Primary 6 Mathematics tuition for Li Hwan Walk families can begin with two fictional groups in the ratio 3:5. Each group then receives twenty-four additional counters from outside, and the ratio changes to 5:7. The original groups contained thirty-six and sixty counters, while the final groups contain sixty and eighty-four.
The important invariant is the difference between the groups: adding the same number to each preserves the gap, although it increases the combined total. A weekday or weekend tutorial is useful when it helps a child identify that relationship before reaching for a ratio formula.
eduKateSG’s focused three-pupil teaching should diagnose what each ratio unit counts, differentiate an external addition from an internal transfer and verify the solution independently. These examples are entirely fictional and do not announce an eduKate branch on Li Hwan Walk.
Guide contents: Equal additions and changing ratio parts
A ratio can change even when both groups gain the same number · Ratio and algebra in the current Primary 6 scope · What the original 3:5 tells us and what it does not · Why equal additions keep the difference unchanged · Worked algebra: solve the original unit value · Check the new 5:7 ratio independently · The fixed gap can be used without full algebra · The difference is a useful separate target
A ratio can change even when both groups gain the same number
Two imaginary groups have counters in the ratio A:B = 3:5. Each group receives twenty-four new counters from outside. The new ratio becomes 5:7. Find the original numbers in the two groups.
The present story is not a transfer between A and B. The new counters are supplied from outside, so A gains twenty-four and B independently gains twenty-four.
Because both gain the same absolute amount, B remains ahead of A by exactly the same numerical difference as before. Yet the proportional ratio changes toward equality.
For Li Hwan Walk parents considering weekday versus weekend Primary 6 tuition, ask whether the pupil recognises the unchanged difference or incorrectly assumes the total remains fixed as in a transfer problem.
The original counts are thirty-six and sixty. After adding twenty-four to each, the counts become sixty and eighty-four, whose simplest ratio is indeed five to seven.
Ratio and algebra in the current Primary 6 scope
The updated MOE Primary Mathematics syllabus places ratio in Primary 6 Standard Mathematics, including equivalent ratios, part-to-part comparisons and dividing quantities according to ratios.
Simple algebraic representations can support a word problem when one unknown unit value must satisfy two conditions. An algebraic equation should be an expression of the story, not a replacement for understanding its quantities.
Parents should check the child’s specific course and school sequence, since upper-primary Standard and Foundation expectations can differ. A student uncertain about equal-part bars should first strengthen that foundation.
Different change stories preserve different quantities: an internal transfer preserves a combined total, equal additions preserve the difference, and a removal from the overall collection reduces its total.
Good teaching asks what happens physically before selecting the numerical relationship. A memorised ratio formula chosen for the wrong action will fail even when its manipulation is accurate.
What the original 3:5 tells us and what it does not
The original ratio A:B = 3:5 describes three equal parts in A and five equal parts in B. The difference is two original parts.
Without an additional condition, the ratio alone does not tell us the absolute count. A pair of 3 and 5, 30 and 50, or 36 and 60 all simplify to the same relative relationship.
Let one original part be u counters. Then A = 3u, B = 5u and the original gap is 2u.
The fixed addition of twenty-four to each creates the second relationship needed to discover u. Those added twenty-four are physical counters, not twenty-four abstract ratio units.
Ask the learner to draw the original three and five equal-part bars and label the difference, rather than immediately guessing a convenient u.
Why equal additions keep the difference unchanged
Before the change, B − A = 5u − 3u = 2u. Afterwards, the counts are B = 5u + 24 and A = 3u + 24.
Subtracting the updated A from B gives (5u + 24) − (3u + 24) = 2u. The two added twenty-four terms cancel because they are equal.
The combined total behaves differently. Original total is 8u; after additions it becomes 8u + 48, because two separate groups each receive twenty-four.
A pupil who says the total stays unchanged may be importing a rule from a transfer problem. Draw two arrows from outside into the two groups to show that the counted system grew.
The distinction between fixed difference and changing total is the reason this particular ratio can be solved efficiently.
Worked algebra: solve the original unit value
The required final ratio is 5:7. Use the updated expressions A = 3u + 24 and B = 5u + 24.
Write the ratio equality as 7(3u + 24) = 5(5u + 24). Expand both sides: 21u + 168 = 25u + 120.
Subtract 21u from both sides and subtract 120: 48 = 4u. Thus u = 12 counters per original part.
Original A equals 3 × 12 = 36 counters. Original B equals 5 × 12 = 60 counters.
Check the starting ratio: 36:60 reduces by twelve to 3:5, matching the first condition. The equation is not finished until the actual group counts are identified.
Check the new 5:7 ratio independently
Add the stated twenty-four counters to both recovered original groups. Group A becomes 36 + 24 = 60; Group B becomes 60 + 24 = 84.
Simplify the final ratio 60:84 by dividing both terms by twelve. The result is 5:7, exactly the stated new ratio.
The original difference is 60 − 36 = twenty-four, while the final difference is 84 − 60 = twenty-four. The invariant holds.
The original total is ninety-six; the new total is 144, an increase of forty-eight from two independent additions.
These three checks—initial ratio, final ratio and changed total with fixed gap—test distinct conditions. A candidate solution satisfying only one is not sufficient.
The fixed gap can be used without full algebra
The initial 3:5 ratio differs by two old ratio parts, while the final 5:7 ratio differs by two new ratio parts. Both part differences represent the same physical gap.
Therefore one old ratio part and one new ratio part have the same numerical size in this specially chosen example. Let that shared size be u.
The initial counts are 3u and 5u. After both gain twenty-four they become 5u and 7u, meaning the increase in A corresponds to two units: 5u − 3u = 2u = 24.
Thus u = twelve, giving original groups thirty-six and sixty. The bar method agrees with the equation because the two ratio differences happen to use two parts each.
Do not teach that ratio parts always retain their numerical size across changes. This equality follows from the specifically equal two-part difference in both given ratios.
The difference is a useful separate target
In the solved example, B originally exceeds A by sixty minus thirty-six, or twenty-four counters. After both receive twenty-four, B still exceeds A by eighty-four minus sixty, again twenty-four.
A question asking for the final combined total would have answer 144, while one asking for the final group difference would have answer twenty-four.
The numerical appearance of twenty-four in both the added amount per group and the fixed difference is coincidental for these chosen ratios.
Ask the pupil to identify what each twenty-four represents rather than assume the same printed number always denotes the same quantity.
The mathematics becomes clearer when every answer is labelled as an original count, final count, addition per group, difference or combined total.
Further applications and changed examples
Worked variation: 2:5 becomes 1:2 after equal additions · Equal removals preserve a difference as well · An internal transfer preserves the total instead · Translate old and new ratios into fractions of their totals · Find the equal addition amount from known original counts · Adding to only one group does not preserve their difference · Ratio parts are not physical counters until scaled · When the data do not determine a unique answer
Worked variation: 2:5 becomes 1:2 after equal additions
An original counter puzzle has A:B = 2:5. Both groups receive eighteen new counters and the ratio becomes 1:2. Find their starting counts.
Let one original part be u. The new amounts are 2u + 18 and 5u + 18. The 1:2 relationship means 2(2u + 18) = 5u + 18.
Expand to 4u + 36 = 5u + 18, which gives u = 18. The original groups had A = 36 and B = 90.
After the additions, the counts become fifty-four and 108, a ratio of 1:2. Their difference stays fifty-four while the total rises from 126 to 162.
This changed example has a different original and final ratio, so a pupil must reconstruct the invariant rather than repeat the 36-and-60 answer.
Equal removals preserve a difference as well
A different fictional pair begins in the ratio 5:7. Twenty counters are removed from each group, and the remaining ratio becomes 3:5. Find the original counts.
Let one old ratio part be u. The updated counts are 5u − 20 and 7u − 20. The final 3:5 relationship gives 5(5u − 20) = 3(7u − 20).
Expand to 25u − 100 = 21u − 60, so 4u = 40 and u = 10. The starting counts were fifty and seventy.
After removing twenty from both, the counts become thirty and fifty, ratio 3:5. The gap is twenty at both times, while total falls from 120 to eighty.
Removing equal amounts from both preserves the gap for the same algebraic reason as adding equal amounts, but the combined total decreases instead of increasing.
An internal transfer preserves the total instead
Take A = 36 and B = 60 from the main Li Hwan Walk example. Suppose twelve counters move from B to A, rather than twelve new counters entering each group.
The new counts are A = 36 + 12 = 48 and B = 60 − 12 = 48. Their new ratio is 1:1.
The original total 36 + 60 = ninety-six remains unchanged after the transfer. But the gap between the groups falls from twenty-four to zero.
Equal external additions would increase both groups and keep the twenty-four-counter gap; internal transfer changes the gap while conserving total.
Ask the pupil to draw arrows showing sources and destinations. The appropriate invariant follows the action, not merely the presence of the word counters.
Translate old and new ratios into fractions of their totals
Originally A:B = 3:5, with combined total ninety-six. A is 36/96 = 3/8 of the original whole, while B is 5/8.
After each receives twenty-four, the total becomes 144. A contains sixty, so its fraction of the new whole is 60/144 = 5/12. B’s fraction is 84/144 = 7/12.
The numerator and denominator both change. A’s count rises by twenty-four, while the overall total rises by forty-eight.
A pupil who expresses the final A count as sixty out of ninety-six has used the updated numerator with the old reference whole.
Ask which groups and which time point define the whole before simplifying any fraction.
Find the equal addition amount from known original counts
Suppose the original group counts of thirty-six and sixty are given explicitly, but the number added equally to each is unknown. After the additions the ratio is 5:7.
Let k be the common addition. The required ratio gives 7(36 + k) = 5(60 + k).
Expand to 252 + 7k = 300 + 5k, so 2k = 48 and k = 24 counters.
Check by adding twenty-four to both counts, producing sixty and eighty-four, which simplify to 5:7.
This inverse problem differs from recovering the original counts: now the original scale is known and the common change is missing. The learner should define the correct unknown.
Adding to only one group does not preserve their difference
Start again with A = 36 and B = 60. Suppose twelve new counters are given only to A, leaving B unchanged.
The resulting counts are A = 48 and B = 60, a ratio of 4:5. Their combined total rises from ninety-six to 108.
The initial gap of twenty-four falls to twelve because only the smaller group gains twelve. There are not equal changes in both groups to cancel when taking the difference.
Do not use the unchanged-difference shortcut here. It depends specifically on the same absolute addition or removal applying to both counted groups.
A tutor can ask learners to compare three small arrows: outside to both groups equally, outside to A only, and B to A. Each has a different invariant.
Ratio parts are not physical counters until scaled
The original 3:5 describes eight equal parts, but each old part happens to be twelve counters only after the new ratio and equal addition supply enough conditions to determine the scale.
After the equal additions, the 5:7 ratio has twelve parts. Each final part is also twelve counters in this specially chosen example; that equality is a consequence, not a general rule.
In the alternative 2:5 to 1:2 example, the old part size is eighteen and the new part size is fifty-four. It changes substantially even though the group difference stays fixed.
A child who insists that a ratio part can never change numerical size is likely to misinterpret a changed-total problem.
Ask the learner to calculate actual counts before simplifying the final ratio and to label the scale associated with each diagram.
When the data do not determine a unique answer
The starting ratio A:B = 3:5 alone permits many pairs: 3 and 5, thirty and fifty, or 36 and 60 all satisfy it.
If the story merely says both receive some equal unknown amount but gives no final ratio, no single original scale or addition amount follows.
The fixed addition of twenty-four plus the new 5:7 ratio provides the extra numerical relationship needed in the main worked example.
Recognising insufficient information is a valid mathematical conclusion. The tutor should ask what extra condition would make the exact quantities determinable.
This habit prevents children from inventing convenient counts simply because a worksheet appears to expect one final number.
More original problems, lesson choices and parent questions
Worked variation: equal additions followed by a transfer · Recover the common addition from the known ending · The two group shares can be expressed as percentages · A ratio is not an average · Predict whether the groups become more or less unequal proportionally · Name the counted system before applying an invariant · Five ratio misconceptions and their targeted repairs · When weekday Primary 6 Mathematics tuition may help · When a weekend tutorial allows clearer bar modelling · How 3-pax tutoring can handle different ratio needs · Four checkpoints for changing-ratio reasoning · Frequently asked questions about changing ratios · MOE curriculum and connected Li Hwan Walk learning
Worked variation: equal additions followed by a transfer
Start from the main Li Hwan Walk counts: A has thirty-six and B sixty. Suppose twelve new counters are first added to each group from outside. A becomes forty-eight and B seventy-two.
Then six counters transfer internally from B to A. The final A is 48 + 6 = 54, and final B is 72 − 6 = 66.
The final ratio 54:66 simplifies by six to 9:11. The original total ninety-six rose by twenty-four during the external additions and stayed at 120 during the internal transfer.
The gap originally was twenty-four. Equal additions kept that gap unchanged, while the six-counter transfer reduced it by twice six to twelve. Check the final gap 66 − 54 = twelve.
This two-action example demands attention to the order and source of each movement, not just a remembered ratio formula.
Recover the common addition from the known ending
Suppose the starting counts are A = 36 and B = 60, and after the same unknown number is added to each, their ratio becomes 5:7. Let the amount added to each be k.
The equation 7(36 + k) = 5(60 + k) gives 252 + 7k = 300 + 5k, so 2k = 48 and k = 24.
Check by adding twenty-four to the original numbers, obtaining sixty and eighty-four. Their ratio simplifies to five to seven.
The problem asks for a change amount, not the earlier ratio unit. The original unit in 3:5 is twelve; the common addition twenty-four is twice that original part.
A tutor should ask the pupil to define the unknown before solving; a correct number attached to the wrong meaning is not a complete answer.
The two group shares can be expressed as percentages
Initially A:B = 3:5, so A represents three eighths of the combined collection and B five eighths. Three eighths is 37.5% of the original ninety-six counters.
After equal additions the ratio becomes 5:7, so A represents five twelfths of the new 144 counters, about 41.67% when rounded to two decimal places.
A’s absolute count rises from thirty-six to sixty, yet its fraction of the combined total changes by a smaller relative proportion because the whole also grew.
The percentage-point difference between five twelfths and three eighths is 5/12 − 3/8 = 10/24 − 9/24 = 1/24 of a whole, or approximately 4.17 percentage points.
Ask the pupil whether a question requests counters, a part-to-part ratio, a fraction of the total or a change in percentage points. These are distinct measures.
A ratio is not an average
The initial group counts are thirty-six and sixty. Their arithmetic mean is (36 + 60) ÷ 2 = 48 counters per group.
The ratio 3:5 expresses the relative counts, not the mean. The number forty-eight is neither one old ratio part nor the total across the two groups.
After each receives twenty-four, their mean increases to (60 + 84) ÷ 2 = 72 counters per group, while their ratio changes to 5:7.
A pupil who divides the total by the number of ratio parts to find a mean has confused two different kinds of ‘equal units’. Ratio units are modelled equal portions of the counts; an average divides the total by observations or groups.
Comparing the ideas helps a confident Primary 6 learner without pretending they are interchangeable procedures.
Predict whether the groups become more or less unequal proportionally
The original larger group B contains sixty while A has thirty-six. Their ratio B:A is sixty to thirty-six, or five to three, about 1.67.
After both receive twenty-four, B has eighty-four and A sixty. The ratio B:A is seven to five, or 1.4.
The numerical gap stays twenty-four, but the larger group is now only forty percent larger than A, rather than about sixty-seven percent larger before.
That movement toward a one-to-one ratio is expected when the same positive amount is added to both unequal positive groups.
An answer giving a more extreme ratio after equal additions should be inspected, since it may have used different additions or the wrong starting scale.
Name the counted system before applying an invariant
When both groups receive new counters from outside, the combined total grows. When counters move only between A and B, their combined total stays fixed.
When the same number is removed from each group to outside, the difference stays fixed but total decreases. When only one group changes, neither the original sum nor the original gap is generally conserved.
Draw arrows and a boundary enclosing A and B. Arrows across that boundary change the combined count; an arrow solely inside the boundary does not.
This representation prevents the unsafe shortcut that all changing-ratio problems preserve the total, or that all ratio parts remain numerically equal across stages.
Good tuition begins with the action and invariant before choosing bar diagrams or symbolic equations.
Five ratio misconceptions and their targeted repairs
A pupil who treats 3:5 as exactly three and five counters needs a unit-scale model. Another who adds twenty-four to the ratio numerals rather than the actual group counts is confusing abstract parts with physical quantities.
A third assumes the total remains ninety-six after two outside additions. That requires a system-boundary and source-of-counters explanation.
A fourth correctly forms the ratio equation but forgets to check the initial 3:5 condition. A fifth takes the final 5:7 unit size as the original part value without justification.
These errors differ in interpretation, arithmetic and verification. Record the first unsupported step in the uncorrected work rather than labelling every wrong answer careless.
A changed problem, such as the 2:5-to-1:2 example, should be attempted independently after the correction.
When weekday Primary 6 Mathematics tuition may help
A weekday lesson can build on a freshly taught school ratio problem while the pupil still remembers why they believed the original total or gap remained fixed.
Bring the first uncorrected attempt, not only a corrected model. The tutor should ask which quantities were added from outside and which groups were counted.
Consider school dismissal, meals, co-curricular commitments, homework, the real journey to a confirmed teaching venue and sufficient sleep. A mathematically useful session needs attention for more than a copied equation.
An illustrative tutorial could begin with the 3:5-to-5:7 problem, compare equal external additions with a transfer, and finish with the 2:5-to-1:2 example privately.
After several days, the child should reconstruct the fixed-difference reasoning without the original bar model visible.
When a weekend tutorial allows clearer bar modelling
A weekend lesson may give children time to lay out three and five equal part blocks, scale them to actual counts and add the same number of physical counters beside both bars.
That visual distinction helps learners who know equivalent ratios but cannot explain how the same addition changes the proportions.
Weekend sport, family visits, other classes and preparation for Monday can nevertheless reduce available energy. The day is useful when the learner can concentrate and recover.
Start with an individual first attempt, model only the missing relationship, then give a different scenario involving equal removals or an internal transfer.
An independent follow-up a few days later reveals whether the correct invariant is recognised without prompts.
How 3-pax tutoring can handle different ratio needs
In a group of three, one pupil may struggle to find the value of a ratio part, another may confuse an outside addition with a transfer, and a third may understand the story but lose track when solving the equation.
These students need different support: an equal-part bar, a source-and-destination arrow or a balance model for the unknown.
An illustrative ninety-minute session could include independent diagnostics, a shared 36-and-60 worked model, a contrast with equal removals, and private exit questions.
Peer discussion helps only when every child demonstrates their own mathematical reason. One confident classmate’s solution cannot establish mastery for the others.
Parents can ask what first hint was needed and whether it was still required on an unfamiliar example days later.
Four checkpoints for changing-ratio reasoning
Checkpoint one identifies the actual group quantities behind a ratio and distinguishes equal parts from physical item counts.
Checkpoint two classifies the change: equal external additions, equal removals, one-sided changes or internal transfers. State which sum or difference, if any, is invariant.
Checkpoint three forms a valid bar model or equation, solves the original counts and checks both original and final ratios.
Checkpoint four changes the story and asks for a missing addition, a reverse transfer or a new fraction of the whole. The learner should solve without a completed model.
These are flexible diagnostic stages, not a guarantee of four lessons or a particular PSLE grade.
Frequently asked questions about changing ratios
What were the original counts? In the main 3:5-to-5:7 story, A had thirty-six and B sixty counters.
What happens after twenty-four are added to each? The counts become sixty and eighty-four, a ratio of five to seven.
What stays fixed under equal additions? The difference of twenty-four counters. The combined total increases by forty-eight.
Is equal addition the same as an internal transfer? No. A transfer preserves the combined total while generally changing the gap.
Are old and new ratio units always equal in size? No. They happen to match in the main chosen example but differ in many others.
Which tuition day is best? Choose attentive teaching, targeted correction, delayed independent practice and adequate rest.
Does Li Hwan Walk name an eduKate classroom? No. Confirm the actual venue, fees and class availability directly.
MOE curriculum and connected Li Hwan Walk learning
The MOE Primary Mathematics syllabus updated October 2025 provides the official Primary 6 ratio scope. The figures and counter movements here are fictional original examples.
Read the eduKate Mathematics Learning Hub and Li Hwan Walk Primary 4 decimals guide for earlier numerical foundations. The Primary 3 budget guide provides a different multi-step context.
For a suitable class, arrange a parent–student consultation with eduKate Singapore. Bring a pupil’s first uncorrected ratio problem, current school course and feasible weekday/weekend windows. Confirm actual venue, group size, fees and travel.
Continue through Li Hwan Walk Mathematics: Primary 1: Picture Graphs · Primary 2: Length Comparisons · Primary 5: Rate Problems · PSLE: Working Backwards · Exam Prep: Error Diagnosis. These are separate grade-level learning guides.