proportion equation is 15/6 = C/10. These are three representations of the same constant price per folder.
Cross-multiplication can solve the equation, but first explain why the fractions compare corresponding quantities in the same order. Dollars per folder equals dollars per folder. Reversing one fraction but not the other would compare different rates.
Why cross-products agree
For non-zero denominators, a/b = c/d implies ad = bc because multiplying both sides by bd clears the denominators. The equation is not made true by drawing diagonal arrows. It is true because equal ratios have been established and the same legal multiplication is applied to both sides.
For simple values, the unitary method can be more transparent. For awkward algebraic values, an equation may be more efficient. Choose the route that exposes the relationship and keeps errors detectable.
13. Extension: inverse proportion needs a fixed product
Inverse proportion is not simply “one goes up while the other goes down”. It means the product of the two quantities stays constant, giving a relationship such as xy = k or y = k/x for positive values.
For a fixed distance of 120 km travelled at a constant speed, the travel time is t = 120/v. At 40 km/h, the time is 3 hours. At 60 km/h, it is 2 hours. Speed multiplied by time remains 120 km.
Doubling the speed halves the travel time only when the distance and the relevant conditions are held fixed. A journey with waiting time or changing speeds may need a more detailed model.
Equal-worker model and its boundary
In a constructed task, four identical machines take six hours to complete a fixed job. If each works at the same constant rate and the work can be shared perfectly, the job requires 24 machine-hours. Eight such machines would take three hours.
The result depends on the assumptions. Adding workers or machines does not always halve a real completion time: some tasks cannot be divided, and shared equipment may limit output. This guide uses the ideal model as mathematics, not as a claim about every workplace.
Keep inverse proportion as a labelled extension when it has not yet been taught. It is useful because it shows that different “things change together” problems preserve different quantities: a quotient in direct proportion, a product in inverse proportion.
14. When a ratio changes, look for what does not
A changing-ratio problem needs a connection between the earlier and later states. That connection may be an unchanged quantity, an unchanged total, an unchanged difference or a specified addition or removal.
Do not assume that one part in the first ratio has the same value as one part in the second. Simplified ratio parts describe relative shares within each state. Their actual sizes may differ after a change.
Worked example: add to one category
Initially, red:blue counters = 2:3. Ten red counters are added and no blue counters are changed. The new ratio is 4:3. Find the original numbers.
The blue count is unchanged. Since its ratio term is 3 in both states, the ratio part size is the same here. Red increases from two parts to four parts, so ten counters correspond to two parts. One part is 5. The original numbers are 10 red and 15 blue.
Check: after adding ten red counters, the counts are 20 and 15, giving 4:3. Both the original ratio and the stated change have been satisfied.
A more general version
If the unchanged category has different displayed ratio terms, match those terms before comparing the changing category. Alternatively, let the original quantities be ak and bk and write the new ratio as an equation. The algebraic route avoids assuming equal part sizes prematurely.
15. Equal additions preserve a difference, not usually a ratio
A collection begins with red:blue = 2:5. Twelve counters are added to each colour. The new ratio is 4:7. Find the original counts.
Let the original counts be 2k and 5k. After the additions, they are 2k + 12 and 5k + 12. Therefore (2k + 12)/(5k + 12) = 4/7. Multiplying gives 14k + 84 = 20k + 48, so 36 = 6k and k = 6.
The original counts are 12 and 30. After the additions, they are 24 and 42, giving 4:7. The difference is 18 both before and after, as expected when equal amounts are added to both quantities.
A non-algebraic viewpoint
The original ratio difference is three parts and the new ratio difference is also three parts. Since the actual difference is unchanged, the part sizes match in this particular example. Each colour gains two parts, so two parts correspond to twelve counters.
That shortcut is valid here because both the part-count difference and the actual difference agree. It is not a rule that the parts in any two ratios are automatically identical. Say what justifies matching them.
The earlier Equations and Equality guide explains the balancing steps used in the algebraic solution.
16. Transfers preserve the total
In an invented token-sharing problem, A:B = 3:7 and the total is 80 tokens. A receives 12 tokens from B. What is the new ratio?
Initially, one part is 80 ÷ 10 = 8, so A has 24 and B has 56. After the transfer, A has 36 and B has 44. The new ratio is 9:11. The total remains 80.
A frequent mistake adds 12 to A but forgets to subtract 12 from B. That creates new tokens which the question did not supply. Another mistake subtracts 12 from the total even though the tokens remain inside the two-person system.
Draw the boundary of the group
If tokens move from A to B, the combined total of A and B is unchanged. If tokens are removed from both and taken outside the group, the total changes. If someone outside contributes tokens, the total increases. “Transfer”, “remove” and “add” are not interchangeable words.
This habit of identifying the group boundary is useful beyond ratio problems. It keeps conservation statements honest: what is unchanged inside the group, and what crosses into or out of it?
17. Scale models preserve corresponding length ratios
A scale of 1:25,000 means one unit of length on the map corresponds to 25,000 of the same units on the ground, within the stated map model. For a map length of 3.6 cm, the ground distance is 3.6 × 25,000 = 90,000 cm = 900 m.
The unit conversion happens after applying the dimensionless scale factor, or equivalently before it if done consistently. The statement is not 3.6 × 25,000 metres. The ratio only cancels units when they are the same.
Reverse scale question
A ground distance of 1.5 km corresponds to how many centimetres on that map? Convert 1.5 km to 150,000 cm, then divide by 25,000. The map distance is 6 cm.
Predict the direction before calculating. A map representation is much shorter than the represented ground distance at this scale. An answer of several kilometres on the page would reveal a reversed factor.
Extension: area does not use the length factor only once
If corresponding lengths of two similar rectangles are in the ratio 2:3, their areas are in the ratio 4:9 because both length and width scale. This extension requires similarity or an equivalent statement that all corresponding linear dimensions scale together. A single pair of lengths alone does not determine an area ratio for arbitrary shapes.
18. A recipe problem is also a resource-limit problem
Consider an invented craft mixture using powder and water in the mass ratio 3:5. To prepare 800 g of mixture under a no-loss model, eight parts correspond to 800 g. The required quantities are 300 g of powder and 500 g of water.
Now suppose only 180 g of powder is available. Keeping the same ratio, three parts correspond to 180 g, so one part is 60 g. The required water is 300 g, and the total mixture is 480 g. The target of 800 g cannot be reached without more powder or a changed ratio.
Do not preserve the total by silently changing the composition
Adding enough water to reach 800 g would give powder:water = 180:620 = 9:31, not 3:5. The total would be correct but the composition would fail.
This is another reason to check more than one condition. A result may satisfy the total while violating the ratio. In practical problems, the relationship may be the most important specification.
Discrete batches
If a package requires three red pieces and five blue pieces, and pieces cannot be divided, available stock may restrict the number of complete packages. With 20 red and 37 blue pieces, red stock permits six complete packages and blue stock permits seven. Only six complete packages can be made, leaving two red and seven blue pieces.
Ratios scale continuously as mathematics, but actual objects can impose whole-number constraints. Interpret the scale factor in the context before rounding or promising a fractional package.
19. Choose the method from the information, not the chapter label
| Information supplied | Useful relationship | First move |
|---|---|---|
| A ratio and a total | The sum of the parts equals the total | Find one part using the sum. |
| A ratio and a difference | The difference of the parts equals the difference | Find one part using the difference. |
| A ratio and one known quantity | That quantity corresponds to its labelled parts | Find one part from that component. |
| Two linked ratios | The shared quantity must match | Put the shared term on a common scale. |
| A constant per-unit model | A quotient is constant | Find the unit value or scale factor. |
| A before-and-after ratio | A stated change connects the states | Identify what is unchanged or write an equation. |
This table is a decision aid, not a replacement for reading. A difficult problem may combine several rows. For example, it may use a total to establish original counts and then a transfer to produce a new ratio.
When a route becomes confused, return to the labels, the state of the quantities and the common unit. Many apparent calculation problems are really mismatched-reference problems.
20. Practice laboratory: establish the relationship first
For every question, write the comparison labels or the meaning of one part before calculating. Questions 13–16 are extension or transfer work; use them when the relevant foundations have been taught.
- Simplify 42:63.
- Find the ratio of 1.5 m to 90 cm in simplest whole-number form.
- Simplify 3/4 : 5/8.
- A collection has red:blue = 4:5, with no other categories. What fraction of the whole collection is red?
- Share 132 cards in the ratio 5:6.
- Two quantities are in the ratio 3:8 and differ by 45. Find both.
- A:B = 7:4 and B = 36. Find A and the total.
- A:B = 3:4 and B:C = 6:5. Find A:B:C.
- Under a constant-price model with no other charges, eight folders cost $20. Find the cost of fourteen folders.
- Is C = 4 + 3n a directly proportional relationship between C and n? Explain.
- Initially red:blue = 3:5 and there are 64 counters. Eight red counters are added. Find the new ratio.
- A:B = 2:3 with 75 tokens in total. A transfers 5 tokens to B. Find the new ratio.
- On a 1:20,000 scale map, a route is 4.5 cm long. Find the corresponding ground distance in metres.
- For a fixed journey, speed increases from 30 km/h to 45 km/h. If the original journey time is 3 hours and each speed is constant, find the new time.
- A mixture uses powder:water = 2:7 by mass. What mass of water is required for 160 g of powder?
- Two quantities are initially in the ratio 1:3. After 10 is added to each, the ratio becomes 1:2. Find the original quantities.
Before looking below, choose one answer to verify using a method different from your original route. A sum check, a difference check, a simplified ratio and a substitution check catch different mistakes.
21. Explained answers
1. Divide both terms by 21: 2:3. The same factor must be applied to both terms.
2. Convert 1.5 m to 150 cm. Then 150:90 = 5:3. Simplifying 1.5:90 without conversion would compare incompatible numerical units.
3. Multiply both fractions by 8: 6:5. The answer is 6:5. This also tells us the first quantity is larger.
4. There are nine parts altogether, so red is 4/9 of the collection. The fraction 4/5 compares red with blue, not with the whole.
5. Eleven parts correspond to 132 cards, so one part is 12. The shares are 60 and 72 cards. Check both their total and their ratio.
6. The difference is five parts. One part is 45 ÷ 5 = 9, giving 27 and 72. Their difference is 45 and their ratio is 3:8.
7. Four parts correspond to 36, so one part is 9. A = 63 and the total is 99.
8. Match B at twelve parts. A:B = 9:12 and B:C = 12:10, so A:B:C = 9:12:10. Both source ratios must survive the combination.
9. One folder costs $20 ÷ 8 = $2.50. Fourteen cost $35. The conclusion uses the explicitly stated constant-price model.
10. No. The fixed 4 means C/n is not constant and doubling n does not generally double C. The variable part 3n is directly proportional to n; the whole expression is not.
11. One original part is 64 ÷ 8 = 8, giving 24 red and 40 blue. After the addition, there are 32 red and 40 blue, so the ratio is 4:5.
12. The original counts are 30 and 45. After the transfer they are 25 and 50, giving 1:2. The total is still 75.
13. Ground distance = 4.5 × 20,000 cm = 90,000 cm = 900 m. The scale compares corresponding lengths in the same unit.
14. The fixed distance is 30 × 3 = 90 km. At 45 km/h, the time is 90 ÷ 45 = 2 hours. Do not multiply the old time by the speed-increase factor.
15. Two parts correspond to 160 g, so one part is 80 g. Seven parts of water have mass 560 g. The total mixture mass would be 720 g under a no-loss model.
16. Let the original quantities be k and 3k. Then (k + 10)/(3k + 10) = 1/2. Therefore 2k + 20 = 3k + 10, so k = 10. The original quantities are 10 and 30; after addition they are 20 and 40.
22. A complete mixed problem: the school display
Problem: A fictional display initially has portrait:landscape cards in the ratio 3:5. There are 96 cards altogether. Twelve landscape cards are removed and twelve portrait cards are added. Find the new ratio. Then determine how many further landscape cards must be added to restore the original ratio, with the portrait count now unchanged.
Initially, eight parts correspond to 96, so one part is 12. There are 36 portrait cards and 60 landscape cards. After the stated changes, there are 48 of each, giving 1:1.
The total has stayed at 96 because twelve were removed and twelve added. But the ratio has changed because the changes affect different categories. An unchanged total does not imply an unchanged composition.
To restore portrait:landscape = 3:5 while portrait remains 48, three parts must correspond to 48. One part is 16. Landscape must therefore become 5 × 16 = 80. Since there are currently 48 landscape cards, add 32 landscape cards.
Check the final ratio: 48:80 = 3:5. Notice that the final total is 128, not 96. Restoring a ratio does not necessarily restore the original amounts or the original total.
Change the question, change the answer
If instead the task asked to restore the ratio by removing portrait cards while keeping landscape at 48, five parts would correspond to 48. Portrait would need to be 28.8 under a continuous-quantity model, which is impossible for indivisible cards. That requested operation cannot exactly restore 3:5 with whole cards under those conditions.
This extension shows why the physical meaning of the quantities matters. An equation can produce a numerical value that does not satisfy a whole-object constraint. The final judgement belongs to the original problem, not merely to the algebra.
23. Error diagnosis: the first wrong relationship
When checking a ratio solution, find the earliest line that no longer represents the question. A later arithmetic error is often easier to notice, but repairing only that line may leave a wrong model untouched.
Wrong whole: the learner treats 2:3 as 2/3 of the combined amount. Repair the reference quantity by drawing two groups and then drawing a boundary around the whole.
Wrong scale: the learner simplifies metres against centimetres without converting. Repair the unit labels before repeating numerical simplification.
Wrong constraint: the learner uses the sum of the parts when the given amount is a difference. Ask exactly which parts the supplied number measures.
Wrong invariant: the learner assumes equal additions preserve the ratio. Compare the fractions before and after the change and identify the difference that actually remains constant.
Wrong proportional model: the learner scales a total charge despite a fixed fee. Separate the fixed and variable components and test a doubled input.
Wrong physical interpretation: the learner accepts a fractional count of indivisible items. Return to the context and decide whether exact fulfilment is possible, or whether the question instead asks for a maximum number of complete sets.
24. Teaching ratio as one connected idea
A proposed teaching sequence begins with a visible comparison, moves to equal parts, then introduces an unknown scale. For example, compare 6 red and 9 blue counters, express 2:3, write 2k and 3k, and ask what extra fact would identify k.
Next vary only the supplied condition. Give a total, then a difference, then one component. The numbers need not be difficult. The teaching target is recognising which group of parts corresponds to the known amount.
Once that is stable, change the representation. Use a table, a sentence, a mixture, a map or an equation. Ask the learner to explain what stayed mathematically the same despite the changed surface. This is the point at which procedure begins to become a transferable relationship.
A lesson can stop before calculation
Give several questions and ask only for the labels, the part equation and the check. For “5:8 with a difference of 27”, a sufficient first representation is “three parts = 27; later check both ratio and difference”. This isolates understanding from arithmetic load.
Then supply a completed but incorrect solution. Ask whether the error is in labels, units, part mapping, arithmetic or interpretation. A student who can locate and explain an error has information that a correct copied solution cannot reveal.
A return task
At a later practice session, ask the learner to construct a ratio problem with a total and another with a difference. Then solve both. Creating a well-defined problem tests whether the learner knows which information fixes the scale and which information leaves it undetermined.
These are suggested teaching moves, not a promise that a fixed number of sessions produces mastery. Use the student’s independent explanations and changed-case performance to decide the next step.
25. Questions students often ask
Can a ratio contain fractions?
Yes. Fractions and decimals can represent the quantities being compared. When asked for simplest whole-number form, multiply every term by a common factor that clears the denominators or decimal places, then simplify.
Must the first ratio number be smaller?
No. A ratio follows the requested order. If the first quantity is larger, the first term can be larger. Reversing the order changes the comparison.
Is cross-multiplication always the fastest method?
No. A known total or difference often makes equal-part reasoning clearer. A constant unit value may be simpler through division to one unit. Cross-multiplication is useful when a correctly formed proportion equation is the cleanest representation.
Does the same ratio mean the same quantities?
No. It means the same relative comparison. A small mixture and a large mixture can have the same composition ratio while containing different amounts.
Can both quantities increase without direct proportion?
Yes. A fixed charge plus a per-item charge is one example. Direct proportion requires a constant multiplier, not merely a common direction of change.
What should I check at the end?
Check the ratio and every condition used to determine the scale: total, difference, known component or change. Then check the units and whether the quantities are allowed by the context.
26. The return path: from ratio to independent modelling
Ratio is a powerful bridge between arithmetic and algebra because it makes an invariant visible. Equivalent ratios preserve a quotient. A total or difference fixes a scale. A transfer preserves a total. Equal additions preserve a difference. Direct and inverse proportional models preserve different relationships.
The learner’s job is to identify the right preserved relationship, not to force every problem through one memorised rule. Before calculating, name the quantities. During calculation, preserve the comparison. Afterwards, return the numbers to the original situation.
Continue with Percentages and Reverse Percentages for comparisons per hundred, or Rate, Speed and Unit Conversion for comparisons between different kinds of quantity. For representation choices, revisit Word Problems and Mathematical Representation.
Sources and teaching boundaries
The MOE syllabus directory is the official curriculum reference. This article is an independent teaching sequence with labelled extensions. It does not assign every example to every subject level or claim an official examination mark scheme.
Supplementary reading: OpenStax, Ratios and Rate, for introductory ratio notation, common-unit comparisons and unit rates. The worked cases and practice questions above are original constructions. Prices, recipes, machines and populations are explicitly mathematical models, not reports about current products or services.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Establish the comparison, test the constraint, check a changed case and return the answer to its labelled quantities.
