PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 5 · GUIDE 19
A bar model is useful when its parts say the same thing as the question. A neat rectangle is not automatically a mathematical model. Every bar, division, bracket and label must correspond to a quantity or a relationship that the problem actually gives.
The central decision is therefore not “How many bars should I draw?” It is “What relationship am I trying to represent?” Are separate parts being combined into a whole? Are two amounts being compared by their difference? Are equal groups being repeated? Is one amount several times another? Is the unknown a whole, a part, a difference, one group or a number of groups?
This extended guide teaches those choices through worked models, contrast pairs, deeper challenges, model repairs and twenty practice questions with full explanations. It is a lesson in selecting and checking a representation, not a demand to draw a bar for every calculation.
Series route: return to the Primary 4 Mathematics Learning Hub. The earlier Multiplicative Comparison and Unknown Units guide provides additional practice with repeated equal quantities.
Learning boundary: this independent lesson applies arithmetic foundations referenced to the MOE Primary Mathematics Syllabus. The model families below are teaching categories, not a claim that every school uses identical terminology or pacing. Optional challenges are clearly identified. All quantities and situations are constructed for practice.
Navigate: what a model promises · parts and wholes · difference · equal groups · model choice · practice · answers · teaching and repair.
1. A model is a claim about relationships
When two bar sections are marked equal, the model claims they represent equal quantities. When one full bar is labelled 720, the model claims that its entire represented quantity is 720. When a bracket spans two separate bars, it claims that the bracket refers to their combined amount.
These claims must be supported by the question. A child cannot divide a bar into four equal parts merely because four boxes fit neatly on the page. The equality needs a reason: equal sharing, identical group sizes, a stated multiplier or a fraction describing equal parts.
Similarly, a small segment labelled 38 has a different meaning from a bracket labelled 38 across two full bars. One might describe an excess; the other might describe a combined total. Moving the same label changes the mathematics.
Before calculating, read the model aloud. For example: “The longer quantity consists of an amount equal to the shorter quantity plus 38.” If that sentence matches the problem, the model has made the relationship visible. If it does not, correct the drawing before doing any arithmetic.
The model should help someone inspect your reasoning without needing to guess what every rectangle means.
2. A drawn bar need not be a ruler
Bar models are often schematic. A section representing 187 objects need not be exactly 187 millimetres long, and a bar representing 244 objects need not be drawn with a perfectly proportional length. The labels and stated relationships carry the exact values.
However, the drawing must not contradict its intended meaning. Equal units should be visibly indicated as equal. A bar representing the larger quantity should not be presented as the smaller one without explanation. An excess segment should extend beyond the matched part of the comparison.
Never measure an unlabelled bar on the page to invent a missing value unless the question explicitly supplies a scale and asks for measurement. An unlabelled bar being twice as long in the drawing does not automatically establish a multiplier of two.
The right balance is to use a clear rough drawing while preserving exact mathematical labels. Precision of meaning is essential; artistic precision is not.
In the text diagrams below, repeated boxes labelled “one unit” represent equal quantities. Their purpose is to record equality, not to reproduce a printed scale drawing.
3. Part-whole models combine separate, non-overlapping parts
A box contains 244 red counters and 187 blue counters. Every counter is one of these two colours. How many counters are there altogether?
The two colour groups are separate parts of one collection. Combine them: 244 + 187 = 431 counters.
Whole collection | 244 red counters | 187 blue counters | Total: 431 counters
The model says that the whole is exhausted by these two parts. The statement that every counter belongs to one of the two colours matters; otherwise another unreported category could exist.
Do not use a difference model simply because the two quantities are unequal. The question asks for their total, not their comparison. A model is selected by the requested relationship, not by the fact that one number happens to be larger.
Check by removing either part from the whole: 431 − 244 = 187 and 431 − 187 = 244. Both checks confirm that the two given parts account for the whole collection.
4. A missing part is found from the whole and the known part
A school has 720 exercise books. Of these, 285 are used and all the others remain unused. How many are unused?
The original whole contains two non-overlapping parts: used and unused. The unknown part is 720 − 285 = 435 exercise books.
Original whole: 720 books | 285 used | unknown unused | Unused = 720 − 285
The subtraction follows from the part-whole relationship. The word “used” does not independently decide the method. If the unused amount and used amount were given instead, we would add them to recover the original whole.
Rotate the unknown deliberately. Give a whole and one part; ask for the other. Then give both parts; ask for the whole. Finally, give the whole and the other part. The drawing may remain almost unchanged while the calculation changes.
This helps the learner understand that a model records a relationship among quantities, not one fixed operation. The position of the unknown determines how that relationship will be used.
5. Check whether the proposed parts overlap
Deeper reasoning check. A group contains eighteen pupils who attend reading club and fifteen who attend art club. Seven pupils attend both. How many different pupils attend at least one of the clubs?
Adding eighteen and fifteen counts the seven shared pupils twice. The number of different pupils is 18 + 15 − 7 = 26.
This example does not require formal set notation. It requires checking whether the supposed parts are genuinely separate. A simple two-part bar labelled eighteen and fifteen would be misleading if it pretended that the clubs contained entirely different pupils.
For core part-whole practice, use clearly non-overlapping categories such as red and blue counters when each counter has only one of those colours. For more complex questions, state how shared membership is handled.
The underlying habit is useful well beyond this particular problem: before adding categories into one total, ask whether any object is counted more than once and whether any category is missing. A model should preserve the collection being counted, not merely make the numbers easy to add.
6. Difference models align equal parts of two quantities
Amir has 146 cards. Bea has 59 fewer cards than Amir. How many cards does Bea have?
Draw Amir’s longer bar and align Bea’s shorter bar beneath it. The part by which Amir extends beyond Bea is 59. Bea therefore has 146 − 59 = 87 cards.
Amir: | amount equal to Bea | 59 extra | Bea: | 87 cards | Amir's complete amount: 146 cards
The difference is the unmatched segment, not one of the full bars. Labelling Bea’s whole bar as 59 would change the statement to “Bea has 59 cards”, which is not what the question says.
The model also supports a reverse question. If Bea has 87 and Amir has 59 more, Amir has 87 + 59 = 146. If both amounts are given, subtract to find the difference.
Ask which amount provides the reference for “more” or “fewer”. A comparison sentence can be read backwards accidentally, especially when the larger amount is not mentioned first.
7. More by an amount is not more by a multiplier
Compare “Tara has five more cards than Jun” with “Tara has five times as many cards as Jun.” If Jun has 24 cards, the first statement gives Tara 29; the second gives Tara 120.
In the additive model, Tara has one amount equal to Jun’s plus a fixed excess of five. In the multiplicative model, Tara has five equal units, each equal to Jun’s whole amount.
The word “five” has a different role. In the first sentence it counts extra cards. In the second it counts how many copies of Jun’s amount make Tara’s amount.
Write the units beside the number when uncertain. “Five cards” can be an additive difference. “Five equal groups of Jun’s cards” describes multiplication. The distinction is semantic, not a matter of spotting one isolated word.
For precise questions, prefer “five times as many” to the potentially ambiguous phrase “five times more”. The model should be built from a clear comparison, not from wording that could support two different interpretations.
8. Equal-group models start with a justified equal unit
Six trays each contain 26 counters. How many counters are there altogether?
Every tray has the same number, so one equal unit represents 26 counters. Six units represent 6 × 26 = 156 counters.
| 26 | 26 | 26 | 26 | 26 | 26 | Six equal groups; total 156 counters
The equality comes from the word “each” together with the stated common group size. It does not arise because the boxes were drawn alike.
If the question instead lists six trays with different amounts, this equal-unit model is not justified. Use a table or a part-whole representation with separately labelled values. Identical containers do not guarantee identical contents unless the question says so.
A good model makes the multiplication sentence explainable: the multiplier counts groups, while the other factor gives the amount in each group. Their product is the combined amount. This distinction will help interpret division when either of those two inputs becomes unknown.
9. Sharing and grouping can use the same division but mean different things
Question A: 156 counters are shared equally among six trays. How many counters are in each tray? The total and number of groups are known, so 156 ÷ 6 = 26 counters per tray.
Question B: 156 counters are arranged with six counters in each tray. How many trays are filled? The total and group size are known, so 156 ÷ 6 = 26 trays.
The arithmetic is identical. The quotient has a different meaning because the known six refers to a different quantity.
For Question A, draw six boxes and put a question mark in each. For Question B, label each group as six and mark the number of groups as unknown. The models show exactly what the division is finding.
This is why “I divided” is not a complete explanation. Ask, “Did the division find the number in one group or the number of groups?” The answer should include the appropriate unit.
When a remainder occurs, use the context to decide whether to report leftovers, complete groups or an additional container. A bar alone does not make that interpretation for the learner.
10. Multiplicative comparison turns one whole amount into a unit
Jun has 24 cards. Tara has five times as many. Jun’s whole amount is one unit. Tara’s whole amount is five units, so Tara has 5 × 24 = 120 cards.
Together they have six units, or 144 cards. Their difference is four units, or 96 cards. The multiplier five describes Tara relative to Jun; it is not automatically the number of units in their total or their difference.
This distinction prevents a common error. If a problem gives the total, divide by the total number of units. If it gives the difference, divide by the unmatched number of units. Do not divide every supplied number by the comparison multiplier.
The same pair of bars can support all three questions. The drawing should show which bracket contains the known information: Tara’s full bar, both full bars together, or only the difference segment.
Once again, the location and meaning of the known quantity determine the operation. The bar model is a relationship map, not an automatic calculation machine.
11. Use a total to recover an unknown unit
Nadia has four times as many beads as Omar. Together they have 210 beads. Find both amounts.
Omar is one unit and Nadia is four units. Their combined total is five units. One unit is 210 ÷ 5 = 42 beads. Omar has 42 and Nadia has 4 × 42 = 168 beads.
Check both conditions: 42 + 168 = 210, and 168 is four times 42. Checking only the total would not be enough because many other pairs also total 210.
An incorrect model might divide the total into four units instead of five. That would treat the larger amount as though it were the whole collection and omit Omar’s separate one-unit amount.
To repair it, ask the child to point to every bead represented by the total. Does the total include only Nadia’s beads, or both people’s? The answer determines the bracket. Once the bracket covers both bars, the fifth unit becomes visible and the correct division follows naturally.
12. Use a difference to recover an unknown unit
Deeper challenge. Hana has six times as many tokens as Iman. Hana has 140 more tokens. Find both amounts.
Iman is one unit and Hana is six units. Align one of Hana’s units with Iman’s whole bar. The unmatched excess is five units, and those five units total 140. One unit is 140 ÷ 5 = 28 tokens.
Iman has 28 and Hana has 6 × 28 = 168 tokens. Check: 168 − 28 = 140 and 168 ÷ 28 = 6.
Dividing 140 by seven would confuse the difference with the combined total. Dividing 140 by six would confuse the difference with Hana’s full amount. The units being counted must match the meaning of the known 140.
A useful repair is to colour or underline only the unmatched units and label their bracket 140. This makes the relationship inspectable before calculation. Do not write the multiplier beside an arbitrary collection of bars and hope the arithmetic will choose the correct interpretation.
13. Total-and-difference problems can be equalised
Deeper challenge. Two children have 174 counters altogether. One has 38 more than the other. Find both amounts.
Represent the larger amount as a copy of the smaller amount plus an excess of 38. Remove that excess from the combined total: 174 − 38 = 136. What remains represents two equal copies of the smaller amount, so the smaller amount is 136 ÷ 2 = 68. The larger is 68 + 38 = 106.
Another route adds 38 to the total instead. That creates two equal copies of the larger amount: 174 + 38 = 212; 212 ÷ 2 = 106. Subtracting the difference gives 68.
Both routes are justified by equalisation. The first removes an excess; the second supplies the missing excess to the smaller bar. Do not teach them as two unrelated formulas.
Check the full result: 68 + 106 = 174 and 106 − 68 = 38. A correct pair must satisfy both conditions at once.
14. Fraction models name equal parts of one reference whole
Three eighths of a collection are 27 counters. Find the whole collection.
Draw eight equal units for the entire collection. The known 27 belongs to three of those units. One unit is 27 ÷ 3 = nine counters; eight units total 9 × 8 = 72 counters.
The model does not say that each small unit is 27. It says that the selected three units together are 27. Place the known label over the correct span.
If the question asks for the remaining counters, five units remain, giving 5 × 9 = 45 counters. Alternatively subtract 27 from 72. These two routes offer a useful check.
Do not assume that equal-looking fraction bars from different problems represent equal physical amounts. Half of 24 is twelve, while one third of 45 is fifteen. The fraction with the larger value can produce the smaller quantity when its reference whole is smaller.
Every fraction model needs a named whole. Equal parts are equal within that specified whole; they are not automatically equal to parts of another collection.
15. Keep nested comparisons on a consistent unit
Optional stretch. B has three times C’s amount. A has twice B’s amount. Their combined total is 260. Find the three amounts.
Choose C as one basic unit. B is three of those units. Since A is twice B, A is six of the same basic units. Altogether there are 1 + 3 + 6 = ten units.
One unit is 260 ÷ 10 = 26. Therefore C has 26, B has 78 and A has 156. Check both comparisons and the total: 78 = 3 × 26, 156 = 2 × 78, and 26 + 78 + 156 = 260.
The common error is to draw C as one unit, B as three and A as two. Those labels use incompatible meanings for “one unit”. A’s two units would each equal all of B, not one copy of C.
A consistent basic unit solves the conflict. This is a useful stretch only after simple one-to-several comparisons are secure; it should not be used to rush a learner who still cannot identify the reference quantity.
16. A bar model must not mix different moments
Suppose A has 84 cards and B has 52, then A gives ten cards to B. A before-and-after table gives A’s final 74 and B’s final 62.
A comparison model for the initial moment describes 84 and 52. A model for the final moment describes 74 and 62. Joining A’s initial bar to B’s final bar would combine quantities from different states and create a misleading total of 146.
Draw separate models for separate moments or use a two-column table. Label “before” and “after” explicitly. Do not keep a unit label unchanged across time unless the corresponding quantity is genuinely unchanged.
For a sequence involving one amount, a timeline may be clearer than several bars. See Before-and-After Problems and Working Backwards for that distinction.
The best representation is the one that preserves the relationships with the least ambiguity, not necessarily the one most recently practised.
17. Choose the model from the unknown and the relationship
| What the problem says | Useful representation | What to label carefully |
|---|---|---|
| Separate parts make a whole | One bar partitioned into parts | The whole bracket and every part |
| One amount is more or fewer by a fixed amount | Aligned comparison bars | The unmatched difference |
| Several groups contain the same amount | Repeated equal units | Group size versus number of groups |
| One amount is several times another | One-unit and multi-unit bars | The reference amount |
| A fraction of a set is known | Equal parts of a named whole | The span represented by the known part |
| One quantity changes through events | Timeline or before-and-after table | The moment belonging to each amount |
| A geometric shape has missing dimensions | Labelled geometric diagram | Actual side relationships and units |
This is a selection guide, not a compulsory classification test. Some questions support more than one valid model. The model earns its place by making the unknown easier to find and the solution easier to check.
A simple calculation such as 24 + 17 may not need any bar at all. Use representation when it clarifies a relationship, not merely to add an extra task.
18. Know when the model cannot supply a missing fact
“A has twenty more counters than B” does not determine either amount. A could have 32 and B twelve, or A could have 47 and B 27. Both pairs have a difference of twenty.
A comparison model can show the fixed excess, but it cannot invent the length of the shared unknown part. A total, one actual amount or another determining condition is needed.
Likewise, a bar drawn twice as long as another does not prove that one quantity is twice the other. Unless the multiplier is given or derived from other facts, the appearance of the sketch is not evidence.
Separate insufficient information from an invalid model. A model can faithfully represent a problem with several possible answers. That is not a failure of drawing; it is an honest record of what is known.
A good final response may therefore identify a missing condition rather than provide one unsupported number. Mathematical representation should reveal uncertainty as clearly as it reveals solvable structure.
19. Audit the drawing before trusting its answer
Read every label back into a sentence. A total bracket should cover everything described by “altogether”. A difference bracket should cover only the excess. Every repeated unit should represent the same quantity. Every bar should belong to a named person, group, object or moment.
Then inspect what the calculation does. Dividing a total by the number of equal units finds one unit. Multiplying a known unit by a unit count finds a larger amount. Subtracting a known part from a whole finds the complementary part.
If the arithmetic is correct but the model is wrong, recopying the arithmetic will not repair the solution. Return to the first label that does not match the words.
Finally, test the resulting quantities in the original problem. A pair found from a total-and-difference model must satisfy both the total and difference. A multiplicative comparison must satisfy the multiplier as well as any total. A fraction model must reproduce the given part of the specified whole.
Models become trustworthy through these checks, not through their neatness or familiarity.
20. Practice laboratory: choose, label and solve
For each question, name the relationship before selecting a model. Draw only what helps. Questions 12–14 and 18–20 include deeper checks. Explain why the chosen representation fits the information.
- A box contains 244 red counters and 187 blue counters, with no other counters. Find the total.
- Of 720 exercise books, 285 are used. Find the unused amount.
- Amir has 146 cards. Bea has 59 fewer. Find Bea’s amount.
- Bea has 87 cards. Amir has 59 more. Find Amir’s amount.
- Two collections have 208 and 147 items. Find the difference.
- Six trays each contain 26 counters. Find the total.
- 156 counters are shared equally among six trays. Find the amount in each.
- 156 counters are placed with six in each tray. Find the number of trays.
- Jun has 24 cards. Tara has five times as many. Find Tara’s amount, their combined total and their difference.
- Nadia has four times Omar’s beads. Together they have 210. Find both amounts.
- Hana has six times Iman’s tokens and 140 more than Iman. Find both amounts.
- Two children have 174 counters altogether, and one has 38 more. Find both amounts.
- Three eighths of a collection are 27 counters. Find the whole and the unselected amount.
- B has three times C’s amount, and A has twice B’s amount. Their total is 260. Find A, B and C.
- A has 84 cards and B has 52. A gives ten to B. Find both final amounts and explain why separate time labels help.
- A has twenty more counters than B. Can either amount be found uniquely from that information alone?
- A child draws one unlabelled bar twice as long as another. Does this establish that the first quantity is twice the second?
- Eighteen pupils attend reading club, fifteen attend art club and seven attend both. Find the number of different pupils attending at least one.
- Which is greater: half of 24 or one third of 45? Explain why the fractions alone do not decide the answer.
- A question says that A has four times B and their total is 210. A student labels the four-unit bar as 210. Explain the model error and repair it.
After solving, identify one question for which a model was especially helpful and one for which a short number sentence was sufficient. Explain the choice. Representation should be a deliberate aid to reasoning, not a compulsory decoration.
21. Explained answers and model checks
1. Use a part-whole relationship: 244 + 187 = 431 counters. The two non-overlapping colour groups account for the whole.
2. The unused books are the missing part: 720 − 285 = 435. Check that used and unused books total 720.
3. Use a difference model. Bea’s shorter amount is 146 − 59 = 87 cards. The 59 labels the excess, not Bea’s full bar.
4. Add the excess to Bea’s known amount: 87 + 59 = 146 cards. This reverses the unknown in Question 3 without changing the relationship.
5. Align the two amounts and find the unmatched segment: 208 − 147 = 61 items.
6. Six equal groups of 26 give 6 × 26 = 156 counters. The repeated units are justified by the equal tray contents.
7. The total is shared into six equal units: 156 ÷ 6 = 26 counters per tray. The quotient is a group size.
8. Each group contains six, so 156 ÷ 6 = 26 trays. Here the quotient is a number of groups.
9. Tara has five units: 5 × 24 = 120 cards. Together they have six units, 144. Their difference is four units, 96.
10. The total represents five units. One is 210 ÷ 5 = 42. Omar has 42 beads and Nadia has 168. Both the total and multiplier check.
11. The difference represents five units, not six. One unit is 140 ÷ 5 = 28. Iman has 28 tokens and Hana has 168.
12. Remove the excess from the total: 174 − 38 = 136. Two equal smaller amounts remain, so the smaller is 68 and the larger is 106. They total 174 and differ by 38.
13. Three units are 27, so one is nine. The whole eight units total 72 counters; the other five units total 45.
14. With C as one unit, B is three and A is six. Ten units total 260, so one is 26. Thus A = 156, B = 78 and C = 26.
15. A ends with 74 and B with 62. Before-and-after labels prevent an initial amount from being combined with another person’s final amount. Their combined total remains 136.
16. No unique amounts are determined. For example, 32 and twelve fit, as do 47 and 27. A further determining condition is needed.
17. No. A rough unscaled drawing does not provide an unstated numerical multiplier. The relationship must come from given or derived information.
18. Adding club counts gives 33 memberships, but the seven shared pupils were counted twice. The number of different pupils is 18 + 15 − 7 = 26.
19. Half of 24 is twelve. One third of 45 is fifteen. Therefore one third of 45 is greater. Different reference wholes prevent the fractions alone from determining the resulting amounts.
20. The 210 is the combined total of A’s four units and B’s one unit. Move the total bracket across all five units. One unit is 42, giving B = 42 and A = 168.
22. Teach model selection before model copying
Begin with a contrast set rather than ten identical questions. Give one part-whole problem, one difference problem and one equal-group problem using manageable numbers. Ask the learner to explain what the unknown represents before drawing.
Then change the unknown while preserving the same relationship. For example, use the same total and parts but ask for a missing part instead of the whole. The student should discover that the model can remain useful while the operation changes.
When a drawing goes wrong, ask the learner to translate it back into words. “This says Bea has 59 cards” may immediately reveal why a difference label was misplaced. Repairing the sentence-model match is more useful than supplying a finished picture to copy.
For equal-group questions, repeatedly distinguish amount in each group from number of groups. For comparisons, identify the reference amount. For fractions, name the whole and the known span. For changing quantities, separate the moments.
A suggested later return is one unfamiliar context that preserves a familiar relationship. Do not announce the model family. The learner’s independent choice is evidence to inspect; merely completing a page of copied models is not.
A useful exit task
Ask the student to invent two different stories for one model, then a third story for which that model would be wrong. This tests both where the representation applies and where its limits lie. The numbers can stay small because the real task is relational understanding.
23. From a good model to more than one valid route
A clear model helps us find an answer, but it can also help us compare methods. A total-and-difference question can be equalised by removing or adding the excess. A missing fraction part can be found by counting unselected units or subtracting from the whole. A division can be checked by rebuilding equal groups.
Continue to Multiple Solution Routes: Choosing, Comparing and Verifying Mathematics Strategies.
Final checkpoint: can the learner explain every label, justify every equal unit, identify the quantity being found and test the answer in all the original conditions? A bar model is successful when it makes those decisions visible.
Source and editorial note
The curriculum reference is the MOE Primary Mathematics Syllabus, updated October 2025. The model-selection categories, examples, practice questions and teaching suggestions are independently written. The optional challenges are not assertions about compulsory school pacing or assessment content.
Editorial approach: Wintour House V1.0 · CivDJ · eduKate Publishing. Choose the representation from the relationship, test the model and return the quantities to the question.
