Primary Mathematics: Invariants, Before-and-After and Unchanged Quantities | Worked Learning Guide

BTT Mathematics / Primary Mathematics Learning Hub / Unchanged Quantities

An invariant is a quantity or property that stays unchanged under a specified action. When counters move from one child to another, their combined total can stay fixed. When both children receive the same number of extra counters, their difference stays fixed. When both amounts are multiplied by the same positive factor, their ratio stays fixed.

Those are three different actions and three different preserved relationships. The most important question is not “Which invariant formula should I use?” It is “What exactly happened, and what does that action leave unchanged?” A quantity does not stay fixed merely because keeping it fixed would make the problem easier.

Consider forty-two counters with one child and eighteen with another. If six counters move from the first child to the second, the new counts are thirty-six and twenty-four. The total remains sixty, but the difference falls from twenty-four to twelve. One counter transferred changes both children’s amounts, so the gap changes by two counters per transfer.

This guide develops before-and-after records, internal transfers, equal additions and removals, common scaling, fixed-quantity ratio problems, area-preserving rearrangements and simple parity arguments. It provides a deeper companion to Word Problems, Bar Models and Checking, concentrating on the relationship that survives a change.

The MOE Primary Mathematics syllabus, page 11, identifies invariance as a mathematical big idea. The examples, problem families and 24 questions here are original teaching material. The word can be introduced through “what stays the same”; the later ratio and puzzle sections are not tasks every younger learner must complete.

Name the action · Transfers and fixed totals · Fixed differences · Changing ratios · Rearrangement and averages · Parity and impossibility · 24 questions · Worked solutions · Teaching and transfer

1. Name the action before naming the invariant

Write a before-and-after record for each quantity. If A gives five counters to B, A decreases by five and B increases by five. If both receive five from an outside box, both increase by five. These actions use the same number five but do not preserve the same relationship.

In the transfer, the combined total changes by minus five plus five, which is zero. In the two gifts, the combined total increases by ten. However, the difference between A and B stays the same under the gifts because both have moved upward by the same amount.

Decide what collection you are counting

If counters move between A and B, the A-and-B total is closed to outside change. If an outside teacher gives counters to each child, that two-child total is no longer fixed. Including the teacher’s supply would create a larger collection whose total might be preserved, but it is not the same total.

This boundary decision is ordinary reading expressed mathematically. Ask where the counters came from and where they went. An object crossing into or out of the named collection changes its total; an internal transfer does not, provided nothing is lost or added.

Only claim what the action preserves

Equal additions preserve a difference but usually change a ratio. Doubling both quantities preserves their ratio but doubles their difference. Cutting and rearranging pieces without overlap or loss preserves area but can change perimeter.

Therefore “the shape changed but something stayed the same” is not a complete solution. Name the object, the allowed action and the specific preserved property. This precision prevents a valid idea from being applied to the wrong situation.

Use a small numerical test to challenge a guess

Suppose a learner claims that adding the same number to both terms preserves a ratio. Test two and four, whose ratio is one to two. Add two to each to obtain four and six, whose ratio is two to three. The claim fails.

One counterexample is enough to disprove an “always” claim. A few successful examples, however, do not prove an unrestricted statement. After testing, explain the action itself: equal additions cancel in a subtraction, which is why the difference rather than the ratio is preserved.

2. Internal transfers preserve the combined total

Example A: Find the equal final amounts

One child has fifty-two counters and another has twenty-eight. Counters are transferred from the first to the second until the amounts are equal. Their combined total is eighty, and no counter enters or leaves the two-child collection. Each final amount must therefore be forty.

The first child gives twelve counters: fifty-two minus twelve is forty. The second receives the same twelve: twenty-eight plus twelve is forty. Reporting twenty-four, the original difference, would overshoot the equal point and reverse which child has more.

Why half the gap is transferred

For every counter passed from the larger collection to the smaller, the larger collection loses one and the smaller gains one. The difference shrinks by two. Closing a gap of twenty-four therefore requires twelve transfers.

The rule depends on the action. If an outside adult added counters only to the smaller collection, closing the same gap would require twenty-four counters. There would be no matching decrease on the larger side. The same original gap can therefore produce two different adjustment amounts.

Example B: A transfer to a stated difference

A has sixty-four cards and B has thirty-two. A gives some cards to B until A has eight more than B. The original gap is thirty-two and the required final gap is eight. The gap must shrink by twenty-four, so twelve cards are transferred.

The new counts are fifty-two and forty-four. Their total is still ninety-six, and their difference is eight. Both checks matter. The goal is not equalisation this time, so automatically halving the whole original gap would solve a different problem.

Example C: Reconstruct the state before a known transfer

After A gives nine cards to B, the final counts are thirty-one and thirty-seven. Restore A’s nine to obtain forty, and remove those nine from B to obtain twenty-eight. The original total is sixty-eight, matching the final total.

This example connects invariants with working backwards. Reversing each person’s change recovers the earlier quantities; the unchanged combined total provides an independent consistency check on the stage labels.

Whole-object equalisation may be impossible

If two children have thirty-one counters altogether and transfers must use whole counters, they cannot finish with equal whole-number amounts. Two equal integers have an even sum, but thirty-one is odd. They could finish fifteen and sixteen, not equal.

The preserved total therefore does more than find an answer. It can show that the requested outcome is impossible under the rules. Splitting a counter would change the allowed actions and the kind of quantities being counted.

3. Equal additions or removals preserve a difference

Suppose two ribbons measure twenty-seven and fifteen centimetres. Extend each by eight centimetres. Their new lengths are thirty-five and twenty-three. The difference is twelve both before and after because the same addition appears in each amount.

Removing the same length from both also preserves the difference, provided the removal is physically possible. A negative ribbon length is not an acceptable result. A general arithmetic relationship still needs the constraints of the objects in the question.

Example D: Find two amounts from the preserved difference

A has eighteen more counters than B. Both receive ten additional counters, and their new total is seventy-four. The difference remains eighteen. Remove that extra eighteen from the total to leave fifty-six split between two equal smaller parts. B’s final amount is twenty-eight, and A’s is forty-six.

Subtract the ten received by each to recover the originals: B had eighteen and A had thirty-six. Check the original difference, thirty-six minus eighteen equals eighteen, and the new total, forty-six plus twenty-eight equals seventy-four.

Age differences in an exact-year model

Two people have stated ages of sixteen and eleven years in a simplified problem. Four years later, they are twenty and fifteen, so the age gap remains five. Both ages increase by the same elapsed time.

Their age ratio changes from sixteen to eleven to four to three. A preserved difference does not imply a preserved ratio. Use the age information exactly as defined by the problem; rounded ages and birthday timing introduce additional interpretation in real situations.

Equal money spending is not the same as equal percentage spending

A has ninety dollars and B has sixty. If each spends fifteen dollars, they have seventy-five and forty-five, retaining the thirty-dollar gap. If instead each spends one-third of their own money, they retain sixty and forty, keeping the ratio three to two but reducing the gap to twenty.

In the first action, the same absolute amount is subtracted. In the second, each amount is multiplied by the same fraction, two-thirds. Those are different operations. A sentence such as “both spend the same” is incomplete unless it says the same money amount or the same fraction.

Use the invariant to choose which quantity to normalise

In a ratio problem, the ratio’s unit difference may represent a fixed real difference before and after an equal addition. The total number of ratio units might change. Normalise the differences, not the totals, when the action preserves the difference.

This decision belongs before any least-common-multiple calculation or bar-unit scaling. Choosing a common unit is helpful only when it aligns quantities that genuinely stay equal across the two stages.

4. Changing ratios: find the quantity that connects the stages

A ratio describes a relationship at one stage. A simplified ratio unit before a change need not have the same actual value as a simplified ratio unit afterwards. The reliable connection is the quantity that remains fixed: a total, a difference or one unchanged person’s amount.

Example E: Fixed total, changing ratio

A and B have counters in the ratio two to five, with seventy counters altogether. Before any transfer, one ratio unit is ten counters. A has twenty and B fifty. B transfers counters to A until the ratio becomes four to three.

The total remains seventy. In this particular example, the new ratio also has seven total units, so one new unit is ten. A now has forty and B thirty. Twenty counters moved from B to A. Check both before and after totals and the new ratio.

Example F: The number of ratio units changes

A and B begin with ninety-six counters in ratio five to three. Their amounts are sixty and thirty-six. A gives counters to B until the ratio is one to one. The final forty-eight each comes from dividing the fixed ninety-six by two, not by the original eight ratio units.

The old ratio unit was twelve counters; the new simplified ratio unit is forty-eight. The transfer is twelve. Treating the two kinds of units as automatically equal would make the before-and-after comparison meaningless. Labels such as “before units” and “after units” can prevent that mistake.

Example G: Only one quantity changes

The ratio of red to blue counters is three to four. Fifteen red counters are added, and the ratio becomes six to four. Blue does not change, and it occupies four units in both descriptions. The before and after units can therefore be taken at the same scale.

Red increases by three units, which represent fifteen counters. One unit is five. Originally there were fifteen red and twenty blue counters; afterwards there are thirty red and twenty blue. The total is not fixed because red counters entered from outside.

Example H: The unchanged quantity needs matching units

Red to blue is initially one to two. Ten red counters are added and the ratio becomes three to four. Blue is unchanged, but its ratio terms are two and four. Rewrite the original ratio as two to four so that the blue terms match.

At this matched scale, red rises from two units to three while blue stays at four. One unit represents the ten added. The original amounts were twenty red and forty blue; the final amounts are thirty red and forty blue. The original one-unit label corresponded to twenty counters, showing why direct subtraction of unmatched ratio terms would fail.

Example I: Fixed difference after equal additions

A to B is initially five to two. Both receive twelve counters, and the ratio becomes seven to four. The difference is three units in both ratios. Because the real difference is preserved by equal additions, those units can be aligned.

A rises from five units to seven, an increase of two units representing twelve. One unit is six. The original counts are thirty and twelve; the final counts are forty-two and twenty-four. Check: both received twelve, the original ratio is five to two and the final ratio is seven to four.

Example J: Match different unit gaps

A to B is initially three to one. Both receive eight counters, after which the ratio is five to three. The real difference stays fixed. The ratio gaps are two units before and two after, so one aligned unit is four counters: each person’s ratio term increases by two units.

Originally A has twelve and B four. Afterwards they have twenty and twelve. A more complicated pair of ratios may have different unit gaps; then multiply the ratio descriptions so that the gaps match before treating their units as equal. The preserved real difference justifies that scaling.

A method-selection question

Ask which action occurred. Internal transfer suggests a fixed combined total. The same absolute addition to both suggests a fixed difference. A change only to A suggests using B’s unchanged amount. Common multiplication suggests a preserved ratio. These are conditional connections, not rules triggered by isolated keywords.

When both amounts change by unrelated quantities, none of those simple invariants may apply. Use the stated changes directly in a before-and-after table or solve the equations the story establishes. Do not invent an invariant to avoid handling the actual conditions.

5. The same idea appears in shapes, measures and averages

Cutting and rearranging preserves area under stated conditions

Cut a two-by-two square into two one-by-two rectangles. Put those rectangles end to end to make a one-by-four rectangle. No material is added, lost or overlapped, so the area remains four square units.

The perimeter changes. The square boundary is eight units, while the one-by-four rectangle boundary is ten. Rearrangement has preserved area, not every property of the figure. A missing-piece or overlap arrangement would require a different area calculation.

Moving unit cubes preserves their total volume

Eight unit cubes can form a two-by-two-by-two block or a one-by-one-by-eight row. Both contain eight cubic units of material. Their exposed surfaces differ because different faces are hidden where cubes touch.

This provides a spatial comparison with transfers: the same pieces are redistributed, so their combined amount stays fixed. Do not infer that the outside shape, surface area or dimensions also stay fixed. Each property must be tested separately.

A unit conversion preserves the measured quantity

One and a half metres is one hundred fifty centimetres. The numerical label changes while the length stays the same. What is invariant is the physical length, not the number written before the unit.

This is why an answer to “what stayed unchanged?” must include a quantity, not only a numeral. Moving to a smaller measuring unit generally increases the numerical count required to describe the same amount.

Redistributing a fixed total across a fixed number preserves the mean

Four values are eight, twelve, sixteen and twenty-four. Their total is sixty and their mean is fifteen. Move four units from the last value to the first to obtain twelve, twelve, sixteen and twenty. The total and the number of values remain fixed, so the mean stays fifteen.

The distribution has changed. The largest and smallest values are now closer together. An unchanged mean therefore does not imply unchanged individual values or unchanged spread. This distinction connects invariance with the data and average guide.

If the number of values changes, reconsider the mean

Adding a new observation changes the divisor in an average calculation. A preserved old total alone does not preserve the mean after a zero is added. Conversely, adding a new value equal to the existing mean preserves the mean while increasing the total.

The point is not to memorise another catalogue of tricks. Identify the formula’s relevant quantities and check which ones the action changes. For a mean, both total and number of observations matter.

6. A preserved property can prove that a target is impossible

Parity means whether a whole number is odd or even. Removing two counters from an odd collection leaves an odd collection. Repeating that action cannot produce zero, which is even. Starting with seventeen counters and removing pairs will finish with one if removals continue as far as possible.

This argument does not need every intermediate total, though listing seventeen, fifteen, thirteen and so on can make the pattern visible. The proof is that every allowed action preserves oddness while the proposed target has the wrong parity.

The action must preserve the property every time

Removing three counters does not preserve parity: odd minus odd is even, and even minus odd is odd. A puzzle allowing removals of either two or three cannot be analysed as though all moves remove pairs. Check every allowed move before using a parity argument.

A single permitted move that changes the claimed invariant is enough to invalidate the proof. It does not prove the target is reachable; it shows that this particular impossibility argument is not justified.

Matching an invariant does not guarantee reachability

Start at two and allow only additions of four. The reachable values are two, six, ten, fourteen and so on. Every value is even, but four is not reachable. It has the right parity but does not belong to the more restricted sequence generated by the action.

An invariant can rule some targets out without ruling all remaining targets in. To show that a target is reachable, construct a valid sequence of moves or give another argument that establishes existence. Distinguish “not yet ruled out” from “proved possible.”

Use small cases to discover, then explain

A learner may notice a preserved total, gap or parity after trying several examples. That observation suggests a conjecture. The next step is to explain why every allowed action preserves it, not simply to add more examples and declare the pattern universal.

For Primary learners, a verbal explanation with counters is often sufficient: taking one from here and adding the same one there leaves the combined collection unchanged. Formal symbolism can follow later when it clarifies rather than conceals the action.

7. Practice: 24 questions

State the action, the quantity or property that stays fixed, and any other condition needed. All counters and cards are whole objects. Money and measurement examples are hypothetical. Keep worked solutions covered. Use the ratio and parity questions only when their prerequisite ideas are familiar.

Questions 1–8: Totals and differences

1. A has thirty-four counters and B eighteen. A gives seven to B. Find the final counts, total and difference.

2. A has forty-six counters and B twenty-two. How many must A give B so that both have equal amounts?

3. A has twenty-six more counters than B. A gives eight to B. How many more does A now have?

4. A has eighteen more counters than B. Both receive eleven from outside. What is their new difference?

5. Two ribbons are 72 cm and 48 cm long. Fifteen centimetres is removed from each. Find both new lengths and their difference.

6. Two quantities are twenty-four and thirty-six. Both double. Find the new amounts, their simplified ratio and their difference.

7. A and B have eighty counters in ratio five to three. How many must A give B to make the amounts equal?

8. A and B have sixty-four counters in ratio three to five. B gives some to A until the ratio is five to three. How many move?

Questions 9–16: Choose the preserved relationship

9. In an exact-year age model, Maya is fourteen and Leo nine. Find their ages and age difference four years later.

10. Ribbons of 9 cm and 15 cm are each extended by 6 cm. Find the new lengths, difference and simplified ratio.

11. Red to blue counters is initially two to three. Twelve red counters are added, producing ratio four to three. Find the original counts.

12. Red to blue is initially two to three. Eight red counters are added, producing ratio four to five. Find the original counts.

13. A to B is initially five to three. Both receive eighteen counters, producing ratio seven to five. Find the original counts.

14. A to B is initially four to one. Both receive fifteen counters, producing ratio three to two. Find the original counts.

15. A has $80 and B $50. Each spends 20% of their own money. Find their remaining amounts, ratio and difference.

16. Two containers hold 1.8 L of water altogether. Four hundred millilitres is poured from one to the other with no spill. What is the combined amount afterwards, in millilitres?

Questions 17–24: Shapes, parity and final checks

17. A 2-by-2 square is cut into two 1-by-2 rectangles and rearranged into a 1-by-4 rectangle without loss or overlap. Compare area and perimeter before and after.

18. Eight unit cubes are rearranged from a 2-by-2-by-2 block into a 1-by-1-by-8 row. What volume stays fixed? Must the external surface area also stay fixed?

19. Start with fifteen counters. The only move removes exactly two whole counters. Can the collection reach zero?

20. Start at four. Each move adds six. Can twenty-two be reached? Give a valid sequence if possible.

21. Start at two. Each move adds four. A learner says four is reachable because both numbers are even. Is that reasoning sufficient?

22. Four values are ten, fourteen, eighteen and twenty-two. Move three units from the last value to the first. Find the new values and compare the means.

23. A and B have sixty counters in ratio three to two. An outside person gives five counters to each. Does the total stay fixed? Does the ratio stay fixed? Find the final amounts.

24. A and B have ninety counters in ratio two to three. B transfers counters to A until the ratio becomes four to five. Find the transfer and explain why the before and after ratio units need not have the same value.

8. Worked solutions

Solutions 1–8

1. A has 27, B has 25; total 52; difference 2. Subtract seven from A and add the same seven to B. The total remains thirty-four plus eighteen, or fifty-two. The original gap sixteen shrinks by fourteen because the transfer changes both amounts.

2. 12 counters. Their total is sixty-eight, so equal final amounts are thirty-four each. A gives 46 − 34 = 12, and B receives 22 + 12 = 34. Equivalently, halve the original gap of twenty-four because each transfer reduces it by two.

3. 10 more. A’s eight-counter loss and B’s eight-counter gain reduce the gap by sixteen. The new difference is twenty-six minus sixteen, or ten. Subtracting only eight would account for only one side of the transfer.

4. 18. Both quantities increase by eleven, so their difference is preserved. Their combined total would rise by twenty-two. The invariant is the difference, not the total, because the counters enter from outside.

5. 57 cm and 33 cm; difference 24 cm. Subtract fifteen from each length. Their original difference was seventy-two minus forty-eight, also twenty-four. Equal feasible removals preserve the gap.

6. 48 and 72; ratio 2:3; difference 24. Common doubling preserves the simplified ratio. The difference doubles from twelve to twenty-four rather than staying fixed. The same action need not preserve all relationships.

7. 10 counters. Eight ratio units represent eighty, so A starts with fifty and B thirty. Equal final amounts are forty each. A transfers ten, preserving the total and closing the twenty-counter gap.

8. 16 counters. Initially one unit is eight, giving A twenty-four and B forty. With the total unchanged, the final ratio gives A forty and B twenty-four. Sixteen moves from B to A. The final amounts satisfy the reversed ratio without changing the total sixty-four.

Solutions 9–16

9. Maya 18, Leo 13; difference 5 years. Both increase by four in the stated model. Equal increases preserve their five-year difference. Their ratio changes, so it should not be treated as an invariant.

10. 15 cm and 21 cm; difference 6 cm; ratio 5:7. Equal extensions preserve the original six-centimetre difference. The original ratio three to five changes to five to seven, showing that equal additions do not generally preserve a ratio.

11. 12 red and 18 blue. Blue is unchanged and has three units in both ratios. Red rises from two aligned units to four, so two units represent twelve and one unit represents six. Original red is twelve and blue eighteen; afterwards red is twenty-four.

12. 40 red and 60 blue. Match the unchanged blue quantity: rewrite 2:3 as 10:15 and 4:5 as 12:15. Red rises by two aligned units, representing eight counters. One unit is four. Original counts are forty and sixty; final red forty-eight gives ratio four to five with sixty blue.

13. A 45, B 27. Equal additions preserve the difference. Both ratios have a gap of two units, allowing their units to align. Each quantity rises by two units, representing eighteen, so one unit is nine. Add eighteen to obtain sixty-three and forty-five, whose ratio is seven to five.

14. A 12, B 3. The original ratio gap is three units. Rewrite the final ratio 3:2 as 9:6, also with gap three. Each quantity rises by five aligned units, representing fifteen, so one unit is three. Original counts are twelve and three; adding fifteen gives twenty-seven and eighteen, ratio three to two.

15. $64 and $40; ratio 8:5; difference $24. Each retains eighty per cent of their own amount. Multiplying both by the same positive fraction preserves the ratio eight to five. The original thirty-dollar gap is also multiplied by 0.8, giving twenty-four dollars.

16. 1,800 ml. The water moves inside the two-container collection and none is lost. The combined amount remains 1.8 litres. Convert the unit using one thousand millilitres per litre. Individual container amounts cannot be found without more information.

Solutions 17–24

17. Area stays 4 square units; perimeter changes from 8 to 10 units. Both arrangements contain the same two pieces without loss or overlap. The square perimeter is four times two. The rectangle perimeter is twice one plus twice four. The rearrangement preserves material area, not boundary length.

18. Volume stays 8 cubic units; surface area need not stay fixed. Both arrangements contain eight unit cubes. The compact block has surface area twenty-four square units, while the long row has thirty-four. Different touching faces are hidden inside, changing the exposed surface.

19. No. Fifteen is odd, and removing two preserves oddness. Zero is even, so it cannot be reached. The sequence eventually reaches one, after which another permitted removal is impossible.

20. Yes: 4 → 10 → 16 → 22. Three permitted additions of six reach the target. Both parity and a valid construction are consistent. The explicit sequence establishes reachability rather than merely failing to rule it out.

21. No; four is not reachable. The permitted sequence is two, six, ten, fourteen and so on. Adding four always jumps over four on the first move and then continues increasing. Shared evenness is a necessary property of reachable values here, but it is not sufficient.

22. 13, 14, 18, 19; both means are 16. Three units move from the last value to the first. The total remains sixty-four and there are still four values, so the mean remains sixty-four divided by four. The individual values and their spread have changed.

23. Final amounts 41 and 29; neither total nor ratio stays fixed. Originally the counts are thirty-six and twenty-four. The two external gifts raise the total from sixty to seventy. The ratio becomes forty-one to twenty-nine, not three to two. The difference twelve is preserved.

24. Four counters move from B to A. Before the transfer, five units share ninety, so one old unit is eighteen and the counts are thirty-six and fifty-four. Afterward nine units share the same ninety, so one new unit is ten and the counts are forty and fifty. The total is invariant; the size of a simplified ratio unit is not.

9. Help the learner find a relationship that survives the action

Begin with two visible collections and make one change. Ask what happened to each collection, then compare total, difference and ratio where appropriate. The learner should identify the invariant from the action rather than choose one first and force the action to fit.

When a transfer is treated as a one-sided change

Use one counter and physically move it from the larger group to the smaller. Record both new counts. Ask how the gap changed. Repeat once and compare. The two-unit change per transfer is visible in the paired loss and gain.

Then change the action: keep the larger collection unchanged and add one counter from outside to the smaller. The gap now changes by one. This paired example shows why “halve the difference” requires an internal transfer, not merely a wish to make two amounts equal.

When the wrong quantity is held fixed in a ratio problem

Ask where objects entered or left. If only red counters were added, blue stayed fixed but the total increased. If counters changed owners inside the collection, the total stayed fixed but neither person’s amount did. The before-and-after table should make that distinction explicit.

Only then align ratio descriptions. Matching the total units when the total changed creates a false scale. Matching an unchanged person’s ratio term is valid because it refers to the same actual amount at both stages.

When before and after units are silently equated

Use a known total with two different ratio sums. Ninety counters in ratio two to three uses eighteen counters per unit. The same ninety in ratio four to five uses ten per unit. Both descriptions are valid, but their simplified units have different sizes.

Ask the learner to calculate or normalise the units explicitly before comparing them. Writing “old unit” and “new unit” is a temporary support that can prevent a large class of otherwise plausible mistakes.

When an invariant is mistaken for a complete solution

An unchanged total may narrow the possible final amounts without determining them. An evenness argument may rule out odd targets while leaving some even targets unreachable. Ask what additional condition or construction completes the argument.

This helps distinguish three statements: a target is impossible; a target is not excluded by this test; a target has been reached by a valid process. They are different conclusions and should not be collapsed into one.

A suggested teaching progression

Begin with an internal transfer, then compare it with equal outside additions. Introduce a simple fixed-total ratio problem once the actions are secure. Follow with a fixed-one-quantity ratio and, later, a fixed-difference ratio. End with a small puzzle in which an invariant rules out a target.

Use the learner’s explanations to decide when to move on. A correct answer produced after the adult names the invariant shows execution with support. A new problem in which the learner identifies the preserved quantity independently gives different evidence. Keep those observations separate in any progress record.

Return the mathematics to the situation

For transfers, verify that the giver loses exactly what the receiver gains. For equal additions, verify that both increases are the stated amount. For ratios, simplify the actual before and after quantities separately. For rearrangements, account for every piece without loss or overlap.

A final number is trustworthy when it satisfies the action as well as the target condition. The invariant supplies a stable relationship through the change; the return check confirms that the calculated change could really be the one described.

Continue through the problem-solving collection

For restoring an earlier stage, use Working Backwards and Reverse Processes. For a fixed number of items with two different contributions, use The Assumption Method and Guess-and-Check. For all permitted arrangements or a complete search after an invariant narrows the cases, use Systematic Listing and Logical Counting.

Return to Ratio, Rate and Percentage for multiplicative comparison, or to the BTT Primary Mathematics Learning Hub to choose another topic.

Original explanations and 24 original questions with separate worked solutions. Conditions such as no loss, equal additions and whole-object transfers are part of each mathematical model. Curriculum reference accessed 6 September 2026. No examination result is promised.

Release Closure: From Preserved Quantity to Independent Equation

The final readiness check for before-and-after Mathematics should force the learner to name the invariant before writing an equation.

Example: two boxes contain 96 counters altogether. Twelve counters are moved from Box A to Box B. The combined total remains 96 because nothing enters or leaves the two-box system. If the final amounts are equal, each box has:

[ 96/2=48. ]

Working backwards, Box A must originally have had 60 and Box B 36. The forward check restores the final equal state:

[ 60-12=48,qquad36+12=48. ]

This release task contains the complete chain:

action → preserved relationship → equation or equalisation → original state → forward check.

Use a changed problem after feedback. If the learner can repeat the invariant reasoning with different amounts, a different transfer and a different target, the method has transferred. If the learner can only reproduce the previous numbers, the understanding is still example-bound.

Final MA-S1-020 standard: identify what changes, identify what stays fixed, convert the invariant into a usable equation, solve, and reconstruct the original situation independently.

Final MA-S1-020 release note: a learner should be able to face a changed before-and-after problem, state the action, name the invariant, form the corresponding equation and reconstruct the original state without copying a previous model. If the action changes the very quantity the learner is trying to preserve, they should reject that invariant rather than force the method. The final check is a forward reconstruction that returns to the stated after-condition.

Release complete: the learner can identify what changes, identify what is preserved, turn the invariant into an equation, solve, reconstruct the original state and verify the after-condition on a fresh problem without copying the earlier model.