BTT Mathematics / Primary Mathematics Learning Hub / Guide 4
To solve a word problem, identify the quantities, state their relationships, choose a representation, calculate and check the result against every condition. A bar model helps when it makes those relationships visible. It does not help merely because a rectangle has been drawn beside some numbers.
The hardest move can come before the first calculation. A child may know how to subtract and still subtract the wrong quantities. Another may produce a correct total but divide it into the wrong number of equal parts. These are not solved by making the same arithmetic faster; the relationship needs to be represented accurately.
This guide moves from one-step stories to comparison, equal units, before-and-after changes and multi-step checking. Consult the MOE Primary Mathematics syllabus for official year-level scope. The progression and difficulty groups here are teaching choices, not official examination bands. All scenarios and questions in this guide are original illustrations.
Read the relationship · Choose a model · Track changes · Check and repair · 24 questions · Worked answers · Teaching and next steps
1. Read for quantities, not just numbers
A number becomes useful when we know what it measures. “Thirty” might count pupils, dollars, centimetres or packets. “Six” might count groups or the amount in one group. Writing the numbers alone throws away information needed to choose the operation.
Begin by naming each known quantity and the unknown. For a packing problem, a useful record might be “total pencils: thirty; pencils in one packet: six; number of packets: unknown.” That record already suggests grouping thirty into sets of six.
Do not insist on a long table for every simple problem. A short label beside a number can be enough. The purpose is to keep the mathematical roles visible, not to create additional copying work.
The question asks for a particular quantity
Suppose a story says that a class has eighteen red counters and seven blue counters. One question asks for the total. Another asks how many more red counters there are than blue. The given numbers are unchanged, but the questions require different relationships.
For the total, combine the two parts: eighteen plus seven. For the difference, compare the quantities: eighteen minus seven. Copying every number correctly does not settle which operation is appropriate. The requested quantity completes the structure.
Before calculating, ask the learner to finish the sentence “I am finding…” with a quantity, not an operation. “I am finding how many packets can be filled” is more informative than “I am doing division.” The operation should serve the quantity we need.
Why keywords are not enough
“Jo has twelve more counters than Kim. Jo has thirty-one counters. How many does Kim have?” The word more appears, but adding twelve to thirty-one would move away from the smaller quantity being asked for. Jo’s amount consists of Kim’s amount plus twelve, so Kim has nineteen.
Likewise, the word altogether does not always mean that the next written step is addition. “The children have sixty-four counters altogether; one has twelve more than the other” gives a known total. Solving for the individual amounts requires removing the difference and splitting the equal part.
Keywords can draw attention to a possible relationship, but they cannot identify which quantity is missing by themselves. Read the sentence as a statement about two amounts. Then ask which direction we must travel to reach the unknown.
Retell without changing the mathematics
Ask the learner to retell a problem in one or two short sentences. A good retelling preserves the quantities, the comparison and the question. It may remove descriptive details that do not affect the calculation, but it must not remove conditions such as equal, remaining, at first or each.
For example, “There are seven equal boxes of eight counters, and nineteen counters are removed” retains the structure needed to find the remainder. Retelling it as “There are seven boxes and nineteen counters are removed” loses the amount in a box and leaves the total unknown.
If a child cannot retell the relationship, do not immediately supply an operation. Ask which amount belongs to which object or stage. The first repair may be a label or a clarification of a pronoun, not another arithmetic demonstration.
2. Choose the model that makes the missing quantity visible
Part-whole: two parts make a total
A part-whole model places two or more non-overlapping parts inside a known or unknown total. For eighteen red counters and seven blue counters, one bar can contain a segment labelled eighteen and another labelled seven. The whole bar represents all twenty-five counters.
If the total and one part are known, the other part is found by subtraction. If both parts are known, the total is found by addition. These are not different story types to memorise separately; the same model has a different missing label.
Check that the parts do not overlap. If a question counts children who like football and children who like swimming, some may belong to both groups unless the conditions exclude that. Adding the counts without checking would not necessarily count distinct children. The model must fit the conditions actually given.
Additive comparison: one amount has an extra part
For “A has fourteen more than B,” draw B’s bar and then draw A’s bar as the same length plus an extra segment labelled fourteen. The shared part represents B’s amount. The extra segment represents the difference, not A’s complete amount.
In words, the relationship is larger amount = smaller amount + difference. From it, we can find the larger amount by addition, the smaller amount by subtraction, or the difference by comparison. The representation stays stable while the unknown changes.
A careful model labels the entire larger amount separately from the extra segment. A common mistake is to attach a number describing the whole larger bar to only the extra part. The picture then looks familiar but says something different from the problem.
Worked example A: Total and difference
Two children have fifty-eight counters altogether. The first has fourteen more than the second. How many counters does each have?
Remove the extra fourteen from the total. What remains is two equal copies of the smaller amount: 58 − 14 = 44. Divide by two to find one smaller amount: 44 ÷ 2 = 22. Add fourteen to obtain the larger amount: 22 + 14 = 36.
The amounts are 22 and 36. Check both conditions: 22 + 36 = 58 and 36 − 22 = 14. Checking only the total would not establish the required difference.
The division by two is justified because removing the extra part leaves two equal bars. It is not a rule to apply whenever two people appear in a story. If the quantities do not become equal after the stated adjustment, this model has not been established.
Multiplicative comparison: one amount contains equal copies of another
“A has three times as many as B” is different from “A has three more than B.” The first says that A contains three equal copies of B’s amount. The second says that A contains one copy of B’s amount and an extra three.
Use a single unit to represent B. A then contains three equal units. If the amounts are combined, there are four units altogether. If they are compared, A has two units more than B. The known number must be attached to either the total or the difference, depending on the wording.
The word unit here means one repeated equal amount in the model. It does not necessarily mean one counter or one dollar. A unit might later turn out to contain sixteen counters. Keeping that distinction prevents an unlabeled bar from becoming a source of confusion.
Worked example B: A multiple and a total
A has three times as many counters as B. Together they have sixty-four counters. The bars contain three units for A and one for B: four units altogether.
One unit is 64 ÷ 4 = 16. B has sixteen counters, and A has 16 × 3 = forty-eight. Check the total, 48 + 16 = 64, and the comparison, 48 = 3 × 16.
Dividing sixty-four by three would incorrectly label the total as A’s three units. The total includes B’s unit too. The most important line is therefore “four units represent sixty-four,” not the later arithmetic.
Worked example C: A multiple and a difference
A has four times as many counters as B, and A has twenty-four more counters than B. A’s four units exceed B’s one unit by three units. The difference twenty-four therefore represents three units.
One unit is 24 ÷ 3 = 8. B has eight counters; A has thirty-two. Check 32 − 8 = 24 and 32 = 4 × 8. The total would be forty, but the total was not the given twenty-four.
Compare this with the previous example. Both use repeated equal units, but one labels the sum of the bars and the other labels their gap. Identifying that distinction is more valuable than memorising two separate strings of operations.
A model is a statement, not a decoration
After drawing, ask the learner to read the model back into words. Which bar represents which person? What does one equal unit stand for? Which collection of units is labelled by the known number? Where is the unknown?
If the model cannot reproduce the sentences in the question, do not calculate from it yet. Neat lines, equal spacing and a correct-looking final number do not compensate for a mislabeled quantity. A rough but faithful model is more useful than a polished model of the wrong problem.
Drawn lengths are usually schematic rather than measured to scale. Labels and stated equalities carry the mathematical information. Do not infer a numerical value by measuring a bar that was drawn only to represent a relationship.
3. Track before-and-after changes without mixing stages
Worked example D: A transfer changes two amounts
Nila has forty counters and Omar has twenty. Nila gives some counters to Omar until they have equal amounts. How many counters are transferred?
The total remains sixty because counters move between the two people rather than entering or leaving the combined collection. Equal final amounts must therefore be thirty each. Nila gives 40 − 30 = ten counters.
Another explanation uses the original difference of twenty. Each counter transferred reduces Nila’s amount by one and increases Omar’s by one, shrinking the gap by two. A transfer of ten closes a gap of twenty.
Giving twenty would not merely remove the difference: it would reverse the positions, leaving Nila with twenty and Omar with forty. This is a useful counterexample because it shows why the difference and the required transfer are not the same quantity.
Use a stage record when the quantities change
A before-and-after table can be more useful than an elaborate bar diagram. For the transfer, record Nila and Omar before, the amount moving from one to the other, and their final amounts. The combined total provides an additional check at each stage.
For spending, the total may decrease. For receiving an external gift, it may increase. Before using “the total stays the same,” ask whether the action merely transfers an amount inside the system or moves an amount across its boundary.
This is ordinary reading expressed mathematically. A gift from one named child to the other preserves their combined total; a purchase from a shop removes money from the two-child total. The story determines what can be held fixed.
Worked example E: Reverse a sequence in the opposite order
A child spends nine dollars, then spends half of the money remaining. Eighteen dollars are left. How much money was there at first?
Immediately before spending half of the remainder, the child had twice the final eighteen dollars: thirty-six. Before the earlier nine-dollar spending, there were 36 + 9 = forty-five dollars.
Check forward: forty-five minus nine leaves thirty-six; spending half of thirty-six leaves eighteen. The reverse order matters. Adding nine to eighteen before doubling would undo the two actions in the wrong sequence and produce a different starting amount.
Write the stages as words if symbols obscure the meaning: at first; after spending nine; after spending half of what remained. Each arrow should describe one action, and the backwards solution should undo the last action first.
Worked example F: A fraction of the remainder uses a new reference amount
A seller starts with ninety-six pieces of fruit. One-quarter is sold in the morning. One-third of what remains is sold in the afternoon. How much remains after both sales?
The morning sale is 96 ÷ 4 = 24, leaving seventy-two. The afternoon fraction applies to seventy-two, not ninety-six. One-third of seventy-two is twenty-four, leaving forty-eight pieces of fruit.
Subtracting one-quarter and one-third of ninety-six would assume both sales referred to the original stock. The phrase “of what remains” changes the reference quantity. Label the stage before applying the next fraction.
Do not make every problem into a bar
A rectangular garden that is twelve metres long and eight metres wide is more naturally represented by a rectangle with labelled dimensions. Its perimeter is the distance around the boundary; its area is the surface enclosed. A bar model would not automatically make that distinction clearer.
A timeline may clarify a duration problem. A table may clarify a repeated rate. A list of possible cases may clarify a small counting problem. Choose the representation that exposes the relationship, rather than forcing every story into the same shape.
There is no reward for using a complicated representation where one clear sentence would suffice. There is also no reward for doing everything mentally when a simple record would prevent stages or units from being mixed.
4. A solution must satisfy every condition
Check the total and the comparison separately
Suppose a problem says two quantities total eighty-four and the first is three times the second. A proposed pair of fifty-six and twenty-eight has the correct total, but fifty-six is only twice twenty-eight. Passing one check does not make the pair a solution.
The correct pair is sixty-three and twenty-one. Their sum is eighty-four, and sixty-three is three times twenty-one. Each sentence in the original question becomes a condition the answer must satisfy.
This approach is particularly useful when an incorrect route produces plausible-looking numbers. A child may feel that an answer is right because the numbers are tidy. Reconstructing the original statements replaces that feeling with explicit evidence.
Check the kind of quantity and its unit
For a twelve-by-eight-metre rectangle, 12 × 8 = 96 gives an area of ninety-six square metres. It does not give the length of fencing needed around the boundary. The perimeter is 12 + 8 + 12 + 8 = forty metres.
The unit can reveal a wrong interpretation. Metres measure length; square metres measure area. Writing a correct unit after the wrong formula does not repair the model. The calculation and the quantity it claims to measure must agree.
Check whether there is enough information
A box contains twelve red counters and eighteen blue counters. How many green counters does it contain? Nothing in those two counts determines the green count unless an additional condition gives the total or a relationship involving green counters.
Zero green counters is possible, but it is not established. Ten green counters is also possible under the stated information. A responsible mathematical answer is that there is insufficient information, not that every number in the question must be combined somehow.
This habit is worth practising. Some educational questions deliberately supply incomplete or irrelevant information, and real situations often do. The learner should distinguish “I have not found the route yet” from “the stated conditions do not determine one answer.”
Repair the earliest incorrect relationship
When a solution fails, find the first line that no longer follows from the question. If “four units” should have been written but “three units” was used, repair that line before redoing division. If a fraction was applied to the original amount instead of the remainder, relabel the stage before recalculating.
A final answer can be wrong because of a small arithmetic slip inside a valid model. It can also be wrong because every later calculation faithfully followed an invalid model. These cases need different corrections, and the written working helps distinguish them.
5. Practice: 24 questions
For each problem, name the unknown and show the relationship before calculating. A bar, labelled sentence, table or other suitable representation is acceptable. Keep the answers covered. The final questions include several-step reasoning and interpretation; choose them only when the component operations are familiar.
Questions 1–8: One relationship at a time
1. A shelf has fourteen books. Nine more books are placed on it. How many books are on the shelf now?
2. A box contains thirty-five counters. Twelve are removed. How many remain?
3. A bag contains eighteen red counters and seven blue counters, and no others. How many counters are in the bag?
4. Mei has twenty-six cards. Ari has nine fewer cards than Mei. How many cards does Ari have?
5. Four packets contain six pencils each. How many pencils are there altogether?
6. Thirty pencils are shared equally among five children. How many pencils does each child receive?
7. Jo has twelve more counters than Kim. Jo has thirty-one counters. How many counters does Kim have?
8. A bag contains forty-two counters, all red or blue. Seventeen are red. How many are blue?
Questions 9–16: Combine relationships
9. Mina has forty-eight cards. Noah has sixteen more than Mina. How many cards do they have altogether?
10. Seven boxes contain eight counters each. Nineteen counters are removed from the collection. How many remain?
11. Two children have sixty-four counters altogether. One has twelve more than the other. How many does each have?
12. A has three times as many counters as B. Together they have seventy-two counters. How many does each have?
13. A has four times as many counters as B and twenty-seven more counters than B. How many does each have?
14. After giving away seven cards, a child has twenty-eight cards left. How many cards were there at first?
15. A set contains ninety counters. Two-fifths are blue. How many are blue, and how many are not blue?
16. Five notebooks cost $3 each and one pen costs $4. A child pays $20 for all the items. How much change should be received?
Questions 17–24: Track, select and verify
17. Arun has twice as many counters as Bela. Together they have fifty-four counters. How many counters must Arun give Bela so that they have equal amounts?
18. A child spends $12, then spends half of the money remaining. There is $18 left. How much money was there at first?
19. A child has three-quarters of the original money left. This remaining amount is $36. How much was there at first, and how much was spent?
20. A seller starts with ninety-six pieces of fruit. One-quarter is sold in the morning. One-third of the remainder is sold in the afternoon. How many pieces remain?
21. A rectangular garden is twelve metres long and eight metres wide. Find the total length of its boundary and its area. Give the correct unit for each.
22. A container holds 2,400 millilitres of juice. Each full serving uses 300 millilitres. How many full servings can be poured?
23. A box contains twelve red counters and eighteen blue counters. The number of green counters is not stated. Can the green count be determined from this information? Explain.
24. Two children have eighty-four counters altogether. A has three times as many as B. A learner proposes fifty-six for A and twenty-eight for B because these add to eighty-four. Explain why the proposal fails and find the correct amounts.
6. Worked answers
Answers 1–8
1. 23 books. The final amount consists of the starting fourteen books plus nine added books: 14 + 9 = 23. The answer must exceed fourteen because books entered the shelf collection.
2. 23 counters. The remaining amount is the original thirty-five minus the twelve removed: 35 − 12 = 23. Check by recombining removed and remaining counters: 12 + 23 = 35.
3. 25 counters. Red and blue are the only two non-overlapping parts of the bag. Their total is 18 + 7 = 25. The phrase “and no others” makes the complete part-whole relationship explicit.
4. 17 cards. Ari’s amount is nine less than Mei’s twenty-six. Subtract: 26 − 9 = 17. Check the comparison in its original direction: Mei’s twenty-six is nine more than Ari’s seventeen.
5. 24 pencils. Four equal packets contain six pencils each, so 4 × 6 = 24. Dividing twenty-four by four gives six per packet again, confirming that all four packets have been counted.
6. 6 pencils per child. The total thirty is split into five equal shares. Calculate 30 ÷ 5 = 6. Five shares of six account for all thirty pencils without a remainder.
7. 19 counters. Jo’s thirty-one consists of Kim’s unknown amount plus twelve. Remove the extra part: 31 − 12 = 19. Adding despite the word more would find an amount larger than Jo’s, contrary to the relationship.
8. 25 blue counters. Red and blue together total forty-two. Remove the seventeen red counters: 42 − 17 = 25. The answer is a missing part, not a difference between two known colour counts.
Answers 9–16
9. 112 cards. Noah has 48 + 16 = 64 cards. Combine both children’s amounts: 48 + 64 = 112. The intermediate sixty-four belongs only to Noah; it is not the requested combined total.
10. 37 counters. The starting collection contains 7 × 8 = 56 counters. After nineteen are removed, 56 − 19 = 37 remain. Check that thirty-seven remaining plus nineteen removed restores the fifty-six counted in the boxes.
11. 26 and 38 counters. Remove the extra twelve from sixty-four, leaving fifty-two split between two equal smaller bars. Each smaller bar is 52 ÷ 2 = 26. The larger amount is 26 + 12 = 38. Check both sum sixty-four and difference twelve.
12. A has 54; B has 18. Three units for A and one for B give four units altogether. One unit is 72 ÷ 4 = 18. A has three units, or fifty-four. Check 54 + 18 = 72 and 54 = 3 × 18.
13. A has 36; B has 9. The gap between four units and one unit is three units. These three units represent twenty-seven, so one unit is nine. A has four units: thirty-six. Check difference twenty-seven and the four-times relationship separately.
14. 35 cards. The original amount must include both the twenty-eight remaining and the seven given away: 28 + 7 = 35. Check forward by removing seven from thirty-five to recover twenty-eight.
15. 36 blue; 54 not blue. One fifth of ninety is eighteen, so two fifths contain thirty-six blue counters. Subtract from the complete set: 90 − 36 = 54 not blue. The two amounts must sum to ninety.
16. $1. The notebooks cost 5 × $3 = $15. Add the pen to obtain a total cost of $19. Subtract this from the $20 paid: $20 − $19 = $1 change. Multiplying the pen cost by five would invent four additional pens.
Answers 17–24
17. 9 counters. The total contains two units for Arun and one for Bela. One unit is 54 ÷ 3 = 18, so Arun has thirty-six and Bela eighteen. Equal final amounts are twenty-seven each. Arun gives 36 − 27 = 9. Check: 36 − 9 = 18 + 9 = 27.
18. $48. Immediately before spending half of the remaining money, the child had 18 × 2 = $36. Restore the earlier $12 spent to obtain $48. Forward check: $48 − $12 = $36, and half of $36 remains as $18.
19. $48 at first; $12 spent. Three equal quarter-units represent $36. One quarter is $12, so the original four quarters total $48. The spent amount is the missing quarter, or $48 − $36 = $12.
20. 48 pieces. Sell 96 ÷ 4 = 24 in the morning, leaving seventy-two. The afternoon sale is 72 ÷ 3 = 24, leaving forty-eight. The two sales total forty-eight, and sold plus remaining returns the original ninety-six.
21. Boundary 40 metres; area 96 square metres. The boundary contains two twelve-metre sides and two eight-metre sides: 12 + 8 + 12 + 8 = 40. The area is 12 × 8 = 96. Length and surface coverage require different quantities and units.
22. 8 full servings. Count groups of three hundred millilitres inside twenty-four hundred millilitres: 24 ÷ 3 = 8. Check 8 × 300 = 2,400 millilitres. Every serving has the stated size and all the juice is accounted for.
23. No; there is insufficient information. The red and blue counts give thirty counters in those two colours, but no total or relationship fixes the green count. Both zero green counters and ten green counters are compatible with the information given, so there is no unique answer.
24. A has 63; B has 21. The proposed amounts pass the total check, but fifty-six is twice twenty-eight, not three times. A correct model contains four equal units altogether. One unit is 84 ÷ 4 = 21. Three units give sixty-three. Check both required conditions: 63 + 21 = 84 and 63 = 3 × 21.
7. Turn a worked solution into independent problem solving
After reading a solution, close it and ask the learner to reconstruct the first relationship rather than recite the arithmetic. For a total-and-difference problem, ask what remains after the extra part is removed. For a multiple-and-total problem, ask how many units the whole collection contains.
Then change one feature. Keep the structure and use different numbers. Or keep the numbers and change the unknown. A learner who can solve “three times as many, total known” should also be invited to explain what changes when the difference is known instead.
Do not treat a failed transfer question as proof that the earlier work was worthless. It identifies the part of the understanding that still depends on a familiar wrapper. Return to the representation and make the changed relationship explicit.
A short lesson can have one precise job
Choose a job such as distinguishing total from difference. Use one additive example and one repeated-unit example. Ask the learner to label the known amount before calculating. Stop when the labels reveal a different difficulty, such as not understanding “three times as many,” and address that prerequisite directly.
A useful lesson need not contain every problem type. The amount of work should follow the evidence from the attempt. If the learner already represents the relationship accurately, more drawing instruction may add effort without repairing the arithmetic that actually failed.
Use prompts that leave the decision with the learner
Ask “What does this number label?” rather than “Divide by four.” Ask “Which amount is the fraction of?” rather than “Use the remainder.” Ask “What would make the two final amounts equal?” rather than “Halve the difference.”
The more direct prompts may be needed during initial teaching, but they should not be mistaken for independent method selection. Record how much help was necessary. A correct calculation after the method has been supplied is evidence of execution, not yet evidence that the learner can choose that method alone.
When the learner gets stuck
Return to the smallest stable facts. Name the unknown, copy one relationship faithfully and choose a simple representation. A learner who cannot solve a long problem may still correctly identify that a total contains four equal units. That line can become the starting point for the next step.
Do not add invented information to make a problem easier. If the green count is unknown, leave it unknown. If a diagram is not to scale, do not measure it. If the problem does not say two groups are equal, do not draw them as equal merely because the calculation would then be convenient.
Use an error record that preserves the reason
Instead of “I forgot to divide,” write “The total contained A’s three units and B’s one unit, so I needed four units altogether.” Instead of “Careless fraction mistake,” write “The second fraction applied to the remaining amount, not the starting amount.”
Such explanations give the learner something specific to notice in a future question. They also help the adult avoid repeating an entire lesson when a small relationship is the real point of repair.
Return the answer to the story
The final line should answer the original question in ordinary language. How many vans are needed? How many dollars were there at first? Which jar has more red counters? What length of boundary must be covered?
A string of correct equations can still stop one step before the requested quantity. Reading the final sentence against the question is a small but valuable finishing habit. It checks that the mathematical work has returned to the purpose for which it was started.
Continue through the Primary Mathematics series
When a sound model breaks during calculation, use Place Value and Regrouping or Equal Groups, Division and Remainders. When the model uses a part, a whole or a changing reference amount, return to Fractions, Decimals and the Same Whole.
The BTT Primary Mathematics Learning Hub connects all four worked guides. For the broader diagnostic question, read Why Can My Child Calculate but Not Solve Mathematics Word Problems?. That explanation and this practice guide serve different purposes: one identifies possible causes; the other provides worked teaching and independent attempts.
For a difficulty that remains unclear after examining the working, use Mathematics Diagnosis. Families considering guided support can use the separate Primary Mathematics Tuition route.
Original learning guide. Curriculum reference checked 6 September 2026. These scenarios are teaching illustrations, not reported student cases, official school questions or examination predictions.