To find an original amount when a percentage of it is known, identify which quantity is 100%, express the known amount as its correct percentage, then scale back to the whole. If 30% of a fictional Punggol reading club’s book budget is $42, the entire budget is $42 ÷ 0.30 = $140. You are not ‘undoing 30 by subtracting 30’; you are reconstructing the 100% base.
Reverse percentage problems feel harder than ordinary ‘find 20% of 200’ questions because the whole is hidden. The printed number might be only 30% of the whole, 80% after a discount, or 115% after an increase. Before any calculation, the pupil has to identify which of those percentages the given amount represents.
This Primary 6 guide develops that exact decision through twelve original worked examples, before-and-after bars, explanations of why 100% must have a reference, parent-friendly error diagnosis, a four-week practice plan and a few deliberately labelled enrichment tasks. The prices, budgets and Punggol scenarios are fictional and are not actual offers or local price data.
Where reverse percentages sit in Singapore’s 2026 PSLE curriculum
The MOE Primary Mathematics syllabus, 2021 edition updated October 2025, specifies at Primary 6 finding the whole given a part and the percentage and percentage increase or decrease. This is a better guide to scope than memorising an old tuition worksheet index. The pupil is expected to understand what the percentage refers to, not merely apply a single shopping-discount formula.
For Standard PSLE Mathematics 0008, the examination format was revised from 2026. Paper 1 lasts 1 hour 10 minutes, is worth 50 marks and does not allow a calculator. Paper 2 lasts 1 hour 20 minutes, is also worth 50 marks and allows a calculator. A pupil who can estimate the answer and justify the base before computing is better prepared for both forms of assessment.
Foundation Mathematics has a different syllabus and paper arrangement. Use the requirements of the registered paper. The examples here are for the Standard Primary 6 learning outcome and are original practice, not reproduced SEAB questions or a forecast of any later PSLE sitting.
The sentence to ask before touching a calculator
Ask ‘What amount is 100% in this story?’ If a bicycle helmet originally costs $100 and is discounted by 20%, the original $100 is 100%, the discount is 20%, and the sale price is 80%. If a club originally has 200 members and grows by 15%, the original 200 is 100% and the new membership is 115%. Those are different relationships, even though both contain the word ‘percentage’.
The base is not necessarily the largest printed number. In the helmet story, 100% is more than the sale price. In the growing club, 100% is less than the new membership. Pupils who interpret 100% as ‘the biggest number’ often get lost as soon as a question discusses an increase.
A reliable reading routine labels original, change and final. Write the percentages before the dollars or people. When a question says ‘after a 20% reduction’, write ‘final = 80% of original’ rather than subtracting 20% from the final figure.
Three common models, three different questions
| Story | What the given amount represents | Set up for the original 100% |
| 42 dollars is 30% of a budget | 30% of original | 100% = 42 × 100/30 |
| Sale price 96 dollars after 20% off | 80% of original | 100% = 96 × 100/80 |
| New total 230 after 15% rise | 115% of original | 100% = 230 × 100/115 |
| After spending 40%, 72 dollars remains | 60% of original | 100% = 72 × 100/60 |
The same arithmetic structure is used in each row, but the percentage assigned to the given amount is different. This is why formula memorisation alone is fragile. The pupil must interpret the relationship first, then choose a proportion or unitary method.
Before-and-after bars: why 80% is not 20%
For a 20% discount, draw a bar of ten equal parts for the original 100%. Shade two parts as the reduction and leave eight as the sale price. If the eight remaining parts correspond to $96, each part is $12. Ten original parts are therefore $120. The bar makes the difference between the lost 20% and the remaining 80% visible.
For an increase, the final bar is longer than the original. Start with the 100% original, then attach an additional 15% segment. The total is 115% of the original. If 115% corresponds to 230, each one-percent unit is 2 and the original 100% is 200. A pupil who can explain both drawings without multiplying first has understood the percentage base.
Once the picture is stable, it can be replaced by fraction or decimal language: 80% = 4/5, so a sale price of $96 is four-fifths of the original and the whole is 96 × 5/4 = 120. This gives the pupil a second method for checking the answer.
Twelve original reverse-percentage worked examples
Worked example 1: the whole given its part
Question. A fictional family reading budget uses $42 for new books. This is 30% of the whole budget. What is the budget?
Solution. 30% = $42, so 10% = $14 and 100% = 10 × $14 = $140.
What to notice. Working from 10% is convenient because 30% divides into three equal ten-percent parts.
Verify. 30% of $140 = $42.
Worked example 2: the original price after a discount
Question. A fictional learning kit sells for $96 after a 20% discount. Find its original price.
Solution. 100% − 20% = 80%. Thus 80% = $96. One percent = $1.20; original = $120.
What to notice. The $96 is the remaining 80%, not the removed 20%.
Verify. 20% of $120 is $24; $120 − $24 = $96.
Worked example 3: recover the number before growth
Question. A hypothetical group now has 230 members after a 15% increase. How many were there before?
Solution. New total = 115% of original. 115% = 230, so 1% = 2; original = 200 members.
What to notice. The final amount can be 115%, so 100% is not necessarily the larger number.
Verify. 15% of 200 = 30; 200 + 30 = 230.
Worked example 4: interpret a published fraction-like statement
Question. Seventy reading cards are 35% of a collection. Find the number of cards in the full collection.
Solution. 35% = 70, so 5% = 10 and 100% = 200. Total = 200 cards.
What to notice. The whole is the entire collection. The part given is 70; the rest need not be separately known.
Verify. 35% of 200 = 70 cards.
Worked example 5: a total that is 125%
Question. An activity initially had some participants. After registration, there are 500, which is 125% of the initial figure. Find the initial figure.
Solution. 125% = 500. As 125% = 5/4, 100% = 500 × 4/5 = 400 participants.
What to notice. Think of 125% as five quarter-parts and the original as four.
Verify. 25% of 400 is 100; 400 + 100 = 500.
Worked example 6: one-quarter reduction
Question. A stack contains 180 booklets after one-quarter of the original stack was removed. How many booklets were there originally?
Solution. One-quarter is 25%. Remaining = 75% = 3/4. Original = 180 × 4/3 = 240 booklets.
What to notice. Fraction and percentage statements can express the same relationship; label what remains.
Verify. 25% of 240 = 60, and 240 − 60 = 180.
Worked example 7: finding 100% from 12%
Question. Twelve percent of a fictional club’s savings is $36. What is the total savings amount?
Solution. 12% = $36, so 1% = $3. Hence 100% = $300.
What to notice. The unitary method is clear when the percentage divides the amount exactly.
Verify. 12% of $300 = $36.
Worked example 8: percentage increase is based on the original
Question. A count rises from 160 to 184. Find the percentage increase.
Solution. Increase = 184 − 160 = 24. Percentage increase = 24/160 × 100% = 15%.
What to notice. The denominator is 160, the original count, not the new count of 184.
Verify. 15% of 160 = 24, and 160 + 24 = 184.
Worked example 9: recover the whole from what remains
Question. After a pupil spends 40% of fictional savings, $72 remains. What was the original sum?
Solution. Remaining = 60% of original. 60% = 72, so 10% = 12 and 100% = $120.
What to notice. The word ‘remains’ changes the given percentage from 40 to 60.
Verify. 40% of $120 is $48; $120 − $48 = $72.
Worked example 10: percentage of the remainder: guided extension
Question. A fictional student starts with $400, spends 25%, then spends 20% of what remains. Find how much was spent altogether and what is left.
Solution. First spending = 25% of $400 = $100; remaining = $300. Second spending = 20% of $300 = $60. Total spent = $160; remaining = $240.
What to notice. The second 20% is based on $300, not $400. This is a useful optional transfer task when the basic whole-finding idea is secure.
Verify. $100 + $60 + $240 = $400.
Worked example 11: discount described without a percentage symbol on the final line
Question. A fictional bag costs $171 after 10% off. Find the undiscounted price.
Solution. After 10% off, 90% remains. 90% = $171; 10% = $19; 100% = $190.
What to notice. Do not add 10% of the sale price; that would calculate 10% of the wrong base.
Verify. 10% of $190 is $19; $190 − $19 = $171.
Worked example 12: a before-and-after amount over 100%
Question. A school project collected 360 counters, which was 20% more than its original target. Find the original target.
Solution. Final = 120% of original. 120% = 360, so 20% = 60 and 100% = 300 counters.
What to notice. The final number represents six equal 20% parts; five parts make the original.
Verify. 20% of 300 = 60, and 300 + 60 = 360.
Why ‘divide by 0.8’ works—and when it does not
If $96 is 80% of a whole, the equation is 0.8 × whole = 96. Division by 0.8 gives the whole: 96 ÷ 0.8 = 120. That is valid because 0.8 refers to the fraction of the original that remains. If you mistakenly divide the same $96 by 0.2, you are pretending the sale price is the discount itself.
A useful check is to ask whether the result should be larger or smaller than the given. After a discount, the original should be larger than the sale price. After an increase, the original should be smaller than the final amount. This does not replace calculation, but it catches a strikingly common type of wrong-base error.
For a 20% rise, simply dividing by 1.2 works when the given is the new 120% amount. If the problem instead states that the increase amount is $24, then $24 is only 20%, and dividing by 1.2 would be wrong. What is printed matters less than what the amount represents.
Eight recurring errors and their likely causes
- Dividing by the announced discount instead of the remaining percentage. Mark the new price as 80% after a 20% discount.
- Calculating 20% of the final price to undo a 20% reduction. The final price is a different base; reconstruct the original 100%.
- Treating 115% as 15% after an increase. Draw an original 100% bar plus the added 15%.
- Adding a percentage to a dollar amount without calculating the percentage of a base. Convert the rate into a value first.
- Using the new number in the denominator of percentage change. Circle the original amount as the reference.
- Applying two successive percentages to the same whole when the second refers to what remains. Write the new base before the second step.
- Trusting calculator output without size sense. Predict larger or smaller before typing.
- Confusing a unitary method with a formula. Ask the child to explain what ‘one percent’ stands for in this story.
A five-question diagnostic for Punggol parents
Keep the test short enough for the pupil to explain every decision. Ask: 10% of a whole is 7; what is 100%? A price of $80 is after a 20% discount; what percentage does $80 represent? A club grows by 10%; is the new number 90%, 100% or 110% of the old? A count increases from 50 to 60; is the percentage increase based on 50 or 60? Finally, ask why a 20% reduction followed by a 20% rise does not necessarily return to the original amount.
| Response | What it reveals | Immediate next activity |
| Cannot find 100% from 10% | Unitary concept is weak | Use ten equal boxes and one visible portion. |
| Says $80 is the 20% discount | Part and remainder confused | Colour the removed 20% separately from the remaining 80%. |
| Says a grown total is 90% | Direction of change unstable | Draw before-and-after bars for rise and fall. |
| Uses 60 as denominator for 50 to 60 | Wrong percentage base | Mark ‘compared with the original 50’ in the sentence. |
| Thinks 20% down then 20% up cancels | Base changes not tracked | Try 100 → 80 → 96 and explain each new base. |
A child may know how to press percentage keys but fail all five reasoning prompts. The correction then should begin with language and diagrams, not bigger calculators or longer worksheets.
What four careful weeks of teaching might achieve
| Week | Core question | Practice and evidence |
| 1 | What is 100% and what does each part represent? | Use fraction bars, 10%, 25%, 50% and exact unitary examples. |
| 2 | How do I recover the whole from 30%, 60% or 80%? | Mix part-given and remainder-given questions; explain each denominator. |
| 3 | What changes after a rise or reduction? | Use 115%, 125%, 90%, 75% models and two independent checks. |
| 4 | Can I solve an unfamiliar story without being told the method? | Mix discount, budget, membership and saving contexts under gentle time limits. |
If a pupil still struggles with fractions such as three-quarters or with division, repair that prerequisite before adding further percentage stories. Four weeks is a useful review cycle, not a guaranteed time to a particular grade. The goal is visible independence: the child names the base, writes the relationship and checks the reasonableness unaided.
Paper strategy: do not confuse permission to calculate with permission to guess
Paper 1 is non-calculator and accounts for half of the revised Standard Mathematics total. Learn friendly relationships: 10% as one tenth, 25% as one quarter, 50% as one half, 75% as three quarters and 125% as five quarters. These provide efficient mental routes without sacrificing meaning.
In Paper 2, a calculator is allowed, yet the story may be harder than the arithmetic. The pupil should still show why 80%, not 20%, represents the sale price, write enough working to make the method understandable, and label units where appropriate. The official SEAB examination document defines the question types; tutors should not imply these particular contexts will appear.
A simple reliability ritual is to predict the scale, compute, and then apply the story in reverse. If you recover a $120 original after a 20% discount, calculate 20% of $120 and verify that subtracting it leaves $96. This habit works across subjects because it checks whether the mathematics actually answers the question.
Choosing support: target the faulty comparison, not the latest score
When a child loses marks on percentage questions, a tutor should distinguish arithmetic slips, misunderstanding of percent-of, wrong starting base, misreading of ‘remaining’, and difficulty transferring to a changed story. These produce similar final wrong answers but need different teaching responses. The useful evidence is in the working line where the base was chosen.
In a small-group Mathematics lesson, the tutor can begin with one clear demonstration and then give pupils different tasks: one learner constructs a 100% bar, another practises fraction equivalents, and another explains percentage change in a new context. Each should finish with an independent unseen question and a delayed review to test retention.
For families in Punggol, school pace, other homework, CCAs and tiredness should influence how much practice is realistic. Additional tuition is not automatically necessary if the child is secure and learning independently. If support is needed, ask how the tutor will demonstrate improvement beyond simply completing more worksheets.
Extended reverse-percentage workbook: the original amount must be identified
The first twelve problems establish a useful question: what quantity is 100%? These additional original exercises deliberately change the way the current amount is described. Sometimes the problem gives a percentage part directly, sometimes a discounted remainder, and sometimes a new value after an increase. The arithmetic becomes reliable when a pupil labels the original whole before calculating.
For the revised Primary 6 syllabus, finding the whole given a part and its percentage is a core percentage relationship; percentage increase and decrease are also specified. The learner should not use the latest or smaller number as the denominator merely because it appears nearer to the question. A percentage change is ordinarily calculated against the original quantity unless the question explicitly defines a different reference.
The distinction becomes visible with a $100 starting price. A 20% increase produces $120. A 20% decrease from the new $120 produces $96, not $100, because the second calculation uses a different 100%. Reversing a 20% increase from $120 requires division by 1.20, not subtraction of 20% of the new price. Pupils who can explain the changed base understand why the two operations are not simple opposites.
Use bar models and a 100-part diagram whenever the language is confusing. If a product is reduced by 25%, the sale price is 75% of the original. If an amount increases by 25%, the new amount is 125% of the original. The direction and base determine whether the multiplier is below or above one. The same reasoning can be expressed as 75/100 and 125/100 without calculators.
Additional worked percentage example 13 — one quarter given
Question. 25% of an imaginary amount is $45. Find the full amount.
Working. One quarter equals $45, so four quarters equal $180.
Why it works. The $45 is only 25%, not the original 100%.
Check. 25% of $180 equals $45.
Additional worked percentage example 14 — recover a discounted price
Question. A fictional item sells for $64 after a 20% discount. Find its original price.
Working. The sale price is 80% of original. Original = 64 ÷ 0.8 = $80.
Why it works. Discount percentage refers to the original price, not to the sale price.
Check. 20% of $80 is $16; $80 − $16 = $64.
Additional worked percentage example 15 — reverse a ten-percent increase
Question. A quantity rises by 10% to 132. What was its original value?
Working. 132 represents 110% of the old value. Original = 132 ÷ 1.1 = 120.
Why it works. The new value is larger than the original; use 110% rather than 90%.
Check. 10% of 120 is 12, and 120 + 12 = 132.
Additional worked percentage example 16 — reverse a fifteen-percent decrease
Question. After a 15% decrease, a quantity is 170. Find its original value.
Working. 170 is 85% of the original, so original = 170 ÷ 0.85 = 200.
Why it works. The given amount is the remaining proportion after a reduction.
Check. 15% of 200 is 30, and 200 − 30 = 170.
Additional worked percentage example 17 — value above one hundred percent
Question. The final quantity is 250 and is 125% of its original. Find the original.
Working. Original = 250 ÷ 1.25 = 200.
Why it works. 125% means all of the original plus one additional quarter.
Check. 125% of 200 = 250.
Additional worked percentage example 18 — a large reduction
Question. A fictional supply falls by 30% and the remainder is 84. Find the earlier amount.
Working. 84 is 70% of the original, so original = 84 ÷ 0.7 = 120.
Why it works. The phrase ‘30% fell’ refers to what was lost; the 84 is what remains.
Check. 30% of 120 = 36; 120 − 36 = 84.
Additional worked percentage example 19 — two fifths of the whole
Question. 40% of a quantity is 96. Find the whole.
Working. 96 corresponds to 40%, so 10% = 24 and 100% = 240.
Why it works. A ten-percent step is convenient because 40% contains four such parts.
Check. 40% of 240 = 96.
Additional worked percentage example 20 — a twelve-percent part
Question. 12% of a sum of money is $18. Find the full sum.
Working. 1% is $18 ÷ 12 = $1.50. 100% is $150.
Why it works. Use a percentage-unit method when the supplied percentage is not a convenient simple fraction.
Check. 12% of $150 = $18.
Additional worked percentage example 21 — an eight-percent fee
Question. A hypothetical fee equal to 8% of a base amount is $24. What is the base?
Working. 1% = $3; therefore 100% = $300.
Why it works. The base amount is the whole to which the fee rate applies.
Check. 8% of $300 equals $24. This is not a claim about a real tax rate.
Additional worked percentage example 22 — three quarters remains
Question. After a 25% reduction, 90 tokens remain. How many were present before the reduction?
Working. 90 represents 75% or three quarters. One quarter is 30, so the whole is 120 tokens.
Why it works. The reduced part is 25%; the known remainder is 75%.
Check. 25% of 120 is 30; 120 − 30 = 90.
Additional worked percentage example 23 — a sixty-percent remainder
Question. A fictional tank contains 72 litres after losing 40% of its original water. Find the original water volume.
Working. The remaining 60% is 72 L. Original = 72 ÷ 0.6 = 120 L.
Why it works. We must not divide by 40%, because 72 L is not the lost amount.
Check. 40% of 120 is 48; 120 − 48 = 72 L.
Additional worked percentage example 24 — half again as much
Question. A new amount of 300 is 150% of the previous amount. Find the previous amount.
Working. Original = 300 ÷ 1.5 = 200.
Why it works. 150% means one and a half times the original quantity.
Check. One half of 200 is 100; 200 + 100 = 300.
Additional worked percentage example 25 — percentage increase from two values
Question. A count rises from 80 to 100. Find the percentage increase.
Working. Increase = 20. Percentage increase = 20 ÷ 80 × 100% = 25%.
Why it works. The denominator is the old amount 80, not the new amount 100.
Check. 25% of 80 is 20; 80 + 20 = 100.
Additional worked percentage example 26 — percentage decrease from two values
Question. A quantity falls from 150 to 120. Find its percentage decrease.
Working. Decrease = 30. Percentage decrease = 30 ÷ 150 × 100% = 20%.
Why it works. The 150 is the original baseline for measuring change.
Check. 20% of 150 is 30; 150 − 30 = 120.
Additional worked percentage example 27 — one more discount reversal
Question. A fictional bag costs $96 after a 20% markdown. Find its original price.
Working. $96 is 80% of original. Divide by 0.8 to get $120.
Why it works. Adding 20% of $96 is incorrect because that percentage is based on the reduced price.
Check. 20% of $120 = $24; $120 − $24 = $96.
Additional worked percentage example 28 — reverse a twenty-five-percent rise
Question. After a 25% increase, a club has 250 tokens. Find the original number.
Working. 250 is five quarters of the old quantity. One quarter is 50, so four quarters equal 200.
Why it works. Percentage-unit bars give a calculator-free route: 125% consists of five 25% units.
Check. 25% of 200 = 50; adding it gives 250.
Additional worked percentage example 29 — a ninety-percent remainder
Question. After a 10% reduction, a quantity is 117. Find its original.
Working. 117 represents 90%; 10% is 13, so original is 130.
Why it works. Nine ten-percent units make 117; one ten-percent unit is thirteen.
Check. 10% of 130 is 13; 130 − 13 = 117.
Additional worked percentage example 30 — why increase and decrease differ
Question. An amount grows from 100 to 120, then falls from 120 to 100. Find the percentage decrease in the second step.
Working. Decrease is 20 on an original-for-that-step base of 120; percentage = 20/120 × 100% = 16 2/3%.
Why it works. The earlier rise was 20%, but the reverse change uses a new denominator; the percentages are not equal.
Check. One sixth of 120 is 20, so the second decrease is one sixth or 16 2/3%.
Additional worked percentage example 31 — recover a quantity after a ten-percent fee
Question. A fictional service subtotal plus a 10% additional charge totals $165. Find the subtotal.
Working. Total is 110% of subtotal. Subtotal = 165 ÷ 1.1 = $150.
Why it works. The extra charge is proportional to the subtotal, not the final total.
Check. 10% of $150 is $15; $150 + $15 = $165.
Additional worked percentage example 32 — a ten-percent sale
Question. A sale price is $135 after a 10% discount. Find the original tagged price.
Working. The sale price is 90% of original. Original = 135 ÷ 0.9 = $150.
Why it works. Do not add ten percent of $135, because that does not restore the correct discount base.
Check. 10% of $150 is $15; $150 − $15 = $135.
Additional worked percentage example 33 — the percentage part rather than the remainder
Question. A charity goal has already received $72, representing 40% of its target. Find the full target.
Working. 40% = 72, so 10% = 18 and 100% = $180.
Why it works. The given amount is the achieved part, not the amount remaining.
Check. 60% still needed is $108; $72 + $108 = $180.
Additional worked percentage example 34 — two successive changes, labelled extension
Question. A fictional price of $200 is first reduced by 10%, then the reduced price is discounted by another 20%. Find the final price and overall percentage reduction.
Working. First price $180; second price 80% of $180 = $144. Total reduction is $56, which is 28% of $200.
Why it works. The two discounts have different starting bases; they cannot simply be added to give a 30% discount. This is a deliberate multistep extension.
Check. $200 × 0.9 × 0.8 = $144, and $56 ÷ $200 = 28%.
A quick reverse-percentage decision table
Classify the given amount before selecting an operation. A sale price after a 20% discount represents 80% of the original; a new amount after a 20% increase represents 120%; a stated 20%-part represents only 20%. Divide the known amount by 0.8, 1.2 or 0.2 respectively, only after confirming which interpretation the words support. The digits 20% alone do not determine the formula.
Once the original whole is found, move forward through the story and reproduce the supplied final amount. A correct original price should return the exact discounted price when the reduction is applied. A correct original count should produce the stated increased count. This reverse-forward check is the surest way to distinguish a valid answer from an accidental numerical coincidence.
A parent exercise that exposes the wrong denominator
Present a price rising from $100 to $125 and ask for percentage increase. The increase is $25 compared with the original $100, so it is 25%. Then ask how much percentage decrease takes $125 back to $100. The decrease is $25 compared with the new starting point of $125, so it is 20%. The two percentages differ despite the identical $25 change.
This experiment shows why percentages need reference wholes. If a child repeatedly divides by whichever number is smaller, the correction begins with labelling ‘before’ and ‘after’ above a bar model, then writing the correct denominator. More practice without changing that interpretation often makes the wrong habit faster.
A focused small-group correction route
In a group of three, one pupil may misunderstand what 100% refers to, another may use the correct model but calculate with decimals incorrectly, and a third may reverse a discount correctly but not an increase. The tutor should assign a distinct repair to each pupil: reference-whole identification, equivalent-fraction calculation, or the complementary multiplier. Everyone then attempts a mixed unseen problem to verify independent transfer.
Parents should judge progress by the student’s ability to explain a new question and check it without prompting, not by the completion of a predetermined number of percentage worksheets. The updated Primary 6 syllabus, the revised 2026 paper format and the learner’s present mastery should determine pacing. Unnecessary complication should not be confused with advanced education.
Questions parents frequently ask
Why is adding 20% to the sale price not the way to reverse a 20% discount?
The discount was 20% of the original, not 20% of the reduced price. After a 20% reduction, the new price represents 80% of the original. Divide by 80% or use a fraction bar to reconstruct the whole.
Is ‘percentage of a percentage’ part of the same idea?
It is a useful reasoning extension when a story changes its base twice. First identify the original for the first percentage, then the remainder or new amount for the next. Do not mechanically add two percentages with different reference amounts.
My child gets the calculator answer but cannot explain it. Is that sufficient?
No. A correct number obtained by guessing a calculation is fragile. Ask the pupil to label the 100% base, name what the given figure represents and demonstrate a reverse check.
What does ‘percentage increase’ divide by?
Percentage increase compares the increase with the original value: increase/original × 100%. The same denominator principle applies to percentage decreases.
What is a sensible way to check a discount question?
After finding the original price, calculate the stated discount from that original amount and subtract. The result must match the given sale price. Also check that the original is greater than the sale price.
Does this apply to Foundation PSLE Mathematics?
Follow the registered Foundation Mathematics 0038 syllabus and its own revised paper structure rather than importing all Standard-paper expectations. Speak to the school about which subject is appropriate for the child.
Official references and helpful next pages
- MOE Primary Mathematics curriculum, October 2025 update — whole-given-part and percentage increase/decrease at P6.
- SEAB: PSLE formats examined in 2026 — subject 0008 and revised arrangement.
- Ratio and fraction links for Primary 6 — identifying part, whole and correct denominator.
- Simple linear equations for Primary 6 — giving the original amount a letter and solving by balance.
- Primary 6 PSLE Mathematics Punggol companion — wider mixed-topic worked problems.
Once a pupil can say what 100% is, the arithmetic becomes a consequence of a clear model. Reverse percentage mastery is the ability to reconstruct the hidden whole, recognise a changed base, and check that the recovered original produces the reported outcome. That skill lasts longer than any one discount formula.
