PSLE Math Question Types: A Complete Guide with Worked Examples

Many parents describe the same frustration: "My child understands the concept when I explain it at home, but when they see a new question in the exam, they don't know which method to use." That gap between knowing a method and recognising when to apply it is exactly what this guide addresses.

Every major PSLE math question type is covered here, with a worked example for each. The guide is structured around the MOE heuristics framework, so your child learns not only how to solve each type of problem but also how to identify the right approach the moment they read the question. Written with input from MOE-experienced teachers, it covers the MOE heuristics framework and the question patterns that repeat across Paper 2.

What the PSLE Math Exam Actually Looks Like

Understanding the exam structure helps your child allocate time and effort correctly. Both papers carry 50 marks each, but Paper 2 contains the harder multi-step problems. Knowing which paper tests which question types shapes how you should prioritise revision.

Paper 1: No Calculator

Paper 1 is divided into two booklets.

Booklet A contains 18 multiple-choice questions: 10 questions worth 1 mark each (10 marks) and 8 questions worth 2 marks each (16 marks). Booklet A total: 26 marks.

Booklet B contains 12 short-answer questions worth 2 marks each. Booklet B total: 24 marks.

Paper 1 total: 50 marks. Time allowed: 1 hour 10 minutes. Calculators are not permitted.

Paper 2: Calculator Allowed

Paper 2 is one booklet with two sections.

The short-answer section contains 5 questions worth 2 marks each. Total: 10 marks.

The structured/long-answer section contains 10 questions worth 3, 4 or 5 marks each. Total: 40 marks.

Paper 2 total: 50 marks. Time allowed: 1 hour 20 minutes. Calculators are permitted for all questions. (Source: SEAB 2026 PSLE Mathematics Syllabus)

Paper 1 vs Paper 2: Format Comparison

	Paper 1	Paper 2
Calculator	Not allowed	Allowed
Question formats	MCQ (Booklet A) and short answer (Booklet B)	Short answer and structured/long-answer problem sums
Total marks	50	50
Time allowed	1 hour 10 minutes	1 hour 20 minutes
Share of total score	50%	50%
Where most marks are lost	Careless errors in MCQ	Multi-step word problems
Heuristics tested	Mainly mental strategies	All heuristics: model drawing, ratio, fractions, geometry

How AL Scoring Works (and Why Question Types Matter)

PSLE Math is graded on the Achievement Level (AL) scale, from AL1 (highest) to AL8. The AL cut-off scores vary by year based on cohort performance. Generally, AL1 requires a high raw score on both papers, and the cut-offs are not published in advance.

Because Paper 2 contains the hardest multi-step problems — structured and long-answer questions worth 3 to 5 marks each — a child who cannot handle these will struggle to achieve a strong AL band regardless of Paper 1 performance. The worked examples throughout this guide focus accordingly on Paper 2-level problem sums: fractions, ratio, rate, and geometry.

Paper 2 problem sums are where the most marks are won or lost at the upper AL bands.

The 8 Heuristics MOE Teaches for PSLE Math

The MOE Primary Mathematics Syllabus formally names eight heuristics that students are expected to apply when solving non-routine problems. These are systematic thinking tools, not shortcuts. The PSLE tests whether a child can select the correct heuristic independently. That selection skill is what separates students at the upper AL bands from those who know the methods but cannot deploy them.

The eight MOE heuristics are:

Model drawing (bar model): Represent quantities as labelled bars to show relationships between parts and wholes. This is the most frequently used heuristic in PSLE Math and applies across fractions, ratios, percentages, and word problems.
Guess and check: Make a systematic first estimate, calculate the result, and adjust based on the error. Used when the number of unknowns matches the number of conditions given.
Working backwards: Start from the known final result and reverse every operation to find the starting value. Used when the end state is given and the starting value is unknown.
Making a list (systematic listing): List all possibilities in a structured order to ensure none are missed. Used in counting problems, coin combinations, and arrangements.
Looking for patterns: Identify the rule governing a sequence of numbers or shapes, then apply the rule to find any term. Used in number sequences and repeating shape problems.
Assumption method: Assume all items belong to one category, calculate the result, compare it to the actual total, and adjust. Used in two-variable problems with a known total count and total value.
Before and after (before-change-after table): Record quantities in a three-column table showing the state before a change, the change itself, and the state after. Used in ratio and fraction problems where a situation changes.
Simplify the problem: Replace large or complex numbers with smaller ones to identify the method, then apply the same method to the original numbers. Used when the structure of a problem is difficult to see through complex values.

Each heuristic is covered with a full worked example in the sections below. The worked examples in this guide do not label the question type for you, because the PSLE does not label it either. Practise identifying the type before checking the solution.

Fraction and Remainder Question Types

Fraction problems are among the most frequently tested question types in PSLE Paper 2. They appear in short-answer questions and multi-step long-answer questions. The three main fraction structures are the remainder concept, the equal fractions concept, and fractions with units.

The Remainder Concept (Branching)

When to use this method: The question tells you a fraction of a total, then gives you a fraction of what remains. The phrase "the remainder" or "what was left" is the signal.

Common mistake: Applying the second fraction to the original total instead of to the remainder. If 3/8 of the stickers are given away, the second fraction applies to the remaining 5/8, not to the full original amount.

Example 1:

Priya had some stickers. She gave 3/8 of her stickers to her sister. Then she gave 2/5 of the remaining stickers to her friend. She was left with 36 stickers. How many stickers did Priya have at first?

Step 1: After giving 3/8 to her sister, the remainder is 5/8 of the original total.

Step 2: She gave 2/5 of the remainder to her friend, so she kept 3/5 of the remainder.

Step 3: Amount left = 3/5 of 5/8 of the original = 3/5 × 5/8 = 15/40 = 3/8 of the original.

Step 4: 3/8 of the original = 36. So 1/8 = 36 ÷ 3 = 12 stickers.

Step 5: Original total = 12 × 8 = 96 stickers.

Check: 3/8 of 96 = 36 given to sister. Remainder = 60. 2/5 of 60 = 24 given to friend. Left = 60 - 24 = 36. Correct.

The Equal Fractions Concept

When to use this method: A fraction of one group equals a fraction of a different group. You are given the fraction and either the actual value or the size of one group, and need to find the other. The signal is "the number of A equals the number of B."

Common mistake: Treating both groups as having the same total. They do not. Find the actual equal value first, then work backwards from each group's fraction to its total.

Example 2:

3/5 of a box of red marbles equals 2/3 of a box of blue marbles. The box of red marbles contains 30 marbles. How many blue marbles are there?

Step 1: 3/5 of 30 red marbles = 18 marbles.

Step 2: The equal value is 18. So 2/3 of blue marbles = 18.

Step 3: Blue marbles = 18 ÷ (2/3) = 18 × 3/2 = 27 blue marbles.

Check: 3/5 of 30 = 18. 2/3 of 27 = 18. Both equal 18. Correct.

Fractions with Units

When to use this method: The question gives you the value of a fractional part and asks you to find the total or another part. The signal is a sentence such as "she spent $36, which was 4/9 of her savings."

Common mistake: Dividing by the denominator rather than finding the value of 1 unit first and then scaling.

Example 3:

Ahmad spent $36 on books. This was 4/9 of his savings. After buying the books, he spent 1/3 of the remaining savings on a game. How much did he spend on the game?

Step 1: 4/9 of savings = $36. So 1/9 of savings = $36 ÷ 4 = $9.

Step 2: Total savings = $9 × 9 = $81.

Step 3: Remaining savings after books = $81 - $36 = $45.

Step 4: Amount spent on game = 1/3 of $45 = $15.

Check: 4/9 of $81 = $36. Remaining = $45. 1/3 of $45 = $15. Correct.

Ratio Question Types

Ratio is tested heavily in PSLE Paper 2. The four main structures are Constant Part, Constant Total, Constant Difference, and Everything Changed. Recognising which type you are dealing with determines the entire approach before you write a single number.

Constant Part

When to use this method: One part of the ratio stays the same across both scenarios. The total changes because the other part changes.

Method: Make the unchanged part equal in both ratios by scaling. Find the LCM of the two values that the constant part takes, then scale both ratios accordingly.

Common mistake: Scaling the part that changed rather than the part that stayed constant. Always identify the unchanged element first.

Example 4:

The ratio of Ali's savings to Ben's savings was 3: 5. After Ali saved another $24, the ratio became 9: 10. How much did Ali have at first?

Step 1: Ben's savings did not change. Ben's share in Ratio 1 = 5 parts; in Ratio 2 = 10 parts.

Step 2: Scale Ratio 1 by 2 so Ben's part becomes 10: Ratio 1 becomes 6: 10. Ratio 2 is 9: 10.

Step 3: Ali increased from 6 units to 9 units, a difference of 3 units = $24. So 1 unit = $8.

Step 4: Ali's original savings = 6 × $8 = $48.

Check: Ben = 10 units = $80. Ali before: $48. Ratio $48: $80 = 3: 5. After Ali saves $24: Ali = $72. Ratio $72: $80 = 9: 10. Correct.

Constant Total (Internal Transfer)

When to use this method: Items move from one person or group to another. Nothing enters or leaves the overall system, so the total stays the same.

Method: Find the LCM of the two ratio sums and scale both ratios so their totals match. The difference between the scaled values shows how much was transferred.

Common mistake: Subtracting the transferred amount from the total. In an internal transfer, the total is unchanged throughout.

Example 5:

The ratio of Mei's stickers to Jun's stickers was 7: 3. After Mei gave some stickers to Jun, the ratio became 3: 2. Together they had 120 stickers throughout. How many stickers did Mei give to Jun?

Step 1: Total = 120 stickers (unchanged). Ratio 1 total parts = 10. So 1 unit = 120 ÷ 10 = 12.

Step 2: Scale Ratio 2 (3: 2) so total = 10 parts: multiply by 2 to get 6: 4.

Step 3: Mei decreased from 7 units to 6 units = 1 unit transferred = 12 stickers.

Mei gave Jun 12 stickers.

Check: Mei before = 84. Jun before = 36. Mei after = 72. Jun after = 48. Ratio 72: 48 = 3: 2. Correct.

Constant Difference (Age Problems)

When to use this method: Age problems where you are told the ratio of two ages now and at another point in time. The age difference between two people never changes.

Method: Use the current ratio to set up expressions for each person's age. Apply the future ratio to write an equation, then solve.

Common mistake: Assuming the ratio of ages stays constant. Only the difference stays constant; the ratio changes every year.

Example 6:

The ratio of Wei's age to his father's age is 1: 4. The difference between their ages is 30 years. In how many years will the ratio of Wei's age to his father's age be 2: 5?

Step 1: Ratio 1: 4 means the difference = 3 parts = 30 years. So 1 part = 10 years.

Step 2: Wei now = 10 years. Father now = 40 years.

Step 3: In n years: Wei = (10 + n), Father = (40 + n). Set up ratio: (10 + n): (40 + n) = 2: 5.

Step 4: Cross-multiply: 5(10 + n) = 2(40 + n). So 50 + 5n = 80 + 2n. Then 3n = 30. So n = 10.

In 10 years, the ratio will be 2: 5.

Check: Wei at 20, Father at 50. Ratio 20: 50 = 2: 5. Correct.

Everything Changed (Ratio with Both Sides Changing)

When to use this method: Both quantities in the ratio change by different amounts. Neither side is constant, and the total also changes.

Method: Assign units to the original ratio. After the change, write new expressions for each quantity in terms of the same units. Use the new ratio to set up and solve an equation.

Common mistake: Assuming a new set of units is needed for the second scenario. One set of units covers both scenarios, connected through the change that is described.

Example 7:

The ratio of boys to girls in a chess club was 3: 4. After 6 boys joined and 2 girls left, the ratio became 3: 2. How many boys were in the club at first?

Step 1: Let boys = 3u and girls = 4u (original).

Step 2: After changes: boys = 3u + 6, girls = 4u - 2. New ratio: (3u + 6): (4u - 2) = 3: 2.

Step 3: Cross-multiply: 2(3u + 6) = 3(4u - 2). So 6u + 12 = 12u - 6. Then 18 = 6u. So u = 3.

Step 4: Boys at first = 3 × 3 = 9 boys.

Check: Boys = 9, Girls = 12. After change: Boys = 15, Girls = 10. Ratio 15: 10 = 3: 2. Correct.

Assumption Method and Whole Number Question Types

The assumption method is one of the most commonly tested heuristics in PSLE Paper 2. It appears in word problems involving two types of items with different values, a known total count, and a known total value.

The Assumption Method

When to use this method: The question gives you the total number of two types of items and the total value. You need to find how many of each type there are.

Method: Assume all items are one type. Calculate what the total would be under that assumption. Compare the assumed total to the actual total. Use the difference to find how many items must be swapped to the other type.

Common mistake: Subtracting the difference in the wrong direction. The excess (or shortfall) tells you how many to swap. Always substitute back into both conditions to verify.

Example 8:

A farmer has chickens and cows. There are 30 animals in total, with 96 legs altogether. How many chickens are there?

Step 1: Assume all 30 animals are chickens. Total legs = 30 × 2 = 60 legs.

Step 2: Actual total = 96 legs. Difference = 96 - 60 = 36 extra legs.

Step 3: Each chicken replaced by a cow adds 4 - 2 = 2 legs. Number of cows = 36 ÷ 2 = 18.

Step 4: Number of chickens = 30 - 18 = 12 chickens.

Check: 12 chickens = 24 legs. 18 cows = 72 legs. Total = 96 legs. Total animals = 30. Correct.

Excess and Shortage

When to use this method: Items are packed or distributed into groups of two different sizes. One size leaves a remainder (excess); the other size leaves you short (shortage). You need to find the total number of items.

Method: Set up two expressions for the same total, one with excess and one with shortage, and solve simultaneously. The key is that the number of groups is the same in both scenarios.

Common mistake: Using a different number of groups for each scenario. Both scenarios describe the same collection of items packed into the same number of groups.

Example 9:

A teacher packs pencils into boxes of 6 and has 4 pencils left over. If she uses boxes of 8, she is short of 6 pencils. How many pencils does she have?

Step 1: Let the number of boxes = b. With boxes of 6: pencils = 6b + 4. With boxes of 8: pencils = 8b - 6.

Step 2: Set equal: 6b + 4 = 8b - 6. So 10 = 2b. Therefore b = 5.

Step 3: Total pencils = 6 × 5 + 4 = 34 pencils.

Check: 34 pencils in boxes of 6: 5 boxes used, 4 left over. 34 pencils in boxes of 8: needs 40 pencils for 5 boxes, which is 6 short. Correct.

Working Backwards and Before-and-After Question Types

Both of these heuristics deal with change over time. Working backwards starts from the end result and reverses to find the start. The before-and-after method tracks how two quantities shift when a defined change occurs.

Working Backwards

When to use this method: The final value is given and you must find the starting value. The signal is any question that tells you what a quantity became after a series of operations.

Method: List every operation that was applied in order. Reverse each operation, starting from the final value and working towards the start. Multiplication becomes division; addition becomes subtraction.

Common mistake: Applying the original operations forward from the final value. The inverse operations must be applied in reverse order.

Example 10:

A number is multiplied by 3. Then 8 is added. Then the result is divided by 2. The final answer is 10. What was the original number?

Step 1: Start from the end. Final value = 10.

Step 2: Reverse "divided by 2": 10 × 2 = 20.

Step 3: Reverse "added 8": 20 - 8 = 12.

Step 4: Reverse "multiplied by 3": 12 ÷ 3 = 4.

Check: 4 × 3 = 12. 12 + 8 = 20. 20 ÷ 2 = 10. Correct.

Example 11:

There were some passengers on a bus. At the first stop, 12 passengers boarded and 5 got off. At the second stop, the number of passengers doubled. There were then 42 passengers. How many passengers were on the bus before the first stop?

Step 1: Final value = 42 passengers.

Step 2: Reverse "doubled": 42 ÷ 2 = 21.

Step 3: Reverse "12 boarded and 5 got off" (net +7): 21 - 7 = 14.

There were 14 passengers before the first stop.

Check: 14. Add 12, subtract 5 = 21. Double = 42. Correct.

Before and After (Table Method)

When to use this method: Quantities are shared between two people or groups, then a change occurs. You are told the ratio before and after the change, along with one actual value.

Method: Draw a three-column table: Before, Change, After. Fill in the ratios for each row. Identify whether the total or the difference is constant, then scale the ratios so they share the same total or difference. Use the relationship between the Before and After values to find the unknown.

Common mistake: Comparing Before and After ratios directly without first scaling them to the same base. The two ratios use different units until you align them.

Example 12:

The ratio of cookies in Box A to cookies in Box B was 5: 3. After 16 cookies were moved from Box A to Box B, the ratio became 3: 5. How many cookies were in Box A at first?

	Before	Change	After
Box A	5u	-16	3u
Box B	3u	+16	5u
Total	8u	0	8u

Step 1: Total is constant (internal transfer). Using the same unit u for both Before and After is valid because the total parts are equal (8u before = 8u after).

Step 2: From Box A: 5u - 16 = 3u. So 2u = 16. Therefore u = 8.

Step 3: Box A before = 5 × 8 = 40 cookies.

Check: Box A = 40, Box B = 24. After moving 16: Box A = 24, Box B = 40. Ratio 24: 40 = 3: 5. Correct.

If your child finds working backwards or the before-and-after method difficult to apply under exam conditions, Ottodot's P5 and P6 Math classes teach each heuristic explicitly and then let students apply it in Roblox games. A trial class is available with no long-term commitment required.

Rate and Speed Question Types

Speed, distance, and time questions, together with work rate problems, are standard in PSLE Paper 2. Both types require students to identify what remains constant and to organise information before writing any equation.

Speed, Distance, Time

When to use this method: Any question involving movement at constant speed. For problems with two objects moving simultaneously, draw a timeline or table showing each object's journey separately before combining.

Key formulas: Speed = Distance ÷ Time. Distance = Speed × Time. Time = Distance ÷ Speed.

Common mistake: Using the total combined distance when the problem is about an overtaking scenario (where the difference in distance matters), or vice versa. Always read whether the objects are moving towards each other or in the same direction.

Example 13:

Car A and Car B are 300 km apart and travelling towards each other. Car A travels at 70 km/h and Car B travels at 80 km/h. They set off at the same time. How long before they meet?

Step 1: When two objects travel towards each other, their speeds combine. Combined speed = 70 + 80 = 150 km/h.

Step 2: Time to meet = Distance ÷ Combined Speed = 300 ÷ 150 = 2 hours.

Check: Car A travels 70 × 2 = 140 km. Car B travels 80 × 2 = 160 km. Total distance covered = 300 km. Correct.

Example 14 (different start times):

Town P and Town Q are 270 km apart. Car A leaves Town P towards Town Q at 60 km/h at 8:00 am. Car B leaves Town Q towards Town P at 90 km/h at 9:00 am. At what time do they meet?

Step 1: By 9:00 am, Car A has already travelled 60 × 1 = 60 km. Gap remaining between the two cars = 270 - 60 = 210 km.

Step 2: After 9:00 am, both cars travel towards each other. Combined speed = 60 + 90 = 150 km/h.

Step 3: Time to close the 210 km gap = 210 ÷ 150 = 1.4 hours = 1 hour 24 minutes.

Step 4: Meeting time = 9:00 am + 1 hour 24 minutes = 10:24 am.

Check: Car A travels from 8:00 am to 10:24 am = 2 hours 24 minutes = 2.4 hours. Distance = 60 × 2.4 = 144 km. Car B travels from 9:00 am to 10:24 am = 1.4 hours. Distance = 90 × 1.4 = 126 km. Total = 144 + 126 = 270 km. Correct.

Rate Problems (Work Rate)

When to use this method: Two or more workers, taps, or machines complete a task at different individual rates. The question asks for the combined time or the time for part of the task.

Method: Find the fraction of the task each completes per unit time. Add these fractions to get the combined rate. Do not add the times.

Common mistake: Adding the times instead of the rates. If Tap A fills a tank in 6 hours and Tap B fills it in 4 hours, the answer is not 10 hours. Add the rates: 1/6 + 1/4 = 5/12 per hour, so combined time = 12/5 hours.

Example 15:

Pipe A can fill a tank in 6 hours. Pipe B can fill the same tank in 4 hours. Both pipes are turned on at the same time. How long will it take to fill the tank?

Step 1: Pipe A fills 1/6 of the tank per hour. Pipe B fills 1/4 of the tank per hour.

Step 2: Combined rate = 1/6 + 1/4 = 2/12 + 3/12 = 5/12 per hour.

Step 3: Time to fill = 1 ÷ (5/12) = 12/5 hours = 2 hours 24 minutes.

Check: Pipe A in 12/5 hours fills (1/6) × (12/5) = 2/5. Pipe B fills (1/4) × (12/5) = 3/5. Total = 2/5 + 3/5 = 1 full tank. Correct.

Example 16:

Worker A can complete a job in 5 days. Worker B can complete the same job in 10 days. They work together for 2 days, then Worker A stops. How many more days does Worker B need to finish?

Step 1: A's rate = 1/5 per day. B's rate = 1/10 per day. Combined rate = 1/5 + 1/10 = 3/10 per day.

Step 2: Work done in 2 days together = 2 × 3/10 = 6/10 = 3/5.

Step 3: Remaining work = 1 - 3/5 = 2/5. B alone at rate 1/10 per day: time = (2/5) ÷ (1/10) = 4 more days.

Check: 3/5 + (4 × 1/10) = 3/5 + 2/5 = 1 full job. Correct.

Geometry Question Types

Geometry questions in PSLE cover area and perimeter, angles in figures, and volume and capacity. They appear in both Paper 1 and Paper 2. The critical first step is always the same: identify which formula or angle property to apply before writing any number.

Area and Perimeter

When to use this method: Any question about the size of a flat shape (area) or the length of its boundary (perimeter). For composite figures, decompose the shape into rectangles, triangles, or semicircles before calculating.

Key formulas:

Area of rectangle = Length × Breadth
Perimeter of rectangle = 2 × (Length + Breadth)
Area of triangle = 1/2 × Base × Height

Common mistake: Confusing area and perimeter, particularly in two-step questions. State which quantity you are finding at each step, and re-read the question before writing the final formula.

Example 17:

A rectangle has a perimeter of 48 cm. Its length is 3 times its breadth. Find the area of the rectangle.

Step 1: Let breadth = b cm. Then length = 3b cm.

Step 2: Perimeter = 2 × (3b + b) = 8b. So 8b = 48. Therefore b = 6 cm.

Step 3: Length = 3 × 6 = 18 cm.

Step 4: Area = 18 × 6 = 108 cm².

Check: Perimeter = 2 × (18 + 6) = 48 cm. Correct.

Angles in Figures

When to use this method: Any question involving unknown angles in diagrams with triangles, straight lines, or parallel lines. Always name the angle property you are using as a reason, because the PSLE marks the reasoning, not only the answer.

Key angle properties:

Angles on a straight line sum to 180°
Angles in a triangle sum to 180°
Vertically opposite angles are equal
Corresponding angles on parallel lines are equal
Alternate angles on parallel lines are equal

Common mistake: Assuming two angles look equal in a diagram and using that assumption in working. State the property that proves them equal instead.

Example 18:

In triangle PQR, angle P = 48° and angle Q is twice angle R. Find the size of angle R.

Step 1: Let angle R = x°. Then angle Q = 2x°.

Step 2: Angle sum of triangle = 180°. So 48 + 2x + x = 180. Therefore 3x = 132 and x = 44.

Angle R = 44°.

Check: Angle Q = 88°. Sum = 48 + 88 + 44 = 180°. Correct.

Example 19:

Two angles are on a straight line together with a third angle. The first angle is 62°. The second and third angles are equal. Find the size of each equal angle.

Step 1: Let each equal angle = c°.

Step 2: Angles on a straight line sum to 180°. So 62 + c + c = 180. Therefore 2c = 118 and c = 59.

Each equal angle = 59°.

Check: 62 + 59 + 59 = 180°. Correct.

For questions involving angles in more complex figures, the Triangle Detective game on the Ottodot Resource Hub gives students timed practice at identifying and applying angle properties.

Volume and Capacity

When to use this method: Any question involving three-dimensional containers, including liquid transferred between containers. When liquid is transferred, the volume of water is conserved.

Key formula: Volume of cuboid = Length × Breadth × Height

Common mistake: Forgetting unit conversion before calculating. If one measurement is in metres and another in centimetres, convert all measurements to the same unit before multiplying.

Example 20:

A rectangular tank measures 30 cm by 20 cm by 25 cm. It is 3/5 full of water. All the water is poured into a second rectangular container with a base area of 200 cm². What is the height of the water in the second container?

Step 1: Volume of first tank = 30 × 20 × 25 = 15,000 cm³.

Step 2: Volume of water = 3/5 × 15,000 = 9,000 cm³.

Step 3: Base area of second container = 200 cm². Height = Volume ÷ Base area = 9,000 ÷ 200 = 45 cm.

Check: 200 × 45 = 9,000 cm³ = volume of water transferred. Correct.

Common Careless Mistakes by Question Type

The table below lists the single most common mistake for each question type covered in this guide and the fastest correction. Use this as a revision checklist before each practice session.

Question Type	Most Common Mistake	How to Avoid It
Remainder concept (fractions)	Applying the second fraction to the original total, not the remainder	Draw a branching bar model first and label each stage
Equal fractions concept	Comparing group sizes without first finding the actual equal value	Find the equal value first, then work backwards through each group's fraction
Fractions with units	Dividing by the denominator instead of finding 1 unit first	Find the value of 1 unit, then multiply up to the required number of units
Constant Part (ratio)	Scaling the changed part instead of the unchanged part	Identify the constant element, then scale to match it across both ratios
Constant Total (internal transfer)	Subtracting the transferred amount from the total	The total never changes; only the distribution changes
Constant Difference (age problems)	Assuming the age ratio stays constant over time	Only the difference is constant; set up the equation using the future ratio
Everything Changed (ratio)	Forgetting to connect the two scenarios through a known quantity	One set of units covers both; the known change is the bridge between them
Assumption Method	Adjusting the count in the wrong direction	Substitute back into both original conditions before writing the final answer
Working Backwards	Applying the original operations in forward order from the final value	Write all operations in sequence first; then reverse starting from the end
Speed problems	Combining speeds when objects are overtaking, not approaching	Draw a timeline; note whether objects move towards or away from each other
Rate problems	Adding times instead of adding rates	Find the fraction completed per unit time for each worker, then add fractions
Area and Perimeter	Using the area formula when asked for perimeter, or vice versa	State what you are finding before writing any formula
Angles in figures	Assuming angles look equal rather than proving it with a property	Name the angle property as a reason in every step of working
Volume	Forgetting unit conversion before multiplying dimensions	Convert all measurements to the same unit before calculating

Practise These Question Types

Ottodot's P5 and P6 Math classes teach each of the eight heuristics explicitly, then give students Roblox-based practice where questions arrive without a label — the same challenge Paper 2 creates. Browse the Ottodot Resource Hub: 64 free Math games covering every P1 to P6 MOE topic, no sign-up needed. A trial class is available with no long-term commitment. View plans and pricing here.

Conclusion

Every PSLE math question type follows a recognisable pattern. "The remainder" signals branching fractions. "Years ago" or "years later" signals a constant-difference problem. "Total number and total value with two item types" signals the assumption method. Learning to read those signals quickly is one of the most transferable exam skills a P6 student can develop.

Knowing the signals is step one. Applying the right method accurately under exam conditions is step two. The worked examples in this guide address step one. Step two comes from practising with questions that do not announce their type.

Bookmark this guide and use it to work through problems with your child. When a question is difficult, return to the relevant section, read the "When to use this method" cue, and work through the heuristic together before looking at the answer. That habit is the closest thing to the actual PSLE that home revision offers.

For interactive practice alongside every question type in this guide, visit the Ottodot Resource Hub. 64 free Math games. Every P1 to P6 MOE topic. No sign-up needed.

For families also helping with Science, see our PSLE Science OEQ guide — it covers all five open-ended question types using the same structured approach.