📅 Last Reviewed: July 2026 · ✍️ By: PSLE Hero Editorial Team
What is the National Mathematical Olympiad of Singapore (NMOS), and how does Olympiad mathematics help Singapore students? Organised annually by the NUS High School of Mathematics and Science, NMOS is a prestigious competition designed for Primary 4 and Primary 5 students. Rather than relying on simple arithmetic or rote memorisation, NMOS challenges students with non-routine problems that demand deep logical thinking, spatial awareness, and mathematical creativity. Developing these problem-solving skills is highly beneficial. By learning how to approach complex, unfamiliar Olympiad problems, students build the cognitive flexibility and resilience needed to tackle challenging Paper 2 Section C questions in the PSLE Mathematics exam.
In this guide, we present 5 original Olympiad-style practice questions inspired by the principles of primary school mathematics competitions. We will break down not just how to solve each question, but how to think through the problem, linking each heuristic back to the core skills tested in the PSLE.
Working backwards is a powerful strategy when a problem describes a series of operations but only gives the final, end state. In the PSLE, this thinking skill is frequently used to solve complex "Before-After" and fraction-of-remainder problems.
Alan, Ben, and Carl have a collection of marbles. First, Alan gives half of his marbles to Ben. Next, Ben gives one-third of the marbles he now has to Carl. Finally, Carl gives one-quarter of the marbles he now has to Alan. In the end, each boy has exactly 24 marbles. How many marbles did Alan have at the start?
Step 1: Start from the final state.
Alan = 24, Ben = 24, Carl = 24. (Total marbles = 72)
Step 2: Undo the final transaction (Carl gives 1/4 of his marbles to Alan).
Before giving 1/4 to Alan, Carl kept 3/4 of his marbles. Thus, Carl's ending count of 24 marbles represents 3/4 of what he had before this step.
Carl's marbles before transaction = 24 ÷ 3/4 = 32 marbles(Carl had 32)
Marbles Carl gave to Alan = 32 - 24 = 8 marbles(Alan received 8)
Alan's marbles before transaction = 24 - 8 = 16 marbles(Alan had 16)
State 3: Alan = 16, Ben = 24, Carl = 32
Step 3: Undo the second transaction (Ben gives 1/3 of his marbles to Carl).
Before giving 1/3 to Carl, Ben kept 2/3 of his marbles. Thus, Ben's count of 24 marbles in State 3 represents 2/3 of what he had before this step.
Ben's marbles before transaction = 24 ÷ 2/3 = 36 marbles(Ben had 36)
Marbles Ben gave to Carl = 36 - 24 = 12 marbles(Carl received 12)
Carl's marbles before transaction = 32 - 12 = 20 marbles(Carl had 20)
State 2: Alan = 16, Ben = 36, Carl = 20
Step 4: Undo the first transaction (Alan gives 1/2 of his marbles to Ben).
Before giving half of his marbles to Ben, Alan kept half. Thus, Alan's count of 16 marbles in State 2 represents half of his starting marbles.
Alan's marbles at first = 16 × 2 = 32 marbles(Alan had 32)
Marbles Alan gave to Ben = 32 - 16 = 16 marbles(Ben received 16)
Ben's marbles at first = 36 - 16 = 20 marbles(Ben had 20)
Carl's marbles at first = 20 (remained unchanged during this step)(Carl had 20)
State 1 (At Start): Alan = 32, Ben = 20, Carl = 20
Why this works: The total number of marbles (72) remains constant because they are only exchanged between the boys. Reversing operations is simple: adding becomes subtracting, and division becomes multiplication.
PSLE Connection: In Paper 2, Section C, students often face questions like "X gave 1/3 of his cards to Y, then Y gave 40% of his new total to Z...". Teaching children to track conservation of total value and build a step-by-step reverse table eliminates the need for messy algebraic equations.
The Assumption Method is a core Singapore Maths heuristic. While standard school problems usually feature two variables (e.g., chickens and rabbits), Olympiad-style questions often introduce a third variable (like unanswered questions or neutral scoring) to test a student's cognitive flexibility.
A mathematics quiz consists of 30 questions. For every correct answer, 5 marks are awarded. For every incorrect answer, 3 marks are deducted. For any unanswered question, 1 mark is deducted. Jerry attempted 28 questions and left the remaining questions blank. If his total score was 90 marks, how many questions did he answer correctly?
Step 1: Determine the number of unanswered questions.
The quiz has 30 questions. Jerry attempted 28, leaving the rest unanswered.
Unanswered questions = 30 - 28 = 2 questions
Marks lost from unanswered questions = 2 × (-1) = -2 marks
Step 2: Adjust Jerry's target score for the attempted questions.
Since the 2 unanswered questions deducted 2 marks, Jerry must have scored the remaining marks from his 28 attempts.
Score from attempted questions = 90 - (-2) = 92 marks
Step 3: Make an assumption. Assume Jerry got all 28 attempted questions correct.
Assumed score = 28 × 5 = 140 marks
Step 4: Find the difference between the assumed maximum score and Jerry's actual score.
Total score difference = 140 - 92 = 48 marks
Step 5: Calculate the score drop when one correct answer is replaced by an incorrect answer.
When you replace a correct answer (+5 marks) with an incorrect answer (-3 marks), you do not get the 5 marks and you lose 3 more marks.
Difference per replacement = 5 - (-3) = 8 marks
Step 6: Find the number of incorrect answers.
Number of incorrect answers = 48 ÷ 8 = 6 questions
Step 7: Find the number of correct answers.
Number of correct answers = 28 - 6 = 22 questions
Why this works: By immediately calculating the exact penalty of the third variable (unanswered questions), we reduce the problem to a standard two-variable scenario. We can then apply the 4-step Assumption Method: Assume all correct → Find total difference → Find individual difference → Divide.
PSLE Connection: A classic mistake in PSLE is using "Guess and Check", which consumes valuable time and increases calculation errors. Master the Assumption Method, and students can solve two-variable (and three-variable) word problems in under 90 seconds.
Cryptarithms (or alphametics) are puzzles where letters stand for digits. They develop deep number sense, place-value comprehension, and logical constraint checking.
In the addition sum below, different letters represent different non-zero digits (1 to 9):
A B A
+ B A B
-------
C D D C
Find the digits represented by the letters C and D.
Step 1: Analyze the maximum possible sum to find the thousands digit (C).
ABA and BAB are 3-digit numbers. Even if we use the largest possible digits (999 + 999), the sum is 1998.
Since the sum is a 4-digit number (CDDC), the thousands digit C must be 1.
C = 1
Step 2: Replace C with 1 and examine the units column.
Our addition looks like this:
A B A
+ B A B
---------
1 D D 1
Looking at the ones/units column (A + B), the sum must result in a units digit of 1. Since A and B are positive digits, A + B cannot equal 1. Thus, A + B must equal 11, carrying over 1 to the tens column.
A + B = 11 (with a carryover of 1)
Step 3: Analyze the tens column to find D.
In the tens column, we add B + A + 1 (the carryover from the units column).
Tens Column Sum = B + A + 1
Since we established that A + B = 11, the sum is:
Tens Column Sum = 11 + 1 = 12
This means the tens digit D must be 2, carrying over 1 to the hundreds column.
D = 2
Step 4: Verify the hundreds column.
In the hundreds column, we have A + B + 1 (the carryover from the tens column).
Hundreds Column Sum = A + B + 1 = 11 + 1 = 12
This yields a hundreds digit of 2 (which is indeed D) and carries over 1 to the thousands column (which is indeed C = 1). The math is fully consistent!
Final Sum = 1221 (e.g., 747 + 474 = 1221 or 838 + 383 = 1221)
Why this works: Instead of guessing digits randomly, we use structural mathematical boundaries (like the maximum limit of adding 3-digit numbers) to establish fixed parameters (C = 1), then cascade the logic through place-value constraints.
PSLE Connection: Non-routine puzzle questions in Section B and Section C of the PSLE Maths paper often test number properties (e.g., prime numbers, factors, divisibility rules). Students who have practiced constraint checking learn to eliminate impossible numbers systematically instead of feeling overwhelmed.
Grid path puzzles are a staple of Olympiad combinatorics. While they look intimidating, they can be solved easily using the Addition Principle (or Pascal's grid labeling), showing students how complex combinations can be broken down into simple addition steps.
A student wants to walk from junction A to junction B along the grid of streets shown below. The student can only walk East (Right) or North (Up). However, junction X is closed due to road construction. How many different paths can the student take from A to B without passing through X?
Step 1: Understand the Addition Principle.
To reach any junction, a student must come from either the junction directly to its West (Left) or the junction directly to its South (Down). Thus, the number of ways to reach a junction is the sum of the ways to reach its left neighbor and its bottom neighbor.
Step 2: Label the starting boundaries.
Junction A is the start, so it is labeled 1. The bottom edge can only be reached by walking straight East, so all junctions along the bottom are labeled 1. Similarly, the leftmost edge can only be reached by walking straight North, so all junctions along the left are labeled 1.
Step 3: Propagate the addition through the grid.
Total routes avoiding X = 11 routes
Why this works: Instead of memorising complex permutations and combination formulas (which fail when road blocks are introduced), the Addition Principle allows students to solve highly customized combinatorics questions using simple, error-free addition.
PSLE Connection: Non-routine pattern drawing or pathway problems appear in both Paper 1 (Section B) and Paper 2 of the PSLE. Students who can list and count systematically using node addition are far more likely to get these correct than those who try to list paths manually in writing.
Geometry questions in primary school exams do not require trigonometry. Instead, they require spatial reasoning, specifically the ability to break complex shapes down into simpler building blocks and analyze overlapping areas.
The figure below shows a square ABCD with a side length of 14 cm. Two quarter circles are drawn inside the square. One is centered at D (covering corner A to C), and the other is centered at B (covering corner C to A). Find the area of the shaded leaf-shaped region in the middle. (Take π = 22/7)
Step 1: Calculate the area of the entire square.
Area of square = 14 cm × 14 cm = 196 cm²
Step 2: Calculate the area of one quarter circle.
The radius of the quarter circle is equal to the side length of the square (14 cm).
Area of 1 Quarter Circle = 1/4 × π × r²
Area = 1/4 × 22/7 × 14 × 14 = 154 cm²
Step 3: Look at the overlapping region.
If we add the areas of the two quarter circles together, the leaf-like overlapping region is counted twice, while the unshaded corners are counted once.
Therefore, the sum of the two quarter circles is equal to the area of the square plus the area of the overlap:
Sum of Quarter Circles = Area of Square + Shaded Overlap
154 cm² + 154 cm² = 308 cm²
Step 4: Subtract the square's area to find the overlap.
Shaded Overlap Area = 308 cm² - 196 cm² = 112 cm²
Why this works: Conceptualizing shapes as overlaps (`A + B - Overlap = Total`) is much faster and mathematically cleaner than attempting to calculate the area of the leaf shape directly using circular segments.
PSLE Connection: Overlapping circle-and-square problems are extremely common in Paper 2 Section C (such as the notorious 2019 "semicircle" question). Students who look for overlapping areas and use rearrangement methods score high marks without getting bogged down in complex decimal arithmetic.
Q: What is the National Mathematical Olympiad of Singapore (NMOS)?
The National Mathematical Olympiad of Singapore (NMOS) is an annual national competition organized by NUS High School of Mathematics and Science since 2006. It is open to Primary 5 students and below in Singapore, and aims to discover and cultivate mathematical talents.
Q: How does Olympiad Maths help with PSLE Mathematics?
While PSLE focuses on the school syllabus, its most challenging Section C word problems are "non-routine." These require students to look at patterns, make assumptions, work backwards, or deduct logically—precisely the thinking heuristics trained in Olympiad Maths.
Q: Is NMOS suitable for all Primary 4 and 5 students?
It is best suited for students who are strong in school mathematics and show a natural curiosity for logic puzzles and number patterns. It is an excellent way to stretch their minds without the pressure of a formal national exam.
📚 Explore More PSLE Guides
Many mathematical thinking skills developed through Olympiad-style problems are also valuable for solving challenging PSLE Maths questions. Continue practising similar problem-solving skills in PSLE Hero.
Practise Heuristics on PSLE Hero→ Privacy Policy | → Data Deletion Request | ← Back to PSLE Hero