How do you know whether a PSLE Mathematics word problem has enough information to solve? First identify the target quantity, then label which facts are needed to determine it. Some story details may be irrelevant; others, such as a group total or perpendicular height, are essential. If two different answers satisfy every stated condition, a unique numerical answer is not determined. This original thirty-six-case clinic teaches pupils to read mathematical relationships rather than blindly use every printed number.
These cases are reasoning exercises informed by the updated MOE Mathematics syllabus, not a claim that SEAB will deliberately set contradictory or impossible questions in the 2027 examination. Real exam questions should be answered according to their stated conditions. This training helps identify where a misunderstanding comes from: wrong target, wrong reference whole, missing scale, unproved geometric condition, overlapping categories or conflicting data.
- Solvable: the stated facts determine the requested answer.
- Extra data: some true facts are not needed for the requested quantity.
- Insufficient data: multiple different numerical answers fit all the information provided.
- Ambiguous data: an unstated order or interpretation can change the outcome.
- Inconsistent data: two or more stated facts cannot all hold simultaneously.
- For actual PSLE questions, read precisely what is given and never invent a missing assumption simply because the diagram looks familiar.
Why data sufficiency matters under the MOE syllabus
The MOE Primary Mathematics syllabus, updated October 2025, emphasises mathematical reasoning, modelling and representation. Students must interpret information and select appropriate strategies. SEAB’s Standard Mathematics 0008 assessment objectives for examination from 2026 likewise mention interpreting, analysing and making inferences. This clinic isolates the reading decisions that must precede arithmetic. It is independent practice and does not represent a new official PSLE item type.
For a learner, the important habit is to ask not ‘How can I use this number?’ but ‘What does this number tell me about the question’s requested quantity?’ A given total may already include a part. A ratio may lack the scale needed for an exact count. A diagram may provide a sloping length where the area formula requires a perpendicular height. A survey may count the same person in two groups. These questions turn apparently easy arithmetic into wrong answers when the relationships are ignored.
Thirty-six original reading and data-sufficiency cases
| No. | Short case | Decision |
|---|---|---|
| 1 | Three colours but only two requested | Solvable; one extra fact |
| 2 | The whole and one part | Solvable |
| 3 | Irrelevant participants in a price story | Solvable; one extra fact |
| 4 | Ratio without a scale | Insufficient information |
| 5 | Ratio with total | Solvable |
| 6 | Ratio with difference | Solvable |
| 7 | Ratio with first group known | Solvable |
| 8 | Percent without a population | Insufficient information |
| 9 | Percent with a complete population | Solvable |
| 10 | Two categories may overlap | Insufficient information |
| 11 | Groups specified as disjoint | Solvable |
| 12 | Perimeter does not fix rectangle area | Insufficient information |
| 13 | Area and one side fix the other side | Solvable |
| 14 | A slanted side is not necessarily triangle height | Insufficient information |
| 15 | Perpendicular height supplied | Solvable |
| 16 | Circle radius and specified pi | Solvable |
| 17 | Tank volume but no base | Insufficient information |
| 18 | Tank base and volume given | Solvable |
| 19 | Minimum boxes for all items | Solvable; context changes rounding |
| 20 | Maximum full boxes | Solvable; same numbers, different target |
| 21 | Average fixes a total but not all values | Solvable |
| 22 | Average does not fix an individual | Insufficient information |
| 23 | One missing average value | Solvable |
| 24 | Two means without group sizes | Insufficient information |
| 25 | Two means with group sizes | Solvable |
| 26 | Pictograph without a key | Insufficient information |
| 27 | Graph intervals without a value scale | Insufficient information |
| 28 | Start and finish time | Solvable |
| 29 | Departure alone cannot determine arrival | Insufficient information |
| 30 | Mixed mass units | Solvable |
| 31 | Fractions of successive remainders | Solvable |
| 32 | Coupon order not specified | Ambiguous information |
| 33 | Union of two categories without overlap stated | Insufficient information |
| 34 | Numerically contradictory ratio | Inconsistent information |
| 35 | Numerically contradictory average | Inconsistent information |
| 36 | Transfer with a final equality | Solvable |
1. Three colours but only two requested
Practice prompt: A class has 12 red pens, 8 blue pens and 5 green pens. How many red and blue pens are there altogether?
Decision: Solvable; one extra fact. Answer, if determined: 20 pens.
Why: Only red and blue are requested, so 12 + 8 = 20. The five green pens are a true fact about the scene but are not relevant to the desired total. A child who includes all three colours would answer twenty-five, an accurate sum of the wrong collection.
Reading repair: Ask the pupil to underline the nouns in the final question and match them to the given counts. A useful retest changes the question to ‘all colours’, at which point the answer becomes twenty-five. The numbers stay the same but the target changes.
2. The whole and one part
Practice prompt: There are 42 books altogether, including 18 red-covered books. The rest are blue. How many blue books are there?
Decision: Solvable. Answer, if determined: 24 books.
Why: The number forty-two already contains the eighteen red-covered books. Blue is the missing part of the whole: 42 − 18 = 24. The phrase ‘including’ signals that adding forty-two and eighteen would count the red books twice.
Reading repair: Draw a bar whose full length is forty-two and mark eighteen within it. Ask whether the blue part plus the red part equals the whole. Check 18 + 24 = 42.
3. Irrelevant participants in a price story
Practice prompt: Five packs cost $6 each. Two volunteers carry the packs. How much do the packs cost in total?
Decision: Solvable; one extra fact. Answer, if determined: $30.
Why: The price relationship uses five packs and six dollars per pack, so 5 × 6 = $30. The two volunteers may explain how the packs are transported but do not alter the charge. Multiplying by two volunteers would introduce a repeated cost never stated.
Reading repair: Have the child label every number with its unit: five packs, six dollars per pack and two volunteers. Only packs and dollars per pack combine to make dollars in this question. Units expose the irrelevant number.
4. Ratio without a scale
Practice prompt: A and B have stickers in the ratio 2:3. How many stickers does A have?
Decision: Insufficient information. Answer, if determined: No unique actual count.
Why: A ratio determines relative parts, not the absolute unit value. A may have two and B three, or A twenty and B thirty, or any common positive whole-number scaling. The first group’s share is two-fifths of the combined total, but the total is not given.
Reading repair: Ask what additional fact would determine the scale, such as the combined total, A’s count, B’s count or the difference. The statement is not a flawed ratio; only the requested absolute count lacks information.
5. Ratio with total
Practice prompt: A and B share 150 stickers in the ratio 2:3. How many does A have?
Decision: Solvable. Answer, if determined: 60 stickers.
Why: The total contains five equal ratio units. Each is 150 ÷ 5 = 30. A owns two units, or sixty stickers. The ratio information becomes sufficient because the total fixes the value of one unit.
Reading repair: Check B has ninety, sum is 150 and 60:90 simplifies to 2:3. This check distinguishes ‘insufficient ratio only’ from ‘solvable ratio plus a scale’.
6. Ratio with difference
Practice prompt: A:B = 2:3 and B has 30 more stickers than A. Find A and B.
Decision: Solvable. Answer, if determined: A=60, B=90.
Why: The difference between three and two ratio parts is one part, so one unit is thirty. A holds two units, or sixty, while B holds ninety. No combined total was required because the difference already sets the scale.
Reading repair: Ask a pupil to draw two bars and mark only the extra unit. A wrong answer obtained by dividing thirty by five would confuse a stated difference with a combined total.
7. Ratio with first group known
Practice prompt: A:B = 3:5 and A has 45 cards. Find the combined total.
Decision: Solvable. Answer, if determined: 120 cards.
Why: Three units represent forty-five, so one unit is fifteen. Both groups together contain eight units, giving 8 × 15 = 120. A child must distinguish the known share from the combined whole.
Reading repair: Check the second group has seventy-five cards and that 45 + 75 = 120. Ratios become fully numeric when an actual corresponding share is supplied.
8. Percent without a population
Practice prompt: In a school survey, 40% chose Science. How many pupils chose Science?
Decision: Insufficient information. Answer, if determined: Exact count cannot be found.
Why: Forty percent is a proportion and does not give a fixed number of pupils. If the survey has fifty respondents, the count is twenty; if it has one hundred, the count is forty. Both satisfy the stated percentage.
Reading repair: Ask for the survey population or an equivalent numerical quantity. The exercise teaches a clear distinction between a percentage share and an absolute count; it is not a prediction that SEAB will deliberately omit required data.
9. Percent with a complete population
Practice prompt: Forty percent of 250 pupils choose Science. How many is that?
Decision: Solvable. Answer, if determined: 100 pupils.
Why: Now the whole is 250, so 0.4 × 250 = 100. The arithmetic is a direct part-of-whole calculation. A quick size check is that forty percent is less than half, so the answer should be below 125.
Reading repair: Compare with the previous example and ask which new sentence made the calculation possible. Identifying one missing datum is more valuable than learning a new multiplication shortcut.
10. Two categories may overlap
Practice prompt: In a club of 100, 60 play chess and 50 play football. How many play at least one?
Decision: Insufficient information. Answer, if determined: Between 60 and 100; exact value unknown.
Why: Some members may play both. Adding sixty and fifty counts those people twice, and gives 110, which exceeds the club’s hundred members. The union is sixty plus fifty minus their overlap. The overlap could range from ten to fifty under these counts.
Reading repair: If the overlap were twenty, the union would be ninety. If it were fifty, everyone who plays football would also play chess and the union would be sixty. The missing overlap prevents one unique answer.
11. Groups specified as disjoint
Practice prompt: Sixty pupils play chess and fifty others play football; no pupil does both. How many participate?
Decision: Solvable. Answer, if determined: 110 pupils.
Why: The words ‘fifty others’ and ‘no pupil does both’ establish that the groups do not overlap. Add sixty and fifty to obtain 110. The problem does not provide a club total, but the question asks only for the combined participants.
Reading repair: A pupil who carries the earlier hundred-member club total into this new independent story has imported information not stated here. Read each problem as its own set of facts.
12. Perimeter does not fix rectangle area
Practice prompt: A rectangle has perimeter 20 cm. Find its area.
Decision: Insufficient information. Answer, if determined: No unique area.
Why: A rectangle whose sides are six and four centimetres has perimeter twenty and area twenty-four square centimetres. Another with sides seven and three also has perimeter twenty but area twenty-one. The same perimeter can belong to different areas.
Reading repair: The missing information is a side length or another relationship restricting the rectangle’s dimensions. The existence of two valid different areas disproves the possibility of a single numerical answer.
13. Area and one side fix the other side
Practice prompt: A rectangle has area 48 cm² and length 6 cm. Find its breadth.
Decision: Solvable. Answer, if determined: 8 cm.
Why: Area equals length multiplied by breadth. Divide forty-eight by six to obtain breadth eight centimetres. The given area and length determine the remaining positive side uniquely under the rectangle model.
Reading repair: Check 6 × 8 = 48 cm². A pupil who stops at forty-eight has restated the given area rather than answering the width question.
14. A slanted side is not necessarily triangle height
Practice prompt: A triangle has base 10 cm and a separate sloping side 7 cm. Find its area.
Decision: Insufficient information. Answer, if determined: Perpendicular height is needed.
Why: Triangle area depends on base multiplied by perpendicular height, divided by two. A sloping side of length seven is not automatically the perpendicular height. Several triangles can have the stated base and sloping side but different heights and areas.
Reading repair: Ask the child to draw a right-angle marker at the base-to-height junction. If the question states that the seven-centimetre line is perpendicular to the base, area would be thirty-five square centimetres; without that condition it is not justified.
15. Perpendicular height supplied
Practice prompt: A triangle has base 10 cm and perpendicular height 6 cm. Find its area.
Decision: Solvable. Answer, if determined: 30 cm².
Why: The required relationship is area = 1/2 × base × perpendicular height. Half of ten times six gives thirty square centimetres. The word perpendicular supplies exactly the geometric condition missing from the previous case.
Reading repair: Check that a corresponding rectangle would have area sixty; the triangle’s area is half that. This is a demonstration of why the nature of the measurement matters, not just its number.
16. Circle radius and specified pi
Practice prompt: A circle has radius 7 cm and the question states π=22/7. Find its circumference.
Decision: Solvable. Answer, if determined: 44 cm.
Why: Circumference = 2πr, so two times twenty-two sevenths times seven equals forty-four centimetres. The supplied radius and pi value are sufficient to calculate the requested length.
Reading repair: Do not write 154 cm², which is the area under the same dimensions. A correct numerical response must match the physical quantity and unit the question requests.
17. Tank volume but no base
Practice prompt: A rectangular tank contains 2 L of water. How high is the water?
Decision: Insufficient information. Answer, if determined: Water depth is not determined.
Why: Two litres equal 2000 cubic centimetres, but the water’s height also depends on the horizontal base area. A tank with base area 200 cm² would have depth ten centimetres, while one with base area 500 cm² would have depth four centimetres.
Reading repair: Both tanks contain the same water volume but give different depths. A base area or length and breadth is needed. This is why a picture of a tank without measurements is not enough to calculate a numerical depth.
18. Tank base and volume given
Practice prompt: A rectangular tank has base 25×20 cm and contains 2 L of water. Find depth.
Decision: Solvable. Answer, if determined: 4 cm.
Why: Convert two litres to 2000 cubic centimetres. The base area is 25 × 20 = 500 square centimetres. Dividing water volume by base area gives four centimetres of water depth.
Reading repair: Verify 500 × 4 = 2000 cm³. The tank’s overall height is not necessary if only the present water level is requested.
19. Minimum boxes for all items
Practice prompt: Sixty-one books must be transported. Each box holds at most ten books. What is the minimum box count?
Decision: Solvable; context changes rounding. Answer, if determined: 7 boxes.
Why: Six complete boxes carry sixty books, leaving one. A seventh box is required to carry all sixty-one. The quotient 61 ÷ 10 is 6 remainder 1, but the final answer counts whole boxes needed.
Reading repair: A child who rounds 6.1 to six using nearest-whole rounding violates the capacity rule. The context is the deciding information, not a special calculator function.
20. Maximum full boxes
Practice prompt: Sixty-one books are packed into complete boxes of exactly ten. What is the maximum number of full boxes?
Decision: Solvable; same numbers, different target. Answer, if determined: 6 complete boxes.
Why: Only six sets of exactly ten fit within sixty-one books. One book is left over and cannot form a seventh complete box. This is the same division as the preceding case but a different real-world question.
Reading repair: Compare the words ‘minimum boxes to carry all’ and ‘maximum complete boxes’. The arithmetic quotient is the same; the response required by the context is different.
21. Average fixes a total but not all values
Practice prompt: Five numbers have an average of 18. Find their total.
Decision: Solvable. Answer, if determined: 90.
Why: Average equals sum divided by count. Multiply eighteen by five to get the total ninety. A total is fixed even though the five individual numbers could be many different combinations.
Reading repair: A pupil should answer only the total requested. The same fact is not sufficient to find a particular individual score unless additional information is provided.
22. Average does not fix an individual
Practice prompt: Five numbers average 18. What is the largest individual number?
Decision: Insufficient information. Answer, if determined: No unique largest value.
Why: The total is ninety, but the distribution is not given. Five numbers all equal to eighteen would have maximum eighteen. Four tens and one fifty would also total ninety but have maximum fifty. Other lists fit as well.
Reading repair: The child should recognise that a group mean constrains the sum, not the full distribution. If only non-negative values are allowed, possible maxima have bounds, but still no unique value is determined.
23. One missing average value
Practice prompt: Five numbers average 18 and four sum to 78. Find the fifth.
Decision: Solvable. Answer, if determined: 12.
Why: The total for five numbers is ninety. Subtract the known sum of the four values, seventy-eight, to get twelve. The newly supplied partial sum makes the individual value identifiable.
Reading repair: Verify 78 + 12 = 90 and 90 ÷ 5 = 18. Do not confuse the mean of the four known numbers with the missing value.
24. Two means without group sizes
Practice prompt: One class averages 20 and another averages 30. What is their combined average?
Decision: Insufficient information. Answer, if determined: No unique mean.
Why: The groups might have equal numbers of pupils, giving a combined mean of twenty-five. Or one group might contain four and the other six, giving a combined mean of twenty-six. Without counts, the displayed means cannot be weighted.
Reading repair: Ask for the number of values contributing to each mean, or their totals. A quick average of twenty and thirty assumes equal group sizes without evidence.
25. Two means with group sizes
Practice prompt: Four pupils average 20 and six pupils average 30. Find combined mean.
Decision: Solvable. Answer, if determined: 26.
Why: The first group’s total is eighty, the second group’s 180. Together 260 points are shared among ten pupils, giving average twenty-six. This differs from a direct average of the two means.
Reading repair: The missing information from the preceding case was the two group sizes. Now both totals and counts can be reconstructed.
26. Pictograph without a key
Practice prompt: A pictograph contains five apple icons. How many apples do they represent?
Decision: Insufficient information. Answer, if determined: Legend or scale is needed.
Why: One icon might represent one apple, three apples or ten apples. The symbol count is five, but that is not necessarily the number of real objects. The pictograph key is part of the mathematical data.
Reading repair: If one icon represents four apples, the total is twenty. Ask pupils to find the key before multiplying; guessing one per picture can be wrong.
27. Graph intervals without a value scale
Practice prompt: A bar reaches four marked intervals above zero. What exact count does the bar show?
Decision: Insufficient information. Answer, if determined: Axis scale is needed.
Why: Four intervals can represent four, twenty, forty or another count depending on the axis labels and interval size. A visible bar height without a numeric scale is not enough to recover the measured quantity.
Reading repair: Give a key of five objects per interval and the bar now represents twenty. Change the key to ten and the same bar represents forty. The picture stayed the same but its meaning changed.
28. Start and finish time
Practice prompt: A bus starts at 2:30 pm and arrives at 4:05 pm. How long was the trip?
Decision: Solvable. Answer, if determined: 1 hour 35 minutes.
Why: From 2:30 pm to 3:30 pm is one hour, then another thirty-five minutes reaches 4:05 pm. The start and finish times determine elapsed duration.
Reading repair: Use a clock timeline rather than treating 4.05 and 2.30 as decimals. The minute place has sixty units per hour, not one hundred.
29. Departure alone cannot determine arrival
Practice prompt: A bus leaves at 2:30 pm. At what time will it arrive?
Decision: Insufficient information. Answer, if determined: Duration or arrival condition is missing.
Why: Departure time by itself does not determine how long the journey lasts. An arrival time requires a travel duration, an independently known finish time, or some equivalent additional relationship.
Reading repair: If a journey duration of one hour thirty-five minutes is supplied, the arrival would be 4:05 pm. That new datum completes the problem.
30. Mixed mass units
Practice prompt: A food box has 1.2 kg of rice and 350 g of flour. Find combined mass in grams.
Decision: Solvable. Answer, if determined: 1550 g.
Why: Convert 1.2 kilograms to 1200 grams, then add 350 grams to get 1550 grams. The unit conversion must precede adding quantities printed in different scales.
Reading repair: An answer of 351.2 grams adds unlike units as bare numbers. Always align the mass units first, and check the result exceeds 1200 g.
31. Fractions of successive remainders
Practice prompt: A pupil begins with $80, spends one-quarter, then one-third of the remaining amount. How much remains?
Decision: Solvable. Answer, if determined: $40.
Why: One-quarter of eighty is twenty, leaving sixty. One-third of sixty is twenty, leaving forty dollars. The phrase ‘remaining amount’ means the second fraction acts on sixty rather than on the original eighty.
Reading repair: Reconcile the three quantities: first spent twenty, second spent twenty and final forty add back to the original eighty. This check is independent of the sequence of subtractions.
32. Coupon order not specified
Practice prompt: An $80 item gets 20% off and a $10 coupon. What is the final cost if the order of reductions is not stated?
Decision: Ambiguous information. Answer, if determined: $54 or $56 depending on order.
Why: If the twenty-percent discount is first, the price becomes $64 and the coupon reduces it to $54. If the ten-dollar coupon comes first, the price becomes $70 and twenty percent off that gives $56. Both are consistent with a vaguely phrased story that does not specify the order or base.
Reading repair: A good word problem should specify the sequence or rule. Do not choose one answer and claim the other calculation is careless when the mathematical conditions genuinely allow both.
33. Union of two categories without overlap stated
Practice prompt: Of 100 children, 55 take art and 50 take music. How many take at least one?
Decision: Insufficient information. Answer, if determined: Overlap count is needed.
Why: Adding gives 105, which double-counts children in both categories. To find the number in at least one category, subtract the overlap from 105. The number in both might be five, twenty or fifty under compatible data, giving different union counts.
Reading repair: If the overlap were twenty, the answer would be eighty-five. A diagram with two overlapping circles helps illustrate why a count from each circle cannot simply be added.
34. Numerically contradictory ratio
Practice prompt: A:B=3:5, total 80. The same statement says A has 60. Can both be true?
Decision: Inconsistent information. Answer, if determined: No solution satisfies every stated fact.
Why: Eight ratio units represent eighty, so one unit equals ten. A must be thirty, while B is fifty. Claiming A is sixty contradicts the original ratio and total. The task is not an underdetermined problem: it supplies incompatible facts.
Reading repair: Ask the learner to calculate the consequence of each condition and show the contradiction precisely. More arithmetic cannot repair incompatible data; the source statement must be corrected.
35. Numerically contradictory average
Practice prompt: Three scores average 10, but their stated sum is 36. Can both be true?
Decision: Inconsistent information. Answer, if determined: No.
Why: A mean of ten across three scores requires sum thirty, because 10 × 3 = 30. The separately stated total thirty-six would instead give average twelve. These claims cannot simultaneously describe the same three values.
Reading repair: This checks the inverse relationship between average, total and count. Distinguish inconsistent from insufficient: there is too much incompatible information rather than too little.
36. Transfer with a final equality
Practice prompt: A and B have 60 counters total. A gives B five, after which each has the same number. Find original amounts.
Decision: Solvable. Answer, if determined: A=35, B=25.
Why: After the transfer each has thirty, because the combined total is sixty and becomes equally divided. Before giving five, A had thirty-five and B had twenty-five. The transfer moves five from one to the other without changing their total.
Reading repair: Check 35 + 25 = 60, and 35 − 5 = 25 + 5 = 30. This pair of independent conditions confirms the solution.
A decision tree for any unfamiliar maths word problem
Start with the last line: is the task asking for a numerical amount, a comparison, an area, a rate, a duration or a proportion? Next list each fact with its unit and role: a total, one part, a relationship between two parts, a change from one state to another, or a measurement. If a numeric question appears to lack a whole, a length or a scale, ask whether that information can be derived from other facts. Do not immediately assume the problem is impossible because one number is not explicitly printed.
Then test the information for completeness and consistency. Could two different actual situations satisfy the given facts while producing different requested answers? If so, the answer is not uniquely determined by the statement. Does one part plus another contradict a supplied total, after accounting for possible overlap? If so, check whether the task has intentionally described different time states or whether the facts are truly inconsistent. Finally, if the relationship is valid and complete, choose an appropriate calculation and check it against the original conditions.
When a number is irrelevant, show why rather than crossing it out blindly
Crossing out irrelevant numbers can help a learner focus, but a number should be excluded only after its role is understood. In a shopping task, an extra object’s price may be irrelevant when the object is not purchased. If the final question instead asks for the cost of everything, that price becomes essential. Similarly, the number of volunteers carrying the packs may be irrelevant to their price, but could matter in a task about sharing the packs equally among those volunteers.
A strong method is to label each number with a noun and state whether it belongs to the requested relationship. This avoids keyword strategies such as ‘more means add’ and ‘each means multiply’ that fail when the unknown is reversed. The goal is mathematical reading, not simply reducing the amount of text on the page.
How to use these cases for a five-day teaching cycle
On the first day, classify six short cases into solvable, insufficient, ambiguous, inconsistent or extra-data groups. On the second, examine a ratio case with and without a total, and a percentage case with and without a population. On the third, use geometry and measurement cases to distinguish a given sloping side from a required perpendicular height. On the fourth, practise charts with missing scales or overlapping categories. On the fifth, mix the categories without headings and ask for one sentence justifying each decision.
If a pupil mistakenly labels a solvable problem impossible, ask whether a missing value can be recovered by an inverse relationship. If they invent a unique count from ratio terms alone, ask for two numerical examples that satisfy the same ratio but have different totals. A small valid counterexample is often more instructive than several pages of arithmetic. Retest after a short gap using changed numbers and wording.
Frequently asked questions
Does PSLE 2027 contain questions with intentionally missing facts?
This practice set makes no such prediction. These are independent reasoning exercises designed to strengthen pupils’ reading and modelling decisions. Follow the conditions actually printed in the official examination question.
Should I use every number in a word problem?
No fixed rule says every printed number must be used. Decide which facts connect to the requested quantity. An apparently extra fact may become relevant if the question target changes.
What is the difference between ambiguous and insufficient information?
Insufficient information means the available relationships do not fix a unique value. Ambiguity means the wording permits different interpretations or sequences, such as a coupon applied before or after a percentage discount. Both can prevent one unique answer, but they require different clarifications.
What if the question’s facts contradict each other?
Check first that you have not confused a before state with an after state or double-counted overlapping categories. If the conditions are genuinely incompatible, no numerical answer satisfies all of them. In ordinary examination practice, discuss questionable items with the school or official source rather than inventing a preferred correction.
Can Foundation Mathematics pupils use this material?
The basic thinking approach is broadly useful, but the examples differ in content. Select questions matching the pupil’s Foundation syllabus; do not treat Standard P6 ratio or circle content as automatically examinable for Foundation.
Related reading and source references
For the words that signal a new reference whole see maths vocabulary and command words. For model selection see the unfamiliar word-problems parent guide. For a non-calculator practice set see 36 exact arithmetic drills. Official sources: MOE Primary Mathematics syllabus updated October 2025 and SEAB Mathematics 0008 format from 2026. Reviewed 13 November 2026.