PSLE Mathematics tuition in Bedok North should help pupils solve changing-ratio problems by tracking the quantities that remain fixed. When Group B transfers fifty-six counters to A, the combined total stays the same but the old ratio 2:3 can become a new ratio 5:4.
The weekday-versus-weekend parent dilemma is best approached with the pupil’s first independent working. The tutor should identify whether the learner confuses abstract ratio parts with actual objects, misreads the transfer direction or fails to check both ratios.
At eduKateSG, focused three-pupil tutorials aim to build multiple valid methods, clearly labelled quantities and an independent forward check. The examples are original teaching questions, not predictions of a PSLE examination item.
Guide contents: Changing ratios and conserved totals
A transfer can change a ratio without changing the whole · What the revised 2026 PSLE Mathematics format tests · The starting ratio 2:3 is not a literal count of five · Label the movement from B to A · Worked algebraic solution from the final 5:4 ratio · Reconstruct the change using actual counts · A second solution using changing fractions of the same total · The difference between groups changes by twice the transfer · The original and final ratio units are different
A transfer can change a ratio without changing the whole
Two fictional groups, A and B, initially have counters in the ratio 2:3. Group B transfers exactly fifty-six counters to Group A. After that movement, their ratio becomes 5:4. How many counters did each group have before the transfer?
For Bedok North parents weighing weekday and weekend PSLE Mathematics tuition, a good first diagnostic is to ask what the transfer changes. A gains fifty-six, B loses fifty-six, and their combined total stays the same. The original and final ratio parts, however, need not be the same size.
The correct initial quantities are 144 in A and 216 in B, totalling 360. After the transfer the counts become A = 200 and B = 160, which simplify to 5:4.
A pupil who adds fifty-six directly to the ratio numeral two is confusing an abstract equal-part representation with an actual count of objects. A child who adds fifty-six to A without subtracting it from B violates the conserved total.
These counters are original teaching data, not actual school records, income or an examination prediction. The value of the problem is the reasoning that can transfer to a different ratio change.
What the revised 2026 PSLE Mathematics format tests
SEAB’s published 2026 Standard Mathematics syllabus sets out assessment objectives involving mathematical knowledge and procedures, interpretation and application, and reasoning with appropriate strategies.
The revised 2026 format has two written papers and three booklets, containing forty-five questions worth 100 marks in total. Paper 1 lasts one hour ten minutes without calculator use; Paper 2 lasts one hour twenty minutes with approved calculators permitted.
Ratio, part-whole comparisons and multi-stage problems provide appropriate practice in choosing a mathematical relationship, but no original example guarantees the content of a particular national examination.
A calculator can divide 360 by five or nine, but it cannot determine that fifty-six counters moved internally rather than entering from outside. That interpretation must come from the story.
Families should check official documents for the child’s actual examination year. Good tuition develops accurate model selection and written checks rather than claims about secret or predictable questions.
The starting ratio 2:3 is not a literal count of five
The initial ratio means Group A has two equal proportional parts for every three such parts in B. Together the groups contain five original ratio parts.
Without a known total, known difference or another independent condition, the original ratio alone does not identify a unique pair of counts. A2 B3, A20 B30 and A144 B216 all satisfy the same ratio.
The fifty-six-counter transfer and resulting final ratio supply the missing relationship that fixes the original scale. The part size cannot be assumed arbitrarily just to make the numbers convenient.
Let u represent the count in one original ratio part. Then A begins with 2u and B begins with 3u. This notation keeps the abstract ratio unit separate from the actual transfer of fifty-six counters.
A tutor should ask which quantity u measures before forming an equation. A correct symbol without a clear meaning can lead to the wrong sign when the transfer is applied.
Label the movement from B to A
Write the initial state as A = 2u and B = 3u. The arrow from B to A carries fifty-six counters in one direction.
After the arrow, A has 2u + 56 and B has 3u − 56. The updated combined total is still 5u because the addition and subtraction cancel.
Check direction: A is expected to grow, while B shrinks. Since the final ratio is 5:4, A is larger than B after the movement, even though A started as the smaller group.
An incorrect model might place plus fifty-six on both sides or subtract from A instead. Such expressions describe different stories and need to be rejected before arithmetic.
A physical pair of trays with fifty-six tokens moving from one to the other can make the conserved total visible to a learner who has trouble interpreting an arrow diagram.
Worked algebraic solution from the final 5:4 ratio
The final ratio A:B = 5:4 means four times A’s new amount equals five times B’s new amount. Substitute the expressions from the transfer:
4(2u + 56) = 5(3u − 56). Expand both sides to get 8u + 224 = 15u − 280.
Rearrange without losing equality: add 280 and subtract 8u, giving 504 = 7u. Divide by seven to obtain u = 72 counters for one original ratio part.
The original counts are 2 × 72 = 144 in A and 3 × 72 = 216 in B.
Check those starting counts satisfy 144:216 = 2:3. Applying the movement gives 144 + 56 = 200 and 216 − 56 = 160, and 200:160 simplifies to 5:4.
The calculation is complete only when the original and final conditions both hold. Solving an equation formed from the wrong transfer signs could produce a number but not a valid solution to the story.
Reconstruct the change using actual counts
Starting with A144 and B216, their combined total is 360 counters. Move fifty-six from B into A.
After the move, A contains 200 and B contains 160. Their combined count remains 360, independently confirming conservation.
To check the new ratio, divide both actual counts by forty: 200:160 becomes 5:4. To check the old ratio, divide 144:216 by seventy-two to obtain 2:3.
The amount gained by A is 200 − 144 = fifty-six, and the amount lost by B is 216 − 160 = fifty-six. Both match the precise transfer.
These are four separate checks: initial ratio, final ratio, transfer amount and combined total. A solution matching only the sum but not the stated ratios is incomplete.
A tutor can ask the child to conduct one check independently after hiding the algebra. That is more informative than repeating the same equation manipulation.
A second solution using changing fractions of the same total
Before the movement, A represents two of the five original ratio parts, or 2/5 of the combined total.
After the movement, A represents five of the nine final ratio parts, or 5/9 of the same total. The total is unchanged because the action is an internal transfer.
The increase in A’s share of the combined total is therefore 5/9 − 2/5 = 25/45 − 18/45 = 7/45.
That increase corresponds to the fifty-six actual counters transferred into A. One forty-fifth of the total therefore equals fifty-six divided by seven, or eight.
All forty-five forty-fifths equal 45 × 8 = 360 counters. The original A count is 2/5 × 360 = 144, and B is 216.
This route uses fraction relationships instead of the first algebraic equation. Both approaches agree because they represent the same conserved total and the same change in A’s share.
The difference between groups changes by twice the transfer
Initially the ratio 2:3 means B exceeds A by one original ratio part. With the solved part size seventy-two, the initial difference is 216 − 144 = 72 counters.
After fifty-six move from the larger B to the smaller A, B loses fifty-six while A gains fifty-six. Their difference shrinks by twice fifty-six, or 112 counters.
The final difference is therefore seventy-two minus 112, which is negative forty when measured as B minus A. In other words, A now exceeds B by 40 counters.
The final ratio 5:4 has a one-part A-over-B difference. Hence one new ratio part equals forty, consistent with final counts of five times forty and four times forty.
This is a useful independent reasoning check. It also shows why a transfer need not equal one old or new ratio part: fifty-six is a separate specified count.
A learner who subtracts only fifty-six from the original difference has overlooked that a transfer changes both sides of the comparison.
The original and final ratio units are different
The original total of 360 is divided into five ratio parts, so each old part represents 360 ÷ 5 = 72 counters.
The same actual 360 is divided into nine final ratio parts, so each new part represents 360 ÷ 9 = 40 counters.
Both ratios therefore describe the same conserved total using different equal-unit scales. The old 2:3 is 144:216, while the new 5:4 is 200:160.
A pupil who uses seventy-two as the numerical size of every final ratio part would report a new total of 648, contradicting conservation.
Ask the child to label each bar model ‘before’ or ‘after’ and state what one unit represents at that time. The part count is a representation, not a fixed physical piece carried unchanged through every story.
Worked variations, teaching decisions and parent questions
Second worked example: equalise a 3:5 ratio by transferring thirty · Find the transfer when the original and final ratios are given · Moving fifty-six in the opposite direction changes everything · A new supply is not an internal transfer · Removing counters entirely has another effect · Third worked example: 3:4 becomes equal after eighteen move · The same counts answer part-to-part and part-to-whole questions · Express the transfer as a percentage of the whole · Recognise when the ratio conditions are insufficient · Six wrong methods with six different repairs · When weekday PSLE Mathematics tuition could help · When weekend tuition offers more bar-model time · Individual diagnosis matters inside a three-pupil class · Four checkpoints for transferable ratio reasoning · Frequently asked questions from Bedok North families · Official PSLE references and next useful Mathematics learning
Second worked example: equalise a 3:5 ratio by transferring thirty
A separate fictional pair of groups starts in the ratio 3:5. The larger Group B transfers exactly thirty counters to the smaller A. After the transfer, both groups contain the same number. Find the original counts.
Because the two groups end equal, their original difference must be twice the transfer, or sixty counters. The starting ratio has two ratio parts of difference, since five minus three equals two.
Therefore two original ratio parts represent sixty, so one part equals thirty counters. A originally has 3 × 30 = 90 and B has 5 × 30 = 150.
After B gives thirty to A, both groups have 120 counters. Their combined total remains 240 throughout.
Check both conditions: 90:150 simplifies to 3:5, and 120:120 simplifies to 1:1. A pupil who transfers the full sixty-counter difference would simply reverse the unequal counts instead of making them equal.
Find the transfer when the original and final ratios are given
Suppose the original ratio is 2:3 and the combined total is specified as 360 counters. After a transfer from B to A, the final ratio is 5:4. How many were moved?
The initial five parts imply one part equals 360 ÷ 5 = seventy-two. Original A therefore has 144 and B216.
The final nine parts imply one part equals 360 ÷ 9 = forty. Final A has 200 and B160.
Group A must have gained 200 − 144 = 56 counters. Independently, Group B must have lost 216 − 160 = 56.
Check the change acts inside the system, so the total remains 360. This reversed question is useful because the learner must identify the correct earlier and later part scales instead of simply repeating the algebra.
Moving fifty-six in the opposite direction changes everything
Start again with A144 and B216 but imagine fifty-six counters move from A to B, the opposite direction from the main problem.
The new counts become A = 144 − 56 = 88 and B = 216 + 56 = 272.
The new ratio is 88:272, simplifying by eight to 11:34. The total still equals 360 because the transfer remains internal.
This result does not satisfy the original final 5:4 condition. The direction of the arrow is essential, even when the magnitude of the transfer is unchanged.
Ask the learner to draw the arrow before writing plus or minus signs. Correct arithmetic after a reversed arrow still solves a different mathematical story.
A new supply is not an internal transfer
Using the original A144 and B216, suppose fifty-six new counters are added to A from outside while Group B keeps all its counters.
Now A has 200 and B remains 216, for a combined total of 416.
The new ratio 200:216 simplifies by eight to 25:27, not the 5:4 result of the internal transfer.
A child who insists the total remains 360 has ignored the external source. The preservation rule is valid only when no objects enter or leave the counted system.
Draw a boundary around both groups. A B-to-A arrow stays inside it, while an arrow from outside into A increases the system’s total. This physical distinction determines which invariant the solution should use.
Removing counters entirely has another effect
Suppose fifty-six counters are instead removed from Group A and taken out of the two-group system. B remains unchanged at 216.
The new counts are A = 144 − 56 = 88 and B = 216. Their sum is 304, a decrease of fifty-six from the original.
The new ratio 88:216 simplifies by eight to 11:27.
Compare this with the opposite-direction transfer, which also leaves A with eighty-eight but sends the removed counters to B, creating B272 and preserving the 360 total.
Tracking destination matters as much as tracking source. A tutor can ask what total should change before allowing the child to calculate the new ratio.
Third worked example: 3:4 becomes equal after eighteen move
A different original puzzle starts with A:B = 3:4. Eighteen counters move from larger B to smaller A, after which the quantities become equal.
The initial difference is one original ratio part because four minus three equals one. A transfer of eighteen reduces the difference by twice eighteen, or thirty-six.
Therefore one original ratio part equals 36 counters. A starts with 3 × 36 = 108 and B with 4 × 36 = 144.
After the transfer A becomes 126 and B becomes 126. Their combined total remains 252, matching the original 108 + 144.
The solution follows from a difference invariant rather than a guessed pair of numbers. A student who transfers thirty-six instead of eighteen would overshoot equality.
The same counts answer part-to-part and part-to-whole questions
Initially A:B = 2:3. A represents two fifths of the total, whereas B represents three fifths. These fractions describe their shares of the combined five original ratio parts.
After the transfer A:B = 5:4. A’s share of the unchanged combined total is now five ninths and B’s share is four ninths.
That is why A’s increase can be calculated as 5/9 − 2/5 = 7/45 of the total, corresponding to the fifty-six counters moved.
The original part-to-part ratio 2:3 is not the same notation as A’s part-to-whole fraction 2/5. The denominator of the latter includes both groups.
Ask the child to name which quantities are compared before simplifying a ratio or fraction. A correct numerator with the wrong reference whole can answer a completely different question.
Express the transfer as a percentage of the whole
The actual conserved total is 360 counters and the transfer moves fifty-six from B into A.
The transferred amount as a share of the whole is 56/360. Simplify by eight to obtain 7/45, or approximately 15.56% of the total.
A’s share before the transfer is forty percent because 2/5 = 0.4. Its share after the transfer is 5/9, approximately 55.56%.
The approximate increase of 15.56 percentage points is consistent with the transferred 7/45 share of the constant total.
The decimal values are approximations, while the fractions 2/5, 5/9 and 7/45 are exact under the teaching model. A pupil should keep that distinction clear when a question requests an exact fraction.
Recognise when the ratio conditions are insufficient
Suppose a question gives only the initial ratio A:B = 2:3 and states that B will transfer some counters to A, without giving either the transfer amount, the final ratio or a fixed total.
Many distinct initial scales and transfers are possible. The ratio alone does not determine one unique original pair.
Even a known transfer amount of fifty-six would generally be insufficient unless another condition fixes the scale. Different original 2:3 collections can accommodate that movement and end with different ratios.
The main problem is solvable because it provides the transfer amount and a specific required final ratio of 5:4. Those independent conditions fix the original unit value.
A pupil who guesses a convenient total without enough information has introduced an assumption that the question never states. Identifying missing conditions is part of sound reasoning.
Six wrong methods with six different repairs
One learner adds fifty-six to the ratio numeral two, mistaking abstract parts for counters. A bar model should first connect a ratio part to an actual count.
Another adds fifty-six to both groups, creating objects despite a pure internal transfer. The tutor should draw a source-to-destination arrow and check the total.
A third solves the algebra with minus fifty-six on A and plus fifty-six on B, reversing the transfer direction. A fourth uses the old part size seventy-two for the final 5:4 ratio.
A fifth reports the original total 360 correctly but fails to recover the original individual counts, even though the question asks for them.
A sixth gives a pair of counts that has the right total but does not satisfy the new 5:4 relationship. The solution must be checked against all stated conditions.
Record the first unsupported step and assign a new variation that specifically tests its repair. A generic ‘careless’ label provides no instruction about which mathematical idea is missing.
When weekday PSLE Mathematics tuition could help
A weekday lesson may be useful after a school question exposes a confusion between ratio units and actual transferred objects. Bringing the original uncorrected working lets the tutor see where the learner changed the meaning of a quantity.
Parents should consider dismissal, CCAs, meals, homework, the real journey to the confirmed class venue and sufficient sleep. A late weekday session might not provide enough attention for multistage reasoning.
A focused ninety-minute tutorial might begin with a private diagnostic, develop the 2:3-to-5:4 solution, check the transfer and finish with the 3:5 equalisation variation.
After several days, set an unseen task: A:B = 3:4 and eighteen move from B to A until equal. The child should recover A108 and B144 without the original explanation visible.
The useful outcome is a correct independent model and fewer prompts, not simply a high number of completed ratio worksheets.
When weekend tuition offers more bar-model time
A weekend session may offer enough space for two bar diagrams labelled before and after, showing that five original parts and nine final parts represent the same actual total.
The model is useful when a child can simplify ratios but cannot explain why the numerical size of a part differs after the transfer.
Weekends may also be crowded with sports, family commitments, other tuition and school preparation. A nominally long session is beneficial only when the child is alert and has recovery time.
An effective tutor should begin with an independent attempt, offer the minimum representation needed and finish with a changed example involving an external addition rather than an internal transfer.
Follow with one brief weekday retrieval question. If the child now names the correct invariant without prompting, the weekend model has supported reusable reasoning.
Individual diagnosis matters inside a three-pupil class
In one eduKateSG group, one learner may divide by the wrong number of ratio parts, another may fail to subtract the transferred objects from B, and a third may find the counts correctly but present an unsimplified or reversed ratio.
Those pupils need different prompts: count equal parts, trace the movement arrow or compare the actual counts with the named ratio order.
An illustrative ninety-minute tutorial can include separate first attempts, one shared worked model, an independent fraction-based alternative and changed private exit questions.
Group discussion helps students learn how different correct routes connect, but a fluent answer from one confident pupil does not establish that the others can solve the problem alone.
Parents can ask what first hint was needed and whether the same hint was necessary after several days. That evidence is more meaningful than worksheet volume.
Four checkpoints for transferable ratio reasoning
Checkpoint one is interpretation. State the before and after ratios, identify the direction and amount of transfer, and decide which total or difference should remain fixed.
Checkpoint two is a numerical model. Express original counts through a valid ratio unit and apply the transfer to the correct group before writing the final relation.
Checkpoint three is verification. Reconstruct initial and final ratios, show that the combined total is conserved, and confirm both the source loss and destination gain.
Checkpoint four is transfer to an unfamiliar story. Reverse the direction, use a different ratio, or introduce an outside supply. The learner should choose the new invariant independently.
These are flexible teaching stages, not a guaranteed four-week programme or a promised examination grade. Some children need stronger fraction or multiplication foundations before advanced ratio changes.
Frequently asked questions from Bedok North families
What were the original group counts? A held 144 counters and B held 216, forming the ratio 2:3.
What are the counts after the transfer? A receives fifty-six and reaches 200; B gives fifty-six and falls to 160, forming 5:4.
What stays fixed? The combined total of 360 because the movement occurs inside the two groups.
Why do the old and new ratio units differ? The same 360 is split into five original parts of seventy-two and nine final parts of forty.
Can the problem be solved without an equation? Yes. A’s fractional share increases by 7/45 of the fixed total, which equals the transferred fifty-six.
Is this a guaranteed PSLE question? No. It is original practice in transferable ratio reasoning.
Is weekday or weekend tuition always better? Neither. Choose a sustainable time for attentive teaching, independent checking and rest.
Does this article announce an eduKate teaching outlet in Bedok North? No. Confirm the actual venue, fees and current three-pupil class availability.
Official PSLE references and next useful Mathematics learning
For the published examination format consult SEAB’s PSLE formats examined in 2026 and the official 2026 Standard Mathematics syllabus (0008). For curriculum scope see the MOE Primary Mathematics syllabus updated October 2025.
The eduKate Mathematics Learning Hub connects fractions, ratios and problem solving. For local geometry see Bedok North Primary 5 triangle area and Bedok North Primary 6 quarter-circle geometry.
To discuss a suitable real three-pupil tutorial, arrange an eduKate Singapore parent–student consultation. Bring the child’s uncorrected first attempt and feasible weekday/weekend windows; confirm venue, fees, travel and current group availability directly.
Continue through Bedok North Mathematics: Primary 1 · Mathematics Foundations · Primary 2 · Number Skills · Primary 3 · Time and Units · Primary 4 · Decimal Division · Exam Prep · Decimal Division and Containers. Read the guide for the pupil’s current school level.