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A sign says twenty per cent off, followed by another ten per cent off the reduced price. Is that thirty per cent off altogether? No. The second discount is taken from a smaller amount, so it removes fewer dollars than ten per cent of the original price would remove.
This guide gives you the short reason first, then builds a method you can use with sale prices, changing quantities and reverse questions. All shop offers below are invented teaching examples, not current promotions or purchasing recommendations. Their conditions are stated so that you can check the mathematics honestly.
The 2027 G1 Mathematics syllabus provides the curriculum reference for practical number and percentage work. Use your school’s sequence to decide how much of this guide to study now. The changing-base idea is useful well beyond a single subject level.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–30
Sections 31–31
For an original price of $150, a twenty per cent discount removes $30 and leaves $120. A further ten per cent discount on that reduced price removes $12 and leaves $108. The total saving is $42, which is twenty-eight per cent of $150, not thirty per cent.
The quickest reliable calculation is 150 × 0.80 × 0.90 = 108. Each multiplier tells you what fraction of the current price remains. If the question asks for the overall percentage discount, compare the final price with the original price: 42 ÷ 150 × 100% = 28%.
Before calculating, ask what each percentage is a percentage of. If both discounts are explicitly based on the original marked price, the model is different and addition may be correct. The wording decides the method; the presence of two percentage symbols does not decide it.
A percentage is a comparison with a reference amount. Twenty per cent means twenty out of every hundred of that reference amount. Without the reference, the percentage alone cannot tell you how many dollars are involved. Twenty per cent of $50 and twenty per cent of $200 are different amounts.
In a first discount, the reference is normally the price immediately before that discount. In a successive discount explicitly applied to the reduced price, the reference changes. That change is the entire reason the two advertised rates cannot simply be added. There is no special shopping exception to ordinary arithmetic.
Write a short label beside each amount: original price, first reduced price, final price. These labels are useful working, not decoration. A student who writes three unlabelled numbers may calculate correctly and still use the wrong one in the final percentage comparison. Meaning needs to survive every line.
Imagine an original price of $100. A twenty per cent discount removes $20, leaving $80. A further ten per cent of $80 removes $8. The final price is $72. You have saved $28 out of the original $100, which makes the combined discount twenty-eight per cent.
Adding twenty and ten would predict a saving of $30. The missing $2 has a clear explanation: the second ten per cent is not applied to the $20 already removed. Ten per cent of that $20 is $2. The addition shortcut counted a discount on money no longer in the price.
Using $100 makes the comparison easy to see, but it is not a restriction on the rule. The same retained fraction applies to any positive original price under these same conditions. Once the reference amounts are understood, you can move to prices that are not convenient multiples of one hundred.
If the discount is fifteen per cent, the customer pays eighty-five per cent of the relevant price. The discount amount is found by multiplying by 0.15. The remaining price is found by multiplying by 0.85. Both calculations are valid, but they answer different questions.
For an $80 item, fifteen per cent of $80 is $12. That is the saving, not the amount to pay. The amount to pay is $68. You can get it by subtracting $12 from $80 or by calculating 80 × 0.85. The two methods describe the same change.
When another discount follows, carry forward the remaining amount. Do not carry forward the first saving as though it were the new price. A useful sentence is: “The first reduction leaves this amount, and the next reduction acts on this amount.” Saying it once can prevent an entire chain of errors.
One hundred per cent of the current price is the whole current price. Removing a discount of twenty per cent leaves eighty per cent. Converting eighty per cent to a decimal gives 0.80. That is why the new price is the old price multiplied by 0.80.
For a discount of seven per cent, ninety-three per cent remains, so the multiplier is 0.93. For a discount of thirty-five per cent, sixty-five per cent remains, so the multiplier is 0.65. Ask yourself whether the multiplier should be below one: a genuine reduction of a positive price should make it smaller.
This reasoning is safer than remembering isolated calculator entries. It also helps you spot an incorrect multiplier such as 1.20 for a twenty per cent discount. A multiplier of 1.20 would increase the price by twenty per cent. The operation must agree with the direction described in the question.
A school bag has an invented marked price of $240. It receives a twenty-five per cent discount. A further eight per cent discount is then applied to the reduced price. Find the final price and the overall percentage discount. Assume no other fees and no intermediate rounding.
After the first reduction, the price is 240 × 0.75 = $180. The second discount is eight per cent of $180, which is $14.40. The final price is therefore $165.60. The compact calculation is 240 × 0.75 × 0.92 = 165.60, with the same result.
The total saving is 240 − 165.60 = $74.40. Relative to the original $240, that is 74.40 ÷ 240 × 100% = 31%. The two advertised rates add to thirty-three per cent, but the actual successive reduction is thirty-one per cent. Always state which percentage your final sentence describes.
The table below follows the $240 example. Notice that the second row’s starting price is the first row’s ending price. That connection is more important than the appearance of the table. It shows exactly where the reference amount changes and which dollars continue to the next stage.
| Stage | Price before reduction | Discount amount | Price remaining |
|---|---|---|---|
| First discount of 25% | $240.00 | $60.00 | $180.00 |
| Further discount of 8% | $180.00 | $14.40 | $165.60 |
| Whole sequence | $240.00 | $74.40 | $165.60 |
Do not put thirty-three per cent in the final row merely because the two earlier rows contain twenty-five and eight. The total row describes a comparison between the first starting price and the last remaining price. It needs its own calculation. Table layout cannot make an invalid addition valid.
If the second discount amount is larger than the first despite a much smaller rate and a smaller base, pause and inspect the entries. Such comparisons are useful warning signals. They are not complete proofs, but they often reveal a misplaced decimal or the accidental reuse of the original price.
After a twenty per cent discount, eighty hundredths of the original price remain. Keeping ninety hundredths of that remainder gives ninety hundredths of eighty hundredths of the original. Multiplication combines these two “of” relationships: 0.90 × 0.80 = 0.72. Seventy-two per cent remains, so twenty-eight per cent has been removed.
You can picture the original whole as one hundred equal units. The first stage leaves eighty. The next stage keeps nine tenths of those eighty units, which is seventy-two. The model is not claiming that every item can physically be divided into a hundred pieces; it represents proportional value.
This is the same structure as a fraction of a remainder. The calculations become easier when you see that the second proportion acts inside what is left, rather than alongside the original whole. If this still feels unclear, practise with ten counters before returning to decimal multipliers.
Suppose a fictional offer explicitly deducts twenty per cent of the original marked price and then deducts another ten per cent of that same original marked price. On a $100 item, those deductions are $20 and $10. The final price is $70. In that different model, the combined discount is thirty per cent.
The rates can be added because their reference amounts are the same and their discount amounts are being combined. That is not the condition stated by “a further ten per cent off the reduced price”. Those two phrases describe different operations, even though the numerals twenty and ten are identical.
Avoid learning the oversimplified rule that percentages can never be added. The better rule is to preserve their bases. Ask whether the percentages refer to the same whole and whether their categories or changes can appropriately be combined. Here the question is about discount calculations, not a general ban on addition.
Two successive discounts can be expressed as one overall reduction from the original price. First multiply the retained fractions. Then subtract that product from one. For discounts of ten per cent and fifteen per cent, the retained fraction is 0.90 × 0.85 = 0.765. The overall discount is 0.235, or 23.5%.
This equivalent single discount is useful when comparing clearly specified offers. It does not tell you whether an item is worth buying, and it does not establish that two shops started with the same original price. It is simply a concise description of a mathematical change from one stated reference amount.
Check the result against the individual discounts. With two genuine positive discounts smaller than one hundred per cent, the combined discount should exceed either single discount but be smaller than their sum. Here 23.5% is larger than 15% and smaller than 25%, which fits that expectation.
Shop A’s fictional marked price is $160 with successive discounts of twenty per cent and ten per cent. Shop B’s fictional marked price is $150 with one discount of twenty-two per cent. Assume identical goods, identical quantities, no fees and that both offers can actually be used.
At A, the final price is 160 × 0.80 × 0.90 = $115.20. At B, the final price is 150 × 0.78 = $117. Shop A is cheaper by $1.80 under these assumptions. A’s twenty-eight per cent combined discount is useful information, but the final prices decide this cost comparison.
If the marked prices or included quantities differ, a larger percentage reduction does not automatically mean a lower bill. This connects with the series guide on unit prices and total costs. Read conditions first, then compare compatible totals rather than the largest advertised numeral.
For two pure percentage discounts applied successively without rounding or other changes, their order does not change the final price. Starting from $200, a ten per cent discount followed by twenty per cent gives 200 × 0.90 × 0.80 = $144. Reversing the discounts gives the same product.
The intermediate prices do change. One route passes through $180, while the other passes through $160. Therefore the individual discount amounts change too. The final price agrees because multiplication is commutative, not because every stage in the two stories is identical. Keep that distinction when explaining your answer.
This conclusion depends on the model. If there is a threshold, a fixed voucher, a cap or a stage-specific rounding rule, order may matter. Do not turn a true statement about pure multiplication into a universal claim about all promotions. The conditions belong in the reasoning, not in small print after it.
Consider an invented $100 price, a twenty per cent discount and a $10 voucher. Applying the percentage first gives $80 and then $70. Applying the voucher first gives $90 and then $72. The final amounts differ because subtraction and percentage multiplication do not generally commute.
The $10 voucher is a fixed amount, not a ten per cent discount. On the $80 reduced price, $10 represents twelve and a half per cent. On the original $100, it represents ten per cent. A fixed monetary amount does not keep the same percentage meaning as the base changes.
When a question includes both kinds of change, write the required order in words before entering a calculator expression. For example: “Take twenty per cent off the original price, then subtract ten dollars.” That sentence leads to 100 × 0.80 − 10, rather than (100 − 10) × 0.80.
A fictional voucher might say ten per cent off, capped at $6. On an eligible reduced price of $50, ten per cent is $5, so the cap does not operate. On an eligible reduced price of $90, ten per cent is $9, but the deduction is limited to $6.
You cannot represent that second deduction by multiplying by 0.90 in every case. You must first calculate the percentage amount, compare it with the cap, and use the permitted deduction. A threshold requiring a minimum eligible spend introduces another condition. These are changes to the mathematical model, not optional details.
In a school question, use the stated conditions rather than assumptions about how real shops usually behave. If a real offer is unclear, obtain its terms before making a cost claim. Mathematics can show the consequences of a rule, but it cannot supply a missing rule merely because one calculation would be convenient.
An item costs $136 after successive discounts of twenty per cent and fifteen per cent. Find its original price, assuming the discounts were applied to the current price at each stage and there was no intermediate rounding. The combined retained fraction is 0.80 × 0.85 = 0.68.
The final price is sixty-eight per cent of the original. Therefore original price × 0.68 = 136. Dividing gives 136 ÷ 0.68 = $200. Check forward: twenty per cent off $200 leaves $160, and fifteen per cent off $160 leaves $136. Both stages agree with the given final amount.
Adding twenty-eight or thirty-five per cent to $136 would not reverse the change. The recovery percentage would be based on the smaller final amount, whereas the overall discount was based on the original amount. Reverse questions are best handled by undoing the retained multiplier, not by guessing an opposite percentage.
Try these without looking at the answers immediately. Question A: an $80 shirt receives twenty per cent off, then a further five per cent off the reduced price. Find the final price. Question B: a $120 item receives ten per cent off and then twenty per cent off the reduced price. Find the overall percentage discount.
Question C: an item costs $153 after successive discounts of ten per cent and fifteen per cent. Find the original price. Question D: a $90 item has deductions of ten per cent and five per cent, each explicitly calculated from the original price. Find the final price and explain why the method differs from A.
Question E: a $100 item receives a twenty per cent discount and a $12 voucher. Calculate the price in each possible order. Assume both orders are allowed and there are no caps or thresholds. Your explanation should identify which operation keeps a fixed amount and which operation keeps a fixed fraction.
For A, the first price is 80 × 0.80 = $64. The second reduction is five per cent of $64, or $3.20, leaving $60.80. The compact calculation 80 × 0.80 × 0.95 agrees. An answer of $60 would incorrectly take twenty-five per cent of the original price.
For B, the retained fraction is 0.90 × 0.80 = 0.72. Thus twenty-eight per cent is removed overall. The final price is $86.40 and the saving is $33.60; dividing 33.60 by 120 confirms twenty-eight per cent. For C, divide 153 by 0.90 × 0.85, using brackets around the whole product. The original price is $200.
For D, deduct $9 and $4.50 from the same original $90, giving $76.50. The combined deduction is fifteen per cent because the bases are identical. For E, percentage first gives 100 × 0.80 − 12 = $68. Voucher first gives (100 − 12) × 0.80 = $70.40. The two models do not have the same result.
Suppose an invented item starts at $250. After a twenty per cent discount and a second discount, its final price is $170. The first reduction leaves $200. The second reduction therefore removes $30 from a base of $200. Its rate is 30 ÷ 200 × 100% = 15%.
The overall discount is thirty-two per cent because the total saving is $80 out of $250. Subtracting twenty from thirty-two would give twelve, but twelve per cent is not the second advertised discount. It describes the second saving as a proportion of the original price, not of the first reduced price.
This question is a useful test of understanding because two correct percentage descriptions coexist. The second saving is twelve per cent of the original $250 and fifteen per cent of the reduced $200. Always include the reference amount when comparing such statements. Different percentages can describe the same thirty dollars.
A quantity of 100 increases by twenty per cent to 120. A later twenty per cent decrease removes 24 and leaves 96. The final amount is four per cent below the original. The increase and decrease use different bases, so equal percentage numerals do not guarantee equal amounts of change.
This is not a separate rule to memorise after learning discounts. It is the same multiplier reasoning: 1.20 × 0.80 = 0.96. A return from 120 to 100 would require a decrease of 20 ÷ 120, or sixteen and two thirds per cent. That is a different rate from twenty per cent.
Keep this as a mathematical illustration rather than a prediction about actual prices or investments. The article is teaching proportional changes. Its examples do not establish current market behaviour, financial returns or the suitability of a purchase. A clear model helps precisely because it makes its own limits visible.
For a purely mathematical question, retain sufficient precision until the requested final rounding. A fictional checkout might instead round the amount payable to cents after each discount. If so, follow that stated process. A staged rounding rule and a final-only rounding rule are different models and can occasionally differ by one cent.
Take $1.07 with two successive ten per cent discounts. With no intermediate rounding, 1.07 × 0.90 × 0.90 = 0.8667, which rounds to $0.87. If the first reduced price of $0.963 is rounded to $0.96 before the second discount, the next result is $0.864, which rounds to $0.86.
Neither result should be presented as the unique real checkout charge without knowing the applicable rule. In an exercise, state the convention given or explicitly assumed. This connects percentage reasoning with rounding and examination accuracy, where the stage of rounding matters as much as the final number of decimal places.
For the $240 example, a direct entry is 240 × 0.75 × 0.92. For a reverse problem, use final price ÷ (0.75 × 0.92). The brackets in the reverse expression tell the calculator to divide by the entire combined multiplier. Without them, a left-to-right calculation may undo one stage and apply another.
If the final price were $165.60, entering 165.60 ÷ 0.75 × 0.92 would give $203.136, not the original $240. The problem is not that the calculator dislikes percentages. It has executed a different expression from the intended one. Translate the relationship correctly before pressing equals.
Write a short equation beside the entry. “Final = original × retained fraction” gives a stable reference for deciding whether to multiply or divide. This is especially useful when the question contains several numbers that could all be entered successfully but only some belong in the correct relationship.
Suppose a student writes 20% + 10% = 30%, then 150 × 0.70 = 105. The multiplication is correct for a thirty per cent discount, but the first line has changed the model. Correcting only the final price misses the useful teaching point. The repair belongs at the percentage combination.
Another student calculates the first reduced price correctly but takes the second ten per cent from $150 again. Ask them to point to the words “of the reduced price”. Then ask which number now represents that price. This connects the language directly to the base without requiring a longer worksheet.
A third student finds $108 correctly but reports the discount as 42 ÷ 108 × 100%. That compares the saving with the final price. To describe the overall reduction from the original, use $150 in the denominator. Each error needs a different correction because each loses meaning at a different stage.
Try this explanation: “The first discount shrinks the amount. The second percentage is calculated on what remains. So we work out the remainder after each stage, then compare the final price with the original.” It is short enough to remember and precise enough to reconstruct the calculation.
If the listener asks why the percentages cannot be added, use the $100 example rather than repeating the rule more firmly. Show $100 becoming $80 and then $72. Ask how many dollars the second discount removes. Eight dollars makes the changing reference concrete in a way that a slogan may not.
If they can explain that example but struggle with $150, ask them to label the stages rather than start again from a formula. The goal is transfer: the structure should remain recognisable when the starting number changes. Confidence built on one convenient example becomes stronger when it survives a second, less tidy one.
On one day, explain the $100 example without notes. On another, calculate two successive discounts on an unfamiliar original price. Later, reverse a final price to find its original value. These tasks revisit the same relationship from different directions, rather than repeating one calculator pattern until it feels familiar.
Include one contrasting question in which both discounts explicitly use the original marked price. The student should explain why addition is appropriate there. Understanding includes recognising when a rule does not apply. A mixed pair often reveals more than ten successive questions with identical wording and different prices.
Keep a correction note in ordinary language: “I reused the original base at the second stage,” or “I divided by the final price when finding the overall reduction.” Such notes are more useful than “careless mistake” because they identify the next action. The repair becomes small, visible and repeatable.
The broader repeated percentage change guide extends the multiplier structure into repeated changes. Use it after you can identify the base in each of the examples here. You do not need to rush into general formulas before a two-stage calculation makes sense.
If the main difficulty is understanding what a percentage describes, return to the percentage change and discount learning guide. If the difficulty is interpreting a story, use the Mathematics learning hub to choose the appropriate foundations rather than opening unrelated advanced topics.
For families considering structured support, the Secondary 1 Mathematics Clementi small-group tutorial page describes the teaching approach. It is a service-information destination, not evidence that a particular lesson guarantees a grade. The useful learning goal remains independent: identify the base, preserve the stages and check the final comparison.
An invented $500 price receives successive discounts of ten per cent, twenty per cent and five per cent. The retained fraction is 0.90 × 0.80 × 0.95 = 0.684. The final price is $342. The overall saving is $158, which is 31.6% of $500, rather than thirty-five per cent.
You can calculate stage by stage if that is clearer: $500 becomes $450, then $360, then $342. The final five per cent removes $18 because its base is $360. This visible sequence is often easier to inspect than a long expression, especially while the method is new.
The compact product is not a more intelligent method than the table. It is a compressed record of the same reasoning. Use the form that lets you preserve the conditions and avoid errors. Fluency means being able to connect the representations, not being forced to use the shortest one every time.
For a positive original price P and successive discount fractions a and b, the remaining price is P(1 − a)(1 − b). Expanding the retained fraction gives 1 − a − b + ab. Therefore the overall discount fraction is a + b − ab. The product ab corrects the overcount in simple addition.
For twenty per cent and ten per cent, a = 0.20 and b = 0.10. Their sum is 0.30, but ab = 0.02. Subtracting gives 0.28. This explains the two percentage-point difference between the naive thirty per cent prediction and the actual twenty-eight per cent discount.
This optional extension is not necessary for every G1 learner’s current work. The numerical sequence is enough to solve the practical questions. The formula is included to reveal the structure for readers ready for it, not to replace the understandable story with letters before those letters have meaning.
If the second discount is zero per cent, its retained multiplier is one and it changes nothing. If a stated discount is one hundred per cent, the remaining price becomes zero; later ordinary percentage reductions of that zero amount still leave zero. These cases help check whether a proposed formula behaves sensibly.
For ordinary discounts between zero and one hundred per cent, the final price should not become negative. A calculation producing a negative amount may have subtracted the advertised percentage numerals as though they were monetary deductions, or applied a fixed voucher beyond its permitted amount. Recheck the model before interpreting such a result.
Boundary checks do not prove that every step is correct, but they expose many implausible expressions. They are especially useful when you devise a general formula. A formula that fails when the second discount is zero cannot be a reliable description of two successive discounts, however neatly it works for one example.
Question F: a $300 item receives twelve per cent off and then twenty-five per cent off the reduced price. Find its final price and combined discount. Question G: a price falls from $400 to $252 after an initial ten per cent discount and a second discount. Find the second rate.
Question H: two successive discounts are both twenty per cent. Is the total reduction forty per cent? Show a calculation using any positive original price. Question I: explain why an offer described only as “twenty plus ten per cent” is not enough to determine the final price without clarifying the reference amounts.
For F, 300 × 0.88 × 0.75 = $198; the saving is $102, or 34%. For G, the first reduction leaves $360, so the next saving is $108 out of $360, which is 30%. For H, retaining 0.80 × 0.80 = 0.64 gives a 36% reduction. For I, distinguish deductions from one original base from a reduction of a remainder.
Return to the $150 item whose final price is $108 after the two reductions. The saving is $42. As a fraction of the original price, that is 42/150 = 0.28, so the discount is twenty-eight per cent. As a fraction of the final price, the same $42 is 42/108 = 7/18, or approximately 38.9%. These are not contradictory calculations. They use different reference amounts and answer different questions.
The second percentage describes how much the final price would have to increase to return to the original price. It does not describe the original discount. If you increase $108 by twenty-eight per cent, you get $138.24, not $150. The increase is being calculated from a smaller base, so the same rate produces a smaller dollar change than the original reduction did.
Try explaining this without decimals. The final price is seventy-two parts when the original is one hundred parts. Recovering the missing twenty-eight parts means adding twenty-eight out of seventy-two, not twenty-eight out of one hundred. The fraction of the current amount needed for recovery is therefore 28/72 = 7/18. This parts model gives another way to see the changing denominator.
For a short check question, suppose an invented price falls from $80 to $60. The reduction is $20 out of $80, or twenty-five per cent. Returning from $60 to $80 requires an increase of $20 out of $60, or thirty-three and one third per cent. The dollars agree in both directions, while the percentage bases differ. Keeping that distinction clear will help with reverse discounts as well as ordinary percentage comparisons.
The forward reduction and backward recovery therefore form a useful matched pair for revision: keep the same two prices, change the direction, and justify the denominator each time.
These questions collect the small distinctions that are easy to lose when working quickly. Read the answer that matches your uncertainty, then return to one example and explain it without copying. A correct sentence about the base is a useful checkpoint before attempting a longer question.
Are two ten per cent discounts the same as twenty per cent off?
No, if the second discount applies to the reduced price. The retained fraction is 0.90 × 0.90 = 0.81, so the overall reduction is nineteen per cent. On $100, the stages leave $90 and then $81. Both discounts use a rate of ten per cent, but their dollar amounts differ.
Can I work out each discount separately?
Yes. Calculate the first saving from the original price, subtract it, then calculate the second saving from the reduced price. This is fully valid. The multiplier method is a shorter version of the same sequence, provided the reference amount is carried forward correctly at each stage.
Must I always calculate the overall discount?
No. Answer the question asked. If it asks only for the final price, a labelled final-price calculation is sufficient mathematically. If it asks for an equivalent single discount, compare the total saving with the original price. Do not add an unnecessary percentage claim that uses the wrong denominator.
What should I remember tomorrow?
Remember the question “a percentage of what?” Then remember that the current remaining amount becomes the next base in a successive reduction. You can reconstruct the calculation from those two ideas. The aim is not to memorise every example here, but to preserve the relationship when the numbers and wording change.
Continue the Mathematics series
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