eduKate Sengkang

Small Group Tutorials

Here to help students catch up, keep up, and move ahead. Book a consultation here.

Chat on WhatsApp

How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0127 | Mathematics: Reverse-Percentage Workshop — Recover the Original Price Before the Discount

G3 SEC Mathematics · Learner’s Guide Vol 0127. Original problems and worked calculations for recovering a value before a percentage change.

The sale label says $72 after a 20% discount. How much did the item cost before it was reduced? Adding 20% of $72 feels reasonable but gives the wrong answer because the original discount was calculated from a different base. The paid amount is 80% of the original, so the missing original is $72 divided by 0.80: $90.

Reverse percentages are not a collection of awkward shopping tricks. They are a way to recover a previous value from a proportional change. Once you understand which quantity the rate refers to, the calculations become ordinary multiplication and division. The main problem is translating the language faithfully.

In this learner workshop, you will build the forward relationship before reversing it, compare a percentage change with percentage points, study several successive discounts, investigate fixed vouchers and tax-inclusive totals, and test when an apparent answer is impossible. A detailed diagnostic casebook provides original problems, not recycled official examination questions.

Chapter index

1. Start by identifying the original 100%

If an item originally costs $90, that is 100% of its original value. A discount of 20% removes $18 and leaves $72. The final payment is therefore 80% of the original. When a question gives you the final payment but hides the original, the proportion still exists.

Let x represent the marked price. “After 20% off, $72 remains” becomes x−0.20x=72, or 0.80x=72. Divide both sides by 0.80 to obtain x=90. This solution is supported by a forward check: 90−20% of 90=90−18=72.

A student might instead multiply the final $72 by 1.20 and obtain $86.40. That operation applies a 20% increase to the reduced price, not the inverse of a 20% reduction from the original. Forward-check the guess: 0.80×86.40=$69.12, not $72.

The useful rule is not “always divide by 80” or “never add a percentage”. It is to ask: what percentage of the starting value does the final amount represent? With a 20% discount, the multiplier is 0.80. With a 15% discount, it is 0.85. With a 20% increase, it is 1.20.

Write the relationship in a sentence before calculating. The correct reference base is more important than the speed of the arithmetic.

2. SEC 2027 G3 syllabus scope

The 2027 SEAB Mathematics K310 syllabus explicitly lists reverse percentages in the N3 Percentage topic. The same section includes increasing and decreasing a quantity by a given percentage, percentages greater than 100% and comparing quantities using percentages. This chapter is a syllabus-aligned practice workshop, not a prediction of a particular exam story.

K310 Paper 1 and Paper 2 each last 2 hours 15 minutes, carry 90 marks and have equal 50% weighting. Paper 1 has approximately 26 compulsory short-answer questions and Paper 2 has 9 to 10 questions, with the final question applying mathematics to a real-world context. An approved calculator may be used. Essential working should still be shown where the assessment requires it.

Reverse percentage can appear as a stand-alone price question or within a multi-part problem about population, attendance, quantities, discounts or rates. Do not restrict your skill to one kind of noun. The mathematical structure transfers to many contexts.

The SEAB 2027 G3 syllabus directory identifies the subject code as K310. Use the Mathematics Hub for the wider Number and Algebra route. For a related but different skill, Volume 0087 develops how the base changes after each movement.

Use exact relationships until the final answer, and follow the rounding instruction in the actual question rather than rounding an intermediate amount because it looks simpler.

3. Convert words into multipliers

“Reduced by 25%” means 75% remains. In decimal notation, that is 0.75. “Raised by 25%” means 125% remains, or 1.25. “Reduced to 25%” is a different expression: it says only one quarter of the starting value remains. The words “by” and “to” are decisive.

A jacket costs $90 after 25% off. Its original is 90÷0.75=$120. If the same jacket costs $90 after being reduced to 25% of its old price, its original would be 90÷0.25=$360. One small phrase changes the answer threefold.

“New price is 120% of old price” and “price increased by 20%” are equivalent under the one-step model. Both use 1.20. “Price increased to 20% of old price” would ordinarily describe a contradictory use of the word increased, and the conditions should be clarified or read carefully.

For a count that has decreased by 12%, the final multiplier is 0.88. A result of 440 means an earlier count of 440÷0.88=500. For a count that has increased by 12% to 5600, the multiplier is 1.12 and the earlier count is 5000.

Before solving, check the direction. After a positive percentage increase, the previous positive amount should be lower than the final amount. After a positive percentage discount, the original positive price should be higher. A result violating that simple expectation suggests the wrong multiplier.

4. Forward checks matter more than a remembered shortcut

The inverse of multiplying by 0.75 is dividing by 0.75, equivalent to multiplying by 4/3. It is not multiplying by 1.25. This explains why a 25% discount needs a 33⅓% increase from the reduced value to restore the original.

Start with $120. A 25% discount gives $90. A 25% increase on $90 gives $112.50, still below $120. An increase of $30 from $90, however, is 33⅓% of the current $90, giving back $120.

This is not mysterious. It is a consequence of percentages always referring to a base. The rate has changed reference. The reduced amount is smaller, so an equal percentage of it represents less money.

A good solution always ends by applying the original change forward. The student claiming that $112.50 was the marked price can be disproved by taking 25% off it: $84.375, not the stated $90. A correct reconstructed price survives the forward test.

You can make this a classroom habit with any discount rate. Write an original value, apply a discount, cover the original, recover it from the final, then check. The two directions should be reversible when the multiplier is non-zero and the model is complete.

5. Percentage points and relative percentages

Imagine a school’s participation rate rising from 40% to 50%. That is a rise of ten percentage points. Relative to the original participation rate, the rise is 10/40×100%=25%. The terms are not interchangeable.

If a pass rate becomes 72% after a relative 20% rise, its previous rate was 72%÷1.20=60%. But if it rose by twenty percentage points to 72%, its previous rate was 52%. The first is a proportional multiplication problem; the second is subtraction of values measured in points.

The problem occurs because the original quantity is itself a percentage. It is easy to see “20%” in the question and select an operation without asking what changed. The correct method follows the full phrase: “20% more than before” or “20 percentage points higher”.

This distinction is useful when reading school statistics, survey shares or business conversion rates. A graph can also exaggerate changes in percentages through its axis choice, a separate topic covered in Volume 0099.

A proper answer names the earlier rate and checks the result in the same terms. The original 60% rises by 20% of itself, which is twelve percentage points, reaching 72%. The original 52% plus twenty percentage points also reaches 72%, but these are different statements.

6. Successive discounts build a chain of bases

Suppose an item first receives 10% off, then a further 20% off the reduced price. The multipliers are 0.90 and 0.80, so the combined multiplier is 0.72. Only 72% of the original remains, meaning a total discount of 28%.

On a $100 starting price, the first discount leaves $90. The second takes 20% of that $90, removing $18 and leaving $72. Simply adding 10% and 20% to announce 30% off would overstate the discount.

When the final price is $144, reverse both multipliers: $144÷(0.90×0.80)=$200. The alternative stepwise reverse is $144÷0.80=$180, then $180÷0.90=$200. Work forward to verify $200→$180→$144.

The wording matters. “A further 20% off the reduced price” is not the same as “twenty percentage points added to the original ten-percent discount”. The first uses multipliers on changing bases. The latter would describe a total 30%-off rate calculated against the original.

When the story has three or more stages, draw an arrow chain and label the amount at each stage. It will help keep the reference quantity attached to the correct percentage.

7. Equal rises and falls usually do not cancel

Start with $100. Raise it by 10%, reaching $110. Then lower the new price by 10%, reaching $99. The combined multiplier is 1.10×0.90=0.99. The final amount is 1% below the starting price, even though the printed percentage rates are equal.

A rise of 25% and a subsequent fall of 20% do cancel because 1.25×0.80=1.00. The second multiplier is the reciprocal of the first. The percentages are different precisely because they apply to different bases.

Similarly, a price rising 30% and then falling 20% has multiplier 1.30×0.80=1.04, leaving a 4% net increase. Do not subtract twenty percentage points from thirty and claim the net change is ten percent.

The right method is to multiply the remaining factors. The left-to-right operation follows the sequence, and a reverse question divides by the combined non-zero multiplier to reconstruct the initial value.

This is also a useful example of why arithmetic rules cannot be applied without context. Percentage labels describe operations on amounts, not freestanding numbers that can always be added together.

8. Fixed vouchers and fees create an order of operations

Suppose a price x is discounted 15%, then a fixed $5 voucher is taken off, leaving $97. The forward equation is 0.85x−5=97. Undo the fixed voucher first: 0.85x=102. Divide by 0.85 to obtain x=$120.

If the voucher had been applied before the percentage discount, the model would be 0.85(x−5)=97. The original would be 97÷0.85+5, a different value. You cannot silently change the order. The exact story determines which expression is appropriate.

A second example includes a $10 delivery fee added after a 20% discount. If the final charged amount is $90, the product’s discounted price was $80. Its original marked price was $80÷0.80=$100. Dividing the full $90 by 0.80 would incorrectly apply the product discount to the delivery charge.

A written arrow diagram can be invaluable: original price → percentage reduction → fixed voucher or charge → final total. Reverse it by undoing the last arrow first.

If a question provides several amounts, identify which are product prices, additional fees and final payments. A correct calculation for the wrong quantity does not answer the task.

9. Tax-inclusive totals are another reverse multiplier

Take a fictional price of $100 before a stipulated 9% tax. The final price is 100×1.09=$109. To recover the pre-tax base from $109, divide by 1.09 and obtain $100.

Subtracting 9% of the final $109 would remove $9.81 and give $99.19, not $100. The original tax was 9% of the pre-tax base, not 9% of the larger tax-inclusive total.

A 10% discount followed by a 9% tax on the reduced price uses multipliers 0.90 and 1.09. Their product is 0.981. A final payment of $98.10 therefore corresponds to an original price of $98.10÷0.981=$100.

Do not guess that a ten-percent reduction followed by a nine-percent increase produces a one-percent decline. The exact combined effect is a 1.9% decline in this model. The percentage changes are multiplicative because each acts on the revised base.

This is a classroom arithmetic scenario, not a statement about actual commercial tax obligations or exemptions. The problem must define what amounts and adjustments the model includes.

10. Edge cases, counts and missing information

A 100% discount has multiplier zero. Under the simple model, any positive original price becomes zero after a complete discount. You cannot recover a unique starting price by dividing by zero. The ordinary inverse method requires a non-zero multiplier.

A count of people must be a whole number. If an exact percentage model produces an original count of 403.57 participants, the data may have been rounded or may not fit an exact whole-person interpretation. Do not report a fractional person without a relevant instruction or explanation.

Likewise, the statement “visits increased by 15%” without either an original or final number does not determine a unique visit count. Many starting values could increase by that proportion. A percentage is a relationship, not enough information to reconstruct every unknown amount.

A decrease exceeding 100% could produce a negative result under a simple formula, which is not a normal positive shop price or number of people. The context provides domain constraints. Check those constraints when the arithmetic produces an unusual result.

Use Volume 0125 on zero denominators for the wider issue of ratios that lack a valid reference. In this workshop, the point is knowing when the reverse multiplier genuinely allows reconstruction.

11. Independent problem set

1. A bag costs $72 after 20% off. Recover the marked price. 2. A jacket costs $102 after 15% off. Recover the marked price. 3. A pair of shoes costs $90 after 25% off. Recover the marked price. 4. A town’s population rises 12% to 5600. Find its old count. 5. A price rises 15% to $276. Find the old price. 6. A count falls 12% to 440. Find its previous value. 7. A price receives 10% off and then 20% off the reduced value, leaving $144. Recover the original. 8. A fictional tax-inclusive total is $109 after 9% tax. Find the pre-tax base. 9. A price rises 25% then falls 20%. Find the overall multiplier. 10. A pass rate increases by 20% relatively and ends at 72%. Find the old rate. 11. Another pass rate rises by twenty percentage points to 72%. Find the old rate. 12. A 15% reduction and $5 voucher leave $97. Recover the initial price. 13. A 10% discount and 9% tax on the discounted amount produce $98.10. Find the original price. 14. A quantity doubles. What is its percentage increase? 15. A quantity becomes 150% of its original. What is its percentage increase? 16. A collection increases 36% to 680 items. Find the previous count. 17. A 100% discount gives price zero. Is a unique starting price recoverable? 18. A 10% increase is followed by 10% decrease. Find the net percentage change. 19. After a 5% decrease, 760 items remain. Find the original. 20. A product’s discount is 20%, yet the recorded final product price exceeds its original price. Identify what additional condition or inconsistency needs checking.

Work through the first six without looking at any solutions. For the mixed cases, write the stages in their given order and reverse them. For questions involving percentage points or missing data, explain the meaning before calculating.

12. Worked answers to the independent set

1. 72÷0.80=$90. 2. 102÷0.85=$120. 3. 90÷0.75=$120. Each original price should be larger than its discounted price.

4. 5600÷1.12=5000. 5. 276÷1.15=$240. 6. 440÷0.88=500. The correct multiplier differs for increases and decreases.

7. 144÷(0.90×0.80)=$200. The combined multiplier is 0.72, not 0.70.

8. 109÷1.09=$100 under the stated pre-tax model.

9. 1.25×0.80=1.00, so the original value is restored.

10. 72%÷1.20=60%. 11. 72%−20 percentage points=52%. These are deliberately distinct wording cases.

12. (97+5)÷0.85=$120. Undo the voucher before the percentage.

13. 98.10÷(0.90×1.09)=$100. The overall multiplier is 0.981.

14. Doubling is a 100% increase. 15. Becoming 150% of the old quantity is a 50% increase.

16. 680÷1.36=500 items. Check 36% of 500 is 180, giving the new total 680.

17. No. The reverse would require division by a zero multiplier, and the original is not unique.

18. 1.10×0.90=0.99, a net 1% decrease.

19. 760÷0.95=800 items. The starting quantity remains a whole number.

20. Under a simple one-step discount, a positive product price should decrease. A final payment greater than the original might include added fees, different quantities or a source error, but none should be invented unless the question supplies them.

Check every result by applying the stated change forward. This protects against a solution that is numerically neat but uses the wrong base.

Worked diagnostic 1: A discounted book and a loyalty voucher

A book is sold for $47 after a 20% discount and then a fixed $1 reduction for loyalty members. First undo the fixed reduction: the percentage-discounted price was $48. Then divide $48 by 0.80 to obtain the original $60. Do not apply 20% to the final $47, because that number includes the later fixed voucher. Verify the stages: $60 becomes $48 after 20% off, then $47 after the voucher. This example teaches the order of inverse operations as well as the changing reference.

Worked diagnostic 2: Membership counts and rounded percentages

A club reports 680 members after a 36% rise from exactly 500. That arithmetic fits. Now imagine the report says the rise was approximately 36%. Recovering 500 by division is a reasonable estimate but not automatically an exact count, because a rounded growth percentage could have arisen from a nearby starting number. In a school question, read whether the percentage is exact or approximate before claiming a uniquely determined whole-person count. The correct relationship should be tested against the statement as written, not a version made more precise by the solver.

Worked diagnostic 3: A relative fall from a relative rise

A transport operator’s fare rises 25%, then returns to its old level through a later reduction. What relative reduction from the higher fare was needed? Its multiplier must be the reciprocal of 1.25, which is 0.80. That is a 20% reduction, not a 25% reduction. You can test with a fictional $4 fare: first it becomes $5, then a 20% reduction removes $1 and restores $4. A learner who simply uses the same percentage in both directions loses the reference change.

Worked diagnostic 4: The difference between ‘by’ and ‘to’

A quantity is reduced by 20% to 40 units. This means the final 40 is 80% of the original; the original is 40÷0.80=50. A different question says the quantity was reduced to 20% of its original and is now 40. That would require an original of 40÷0.20=200. The two sentences share the words reduced, twenty percent and forty, but describe very different final fractions. Before solving, paraphrase the sentence as ‘the final is ___ of the original’. The blank reveals the correct multiplier.

Worked diagnostic 5: Free entry is not a percentage model

A school announcement says admission is free for confirmed participants. A pupil attempts to recover an original ticket price from the zero charge by assuming a 100% discount. That may not follow. The event might always have been free rather than marked down from a previous price. Even if a 100% discount really occurred, zero final price does not recover a unique positive original because the remaining multiplier is zero. The appropriate conclusion is that more information is needed. Not every everyday description of zero cost contains a reversible percentage transaction.

Worked diagnostic 6: Final rate versus percentage points

A participation rate reaches 60%, having risen by ten percentage points. The earlier rate was 50%. If the story instead says that it increased by ten percent relative to the earlier rate, the earlier rate was 60% ÷ 1.10, approximately 54.545%. These are not equivalent descriptions. The percentage-point problem removes ten points directly, while the relative problem divides by a multiplier of 1.10. A careful report names which kind of change is intended rather than leaving the reader to guess from a percent sign.

Worked diagnostic 7: A fixed fee after a percentage change

A fictional admission charge after 20% reduction is $64, and a separate $6 handling fee is added. The customer pays $70. If asked for the original admission charge, subtract the $6 fee from the total payment, then divide $64 by 0.80 to get $80. Dividing $70 by 0.80 would incorrectly treat the handling fee as a discounted portion of the admission charge. The solution should name the exact object being recovered. The final payment and the product’s price are not the same quantity once a fee is added.

Worked diagnostic 8: A high percentage is not automatically an error

An event’s number of registrations rises from 80 to 200. The increase is 120, which is 120/80×100%=150%. A student may reject 150% because the number exceeds 100, but percentage increases greater than 100% are valid. The new level is 250% of the previous level. Reverse the description: if a final count is 200 after a 150% rise, divide by 2.50 to get 80. This case tests whether the solver distinguishes the amount of the increase from the total amount remaining.

Worked diagnostic 9: Why answer direction is a useful check

A receipt says an item costs $75 after 25% off. A learner obtains $60 as its original price. That result is smaller than the reported post-discount price, contradicting the direction of a simple positive-price discount. The equation should be 0.75x=75, so original x=$100. The direction check cannot find every arithmetic error, but it quickly catches an especially common one. After any decrease, the recovered starting positive amount must exceed the final value unless other stated adjustments change the model.

Worked diagnostic 10: One final check is stronger than three guesses

A jacket sells for $90 after a 25% discount. A learner proposes original prices of $110, $115 and $120. Instead of debating which looks reasonable, apply the stated multiplier 0.75 to each candidate. The resulting final prices are $82.50, $86.25 and $90. Only the last reproduces the source. This independent forward test evaluates the entire claim at once. Guessing can be a useful initial estimate, but verification is the step that makes a mathematical answer defensible.

Worked diagnostic 11: A discount applied twice to the wrong base

A flyer offers 20% off the original price and then another 10% off the reduced price. Starting from $200, the correct path is $200→$160→$144, an overall reduction of 28%. A learner who removes 30% of the original would get $140, which is not correct. Another who applies the second 10% to the original would also remove $20 rather than $16. Each discount stage has its own reference. The final multiplier 0.80×0.90=0.72 captures the correct sequence and is easy to reverse.

Worked diagnostic 12: Checking a population count with an integer domain

A report states that an exact 20% increase produced 96 members. The earlier number is 96÷1.20=80 members, a whole number. That is consistent with an ordinary headcount. If the calculation instead produced 80.5 people, it would invite a question about whether the reported percentage was rounded or the quantities were estimated. A calculator’s decimal display is not automatically a valid interpretation. After division, check whether the answer belongs to the allowed mathematical domain, especially when the quantities count people, tickets or separate objects.

Worked diagnostic 13: Reading the reference from a sentence

Compare ‘The sale price is 80% of the marked price’ with ‘The marked price is 80% of the sale price’. In the first, a sale price of $80 implies a marked price of $100. In the second, a marked price of $80 implies a sale price of $100. The mathematical relation changes when the sentence reverses the subject and reference. The word ‘of’ identifies which amount supplies the percentage base. Under exam pressure, underline the quantity after ‘of’ before setting up a multiplier equation.

Worked diagnostic 14: A price increase before a final discount

A fictional product rises by 20% and then receives 10% off the raised price. Its final amount is $108. Combined forward multiplier is 1.20×0.90=1.08, so the original price is 108÷1.08=$100. Forward check: $100 becomes $120 and then $108. Subtracting the printed rates to claim a net ten-percent increase would be wrong: the actual net increase is eight percent. The case illustrates how a two-stage chain can be compressed into one correct multiplier without forgetting the intermediate reference.

Worked diagnostic 15: When the problem names an exact bound

A discount of 20% is advertised on a ticket, but a minimum final charge of $50 is enforced. The simple equation final=0.80×original may apply only when the discounted result is at least $50. If a customer paid exactly $50, many marked prices might be compatible with a minimum-charge rule depending on its wording. The final payment alone may not identify a unique original. A sophisticated reader tests whether the percentage rule operates without an additional floor or ceiling. This is distinct from making an arithmetic error in an otherwise complete model.

14. A one-page diagnostic you can reuse

When you read a reverse percentage question, write four things: the original 100%, the change, the fraction of the original represented by the final amount, and the value to recover. If several changes occur, describe each stage in order and attach the correct reference to every rate. If a fixed fee or voucher appears, keep it separate from a percentage multiplier.

Before using the calculator, estimate whether the recovered original should be larger or smaller than the reported final amount. Then do the algebra and apply the original change forward to verify the answer. This takes little time and helps distinguish a correct calculation from an accidental match.

If a result is a fraction of a person, a negative normal price, or a value inferred from a zero multiplier, do not hide the problem. Examine whether the source is approximate, the domain has been violated or the information is insufficient. A good examination answer respects both mathematics and context.

15. Seven days of repair and transfer

Day 1: Build multipliers from ‘increased by’, ‘reduced by’ and ‘became’ statements. Translate words into one clear equation before calculating.

Day 2: Solve straightforward original-price problems. Forward-check every answer and explain why adding the same percentage to the sale price is not the reverse.

Day 3: Compare percentage points with relative percentage changes using school-rate examples. Identify the reference after the word ‘of’.

Day 4: Work through successive discounts and increases. Multiply forward multipliers and reverse them without confusing the base at each stage.

Day 5: Add fixed vouchers, delivery charges or hypothetical tax-inclusive amounts. Preserve the exact sequence of operations.

Day 6: Study incomplete or impossible models, including a 100% discount, a floor charge or rounded percentage. Explain when a unique original cannot be recovered.

Day 7: Complete the independent questions and the diagnostic casebook without notes. Group mistakes by reference base, multiplier, sequence, domain or unsupported assumption. Revisit the weakest group with new figures.

The most useful question from a parent or tutor is, ‘What amount was one hundred percent?’ If the learner can answer that, the rest is usually an ordinary mathematical operation.

16. The original value is the key to the story

Reverse percentages turn a final amount into evidence about the past. A multiplier records the change, and division by that non-zero multiplier reconstructs the earlier base. Prices, ticket counts, populations and percentage rates all become easier when the quantity and reference remain attached.

Continue through the Mathematics Hub, the G3 SEC Learner’s Guide, Volume 0087 on successive percentage transfers and Volume 0125 on zero denominators.

Official examination scope and format come from the SEAB 2027 G3 syllabus index and K310 Mathematics syllabus. The problems are original practice, not official examination questions.