Secondary Mathematics: Ratio, Percentage and the Correct Base

Contents · choose your starting point

1. 1. Three questions can use the same two numbers and have different answers
2. 2. A ratio describes relative size, not necessarily an actual count
3. 3. Fractions and percentages are two representations of a comparison
4. 4. Percentage change uses the starting value as its base
5. 5. Multipliers make forward percentage changes explicit
6. 6. Reverse percentage problems recover the original whole
7. 7. Successive changes act on successive bases
8. 8. A percentage of the remainder is not a percentage of the original
9. 9. Changing a ratio requires tracking what stays fixed
10. 10. Combining ratios means matching the shared quantity
11. 11. Rates compare quantities with different units
12. 12. Average speed is total distance divided by total time
13. 13. Not every relationship is directly proportional
14. 14. Percentage points and relative percentage change are different
15. 15. Combining groups requires the actual group sizes
16. 16. Capstone: a budget with two changing bases
17. 17. Independent practice
18. 18. Worked answers
19. 19. A repair routine that keeps the denominator visible
20. 20. Continue through the BTT Mathematics library
21. Sources and scope

Secondary Mathematics · Worked Repair Guide 03

The most important question in a percentage problem is not “Should I multiply or divide?” It is “Percentage of what?” In a ratio problem, the corresponding question is “Which quantities are being compared, and what does one part represent?” A correct calculation using the wrong base still answers the wrong question.

This guide connects fractions, ratios, percentages, rates and changing totals. It teaches how to identify the reference quantity, distinguish additive from multiplicative change, reverse a percentage operation, combine groups correctly and check whether a proportional model is justified.

The worked situations are invented for learning. Prices, budgets, attendance figures and journeys below are not current quotations or reports about real people. The purpose is to make the mathematics visible without requiring outside knowledge of a particular business or event.

Begin with the questions that expose the base. Move to calculation only after you can name the whole or comparison quantity in an ordinary sentence.

1. Three questions can use the same two numbers and have different answers

Suppose a box contains 20 red counters and 30 blue counters. There are 50 counters altogether. The ratio of red to blue is 20:30 = 2:3. The fraction of all counters that are red is 20/50 = 2/5. The number of red counters as a fraction of the number of blue counters is 20/30 = 2/3.

Those answers do not conflict. They answer three different comparisons. The ratio 2:3 does not mean that two thirds of the entire collection is red. The whole consists of 2 + 3 = 5 equal parts, so the red share is two fifths.

Now ask how much greater the blue count is than the red count. The difference is 10. Relative to the red count, 10/20 = 50%, so there are 50% more blue counters than red counters. Relative to the blue count, 10/30 = 1/3, so there are 33⅓% fewer red counters than blue counters.

The same difference produces different percentage comparisons because the denominators differ. Words such as “of”, “than”, “original”, “remaining” and “altogether” identify which quantity belongs in that denominator.

Entry check: in a class with 12 students who cycle and 18 who do not, state the cycling-to-non-cycling ratio, the cycling percentage of the class, and the percentage by which the non-cycling count exceeds the cycling count. The answers are 2:3, 40%, and 50%. Explain all three denominators before continuing.

2. A ratio describes relative size, not necessarily an actual count

The ratio A:B = 2:3 says that the two quantities can be represented as 2k and 3k for a common scale k. The value of k depends on the information in the problem. It is not automatically one person, one dollar or one centimetre.

If A and B together equal 75, the five parts total 75, so one part is 15. Therefore A = 30 and B = 45. Check both the total, 30 + 45 = 75, and the relationship, 30:45 = 2:3.

If instead B exceeds A by 18, the one-part difference equals 18. Then A = 36 and B = 54. The same starting ratio gives different actual quantities because a different absolute condition has been supplied.

Worked example: three quantities are in the ratio 2:3:5 and total 180. There are ten parts, so one part equals 18. The quantities are 36, 54 and 90. A request for the difference between the largest and smallest would use 5 − 2 = 3 parts, giving 54, not three tenths of an arbitrary individual quantity.

A ratio can also compare measurements with units. A length ratio of 2 m to 50 cm should be converted to a common unit before simplifying: 200 cm:50 cm = 4:1. Writing 2:50 ignores the different units and changes the comparison.

When a ratio represents counts of indivisible objects, the final values must be whole numbers. A ratio statement alone allows a scale to be symbolic; the context determines which scales are feasible. Keep that feasibility check separate from the arithmetic of the ratio.

3. Fractions and percentages are two representations of a comparison

A percentage means a number of parts per hundred. Thus 25% = 25/100 = 1/4, while 12.5% = 12.5/100 = 1/8. These are exact numerical equalities. Converting a fraction to a percentage multiplies its numerical value by 100 and attaches the percent symbol.

To find 35% of 240, identify 240 as the whole and calculate 0.35 × 240 = 84. To find what percentage 84 is of 240, calculate 84/240 × 100% = 35%. To find the whole when 35% is 84, solve 0.35W = 84, giving W = 240.

These are the three orientations of one relationship:

part = percentage as a decimal × whole.

Learning them as a connected relationship is more useful than memorising three unrelated verbal rules. Write which of the three quantities is unknown before selecting the operation.

Worked example: 42 students represent 30% of a group. The whole is not 42, so taking 30% of 42 would answer a different question. Let the group size be G. Then 0.30G = 42, hence G = 140. A check gives 30% of 140 = 42.

Some fractions do not have terminating decimal representations. One third is exactly 33⅓%, but approximately 33.3% to one decimal place. Keep the fraction exact during a multistep calculation unless the context requires rounding. Do not let an early rounded representation silently become an exact quantity in the next step.

4. Percentage change uses the starting value as its base

For a positive starting value, percentage change is calculated by comparing the change with that starting value:

percentage change = (new value − original value) / original value × 100%.

If a quantity rises from 200 to 250, the increase is 50 and the percentage increase is 50/200 × 100% = 25%. If it then falls from 250 back to 200, the decrease is 50 but the percentage decrease is 50/250 × 100% = 20%.

The return journey is not a 25% decrease, because the base has changed. A 25% decrease from 250 would be 62.5, leaving 187.5.

Worked example: a practice time falls from 80 minutes to 68 minutes. The reduction is 12 minutes, and 12/80 × 100% = 15%. It is not 12/68 × 100%, because 68 is the new value rather than the starting value.

A percentage change from an original value of zero is not defined by this formula: division by zero is unavailable. For example, a count rising from zero to ten has increased by ten, but there is no finite percentage increase obtained from 10/0. State the absolute change instead of inventing a percentage.

Comparisons involving negative baselines require additional interpretive care. In this guide’s ordinary increase and decrease examples, the original quantities are positive. Do not transfer the everyday meaning of “20% growth” to every signed-number situation without defining what is being compared.

5. Multipliers make forward percentage changes explicit

An increase of 15% retains the original 100% and adds 15%, giving 115% of the starting value. The multiplier is 1.15. A decrease of 15% retains 85%, giving a multiplier of 0.85.

For an original amount P:

after a 15% increase: 1.15P;
after a 15% decrease: 0.85P.

This representation separates the final amount from the size of the change. Multiplying by 0.15 finds only the change, not the resulting amount.

Worked example: an illustrative $160 item is reduced by 25%. The reduction is $40, but the new price is $120. Using a multiplier gives 0.75 × 160 = 120 directly. A check subtracts 40 from 160.

A change greater than 100%: an increase of 150% makes a quantity 250% of its original value, so the multiplier is 2.5. If the original is 40, the result is 100. “Increased by 150%” and “increased to 150% of the original” do not mean the same thing; the latter would give 60.

For a positive amount, an ordinary reduction of 100% leaves zero. A reduction greater than 100% would produce a negative value under the arithmetic formula. Whether that is meaningful depends on the context. An object cannot have a negative physical mass merely because a multiplier can be calculated.

Read the final state aloud: “The new amount is 75% of the original.” This sentence provides a clear direction for both calculation and checking.

6. Reverse percentage problems recover the original whole

Suppose an item costs $72 after a 20% reduction. The $72 is 80% of the original price, not 100%. Let the original be P:

0.80P = 72
P = 72/0.80 = 90.

Adding 20% of $72 would give $86.40, which is not the original price. That calculation uses the reduced amount as the base for the restoration. The actual increase required to go from 72 to 90 is 18/72 = 25%.

Worked example: a quantity becomes 345 after a 15% increase. The relationship is 1.15Q = 345, so Q = 300. Check: 15% of 300 is 45, and 300 + 45 = 345.

A reverse question should begin with the sentence “The given final amount represents ___% of the original.” That blank determines the divisor. It is a more dependable prompt than asking whether the word “discount” always requires subtraction.

Sometimes the change itself is given instead. If a 12% increase amounts to 36 units, then 0.12Q = 36 and the original is 300. The final amount is 336. Do not divide 36 by 1.12, because 36 represents the extra 12%, not the complete 112%.

Contrast pair: “After losing 30%, the remaining amount is 56” gives an original of 80. “The lost 30% is 56” gives an original of 560/3, or 186⅔. Similar words can identify different known quantities. Label the known part before touching the calculator.

7. Successive changes act on successive bases

Consider an original amount of 1,000 reduced by 10% and then by 20%. The first change gives 0.9 × 1,000 = 900. The second gives 0.8 × 900 = 720. The combined multiplier is 0.9 × 0.8 = 0.72, so the total reduction is 28%, not 30%.

The percentages cannot simply be added because the second reduction is applied to a smaller base. Its absolute size is 180, not 200.

Now consider an increase of 20% followed by a decrease of 20% on an initial 200:

200 × 1.2 × 0.8 = 192.

The final amount is 96% of the original, a 4% loss. In general, increasing a positive quantity by a decimal fraction r and then decreasing the new amount by the same fraction gives the multiplier (1 + r)(1 − r) = 1 − r². The order of these two pure percentage multipliers does not affect the final product.

That commutativity does not extend to every real instruction. A fixed $10 reduction followed by a 20% reduction gives a different result from a 20% reduction followed by $10 off. Starting with $100, the first route gives 90 × 0.8 = 72; the second gives 80 − 10 = 70.

Identify whether each change is multiplicative or additive. Also inspect which items or amounts each change applies to. The algebra of successive percentages assumes the described multipliers apply to the full successive amounts, without additional thresholds, caps or exclusions.

8. A percentage of the remainder is not a percentage of the original

An invented activity budget contains $800. Forty per cent is used for materials. That is $320, leaving $480. If 25% of the remainder is used for printing, printing costs 0.25 × 480 = $120, not $200.

The remaining money is $360. As a fraction of the original, this is 360/800 = 45%. The multiplier route makes the structure especially clear:

remaining fraction = (1 − 0.40)(1 − 0.25) = 0.60 × 0.75 = 0.45.

The second percentage refers to the remainder. Adding 40% and 25% would incorrectly assume both spending amounts use the same original base.

Reverse version: after spending 30% of a fund and then 20% of the remainder, $280 is left. The remaining fraction is 0.7 × 0.8 = 0.56. Therefore the original fund was 280/0.56 = $500.

A table can help when there are several stages. Its useful columns are “stage”, “amount before”, “operation” and “amount after”. Do not give every row a percentage of the original unless you have actually converted it to that common base.

For a visual representation, imagine the original as a full bar. After removing 30%, 70% remains. Taking 20% of that remaining bar removes 14% of the original. The full spending is therefore 30% + 14% = 44%, leaving 56%. This agrees with the multiplier calculation and explains the changing whole.

9. Changing a ratio requires tracking what stays fixed

Suppose a container has red and blue counters in the ratio 2:3, with 200 counters altogether. The original counts are 80 red and 120 blue. Adding 50 red counters produces 130 red and 120 blue, so the new ratio is 13:12.

The blue count stayed fixed; the total did not. Trying to preserve the original five-part total would misrepresent the operation. After a change, re-establish the quantities before simplifying the new ratio.

Unknown starting scale: A and B have quantities in the ratio 3:5. Adding 12 to A makes them equal. Let their original quantities be 3k and 5k. Then:

3k + 12 = 5k
12 = 2k
k = 6.

The original quantities are 18 and 30. After the addition they are 30 and 30. The fixed quantity was B; the difference between the original amounts identified the two-part gap.

Transfer example: two groups start in the ratio 3:2. Twelve members move from the second group to the first, changing the ratio to 7:3. The total remains fixed. Initially the first group is 3/5 of the total; afterwards it is 7/10. Its increase is 1/10 of the total, which equals 12. Therefore the total is 120. The counts change from 72 and 48 to 84 and 36.

In an addition problem, a total may rise. In a removal problem, it may fall. In a transfer problem, it may remain unchanged. Identify that invariant before choosing a ratio method.

10. Combining ratios means matching the shared quantity

Suppose A:B = 2:3 and B:C = 4:5. The two appearances of B must refer to the same amount before the ratios can be combined. In the first ratio B occupies three parts; in the second it occupies four parts. Match both to twelve:

A:B = 8:12;
B:C = 12:15.

Therefore A:B:C = 8:12:15. Writing 2:3:5 would incorrectly treat different scales as though they were already the same.

A numerical check helps. With the combined ratio, A/B = 8/12 = 2/3, and B/C = 12/15 = 4/5. Both original relationships are preserved.

Worked example: the ratio of notebooks to folders is 3:2, and the ratio of folders to pens is 5:4. Match the folder quantity to ten parts. Notebooks:folders becomes 15:10, while folders:pens becomes 10:8. The three-part ratio is 15:10:8.

The shared label must represent the same collection, time and unit. A morning stock of folders and an afternoon stock after sales cannot be matched merely because both are called “folders”. A ratio is a relationship between specific quantities, not a pattern of numbers detached from their meaning.

This is a useful point of contact with equations. Representing B consistently as a single quantity is equivalent to using the same variable for the same object. Different scales must be reconciled before information can be combined.

11. Rates compare quantities with different units

A ratio such as 4:1 may compare two lengths after conversion to a common unit. A rate such as 60 kilometres per hour compares different kinds of quantity. Its units are part of its meaning.

The relationship distance = speed × time must use compatible units. At 60 km/h for 30 minutes, the time is 0.5 hours, so the distance is 30 km. Multiplying 60 by 30 without converting minutes would produce a number with the wrong interpretation.

To convert 60 km/h to metres per minute, convert both dimensions:

60 km/h = 60,000 m / 60 min = 1,000 m/min.

A “per” quantity is a division. Kilometres per hour means kilometres divided by hours. That meaning helps both in conversion and in rearrangement.

Unit-price example: an imaginary pack of three pens costs $4.50, while a pack of five costs $7. The unit prices are $1.50 and $1.40 per pen. The five-pack has the lower unit price. But someone needing exactly three pens, with no use for extras, would spend less by buying the three-pack. A lower rate is not automatically the best decision under every purchasing constraint.

For word problems, write the required output unit before calculating. If the question asks for time, a distance divided by a speed is plausible because kilometres divided by kilometres per hour gives hours. Units provide an additional check on the selected operation.

12. Average speed is total distance divided by total time

A journey travels 60 km at 60 km/h and then 60 km at 40 km/h. The times are 1 hour and 1.5 hours. Total distance is 120 km and total time is 2.5 hours, so the average speed is 48 km/h.

The arithmetic mean of 60 and 40 is 50, but the traveller spends unequal amounts of time at those speeds. The slower section occupies more time, so an unweighted average of the two speed figures is inappropriate.

If the traveller instead spends one hour at each speed, total distance is 100 km and total time is two hours. The average is then 50 km/h. The simple mean works in that case because the time weights are equal.

Another example: travel 60 km at 40 km/h and 120 km at 60 km/h. The times are 1.5 hours and 2 hours. Average speed is 180/3.5 = 360/7 km/h, approximately 51.4 km/h to one decimal place.

The safest general route does not require memorising separate average-speed shortcuts. Recover the distance and time for each stage, add like quantities, and divide the totals. Include waiting time when the question defines the journey duration to include that wait.

For example, adding a half-hour stop to the first 120 km journey makes the elapsed time three hours, so the journey’s average speed including the stop is 40 km/h. Always identify the time interval the question is asking about.

13. Not every relationship is directly proportional

Direct proportion has the form y = kx for a constant k over the stated domain. Doubling x doubles y, and y/x is constant wherever x is non-zero. This model includes no non-zero fixed starting amount.

Suppose an illustrative printing service charges a fixed $5 setup amount plus $0.20 per page. Its cost is C = 5 + 0.20n. Ten pages cost $7 and twenty pages cost $9. Doubling the page count does not double the total cost because the fixed charge is not doubled.

For a fixed distance, time is inversely proportional to a constant speed: t = d/v, with v > 0. Doubling speed halves the travel time within that idealised constant-speed model. That does not establish that real travel can always be sped up in this way; roads and other constraints belong to the actual situation.

Similarly, doubling the number of workers halves a job’s time only under assumptions such as equal independent rates and perfectly divisible work, without new coordination costs. State those assumptions when using an idealised rate model. A mathematical formula does not guarantee that a real project obeys it.

For geometrically similar shapes, a length scale factor k produces area factor k² and volume factor k³. A length increase of 50% has k = 1.5, so the area factor is 2.25 and the volume factor is 3.375. These correspond to increases of 125% and 237.5%, not 50%. The power follows the dimension of the quantity being scaled.

14. Percentage points and relative percentage change are different

Suppose a success rate rises from 40% to 55%. The difference is 15 percentage points. Relative to the starting rate, the increase is 15/40 × 100% = 37.5%.

Both statements can describe the same change, but they use different units and comparisons. “Fifteen percentage points” subtracts two percentages. “A 37.5% relative increase” compares that difference with the starting percentage.

A test score moving from 45 out of 100 to 60 out of 100 rises by 15 marks and by 15 percentage points of the available marks. Relative to the initial score of 45, it rises by 33⅓%. Those numerical descriptions do not by themselves explain why the score changed, whether the tests were comparable, or whether the improvement will persist.

When reading an assertion about improvement, ask for the starting value, ending value, denominator and observation period. A very large relative increase can arise from a very small starting value. Moving from one successful attempt to two is a 100% increase in the count, but it is still only one additional success.

Conversely, a small percentage of a large base can represent a large absolute quantity. The number of affected cases cannot be understood from the percentage alone. This is not a reason to distrust percentages; it is a reason to insist that their bases remain visible.

15. Combining groups requires the actual group sizes

An invented activity has two classes. Class A has 60 students, of whom 75% attend. Class B has 40 students, of whom 60% attend. The attendance counts are 45 and 24, so total attendance is 69 out of 100, or 69%.

Averaging the two percentages gives 67.5%, which is wrong for the combined group. It gives equal weight to classes with unequal sizes. The correct combined percentage uses total attendees divided by total students.

The same principle applies to combining concentrations, test results or completion rates when the underlying bases differ. Convert each percentage to its corresponding amount, add compatible amounts, then divide by the combined base.

Reverse example: a group of 80 has a 70% completion rate, so 56 people complete the task. A second group of 20 has a 90% completion rate, giving 18 completions. Together they have 74 completions out of 100, or 74%, not 80%.

Before combining, make sure the definitions match. “Completed within a week” and “eventually completed” are not identical outcomes. Even flawless arithmetic cannot make different definitions interchangeable. For mathematical exercises, the shared definition is normally stated; for real data, it must be checked.

This section brings the entire guide back to its opening question. A percentage is never free-floating. Its denominator determines its weight in a combined result.

16. Capstone: a budget with two changing bases

An imaginary student programme begins with a budget of $2,400. It spends 40% of the original budget on materials, then 35% of the remaining money on printing. Find the money left and its percentage of the original budget. Then find the original budget needed to leave $1,170 under the same spending rules.

Materials cost 0.40 × 2,400 = $960. The first remainder is $1,440. Printing costs 0.35 × 1,440 = $504. The final remainder is $936.

The remaining fraction is 0.60 × 0.65 = 0.39, so $936 is 39% of the original $2,400. The spending percentages cannot be added as 40% + 35% because the second uses a different base.

For the reverse problem, let the required starting budget be B. The relationship is 0.39B = 1,170, giving B = $3,000. Check: materials take $1,200, leaving $1,800. Printing takes $630, leaving $1,170.

Now change the printing instruction to “35% of the original budget”. The total spending becomes 75% of the original, so 25% remains. On $2,400, that would be $600. Only one phrase changed, but the mathematical model changed with it.

A strong explanation identifies the changed denominator before calculating. That ability is more valuable than learning a formula keyed to the superficial appearance of the question.

17. Independent practice

For every percentage task, write the base in words. For every ratio task, identify whether a total, a difference or a fixed quantity supplies the scale.

1. A:B = 3:7 and A + B = 150. Find A and B.
2. A:B = 4:9 and B − A = 35. Find A and B.
3. In a collection of 18 red and 42 blue counters, find the red percentage of the whole.
4. A is 24 and B is 30. By what percentage is B greater than A?
5. By what percentage is A less than B in Question 4?
6. Find 12.5% of 360.
7. A quantity rises from 160 to 184. Find the percentage increase.
8. A quantity falls from 250 to 215. Find the percentage decrease.
9. An illustrative price is $84 after a 30% reduction. Find the original price.
10. After a 12% increase, a quantity is 448. Find its original value.
11. A quantity of 500 decreases by 10% and then by 10%. Find the result and overall percentage decrease.
12. After spending 20% of a fund and then 25% of the remainder, $420 is left. Find the original fund.
13. A:B = 2:5 and B:C = 3:4. Find A:B:C.
14. A and B are in the ratio 4:7. Adding 18 to A makes them equal. Find their original values.
15. A vehicle travels 90 km in 1.5 hours. Find its average speed in km/h.
16. A journey covers 40 km at 40 km/h and another 40 km at 80 km/h. Find its average speed.
17. A rate rises from 30% to 36%. State its change in percentage points and its relative percentage increase.
18. A group of 30 has 80% attendance and a group of 70 has 60% attendance. Find the combined attendance percentage.
19. Similar shapes have length scale factor 2.5. Find their area scale factor.
20. A cost model is C = 8 + 0.5n. Find the cost for 12 units and explain why total cost is not directly proportional to n.

18. Worked answers

1. A = 45 and B = 105. Ten parts total 150, so one part is 15. The requested quantities use three and seven of those equal parts.

2. A = 28 and B = 63. The difference is five parts, so one part is 35/5 = 7. Do not divide the difference by the thirteen-part total.

3. Thirty per cent. There are 60 counters altogether, so the red share is 18/60. The denominator is the whole collection, not just the blue count.

4. Twenty-five per cent. The difference is 6, compared with the starting comparison quantity A = 24. Thus 6/24 × 100% = 25%.

5. Twenty per cent. The same difference is now compared with B = 30. Therefore 6/30 × 100% = 20%. The wording changes the base.

6. Forty-five. Since 12.5% = 1/8, divide 360 by 8. Alternatively, 0.125 × 360 gives the same exact answer.

7. Fifteen per cent. The increase is 24, and the original is 160. The multiplier check is 1.15 × 160 = 184.

8. Fourteen per cent. The decrease is 35. Dividing by the original 250 gives 0.14, so 86% of the original remains.

9. One hundred and twenty dollars. The final price is 70% of the original, so divide 84 by 0.70. A 30% reduction of $120 is $36, leaving $84.

10. Four hundred. The final quantity is 112% of the original, so 448/1.12 = 400. Dividing by 0.12 would treat the complete final amount as though it were only the increase.

11. Four hundred and five; a 19% decrease. The multiplier is 0.9² = 0.81. Therefore 500 × 0.81 = 405. The second loss is 45 rather than 50.

12. Seven hundred dollars. The retained fraction is 0.8 × 0.75 = 0.60. The original is 420/0.60. Check the two spending stages separately.

13. 6:15:20. Match the B quantity to fifteen parts. Multiplying the first ratio by 3 and the second by 5 preserves both relationships.

14. Twenty-four and forty-two. The original gap is three parts and equals 18. One part is 6. After the addition, A also equals 42.

15. Sixty kilometres per hour. Average speed is 90/1.5. The units are kilometres divided by hours, matching the requested output.

16. 160/3 km/h, approximately 53.3 km/h. The times are one hour and half an hour. Divide the 80 km total by 1.5 hours, rather than averaging the two speeds equally.

17. Six percentage points; a 20% relative increase. The rate difference is 36 − 30 = 6 percentage points. Relative to the initial rate, 6/30 = 20%.

18. Sixty-six per cent. The attendance counts are 24 and 42. Their sum is 66 out of 100. The unequal group sizes determine the weights.

19. 6.25. Area scales as the square of the length factor, so 2.5² = 6.25. This is a factor, not a 6.25% increase.

20. Fourteen currency units. Substitution gives 8 + 0.5(12) = 14. The fixed term 8 means doubling n does not generally double the total cost. Only the variable part 0.5n is directly proportional to n.

19. A repair routine that keeps the denominator visible

When a solution fails, circle the exact quantity used as the base. Ask whether it represents the original total, a remainder, a comparison quantity, a time interval or a combined group. Many wrong operations become obvious once their implied denominator is spoken aloud.

Keep a record that names the mistake specifically: “Used final price as the original”, “Added percentages with different bases”, or “Averaged speeds with unequal times”. A useful next question changes one relevant feature while retaining the underlying principle.

For reverse percentages, alternate between knowing the final amount and knowing the change itself. For ratios, alternate between total, difference and transfer conditions. For averages, compare equal-time and equal-distance cases. The contrast is the teaching tool: it forces the base to be chosen rather than guessed from a familiar surface pattern.

Before declaring a repair complete, ask for a short explanation without a numerical example. “Why does a 20% increase followed by a 20% decrease not return to the original?” can be answered through changing bases or through the product 1.2 × 0.8. A learner who can explain and apply either route has shown more than recognition of a copied answer.

Use the record to select the next task, not to assign a fixed ability label. These practice routines are suggestions for instruction; progress depends on what the learner can actually do independently in later work.

20. Continue through the BTT Mathematics library

The BTT Mathematics Hub provides the wider learning route. The BTT Mathematical Lab is the next doorway when the same comparison error persists across several topics.

For the algebra underneath a ratio model, use Signed Numbers, Brackets and Algebraic Structure. To solve for an unknown original quantity, use Equations, Balance and Checking. To recognise direct proportion, fixed starting amounts and rates visually, continue to Graphs, Tables and Relationships.

Sources and scope

All worked calculations, invented contexts, counterexamples and practice answers are original teaching material. Their conditions and derivations are displayed rather than delegated to an external authority. Hypothetical costs are not financial advice, actual service prices or tax statements.

Use MOE: Curriculum for secondary schools to check the applicable school and subject-level framework. This guide is organised by mathematical dependencies and recurring comparison errors, not by a claim that all topics occur in the same year for every student. Framework checked on 6 September 2026.