Secondary Mathematics Tuition | How to Solve Math Problems: A Complete Secondary Problem-Solving System

Secondary Mathematics Tuition · From an unfamiliar question to a justified solution

Clara has thirteen sealed boxes on the page in front of her. Some contain six components; the others contain ten. Altogether there are 102 components. The question asks how many boxes of each kind there are.

She knows how to multiply, subtract and solve equations. None of that knowledge is missing. What she does not yet see is how to make those abilities meet this question. Ryan starts trying combinations. Aisha writes two equations. Clara looks at both approaches and wonders how anyone knew which move to make first.

To solve a math problem, identify the exact result required, represent the given relationships, choose a method that connects them to the target, carry out the method while preserving its conditions, and test the conclusion against the original question. When a route stops being useful, inspect the last justified step and change something specific: the representation, the intermediate goal or the method. Do not merely repeat the same unsuccessful move with greater determination.

This is a worked casebook for secondary mathematics. It shows the decisions inside complete solutions, including alternatives that are tempting but wrong, moments when a change of method helps, and answers that must be rejected because they fail a condition. The examples are original. Clara, Ryan and Aisha are fictional recurring learners, not reported student cases or permanent ability types.

Find your starting point: unable to begin? Start with the thirteen boxes. Confused by wording? Read the translation chapter. Unsure which method fits? Open route selection. Stuck midway? Use the recovery chapter. Ready for independent work? Attempt the mixed checkpoint before reading its answers.

School systems place topics in different years. Work only on mathematics you have been taught, and use your course materials for its required methods and assessment rules. Here, math, maths and mathematics name the same subject. The Secondary Mathematics Learning Hub supplies the existing topic and stage routes. For study organisation use How to Study Maths Effectively; for the reasons behind rules use How to Understand Mathematics Instead of Memorising It. This guide concentrates on assembling those abilities into solutions.

1. The first useful move does not have to be the final method

Return to the thirteen boxes. A useful first move is to imagine that all thirteen are the smaller kind. They would contain 13 × 6 = 78 components. The real total is 102, an excess of 24. Replacing one six-component box with one ten-component box adds four components without changing the number of boxes. Six replacements account for the excess, so there are six large boxes and seven small boxes.

The answer checks both pieces of information: 6 + 7 = 13 boxes, and 6 × 10 + 7 × 6 = 102 components. Checking only the component total would not be enough. Two large boxes and many small ones might reproduce a total in another version of the problem while violating the required box count. Every supplied condition has a role.

Why was the imaginary all-small collection useful? It respected one constraint immediately: thirteen boxes. It then converted the remaining difference into repeated changes of a known size. The imagined collection did not need to be physically present. It was a mathematical reference case selected because comparing it with the real collection would isolate the unknown.

Aisha’s equations express the same structure. Let s and l be the numbers of small and large boxes. Then s + l = 13 and 6s + 10l = 102. Multiplying the first equation by six gives 6s + 6l = 78. Subtracting this from the component equation gives 4l = 24, so l = 6 and s = 7. The algebraic subtraction is the same comparison that Clara made with the imagined collection.

Ryan’s systematic trial can also work. If there are no large boxes, the total is 78. One large box gives 82, two give 86, and each further replacement adds four. The list reaches 102 after six replacements. It is not necessary to condemn trial as inferior thinking. The important difference is between a controlled sequence whose change is understood and disconnected guesses that forget what was already tried.

The method can be chosen after the structure begins to emerge. Clara did not need to know the name of an algebraic technique before calculating 78. That calculation was useful because she could state what it represented and how it would help. In contrast, multiplying 13 by 102 would combine two visible numbers without a relationship connecting that product to the target.

Now change the stated component total to 104 while keeping all other conditions. The all-small reference still contains 78, leaving an excess of 26. No whole number of four-component replacements produces 26. The equations give l = 6.5, which is not a permissible box count. The correct conclusion is that the stated conditions cannot all hold, not that six and a half boxes should be rounded to seven.

This variation reveals a second purpose of representation. A good model does not merely manufacture an answer. It can expose inconsistency. Arithmetic that reaches a fractional count is not necessarily faulty; it may be telling you that the supplied story is impossible under its own rules. Return to the model before blaming the last division.

For a general collection of N boxes holding a or b components, with b greater than a, the all-small reference holds aN. If T is the actual total, the large-box count is (T − aN)/(b − a). This expression is useful only when the result is an integer between zero and N. The formula compresses the comparison, including the need to respect the count domain.

That is our first problem-solving lesson. A productive move changes the state of the question: it identifies a relationship, removes an unknown, establishes a bound or reveals a contradiction. The move need not finish the problem. It must make the next decision better informed than it was before.

2. Decide what a finished answer must contain

Before choosing a calculation, decide what would count as answering the question. Is the target one value, every permitted value, a range, the least possible count, a comparison or a proof? These are different mathematical jobs. A page can contain correct calculations and still stop before the required job has been completed.

Suppose a club rents equipment for a fixed charge of 18 units plus 7 units per hour. A student has a budget of 80 units. Solving 18 + 7h = 80 gives h = 62/7, about 8.86. Whether that is the answer depends on the rental rule and the question. If hours are continuously divisible, it is the maximum duration under the stated model. If only complete one-hour sessions are sold, the maximum is eight sessions.

If the question asks which whole session counts are affordable, giving only eight is incomplete. The permissible positive counts are one through eight; zero can be included when the context allows no purchase. If the question asks whether nine sessions are affordable, a direct check gives 18 + 63 = 81, exceeding the budget. One relationship supports several different answer forms.

Write a brief target sentence before working: “Find the largest whole number of sessions whose cost does not exceed 80.” That sentence contains the unknown, the integer condition and the inequality. It is more useful than copying “budget” into the margin. It also supplies a final check: show that eight works and the next whole count does not.

A request for all solutions needs completeness. The equation x² = 49 has real solutions 7 and −7. Showing that 7 works proves only one candidate. Conversely, the expression √49 asks for the principal square root and has the single value 7. The wording and notation decide whether a second answer is required, not a general habit of adding or omitting a ± sign.

A proof has a different finish again. To show that the sum of two consecutive integers is odd, calculating 6 + 7 = 13 is an example. A general argument writes the integers as n and n + 1, giving 2n + 1. For every integer n, this has the form of an odd integer. The finish is the connection that covers the full stated class.

For a comparison, equality is often only a boundary. Two plans costing 18 + 7h and 30 + 5h are equal at h = 6. To answer which is cheaper, compare either side as well: the first is cheaper for nonnegative h below six, and the second for h above six. At six they tie. “Six” alone does not answer the comparison, although it is an important intermediate result.

For an estimate, exact-looking precision can misrepresent the information. If a graph supports only an approximate reading, write an approximation and explain the reading where needed. If a task requests an exact value, rounding an irrational result prematurely can lose the required answer form. The endpoint must match the information and instructions, not the number of decimal places a calculator happens to display.

Aisha uses a simple final question: “Would someone who has not seen my working know what this number refers to?” A bare 8 may mean hours, sessions, boxes or a coordinate. A sentence such as “The largest affordable purchase is eight one-hour sessions” closes the connection to the task. That sentence is not decorative English after the mathematics; it identifies the quantity the mathematics has found.

Do not expand this into a long ritual for every short exercise. In a straightforward substitution, the target may already be obvious. Use the explicit target sentence when a question has several conditions, parts or plausible stopping points. The goal is to prevent the route from ending at a convenient intermediate quantity rather than at the requested conclusion.

3. Read relationships instead of hunting for operation keywords

Words matter, but individual words do not reliably select an operation on their own. “More” can appear in a subtraction question, “altogether” can accompany an equation with unknown groups, and “of” may describe a part, a rate or an item category. Read the relationship between quantities before choosing how to calculate it.

Clara has 17 cards, which is five more than Ryan has. The relationship is C = R + 5. With C = 17, Ryan has 12. Adding five to seventeen because the word “more” appears would answer a different question. By contrast, if Ryan has seventeen and Clara has five more than Ryan, the same relationship gives Clara twenty-two. Which quantity is known changes the direction of use.

Translate the complete sentence before inserting the number. “A is five more than B” becomes A = B + 5. “A is three times B” becomes A = 3B. “A is five less than three times B” becomes A = 3B − 5. This small separation between relation and data reduces the temptation to combine the numbers before deciding what they represent.

Brackets preserve the scope of a phrase. “Four times the sum of a number and two” is 4(x + 2), while “two more than four times a number” is 4x + 2. At x = 3 the values are twenty and fourteen. The difference is not a subtle preference in notation. The first statement multiplies both contributions to the sum; the second adds two only after multiplying x.

Reference quantities create another reading demand. A container is one quarter full and contains 18 litres. The 18 represents a quarter of the capacity, so capacity is 72 litres. If the container has 18 litres of empty space while one quarter full, that empty space represents three quarters, giving capacity 24 litres. The same fraction and the same number can describe two different parts.

Ask what each number measures. In a speed question, is the given time the moving time, the whole elapsed time or a delay before departure? In a price question, is the amount before a discount, after a discount or a fee added separately? In a geometry question, is a marked length a side, a perpendicular height or the total of several segments? A numerical value has no problem-solving meaning until its quantity is identified.

Not every sentence needs a symbol. Sometimes a short labelled note does the job better. “Twelve litres remains after the second transfer” preserves the timing of a quantity. “The two widths together total eight” preserves a count of repeated dimensions. Choose a notation that makes a likely confusion harder to repeat.

Keep conditions while simplifying the story. Names, colours and dates may be irrelevant to a calculation, but “without replacement”, “positive integer”, “constant speed” and “parallel” can determine the model. Removing every word to leave only numbers is not mathematical abstraction. Abstraction preserves the relationships that control the answer.

For example, a bag starts with five red and four blue counters. After a red is drawn without replacement, the next draw comes from four red and four blue counters. Ignoring the two words “without replacement” changes the second probability. They contain no numerical digits, yet they determine which numerical calculation is valid.

A useful translation check runs backwards. Read your equation as a sentence and compare it with the original. If you wrote 3x − 8 = 19, say “three times the unknown, then subtract eight, gives nineteen”. Does that match the stated sequence? This catches a modelling error before later algebra makes the wrong representation look authoritative.

For focused work on notation, use Read Mathematical Notation Like a Language. For a concise explanation of the learner-side process, How Mathematical Problem Solving Works for a Student provides the wider mechanism. The cases here show how to carry those reading decisions through a complete mathematical task.

4. Choose a representation for the uncertainty it removes

A diagram, equation or table is useful when it makes a relationship easier to inspect. It is not valuable merely because a problem-solving checklist says to draw something. Ask which uncertainty the representation will remove. If the answer is unclear, the representation may be adding work without adding information.

Consider a mixture of two kinds of grain. One costs 4 units per kilogram and the other 7 units per kilogram. A 12-kilogram mixture costs 66 units in total, with no extra costs and exact quantities in this model. Find the mass of each kind. The unknowns concern two quantities with one total mass and one total cost.

A compact table can keep the measures aligned. The first row contains mass x, unit cost 4 and total cost 4x. The second contains mass 12 − x, unit cost 7 and total cost 7(12 − x). The cost condition becomes 4x + 7(12 − x) = 66. Expanding gives 84 − 3x = 66, so x = 6. Both types contribute six kilograms.

The table’s purpose is not to create a new arithmetic method. It prevents adding a mass directly to a cost and makes clear why each contribution is a product. A student who writes 4 + 7 + 12 = 66 has not merely chosen an inefficient procedure; the left side combines unlike quantities without representing the stated cost relationship.

An all-expensive reference gives another representation. Twelve kilograms at seven units per kilogram would cost 84. The real mixture is 18 units cheaper. Replacing one kilogram of the expensive grain with the cheaper grain saves three units, so six kilograms must be replaced. This is the same mathematical comparison used for the boxes, now with continuous mass instead of whole box counts.

A line graph could show total cost as the cheaper mass varies from zero to twelve. The equation C = 84 − 3x is decreasing, and C = 66 occurs at x = 6. This view makes bounds visible: the total cost must lie between 48 and 84. A claimed cost of 90 would be impossible under the model, even before the equation was solved.

Would drawing twelve little bags help? Possibly, for a learner who needs to understand the fixed total, but it is not the most direct representation of continuously divisible mass. The appropriate representation depends on the actual obstacle. A picture chosen for one learner’s uncertainty may be unnecessary for another who already controls that relationship.

The What Works Clearinghouse problem-solving guide for Grades 4–8 gives strong-evidence ratings to teaching visual representations and to helping learners monitor and reflect on solving. Its scope does not validate every activity for every secondary course. Here, the practical use is specific: choose a representation that exposes a relationship you need to control. See Improving Mathematical Problem Solving in Grades 4 Through 8.

Translations between representations need checking. If your table says the masses total twelve, the algebra must preserve x + (12 − x) = 12. If your graph uses cheaper mass as the input, its negative gradient describes falling cost as cheaper material replaces expensive material. Changing the input to expensive mass would reverse the direction of the graph. The label is part of the model.

Before leaving a representation, ask what it has accomplished. Has it removed an unknown, shown a bound, revealed an overlap, or made a repeated contribution countable? Then use that information. Spending time beautifying a table after its decisive relationship is already visible may distract from the next mathematical step.

The specialist guide Represent the Problem Before You Calculate is the deeper route for this local skill. In complete problem solving, representations are selected, tested and sometimes replaced. They are working descriptions of the task, not compulsory decorations around an answer.

5. Choose a route by asking what it will make simpler

Knowing several methods is useful only when you can decide what each method offers here. “I know substitution, elimination and graphs” is a list of abilities. Choosing among them requires inspecting the present relationships, the desired precision and the amount of work each route creates.

Take the system 2x + y = 11 and x + 3y = 18. Substitution is attractive because the first equation gives y = 11 − 2x immediately. Replacing y in the second gives x + 3(11 − 2x) = 18, so −5x + 33 = 18, x = 3 and y = 5. The route uses a variable already easy to isolate.

Elimination is equally valid. Double the second equation to get 2x + 6y = 36, then subtract the first to obtain 5y = 25. Again y = 5 and x = 3. This route avoids substituting an expression into a bracket but introduces a multiplied equation. Which is easier may depend on the learner’s current fluency, not on a universal ranking of methods.

A graph locates the intersection at (3, 5), but an inaccurately drawn graph may give only an approximate reading. If the question requires exact values, algebra supplies them directly. The graph remains useful as an interpretation or check: the answer must lie on both lines. A representation can be excellent for understanding without being the most efficient exact calculation.

Try a changed system: y = 4x − 9 and 3x + y = 12. Substitution now requires almost no preparation. In a system with coefficients already equal or opposite, elimination may be the natural first candidate. Look for the feature that makes one route cheap, rather than choosing whichever method appeared on the last worksheet.

Route choice also depends on the target. For x² − 10x + 21, factorisation reveals zeros at three and seven. Completing the square gives (x − 5)² − 4 and reveals the minimum. Both forms describe the same expression, but the question decides which information you need first. Starting with a familiar method is reasonable; continuing with it after it stops serving the target is not automatically sensible.

The algebra practice guide from the What Works Clearinghouse recommends deliberate choice among alternative strategies and rates that recommendation as supported by moderate evidence. Its other recommendations on solved-problem analysis and algebraic structure have minimal-evidence ratings. The recommendations should not be presented as equally tested interventions. See Teaching Strategies for Improving Algebra Knowledge in Middle and High School Students.

For your own work, write a small prediction: “Substitution will leave one equation in x,” or “Factoring will reveal when the product is zero.” This is not a lengthy explanation for every line. It is a way to make route selection explicit enough that you can recognise whether the promised simplification actually occurs.

Do not generate five complete solutions before choosing one. One or two plausible routes are usually enough for a useful comparison in a learning task. If you already see a valid efficient route, solve it. Alternative methods are valuable when they reveal structure, provide a check or repair a difficulty, not when they become a performance of maximum method count.

Ryan chooses elimination because his bracket handling is uncertain. Clara chooses substitution and then makes the bracket step visible on a separate line. Aisha uses the graph afterward to interpret the shared solution. Their work can all be good without becoming identical. The common standard is validity, relevance to the target and an answer that satisfies the original system.

The focused resource Compare Different Methods Until You Can Choose supplies more detailed comparisons. The wider system asks one practical question: what will this route make simpler, and is that simplification useful for the quantity I need?

6. Build a chain of intermediate goals before doing the arithmetic

A multistep problem often feels difficult because the requested quantity is not directly connected to the first available number. The solution needs an intermediate quantity. Finding that quantity is part of the mathematics, not an administrative step before the “real” calculation.

An item receives a 10% discount on its original price. A delivery charge of 18 units is then added, making the total payment 198 units. Find the original price. The final payment is not the discounted price: it contains a separate fee. The first useful intermediate quantity is therefore the item price after the discount but before delivery.

Subtract the fee: 198 − 18 = 180. This is 90% of the original price, so the original is 180/0.9 = 200. A check repeats the actual order: 10% off 200 leaves 180; adding 18 gives 198. The sequence matters because the fee was not stated to receive the discount.

The one-line model is 0.9p + 18 = 198. It contains the same intermediate structure in compressed form. A student who writes 0.9(p + 18) = 198 discounts the delivery fee as well. Solving that equation accurately would answer a different pricing rule. The error occurs before the algebra, in the relationship chosen.

Work backwards from the requested original price: to find it, I need the amount that represents 90% of it. To find that amount, I must remove the added fee from the final payment. Work forwards to verify the chain: original price, discounted item price, total payment. The two directions help establish whether every intermediate quantity has the intended meaning.

Now suppose the question instead says that an 18-unit fee is added first and the combined amount is discounted by 10%, giving a final payment of 198. Then 0.9(p + 18) = 198 is correct, so p + 18 = 220 and p = 202. The two models differ only in the operation order, yet the original prices differ. A memorised “reverse percentage” routine must still read the sequence it is reversing.

Intermediate goals also occur in geometry. To find the area of a triangle, you may first need a perpendicular height; to find that height, you may need a right-triangle relation. In statistics, to combine group means, you first recover group totals. In probability, to calculate an event after a draw, you first update the remaining collection. The useful intermediate quantity depends on the structure, not just the topic name.

Clara writes her chain as three short statements rather than three disconnected computations. “Remove delivery to recover discounted price. Divide by the retained proportion to recover original price. Replay the pricing rule.” When she later checks the work, the purpose of each number remains visible. A naked 180 in the margin would be easier to reuse incorrectly.

Keep units and conditions attached to intermediate results. A time calculated in minutes should not enter a rate formula expecting hours without conversion. A value rounded for one display should not replace an exact quantity needed by a later step. The chain is only as reliable as the meaning passed from one stage to the next.

Sometimes a planned chain exposes missing information. If the delivery fee is only described as “an additional charge” with no amount or rule, the original price cannot be uniquely recovered from the final payment alone. Recognising the missing link is a legitimate result. Do not fill it with an invented assumption merely to keep the calculation moving.

The goal is not to plan every arithmetic detail before writing anything. It is to see enough of the dependency to avoid solving for irrelevant quantities. A useful next subgoal is one that supplies information a later step actually needs. When you cannot explain that connection, pause and reconsider why you are calculating it.

7. In geometry, trace the object the question actually asks you to measure

A rectangular sheet measures 14 units across and 10 units high. A rectangular notch 5 units wide and 4 units deep is cut downward from the middle of the top edge, leaving positive lengths of top edge on both sides. Find the remaining area and the perimeter of the remaining shape.

The area is straightforward subtraction: the original 140 square units loses a 20-square-unit notch, leaving 120. The perimeter needs a different representation. It measures the entire exposed boundary after the cut, including the two vertical sides of the notch and its bottom edge.

The original perimeter is 48. Cutting the notch removes a top segment of length five and adds a bottom segment of the same length, so those horizontal changes cancel. It also adds two new vertical boundary segments of length four. The new perimeter is therefore 48 + 8 = 56 units.

Subtracting the perimeter of the removed rectangle from 48 would give the wrong boundary. The notch’s rectangle is not simply an independent piece of boundary that existed in the original figure. Some new edges are created by the cut, while one old edge segment disappears. Trace the remaining shape rather than treating perimeter as an area-like subtraction rule.

A complete boundary trace gives the same result. The bottom is fourteen, the two outer sides total twenty, the surviving top pieces total nine, the notch bottom is five and the two notch sides total eight. Adding gives 14 + 20 + 9 + 5 + 8 = 56. This route checks every exposed segment and makes the geometry explicit.

Now move the same 5-by-4 cut to a top corner. The remaining area is still 120, but the perimeter becomes 48: lengths removed from the original outer boundary are replaced by equal horizontal and vertical lengths along the cut. Equal removed area does not imply equal change in perimeter. Position matters to the boundary even when it does not change the area of the removed rectangle.

This contrast shows why the representation must match the target. For area, dividing a region into non-overlapping pieces or subtracting a contained region can be convenient. For perimeter, inspect the external and exposed internal edges. A drawing that helps one calculation may not by itself settle the other.

A different geometry task may need an added construction. In an isosceles triangle with equal sides thirteen and base ten, drawing the perpendicular from the apex to the base also bisects the base, giving two right triangles with horizontal leg five. The height is √(13² − 5²) = 12. The original area is 10 × 12/2 = 60 square units.

The bisection is justified by the isosceles structure; it should not be assumed in a general triangle merely because the drawn perpendicular looks central. Without the equal-side condition, the base parts may differ. An auxiliary line is useful because of the relationships it legitimately introduces, not because drawing an extra line automatically makes a geometry problem easier.

Ryan initially uses thirteen as the triangle’s height because it is the only visible long side. Aisha asks which length is perpendicular to the chosen base. That single question identifies the missing subgoal: find the height before calculating the area. The repair is not to memorise another triangle formula, but to distinguish a sloping side from a perpendicular distance.

Before accepting a geometry result, point to the object measured: this region, this boundary, this angle or this distance. State the given or established relationship that permits each key step. When the picture is unclear, redraw only the necessary structure and label it. A precise small sketch is often more useful than an elaborate copy of the original page.

8. Proportion problems become easier when you identify what scales together

A map uses a scale of 1:25,000. Two locations are 7.2 centimetres apart on the map. What is their actual separation in kilometres? A learner who remembers “multiply by the scale” may obtain the right answer, but the units still have to be controlled. The map distance corresponds to 7.2 × 25,000 = 180,000 centimetres in reality.

Now convert the unit: 180,000 centimetres is 1,800 metres, or 1.8 kilometres. Writing the conversion after the multiplication matters because the scale compares like length units. Treating 25,000 as though it directly converts centimetres to kilometres would hide two different operations inside one number.

The scale statement can be read as a ratio: map length / real length = 1/25,000. If the map length is 7.2 centimetres, real length r satisfies 7.2/r = 1/25,000, with both lengths temporarily measured in centimetres. Cross multiplication gives r = 180,000. The proportion works because the map is assumed to use a constant linear scale.

Change the question. An actual road is 3.5 kilometres long. How long should it appear on the same map? Convert 3.5 kilometres to 350,000 centimetres and divide by 25,000, giving 14 centimetres. The same scale relation is now used in the opposite direction. A learner who always multiplies by 25,000 has memorised one direction of one familiar wording rather than the proportion itself.

Aisha asks what would happen to an area shown on the map. If both length dimensions use a factor of 25,000, an area scale uses the square of that factor. A one-square-centimetre region on the map corresponds to 625,000,000 square centimetres in reality. The linear scale cannot be applied directly to an area because area depends on two length dimensions.

This is a common problem-solving theme: identify what is being scaled and how many dimensions participate. Similar figures, currency conversions, recipes and constant-speed models can all involve multiplicative relationships, but their quantities and conditions differ. The word “scale” is not the method. The invariant relationship is.

Consider a recipe using flour and water in ratio 5:3. If 450 grams of flour is used, the water mass is 270 grams because one ratio part is 90 grams. If instead the total mixture is 640 grams, eight ratio parts make 640, so one part is 80 and the ingredient masses are 400 and 240. The same ratio supports two different routes because the given number represents a different part of the relationship.

A careless keyword strategy might divide every number by eight as soon as it sees 5:3. That works only when the supplied quantity is the total of both parts. When the 450 grams refers to the five-part flour amount, dividing by five is the relevant first move. Read what the number measures before choosing the operation.

For direct proportion, a useful algebraic representation is y = kx. If y = 42 when x = 6, then k = 7 and y = 7x. If x later doubles, y doubles. If the model is y = 7x + 10, the constant addition breaks direct proportionality even though the graph is still a straight line. The fixed offset matters.

A problem can also look proportional while failing the constant-ratio condition. A taxi fare with a starting charge plus a per-kilometre rate is linear but not directly proportional to distance. Doubling the distance does not double the total fare because the fixed charge is paid once. A graph or equation can make that distinction easier to see than a memorised list of “proportion words”.

Ryan uses one question to test his model: “If the input were zero, would the output also have to be zero under direct proportion?” If the answer is no because a fixed amount remains, the relationship is not of the form y = kx. This boundary case is quick and often revealing.

When the problem supplies a scaling relationship, preserve the units, identify which quantity is known and check whether the relation really stays constant over the stated range. Proportion solving is not a single cross-multiplication technique. Cross multiplication is one algebraic way to use a multiplicative relationship after that relationship has been justified.

9. Tables are useful when the problem is about systematic change

A theatre has 24 rows. The first row contains 18 seats, and each later row contains two more seats than the preceding row. How many seats are there altogether? One route is to recognise an arithmetic sequence. Another is to construct enough of a table to see the repeated change and then compress it into a formula.

The first few rows contain 18, 20, 22 and 24 seats. The final row contains 18 + 23 × 2 = 64 seats because there are twenty-three increases between the first and twenty-fourth rows. The total is 24(18 + 64)/2 = 984 seats.

The table’s job is not to list all twenty-four values unless that is genuinely useful. It reveals the constant first difference of two and confirms which count belongs to the increment. Once the structure is visible, the arithmetic-sequence formula makes the repeated addition efficient.

Now change the rule: each row has 10% more seats than the previous row, ignoring the practical issue of whole seats for the moment. The first difference is no longer constant. A table now reveals multiplication by 1.1 rather than addition of two. Using the arithmetic-sequence formula would preserve the visual idea of a row-by-row pattern while changing the mathematics.

Tables are particularly useful when several cases need to be compared. Suppose two mobile plans charge A = 12 + 5n and B = 30 + 2n for n units of use. A short table for n = 0, 3, 6, 9 shows A as 12, 27, 42, 57 and B as 30, 36, 42, 48. The equality at n = 6 becomes visible before algebra is used.

The table does not prove the complete cheaper-plan ranges by itself unless the linear models are retained. Algebra compares A − B = −18 + 3n. It is negative below six, zero at six and positive above six. The table suggests the crossing; the expression explains the entire stated domain.

A table can also expose impossible conditions. Imagine integer pairs x and y satisfying x + y = 10 and xy = 40. Listing possible positive integer pairs with sum ten gives 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5. Their products are 9, 16, 21, 24 and 25. No pair works. If the domain is positive integers, that finite table can settle the question completely.

But if x and y are real numbers, the table of positive integers is not enough. Substituting y = 10 − x gives x(10 − x) = 40, or x² − 10x + 40 = 0. Its discriminant is 100 − 160 = −60, so there are no real solutions either. Here the algebra extends the conclusion from a finite integer sample to the whole real domain.

Clara uses tables when a variable changes in regular steps and the effect of that change is itself informative. She does not use them automatically for every problem. For a single linear equation, a table may create unnecessary cases. For a recurrence, comparison or discrete counting process, it may be exactly the representation that makes the structure visible.

When building a table, label the columns with quantities and units. A column headed “3, 5, 7” without a quantity name invites later confusion about whether those values are times, costs, lengths or case numbers. A table is a model. Its headings carry meaning just as variables do in an equation.

Stop extending the table when it has revealed the decision you need. If the pattern can now be represented exactly, switch to a formula or argument. Problem solving is not about loyalty to the first representation chosen. It is about using each representation until its job is done.

10. Break a large problem into subproblems that hand useful information to one another

A water tank is initially 40% full. After 120 litres are added, it is 70% full. Then 15% of the tank’s total capacity is drained. How much water remains? The problem contains several percentage statements, but the useful first unknown is the tank’s capacity.

The increase from 40% to 70% is thirty percentage points of the total capacity. That change corresponds to the added 120 litres. Therefore 30% of capacity is 120, so the capacity is 400 litres. At 70% full, the tank contains 280 litres.

Draining 15% of total capacity removes 60 litres, leaving 220 litres. A common wrong route calculates 15% of the current 280 litres, removing 42 instead. The phrase “15% of the tank’s total capacity” fixes the base at 400, not at the current amount.

The solution can be seen as three subproblems with interfaces. First, convert a percentage-point change and an added volume into capacity. Second, convert the stated fullness into current volume. Third, interpret the drainage base and subtract. Each subproblem produces exactly the quantity needed by the next.

A weak decomposition can create work that does not connect. Calculating 40% of 120 or 70% of 120 produces numbers, but those numbers do not represent quantities named by the problem. Decomposition is not the act of making any small calculation. Each subproblem should answer a question whose output is needed later.

Ryan writes the interfaces explicitly: “120 litres = 30% capacity → capacity 400. Then 70% capacity → 280 litres. Then 15% capacity → 60 litres removed.” This compact chain makes the base visible at each stage. It also makes it easier to check the final answer: 220 litres is 55% of 400, consistent with 70% minus 15 percentage points.

Now change the final instruction to “15% of the water currently in the tank is drained.” The first two subproblems remain identical, but the third changes. Remove 15% of 280, which is 42, leaving 238 litres. The earlier decomposition helps isolate exactly where the changed wording changes the mathematics.

A geometry problem can have the same dependency shape. To find the area of an isosceles triangle, first find half the base, then use a right-triangle relationship to find the perpendicular height, then calculate the area. A statistics problem may first reconstruct totals from means, combine the totals, then calculate a new mean. The topic changes; the handoff principle remains.

A subproblem should preserve the conditions needed later. If a quadratic step generates two possible lengths, do not choose one arbitrarily before returning to the geometry. If a measurement is rounded, retain the appropriate interval when a later bound depends on it. Intermediate work can lose essential information if it passes only a convenient number forward.

When stuck, ask which quantity would make the final step easy. Then ask whether that quantity can be found from the givens. This backward planning can generate a useful subgoal without requiring the entire route to be visible at once. The focused guide Work Backwards When the Route Is Hidden develops that move in more detail.

Good decomposition reduces uncertainty while preserving interfaces. It should make the original problem easier to reassemble, not turn it into several unrelated mini-exercises. At the end, return through the chain and ask whether every intermediate quantity retained the meaning the final answer depends on.

11. Inequalities describe regions of possibility, not only boundary values

A school club must rent a room for 90 units and buy materials costing 6 units per participant. It can spend at most 300 units. How many participants can it support if participant count is a nonnegative whole number? The cost model is 90 + 6n ≤ 300.

Subtracting ninety gives 6n ≤ 210, so n ≤ 35. With the whole-number condition, the maximum is thirty-five participants. Check the boundary: thirty-five costs 300 exactly, while thirty-six would cost 306. The largest valid integer is therefore established from both sides of the boundary.

Now suppose the budget is 302. Algebra gives n ≤ 212/6 = 35⅓. The maximum whole count is still thirty-five. Rounding to the nearest whole number would give the same result here by chance, but the correct reasoning is to choose the greatest integer not exceeding the bound. In another problem, ordinary rounding could violate the inequality.

Contrast a capacity question. Thirty-six people require vans holding at most eight people each. The division 36/8 = 4.5 means five vans are required. Here the answer must be rounded upward because four vans provide only thirty-two places. The direction of the adjustment follows from the constraint, not from a universal instruction to round counts down or up.

Inequalities can have continuous domains too. Solve 3x − 4 < 11 over the reals: x < 5. The answer is an interval, not simply the boundary point five. Testing x = 4 confirms a permitted value; testing x = 6 confirms a rejected one. Those tests support the direction but do not replace the algebraic argument for every real value.

Multiplying or dividing by a negative number reverses order. From −2x ≥ 8, division by −2 gives x ≤ −4. A number-line interpretation explains the reversal: multiplying every number by −1 reflects the line around zero, reversing left and right order. A memorised arrow-flip rule is safer when its relationship can be reconstructed.

A compound condition can describe an overlap of constraints. Suppose a machine operates safely when temperature t is at least 15 and below 40. The domain is 15 ≤ t < 40. If another requirement demands t > 20, the combined condition is 20 < t < 40. Solving a problem can involve intersecting permissible regions, not only isolating a single variable.

Geometry can create inequalities before algebra begins. Three positive lengths form a triangle only if each is less than the sum of the other two. If two sides are 7 and 11, a third side x must satisfy 4 < x < 18. The lower bound comes from 7 + x > 11; the upper from 7 + 11 > x. A positive length alone is not enough.

Clara draws a number line whenever several bounds are being combined. Aisha prefers interval notation after drawing the line. Ryan uses test values to verify each region. These are different representations of the same permitted set. The goal is to preserve whether endpoints are included, not to force one notation before its meaning is clear.

A final answer to an inequality-based problem should return to its domain. “n ≤ 35” is not yet the complete result if n counts people and negative integers are meaningless. “0 ≤ n ≤ 35, n an integer” states the full model. In a shorter context, saying “at most 35 participants” may convey the same practical conclusion.

When a problem asks “maximum”, “minimum”, “at least”, “no more than” or “must be enough”, translate the constraint first. Then decide whether the variable is continuous or discrete. The hardest part is often not the inequality manipulation but preserving what the bound means for an allowable answer.

12. When you are stuck, recover from the last justified state

Getting stuck after three valid steps is different from not knowing how to begin. Preserve the work that is correct. Identify the last line you can justify and the specific decision that is missing next. Starting the entire problem again can erase useful information and recreate the same difficulty.

Consider the equation 1/(x − 1) + 1/(x + 1) = 3/4 over the reals. A learner correctly notes that x ≠ 1 and x ≠ −1, then stalls. The last justified state already contains useful information: the denominators are nonzero, so multiplication by their product is permitted.

Multiplying through by 4(x − 1)(x + 1) gives 4(x + 1) + 4(x − 1) = 3(x² − 1). Simplifying gives 8x = 3x² − 3, or 3x² − 8x − 3 = 0. Factoring gives (3x + 1)(x − 3) = 0, so x = −1/3 or x = 3. Neither violates the domain restrictions.

Check in the original equation. At x = 3, 1/2 + 1/4 = 3/4. At x = −1/3, the two terms are −3/4 and 3/2, again summing to 3/4. Both solutions are valid. The domain note made at the beginning remained relevant until the end.

Suppose the learner instead multiplies only the first denominator away and creates a more complicated expression. Do not necessarily erase the page. Ask whether the new equation is still equivalent under the stated restrictions. If it is, you may continue. If a transformation is invalid or loses a condition, return to the last valid equation rather than to line one.

A recovery question should be local: “What single operation would remove both denominators?” or “Which quantity have I not yet represented?” This is more useful than “Which chapter is this?” when the topic is already clear. The problem is not always memory of a named method; it may be a missing connection between the current state and the next subgoal.

If no local move appears, create a simpler problem with the same obstacle. Replace the equation’s numbers with easier ones or solve a special case. For a counting problem, examine three objects before n objects. For a graph, choose easy coordinates before general parameters. The simpler case should preserve the feature you are trying to understand.

For example, if the obstacle is combining reciprocal terms, first simplify 1/a + 1/b under nonzero conditions to (a + b)/(ab). Then return to x − 1 and x + 1 as the two denominators. The smaller algebraic relationship supplies a component, but the original domain and equation still govern the final answer.

Know when outside help is appropriate. If the missing method has never been taught or a definition is unknown, an explanation is not cheating in ordinary learning. Use the smallest support that restores meaningful work, then attempt a fresh problem without the support when you want evidence of independence.

The SEC Mathematics Recovery route covers getting unstuck in the Singapore-specific SEC system, while Solve a Simpler Problem First develops one transferable recovery technique. This global guide keeps the narrower job: recover the mathematical route without pretending the failed line invalidated everything that came before it.

13. Switch methods when the new route removes the actual obstacle

Changing method is useful when it addresses the reason the current route is failing. Switching because a question feels difficult can lead to a sequence of abandoned beginnings. Before changing, name the obstacle and what the new route is expected to improve.

Suppose a quadratic equation is x² − 6x − 7 = 0. Factoring works quickly: (x − 7)(x + 1) = 0, so x = 7 or x = −1. Using the quadratic formula would also work but creates more arithmetic. In this case, switching away from factorisation would not solve a problem because there is no obstacle to solve.

Now consider x² − 6x − 2 = 0. Searching for integer factor pairs cannot succeed because the real roots are 3 ± √11. Completing the square gives (x − 3)² = 11 directly. The quadratic formula also works. Continuing to search for integer factors after the structure has ruled them out is not persistence; it is ignoring information from the polynomial.

A graph can estimate the roots and help check their locations: one is negative and one positive. But if exact values are required, the graph does not replace the algebraic route. A method switch should preserve the answer standard. Moving from an exact route to an approximate one can be useful for checking while still being insufficient as the final solution.

Geometry creates similar decisions. A coordinate approach can convert a spatial relationship into algebra; a synthetic theorem can avoid lengthy coordinate calculations. If a diagram supplies clear parallel or cyclic structure, geometry may be shorter. If coordinates are already given and the target is a line equation, an analytic route may make the data directly usable.

Aisha tries a similarity approach to a problem but cannot establish a second angle equality. She changes to coordinates because the vertices already have simple coordinates. That switch is justified: the missing similarity condition is the obstacle, while the coordinate data provide an independent route. Clara, by contrast, should not abandon a valid similarity proof merely because coordinate geometry is also available.

Probability offers another example. Direct enumeration is excellent when the sample space is small. For six equally likely outcomes, listing can make overlap visible. For a sequence of ten independent trials, a compact counting or complement method may be easier than listing 1,024 outcomes. The underlying event remains the same while the representation cost changes.

Before switching, ask whether the current method is invalid, inefficient or merely unfinished. Invalid steps must be repaired. Inefficient routes may still be worth completing once during learning because they reveal structure or provide a check. An unfinished route may simply need one intermediate result rather than a complete replacement.

When you do switch, carry forward facts that remain valid. A domain restriction, a calculated length or a proven angle equality does not disappear because the next method changes. Reusing established information prevents the second route from becoming a complete restart with new opportunities to lose earlier conditions.

Ryan writes one sentence beside a switch: “Factoring over integers is not available; completing the square reveals the roots exactly.” That sentence keeps the change purposeful. It also creates a study record: the difficulty was not solving quadratics in general but recognising when one familiar route no longer fits the expression.

Problem-solving flexibility is therefore not maximum novelty. It is the ability to maintain a valid route, diagnose its limits and change representation or method when the change has a mathematical reason. The strongest solver is not the one who uses the most methods, but the one who chooses and controls methods in relation to the problem.

14. In probability, define the event before multiplying fractions

A bag contains five red counters, four blue counters and three green counters. Two counters are drawn at random without replacement, with each remaining counter equally likely at each draw. What is the probability that the two counters are the same colour?

The event is a union of three disjoint possibilities: two red, two blue or two green. Their probabilities are (5/12)(4/11), (4/12)(3/11) and (3/12)(2/11). Adding gives 20/132 + 12/132 + 6/132 = 38/132 = 19/66.

Why add? The three colour-pair events cannot occur together on the same two-draw trial. Why multiply within each branch? The second draw occurs after the first and its probability is conditional on what was removed. The arithmetic follows the event structure rather than a generic instruction to “multiply down and add across”.

A counting approach reaches the same result. There are C(12, 2) = 66 unordered pairs of counters. Same-colour pairs number C(5, 2) + C(4, 2) + C(3, 2) = 10 + 6 + 3 = 19. With equally likely unordered pairs under the stated random draw, the probability is 19/66. Comparing the two routes provides a check and reveals the same combinatorial structure.

Now change the event to “at least one red”. The complement is easier: no red means both counters come from the seven non-red counters. Its probability is (7/12)(6/11) = 7/22. Therefore the desired probability is 1 − 7/22 = 15/22. Listing the red branches directly would also work, but the complement reduces the number of cases.

Change the draw rule to replacement. The same-colour probability becomes (5/12)² + (4/12)² + (3/12)² = 50/144 = 25/72. The second probability in each branch no longer uses eleven counters because the first counter returns. The phrase “with replacement” changes the model before any calculation begins.

Ryan initially writes 5/12 + 4/12 + 3/12 = 1 and says every pair must be the same colour because one of the first-draw colours must occur. The arithmetic statement about the first draw is true but irrelevant to the two-draw event. A correct calculation can still answer the wrong probability question.

Aisha asks a more diagnostic question: “What would count as success after the first counter is red?” The second must also be red, so the probability changes to 4/11 without replacement. That conditional question forces the event to be followed across stages rather than collapsed into a one-draw colour count.

Probability answers should stay between zero and one, but that is only a coarse check. A result of 1/2 could still be wrong while looking plausible. Compare with a bound or another route where possible. Since red is the largest colour group but still less than half the bag, a same-colour probability around 0.288 is plausible; a probability near 0.9 would demand investigation.

Use the Split a Difficult Problem into Cases guide when the main difficulty is organising mutually exclusive cases. The complete problem-solving habit is to define the event, decide whether cases overlap, update changing conditions and then choose counting or probability operations accordingly.

15. A graph can reveal a crossing, a bound or a trend—but it must match the claim

Two reservoirs are monitored under idealised linear models. Reservoir A contains 500 units and loses 12 units per hour. Reservoir B contains 320 units and gains 6 units per hour. When do their amounts become equal, provided both linear models remain valid over that time?

The models are A(t) = 500 − 12t and B(t) = 320 + 6t. Equality requires 500 − 12t = 320 + 6t, so 180 = 18t and t = 10 hours. Both amounts are then 380 units.

On a graph, the two lines intersect at (10, 380). The downward slope of A and upward slope of B make the crossing inevitable if the models continue long enough. The graph also explains the inequality either side: before ten hours A is above B; after ten, B is above A.

A graph drawn from approximate plotting might suggest 9.8 or 10.2 hours. If an exact result is available algebraically, use the exact calculation for the final answer. The graph’s value is interpretation, prediction and checking. It is not automatically the most precise route.

Now add a capacity limit: Reservoir B can hold at most 350 units. It reaches that capacity at 320 + 6t = 350, so t = 5 hours. The original line B(t) = 320 + 6t cannot represent retained volume beyond that point unless the model specifies overflow or a change in inflow. The later intersection at t = 10 lies outside the stated validity of the unchanged B model.

The problem therefore changes. If B remains at 350 after reaching capacity and A continues to decrease, equality with 350 occurs when 500 − 12t = 350, giving t = 12.5 hours. But that answer depends on the newly supplied post-capacity rule. A graph that continues the original B line past capacity would be visually neat and physically wrong.

Clara learns to ask two graph questions separately: “What mathematical relation is the line showing?” and “Over which input values is that relation supposed to represent this situation?” The distinction prevents a formula from being treated as an unlimited physical law merely because the graphing software extends it across the screen.

Graphs also help solve inequalities. To find where x² − 5x − 6 < 0, factor as (x − 6)(x + 1), giving roots −1 and 6. An upward-opening graph is below the horizontal axis between the roots, so −1 < x < 6. The graph supports the sign analysis; the factorisation supplies exact boundaries.

If the quadratic were −x² + 5x + 6, the graph would open downward and the sign regions would reverse. Remembering “shade between roots” without considering the leading sign would answer only one family of cases. The representation makes the sign structure visible when read in relation to the axis.

A graph can show that a proposed answer is implausible. If an equation’s two sides are graphed and the visible intersection lies near x = 2, an algebraic answer of x = 20 should trigger checking. But the graph’s viewing window may hide additional intersections. Visual evidence is a check whose scope depends on the function and display, not a universal proof of completeness.

Use graphs as arguments about relationships: crossing, monotonicity, sign, intercept, rate and domain. Label axes and units. Preserve exact algebra when the problem requires it. A graph is strongest when it clarifies the structure of a solution rather than acting as an unexplained picture beside one.

16. Working backwards succeeds when each reversed step is justified

A number is multiplied by three, then seven is subtracted, then the result is squared to give 400. What are all possible original real numbers? Working backwards is tempting, but the squaring step is not one-to-one over the reals, so the reverse route must branch.

Let the original number be x. The forward process gives (3x − 7)² = 400. Reversing the square gives 3x − 7 = 20 or 3x − 7 = −20. The first gives x = 9; the second gives x = −13/3. Both check in the forward rule.

A learner who takes only √400 = 20 and continues obtains one valid answer but misses the second. The reverse of squaring over the reals requires both signs when solving an equation. This is different from evaluating the principal square root expression √400, which is twenty.

Now consider a process that begins with a positive number, doubles it and takes the positive square root of the result to produce six. The positivity and root operation constrain the route differently. Writing √(2x) = 6 and squaring gives 2x = 36, so x = 18. No ± branch is introduced by squaring both sides here because the original equality already requires the root value to be nonnegative and the right side is six.

Working backwards through addition and multiplication by known nonzero constants is reversible. Working backwards through squaring, absolute value, rounding or many-to-one functions may require cases or may be impossible to do uniquely without additional information. The solver should ask whether the forward operation lost information.

A rounded measurement is a simple example of information loss. If a length rounds to 12 centimetres to the nearest centimetre, the original length is not uniquely recoverable. Under the usual convention, it lies from 11.5 up to but not including 12.5. Working backwards produces an interval, not one exact number.

Absolute value similarly loses sign. |x − 4| = 7 gives x − 4 = 7 or x − 4 = −7, so x = 11 or −3. The two branches are not an arbitrary rule attached to vertical bars. They describe two positions seven units from four on the number line.

Ryan uses a reverse-arrow diagram to mark whether each operation is one-to-one under the current domain. An addition by five gets one reverse arrow. A square over all reals gets two possible reverse branches. A rounding operation gets an interval. The diagram helps him decide what information the reversed problem can legitimately recover.

Work backwards also helps with construction problems. If a final average is required, ask what total would produce it. If a target percentage is required, ask what original base and multiplier create it. If a proof needs a particular factor, ask what earlier algebraic form would make that factor appear. The strategy is not limited to reversing arithmetic instructions.

The focused Work Backwards When the Route Is Hidden page develops the technique. The controlling principle in a full solution is to reverse only relationships whose information loss you understand, and to branch or retain ranges when the forward step was not uniquely invertible.

17. Look for something that stays unchanged while the situation moves

Some problems become manageable when you identify an invariant: a quantity or property that remains unchanged under the allowed moves. The invariant can rule out impossible targets or reduce a large search to a small calculation.

Suppose three boxes contain 7, 12 and 19 counters. A legal move transfers exactly two counters from one box to another. Can the first box ever contain exactly 10 counters? The total number of counters remains 38, but that alone does not settle the first-box target.

Each legal move changes the first box by 0, +2 or −2. It starts with an odd count, seven, so its parity remains odd. Ten is even. Therefore the first box can never contain exactly ten counters under these moves. No long simulation is required.

The parity invariant is stronger for this target than the total invariant. The total still matters globally: counters are only transferred, not created or destroyed. But many states with total 38 are possible in principle, including some with first-box count ten. Parity identifies why those particular states are unreachable by the allowed moves.

Change the target to eleven counters. Parity no longer rules it out. One legal move transferring two counters into the first box reaches nine; another reaches eleven, provided the donor boxes have enough counters. An invariant can establish impossibility, but satisfying the invariant does not always prove reachability. Additional constraints may still matter.

A geometric invariant can appear under rearrangement. Cutting a shape into pieces and translating or rotating the pieces preserves total area. That allows one figure’s area to be related to another without recalculating every small dimension. The invariant is not perimeter: rearrangement can create a different boundary length.

Algebra contains invariants too. Adding the same number to both sides of an equation preserves equality. Multiplying numerator and denominator of a fraction by the same nonzero factor preserves its value. These familiar transformations are small invariant-based arguments: the representation changes while a relationship remains fixed.

In a sequence problem, a remainder class may remain unchanged under allowed changes. If a number begins congruent to 1 modulo 3 and every move adds six, it remains congruent to 1 modulo 3. A target congruent to 2 modulo 3 is impossible. This is the same reasoning pattern as parity with a different modulus.

Aisha asks what makes an invariant worth searching for. Look at the legal moves: what do they change, and what can they not change? Repeated changes by an even number suggest parity. Transfers suggest a conserved total. Similarity transformations suggest angle preservation and proportional lengths. Symmetry may preserve distance from an axis or centre.

Do not force an invariant into every problem. For the thirteen boxes, a difference comparison was more direct. For a straightforward linear equation, equality preservation is already built into the usual method and does not need a separate puzzle-style search. An invariant is useful when it meaningfully narrows the space of possible states or justifies a transformation.

When an invariant proves impossibility, state the contradiction clearly: “The first box remains odd after every legal move, but ten is even.” A conclusion such as “I tried many moves and could not get ten” is empirical evidence about your search, not a proof about every possible sequence of moves.

Recognising invariants turns some problems from route-finding into state-checking. Rather than asking how to reach the target, first ask whether the target belongs to the same invariant class as the start. If not, the best solution may be a proof that no route exists.

18. Split into cases when one assumption cannot describe every possibility

Solve |2x − 5| = x + 1 over the reals. The right side must be nonnegative because it equals an absolute value, so x ≥ −1. The absolute value then creates two algebraic cases according to the sign of 2x − 5.

Case one: 2x − 5 ≥ 0, so x ≥ 2.5. Then |2x − 5| = 2x − 5 and the equation becomes 2x − 5 = x + 1, giving x = 6. It satisfies the case condition and the original equation.

Case two: 2x − 5 < 0, so x < 2.5. Then |2x − 5| = 5 − 2x and 5 − 2x = x + 1, giving 4 = 3x and x = 4/3. It satisfies x ≥ −1 and the case condition. Both solutions are valid.

A student who writes ±(2x − 5) = x + 1 without tracking the corresponding sign conditions may still stumble onto the same candidates, but the reasoning becomes harder to verify in more complicated examples. Case conditions tell you which formula represents the absolute value in each region.

Piecewise rates create a similar need. A parking system charges 5 units for up to two hours and then 3 units for each additional hour, with a maximum stay of six hours in this invented example. One expression cannot describe all durations without a piecewise rule. The boundary at two hours changes the relation.

For t between zero and two, cost is 5. For t above two and at most six, cost is 5 + 3(t − 2) if time is charged continuously under the stated model. To find durations costing 11, use the second case: 5 + 3(t − 2) = 11 gives t = 4. It belongs to the case domain.

Case splitting should be exhaustive and non-overlapping where practical. If the domains are x ≥ 2.5 and x < 2.5, every real x belongs to exactly one. Leaving x = 2.5 out would create a gap; including it in both can create duplicated work. The boundary should be assigned deliberately.

Geometry may require cases when an angle can be acute or obtuse, or when a point can lie on either side of a line. A counting problem may separate even and odd n. A parameter equation may behave differently at the parameter value that makes a coefficient zero. The need for cases comes from a structural change, not from an arbitrary desire to make the solution longer.

Clara writes the trigger above each case: “if 2x − 5 is nonnegative” and “if 2x − 5 is negative”. That note prevents the branch from becoming a detached equation. At the end she filters each candidate through its branch condition and then through the original problem.

Use cases when one formula, sign or condition cannot honestly represent every allowed input. The Split a Difficult Problem into Cases guide develops this skill more fully. In a complete problem-solving system, the case structure is part of the model and must travel with the calculations it generates.

19. Estimate the answer’s shape before calculating it exactly

A cylinder-like container problem, a percentage calculation and a graph intersection all permit different kinds of predictions. Estimation is not always about rounding numbers. It can mean predicting sign, order of magnitude, interval, monotonicity or whether an answer should be above or below a reference value.

Suppose 17.8% of 246 is required. Before calculating, 20% of 250 is 50, so the exact answer should be somewhat below fifty. The calculation 0.178 × 246 = 43.788 fits that expectation. An output of 437.88 would be inconsistent with the scale even if it appeared on a calculator after an entry mistake.

For 7.8 ÷ 0.39, the answer should be much larger than 7.8 because the positive divisor is below one. Since 0.39 × 20 = 7.8, the exact quotient is twenty. A learner who believes division always makes a number smaller may reject the correct answer because the misconception controls the reasonableness check.

In an equation such as 0.52x = 104, x must exceed 104 because multiplying by roughly one half produced 104. An answer near 200 is plausible; an answer near 54 is not. This qualitative prediction can catch an incorrect inverse operation before the final line is accepted.

Geometry provides bounds. A rectangle with sides 9.8 and 20.3 has area slightly below 10 × slightly above 20, so around 200 square units. Exact multiplication gives 198.94. A result of 19.894 likely reflects a misplaced decimal point; 1,989.4 likely reflects another scale error.

Reasonableness depends on the model. If a probability result exceeds one, it is impossible. If an average of positive values lies far outside the minimum and maximum, the calculation is wrong for the ordinary arithmetic mean of those observations. If a length from a triangle violates the triangle inequality, the geometry cannot hold.

Aisha writes one expectation before a long calculation: “positive and between 30 and 50”, “less than the original amount”, or “one root on each side of zero”. She does not try to estimate every intermediate decimal. The note is a reference against which the later exact result can be compared.

Prediction can also choose methods. If an answer should be an integer because it counts objects, an equation producing an awkward decimal may indicate inconsistent data or a modelling error. If exact algebra is expected to produce a simple surd, early decimal approximation may obscure the structure. The anticipated answer form can guide how long exact values are retained.

Do not use an estimate to overwrite exact evidence. If a carefully checked result surprises you, inspect the estimate’s assumptions too. A compound percentage change may behave differently from an intuitive “net percentage” guess. An average speed may not lie at the simple arithmetic mean of two displayed speeds. A reasonableness check is another model, and it can itself need correction.

Estimate early enough to influence checking, then calculate. The Check an Answer Through an Independent Route page develops verification methods. Here the problem-solving role is to establish an expected region before the exact route becomes psychologically persuasive merely because it produced a neat number.

20. Verify the whole problem, not only the last equation

A final numerical answer can satisfy the last equation while failing an earlier condition, a unit, an integer requirement or the actual question. Verification should therefore return through the full problem statement. Ask which claims the answer must satisfy and test those claims directly where possible.

Consider a rectangle with length three units greater than width and area forty square units. Let width be w, so w(w + 3) = 40. Rearranging gives w² + 3w − 40 = 0, or (w + 8)(w − 5) = 0. The algebraic candidates are w = −8 and w = 5.

Both satisfy the quadratic equation, but only w = 5 can represent a positive width. The corresponding length is eight and the area checks as forty. The negative candidate is not an algebra error. It is a solution of the algebraic model before the physical domain is applied.

A different problem might permit negative values, so “discard the negative root” is not a universal quadratic rule. The rejection is justified by the quantity being a length. In a coordinate problem, a negative coordinate may be perfectly valid. The context, not the visual sign of the number, determines admissibility.

Units are another final filter. If a rate calculation returns twelve, state whether that means metres per second, items per minute or currency units per kilogram. Multiplying a length by a length produces square units, while adding quantities with incompatible units is usually a sign that the model has mixed different kinds of measure.

Completeness matters too. If a trigonometric equation over a specified interval has several solutions, verifying one does not establish that the list is complete. If a counting task asks for the minimum number of containers, checking that n containers suffice should be paired with showing that n − 1 do not. A boundary claim needs evidence on both sides.

Use an independent route when it is economical. For simultaneous equations, substitute the pair into both originals. For a factorisation, expand. For an area, decompose the region another way. For a probability, compare counting and complement routes. Independent checks are valuable because they are less likely to reproduce the identical error mechanism.

But do not assume a different-looking calculation is independent if it rests on the same wrong model. If you misread 20% as being of the final amount, both a calculator and algebra can faithfully solve that wrong relationship. Verification should attack the relationship as well as the arithmetic.

Clara reads her answer into the original sentence. Ryan checks a boundary case. Aisha asks whether every stated condition has been used. These three habits catch different failures: interpretation, range and missing information. No single check can replace all others.

The completed solution should be able to answer four questions: Does the value satisfy the mathematics? Does it belong to the allowed domain? Does it carry the right units or object meaning? Does it answer the requested target completely? If one answer is no, the problem is not finished.

Verification is not an apology for distrusting mathematics. It is mathematics applied to the claim you are about to make. A solution becomes stronger when it can survive a deliberate attempt to show that it violates its own conditions.

21. Data problems begin by asking what the summary must preserve

Two classes take the same test. Class A has 12 students with mean score 68. Class B has 18 students with mean score 74. What is the mean score of all thirty students? Averaging 68 and 74 gives 71, but that would give the two classes equal weight despite their different sizes.

Recover the totals. Class A contributes 12 × 68 = 816 score points. Class B contributes 18 × 74 = 1,332. Together they contribute 2,148 points across thirty students, giving a combined mean of 71.6.

The result lies between the two means and closer to 74 because the larger class has the higher mean. That qualitative prediction is a useful check. A combined mean of 76 would be impossible when every group mean lies below it unless the data description had changed.

Now suppose the question asks for the median of the thirty scores. The two group means and class sizes are not enough. Very different score distributions can share those means while having different combined medians. Recognising that information is insufficient is part of solving the problem; no formula can recover data that the summary did not retain.

Aisha constructs two possible Class A score sets with mean 68 and different medians. The exercise shows why a mean cannot answer every question about a distribution. The problem-solving move is not to search harder for a median formula. It is to identify what the given information actually determines.

Weighted averages in rates use the same principle. If a traveller spends two hours at 40 kilometres per hour and one hour at 70 kilometres per hour, total distance is 80 + 70 = 150 kilometres over three hours, giving average speed 50 kilometres per hour. The time weights matter because average speed is total distance divided by total time.

If the traveller instead covers equal distances at the two speeds, the times differ and the same calculation changes. For 120 kilometres at each speed, time is three hours plus 12/7 hours, so average speed is 240 divided by 33/7, or 560/11 ≈ 50.91 kilometres per hour. The visible speeds are the same; the weighting quantities are not.

Data problems often hide a question about what has been aggregated. Combine totals before dividing when the definition requires it. Preserve frequencies when merging distributions. Keep units attached to rates. Do not average already-averaged numbers unless the conditions make that shortcut valid.

For grouped or sampled data, also separate calculation from inference. A perfectly computed sample mean does not prove that an entire population has that mean. The sampling method, sample size and question determine what inference is defensible. A correct arithmetic procedure cannot supply missing design information.

Ryan writes the defining relationship before manipulating summaries: mean = total/count, rate = total output/total input measure, probability = favourable weight/total weight under the stated model. This prevents a visible pair of numbers from being averaged merely because they appear side by side.

A useful check asks whether the summary lies inside a necessary range. A weighted mean of positive weights lies between the smallest and largest group means. A probability lies between zero and one. A median of an ordered finite set lies between its minimum and maximum. These bounds do not prove the answer correct, but they reject some impossible results quickly.

Problem solving with data therefore begins with the definition of the requested summary. Ask what information that definition needs, which information the problem supplies, and whether anything has been compressed away. Sometimes the right conclusion is a number. Sometimes it is that the requested quantity cannot be uniquely determined.

22. Maximum and minimum problems become clearer when you expose the competing effects

A rectangle has perimeter 40 units. What dimensions produce the greatest possible area? Let one side be x, so the other is 20 − x, with 0 < x < 20. The area is A = x(20 − x) = 20x − x².

Completing the square gives A = 100 − (x − 10)². Because a real square is nonnegative, A ≤ 100. Equality occurs at x = 10, so the maximum-area rectangle is a 10-by-10 square with area 100.

This solution reveals the competing effects. Increasing x makes one dimension larger but forces the other smaller because the perimeter is fixed. The product peaks when the two dimensions balance. The completed-square form turns that intuition into an exact bound.

A table could suggest the same result: widths 8, 9, 10, 11, 12 give areas 96, 99, 100, 99, 96. But the finite table does not by itself prove that no noninteger width such as 9.7 gives a larger area. The algebraic bound covers every real x in the allowed interval.

Change the perimeter to P. With side x and other side P/2 − x, area is x(P/2 − x). Completing the square gives P²/16 − (x − P/4)². The maximum occurs when both sides are P/4. The same structural conclusion holds for every positive perimeter under the rectangular model.

Optimization problems often tempt learners to differentiate immediately when calculus is available. Differentiation is a valid route for suitable functions, but it is not always necessary. If a quadratic can be rearranged into a clear square, the structure may provide a shorter exact argument. Route choice should respond to the function, course and target.

For a non-quadratic problem, another method may be required. Suppose a box volume depends on several dimensions with constraints. A table can explore, algebra can reduce variables, and calculus may later locate an optimum. The problem-solving sequence still begins before differentiation: define the variable, express the constraint, identify the objective and establish the allowed domain.

Clara’s first mistake is to maximize one side independently. She chooses x = 19 because it is nearly the largest possible width. The other side then becomes one, giving area nineteen. The problem does not ask for the largest side; it asks for the largest product under a fixed sum. The target function matters.

Aisha checks boundary behaviour. As x approaches zero or twenty, the area approaches zero. A maximum somewhere in the interior is therefore plausible. This does not locate it, but it gives the later algebra a shape to agree with.

Optimization in practical models also needs context. If side lengths must be whole centimetres, the continuous optimum may need comparison with nearby integers. If material thickness or manufacturing constraints exist, the simple perimeter model may not be sufficient. The mathematical optimum is always relative to the model and domain.

The lesson generalises: identify the quantity being optimized, express it using as few variables as possible, preserve the constraints, and choose a method that establishes a global or relevant local bound. A neat stationary point is not automatically the required optimum without domain and boundary checks.

23. A parameter can change the kind of problem, not just its numerical answer

Consider the equation ax = 6, where a is a real parameter. If a ≠ 0, the solution is x = 6/a. If a = 0, the equation becomes 0 = 6 and has no solution. Dividing by a immediately without discussing the zero case silently removes one qualitatively different regime.

Now consider ax = 0. If a ≠ 0, the only solution is x = 0. If a = 0, every real x is a solution because the equation becomes 0 = 0. The parameter value changes the solution set from one point to the whole real line.

A quadratic parameter can change the number of real roots. For x² − 4x + k = 0, the discriminant is 16 − 4k. There are two distinct real roots when k < 4, one repeated real root when k = 4 and no real roots when k > 4. The parameter does more than alter the roots’ values; it changes the problem’s state.

Completing the square gives (x − 2)² = 4 − k, which makes the same classification visible. A real square is nonnegative. If 4 − k is positive, there are two symmetric roots around two; if zero, one; if negative, none. Different methods expose the same structural transition.

In geometry, a parameter might control whether a shape is possible. A triangle with sides 5, 8 and p requires 3 < p < 13. The parameter range is part of the existence condition. Solving later angle or area questions without first enforcing this range can produce algebraic expressions for a triangle that cannot exist.

In probability, a parameter can represent an unknown success probability p, which must satisfy 0 ≤ p ≤ 1. Algebraic manipulation producing p = 1.2 signals an inconsistency with the probability model. The variable’s allowed set is not generic merely because the symbol is unfamiliar.

Ryan writes parameter cases only when a transformation depends on a condition. If an equation contains a/(a − 2), he marks a ≠ 2. If a coefficient might become zero, he asks how the equation changes at that value. This prevents the common mistake of treating a symbolic coefficient as though it were an ordinary known nonzero number.

A parameter problem may ask for the values that cause a particular event: no real roots, tangency, equal solutions, integer output or a target range. Translate that event into a mathematical condition before solving for the parameter. For tangency of a line and a quadratic, for example, equality of the two expressions leads to a quadratic in x whose discriminant must be zero.

Suppose y = x² and y = mx + 1 are tangent. Equating gives x² − mx − 1 = 0. Tangency would require discriminant m² + 4 = 0, which has no real m. The correct conclusion is that no real line of that particular intercept form is tangent to y = x². A solver should not force a real parameter value because the question sounds as though one should exist.

Change the line to y = mx − 1. The equation becomes x² − mx + 1 = 0, and tangency requires m² − 4 = 0, so m = ±2. Small changes in the parameterized family change whether the target condition is achievable.

Parameter problems reward case awareness and condition tracking. Before dividing, cancelling, taking roots or applying a formula, ask whether a parameter value changes the legitimacy of that step. A general symbolic answer is incomplete if it fails precisely where the algebra changed character.

24. Some problems end with a proof rather than a number

Show that the product of three consecutive integers is divisible by six. A few examples—2 × 3 × 4 = 24 and 5 × 6 × 7 = 210—suggest the claim but do not explain why every integer triple works.

Let the three consecutive integers be n, n + 1 and n + 2. Among any three consecutive integers, one is divisible by three. Also, at least one is even. Therefore their product contains a factor of three and a factor of two, so it is divisible by six.

This argument solves the problem by identifying invariants of consecutive residues. The exact integer n never needs to be found. The target is a general divisibility statement, so the useful representation is not a numerical table but a structure that covers every case.

An algebraic expansion n(n + 1)(n + 2) is possible but not automatically helpful. Expanding gives n³ + 3n² + 2n. Divisibility by six is less visible in that form. A representation can be equivalent and still be poorly suited to the target.

A proof problem can also be attacked through cases. To show the same product is even, split n into even and odd. If n is even, the first factor is even. If n is odd, n + 1 is even. This covers every integer n. For divisibility by three, use residues modulo three or observe that one of three consecutive integers must be a multiple of three.

A counterexample solves a different kind of proof problem. The claim “the sum of two irrational numbers is irrational” is false because √2 + (−√2) = 0. One allowed counterexample disproves the universal statement. Searching for an elaborate general proof would be wasted effort because the claim itself fails.

When a statement includes “for all”, “always” or “must”, first test simple and boundary cases. A counterexample can save time. If no counterexample appears and the claim remains plausible, identify what general structure would force it to hold. Examples guide conjecture; proof carries the final burden.

In geometry, proof often means linking givens to a target property through established theorems. If you need to prove two lines parallel, equal alternate angles or equal corresponding angles may be useful routes. If you need to prove triangles congruent, the target suggests which side and angle relationships would be sufficient. Working backwards from the proof target can generate subgoals.

Aisha writes the final claim at the bottom of a proof page and asks what would be enough to justify it. Then she looks for those relationships in the givens. This backward planning does not mean assuming the conclusion. It means using the target to choose which facts to establish.

The Build a Proof from Claims, Reasons and Checks guide develops the local craft. Within problem solving, proof is one possible answer form. The method should be judged by whether it covers the claim’s full scope, not by whether it resembles a calculation-heavy worksheet.

25. Calculators and digital tools should accelerate a route you can still inspect

A calculator can reduce arithmetic load, graph a function and test values. It cannot decide which model the question intended unless you have already represented that model correctly. Entering the wrong expression accurately only automates the wrong problem.

Suppose a learner intends to calculate (18 + 6)/3. Entering 18 + 6 ÷ 3 without brackets gives 20 under standard order of operations, not eight. The machine has not made a mathematical error. The entered expression did not match the intended grouping.

Write the mathematical expression before the key sequence when grouping matters. Estimate the result. Then use the tool and compare. This three-step routine is often enough to catch a missing bracket or decimal-place error without turning calculator use into a separate ritual.

Graphing tools create another risk: the viewing window can hide behaviour. A polynomial may have an intersection outside the default display. A rational function may look continuous if an excluded point is not visible at screen resolution. A nearly flat segment can obscure a turning point. The screen is a sampled representation, not a proof that no other feature exists.

Use zooming or algebraic analysis when completeness matters. If an exact factorisation shows roots at −12 and 3 but the window displays only the positive root, the algebra has revealed a display limitation. Conversely, a graph can alert you when an algebraic root was lost because a division removed a zero case.

A spreadsheet can explore recurrence or data efficiently, but cell formulas should still be interpretable. If a copied formula changes a reference incorrectly, an entire column may look systematic while implementing the wrong relationship. Spot-check a few rows manually and understand which references are intended to vary.

Automated solvers and general AI tools can propose routes quickly. Use them as explanations or checks in permitted learning contexts, not as evidence that you independently solved the task. Ask for a hint, a diagnosis of the first invalid line or a comparison of two methods. Verify every mathematical claim, especially conditions and domains.

Clara asks a digital tool to solve an equation and receives x = 4. She substitutes four into the original and finds it works. Aisha notices the equation was quadratic and asks whether another solution exists. The single verified candidate is not evidence of completeness. Tool output still has to be read in relation to the original question.

Privacy and assessment rules matter too. Do not send unnecessary personal information to an external service, and do not use assistance where an assessment prohibits it. The learning goal here is to strengthen the learner’s mathematical decisions, not to obscure who supplied them.

A useful post-tool note is: “The tool supplied the graph; I interpreted the domain and solved the exact boundary algebraically.” This separates assistance from independent reasoning. Such transparency is valuable for diagnosis and for deciding what still needs practice without support.

Tools are strongest when they remove low-value friction while leaving the mathematical model, selection and checking visible. If the learner can no longer explain what was entered, what came out or why the result answers the problem, the tool has become transportation rather than support.

26. Under time pressure, protect the high-value decisions first

A timed assessment adds a resource constraint: you cannot explore every possible method indefinitely. Efficient problem solving under pressure therefore needs triage. Read enough to classify the task, preserve conditions, choose a plausible route and know when to move on temporarily.

This guide does not prescribe one examination timing rule for every system. Paper structures differ. Use the official instructions and the existing examination owner for your course. The transferable point is that time pressure magnifies the cost of an unexamined wrong model.

On first reading, mark the target and the decisive conditions. If the question asks for a minimum whole count, keep the discreteness visible. If a diagram is not drawn to scale, do not infer lengths from appearance. If a probability is without replacement, mark the changing total. Ten seconds preserving a condition can prevent minutes of solving the wrong problem.

Choose the shortest valid representation you can still control. A full table may be unnecessary when a linear equation is immediate. A sketch may be essential when geometry relationships are hard to hold mentally. Efficiency is not minimal writing at any cost; it is enough external structure to prevent avoidable rework.

If you are stuck, record the useful state before moving on: equations formed, known angle, domain restriction or partial factorisation. Returning later is easier when the earlier work remains interpretable. A page of crossed-out beginnings can cost more time than a short labelled checkpoint.

Do not let one difficult question consume all remaining time simply because you have already invested in it. That is an assessment decision, not a statement about the mathematical value of the problem. Preserve what you can, attempt other accessible work and return if the paper allows.

When time is limited, verification should target likely failure points. Check a sign-changing operation, a domain-sensitive root, a unit conversion or the boundary in a minimum/maximum problem. Recomputing every arithmetic line identically may be less useful than one independent substitution into the original equation.

Ryan reaches a quadratic with two candidates and wants to move on immediately after writing both. Clara notices the question concerns a length. Testing the sign condition takes seconds and removes the negative candidate. That small interpretation step converts algebraic completion into problem completion.

Aisha, by contrast, spends too long proving a result through two methods when one complete proof is enough. Multiple methods are valuable in study, but the performance setting may reward a single clear route. Practice should teach the learner when redundancy is useful and when it merely consumes the clock.

After the assessment, a slower post-mortem can revisit route choices and missed checks. The existing Mathematics Examination Craft route owns that performance layer. This problem-solving guide keeps the core relationship intact: time changes how much exploration is affordable, but it does not make invalid mathematics valid.

27. Independent checkpoint: twenty-four problems that require a route, not only a calculation

The following original problems sample the decisions in this guide. They are not a standardised test, they do not represent every secondary syllabus, and there is no validated pass mark. Attempt only topics you have studied. Keep the worked explanations closed until you have written a route, an answer and a check where one is appropriate.

Record whether you used a hint, example, calculator, graphing tool or full solution. A supported attempt can be an excellent part of learning, but its conditions should remain visible. The purpose of the checkpoint is to discover which problem-solving decisions you can currently make for yourself.

Problem 1 — Two types under one total

A workshop packs 18 containers. Some hold 4 units and the others hold 9 units. The containers hold 107 units altogether. How many containers of each type are there?

Worked explanation: Imagine all eighteen are four-unit containers. They would hold 72 units. The actual total is 35 higher. Replacing a four-unit container with a nine-unit container adds five units, so seven replacements are needed. Therefore there are seven nine-unit containers and eleven four-unit containers. Check: 7 × 9 + 11 × 4 = 63 + 44 = 107 and the counts sum to eighteen. An equation route gives the same result. The useful structural move is the difference from a reference case, not a keyword.

Problem 2 — The order of a percentage and a fee

An item is reduced by 20% and then a fixed service fee of 12 units is added. The customer pays 92 units. Find the original price.

Worked explanation: Remove the fee first: the reduced item price is 80. This is 80% of the original, so 0.8p = 80 and p = 100. Check forward: 20% off 100 gives 80; adding 12 gives 92. The equation is 0.8p + 12 = 92. Writing 0.8(p + 12) = 92 would discount the fee too and model a different order of operations.

Problem 3 — Same ratio, different given quantity

Two quantities are in ratio 5:8. The larger exceeds the smaller by 27. Find the quantities.

Worked explanation: The difference between eight parts and five parts is three parts. Three parts represent 27, so one part is nine. The quantities are 45 and 72. Check: 72 − 45 = 27 and 45:72 simplifies to 5:8. Dividing 27 by thirteen would wrongly treat 27 as the total of both quantities rather than their difference.

Problem 4 — Linear scale versus area scale

A plan uses a linear scale of 1:200. A rectangular room measures 4.5 cm by 3.2 cm on the plan. Find its actual area in square metres.

Worked explanation: The actual lengths are 900 cm = 9 m and 640 cm = 6.4 m. The area is therefore 9 × 6.4 = 57.6 m². Equivalently, the plan area is 14.4 cm² and the area scale factor is 200² = 40,000, giving 576,000 cm² = 57.6 m². Multiplying the plan area by only 200 would use a one-dimensional scale on a two-dimensional quantity.

Problem 5 — Choose substitution or elimination

Solve 3x + y = 17 and 2x − y = 3. Then explain why elimination is especially convenient.

Worked explanation: Adding the equations removes y immediately: 5x = 20, so x = 4. Substituting into 3x + y = 17 gives y = 5. Check both equations: 12 + 5 = 17 and 8 − 5 = 3. Elimination is convenient because the y coefficients are already opposites. Substitution is still valid; method convenience comes from the structure of the current equations.

Problem 6 — Maximum whole count under a budget

A fixed setup costs 32 units and each participant costs another 9 units. The budget is 250 units. What is the maximum whole number of participants?

Worked explanation: Solve 32 + 9n ≤ 250. Then 9n ≤ 218, so n ≤ 218/9 ≈ 24.22. Since n is a nonnegative integer, the maximum is 24. Check: 32 + 216 = 248 fits; twenty-five would cost 257 and exceed the budget. The result is not obtained by ordinary nearest-integer rounding but by respecting an upper-bound constraint.

Problem 7 — Minimum containers under a capacity constraint

Seventy-four identical objects must be packed into boxes holding at most twelve objects each. What is the minimum number of boxes?

Worked explanation: 74/12 = 6 remainder 2, or about 6.17. Six boxes can hold only 72 objects, so seven are necessary. Seven boxes provide enough capacity. This is a lower-bound problem, so the integer adjustment goes upward. It contrasts with the budget problem, where exceeding the real-valued bound is forbidden.

Problem 8 — A notch changes perimeter differently from area

A 16-by-11 rectangle has a 4-by-3 rectangular notch cut downward from the middle of its top edge. Find the remaining area and perimeter.

Worked explanation: Area is 16 × 11 − 4 × 3 = 176 − 12 = 164 square units. Original perimeter is 54. The cut removes a top segment of length four and adds a notch bottom of length four, which cancel, while two new vertical sides of length three add six. New perimeter is 60 units. The boundary must be traced; perimeter is not found by subtracting the removed rectangle’s perimeter.

Problem 9 — A right-triangle subgoal inside an area problem

An isosceles triangle has equal sides of length 17 and base 16. Find its area.

Worked explanation: The perpendicular from the apex to the base bisects the base into two lengths of eight. The height is √(17² − 8²) = √225 = 15. The area is 16 × 15/2 = 120 square units. The equal-side condition justifies the bisection. Using seventeen directly as the height would confuse a sloping side with a perpendicular distance.

Problem 10 — Same colour without replacement

A bag contains six red and four blue counters. Two are drawn randomly without replacement. Find the probability that they have the same colour.

Worked explanation: The disjoint possibilities are red-red or blue-blue. Their probabilities are (6/10)(5/9) = 1/3 and (4/10)(3/9) = 2/15. The total is 1/3 + 2/15 = 7/15. A counting check gives [C(6,2) + C(4,2)]/C(10,2) = (15 + 6)/45 = 7/15. Both routes describe the same event.

Problem 11 — At least one success through a complement

Three independent fair coins are tossed. Find the probability of at least one head.

Worked explanation: The complement is no heads, which means TTT with probability (1/2)³ = 1/8. Therefore the desired probability is 7/8. Counting directly also works: seven of the eight equally likely ordered outcomes contain at least one head. The complement is efficient because it replaces several success cases with one failure case.

Problem 12 — Combine means through totals

A group of 10 observations has mean 14 and a group of 15 observations has mean 20. Find the combined mean.

Worked explanation: The totals are 140 and 300, giving 440 across twenty-five observations. The combined mean is 17.6. Directly averaging fourteen and twenty would give seventeen and incorrectly weight the two groups equally. The result should lie between the two means and closer to twenty because the larger group has the higher mean.

Problem 13 — Average speed with unequal times

A cyclist travels for two hours at 18 km/h and then for three hours at 24 km/h. Find the average speed for the whole five-hour journey.

Worked explanation: Distances are 36 and 72 kilometres, giving 108 kilometres in five hours. Average speed is 21.6 km/h. The arithmetic mean of the two speeds, 21, is not appropriate because the cyclist spends unequal times at the two speeds. Average speed is total distance divided by total time.

Problem 14 — A graph model hits a boundary

A tank starts with 70 litres and receives 8 litres per minute. Its capacity is 150 litres. Under the constant-inflow model, when does it first reach capacity, and why should the formula not automatically be used to describe retained volume after that time?

Worked explanation: Solve 70 + 8t = 150, giving t = 10 minutes. The formula models filling before the capacity constraint changes the physical situation. After ten minutes, continued substitution gives values above 150, which cannot be retained in the tank. A post-capacity model needs information about overflow or stopping the inflow. The algebra remains calculable beyond ten; the physical interpretation does not automatically remain valid.

Problem 15 — Reverse a square without losing a branch

A number is doubled, five is subtracted, and the result is squared to give 81. Find all possible original real numbers.

Worked explanation: Let the number be x. Then (2x − 5)² = 81. Therefore 2x − 5 = 9 or −9. The first gives x = 7; the second gives x = −2. Both check. Taking only the principal square root nine and continuing would miss the branch produced because squaring is not one-to-one over the real numbers.

Problem 16 — An invariant rules out a target

A counter begins at 5. Each legal move adds or subtracts 4. Can it ever reach 22?

Worked explanation: Every reachable value has the same remainder as five modulo four, namely one. Twenty-two has remainder two modulo four. Therefore it is unreachable. Repeated trial may fail to find 22, but the invariant explains why no legal sequence can reach it. The same reasoning can be stated through parity classes only if parity is strong enough; here modulo four carries the decisive information.

Problem 17 — Absolute value creates cases

Solve |3x + 1| = 8 over the reals.

Worked explanation: Either 3x + 1 = 8 or 3x + 1 = −8. The solutions are x = 7/3 and x = −3. Both check. The two cases represent points whose signed expression is eight units from zero. The result is not obtained by placing a ± sign randomly on x; the branching occurs at the value inside the absolute expression.

Problem 18 — A fixed perimeter creates an area maximum

A rectangle has perimeter 52 units. Find the maximum possible area over positive real side lengths.

Worked explanation: If one side is x, the other is 26 − x. Area is x(26 − x) = 169 − (x − 13)². Therefore area is at most 169 square units, achieved when both sides are thirteen. A finite table of integer widths can suggest the result but does not prove the real-valued maximum as directly as the completed-square form.

Problem 19 — A parameter changes the number of solutions

For what real values of k does x² + 2x + k = 0 have two distinct real solutions, one repeated real solution, or no real solutions?

Worked explanation: The discriminant is 4 − 4k = 4(1 − k). It is positive when k < 1, zero when k = 1 and negative when k > 1. Therefore there are two distinct real roots for k < 1, one repeated root for k = 1 and none for k > 1. Completing the square gives (x + 1)² = 1 − k and reveals the same classification through the nonnegativity of a real square.

Problem 20 — A counting domain rejects a valid algebraic value

A fixed charge is 7 units plus 4 units for each complete item. Can the total cost be exactly 30 units for a nonnegative whole number of items?

Worked explanation: Solve 7 + 4n = 30, giving n = 23/4 = 5.75. This is not an allowed item count, so no exact whole-item purchase has cost thirty. Five items cost 27 and six cost 31. Rounding the algebraic value would change the equality rather than solve it.

Problem 21 — The data are insufficient

A set of five real numbers has mean 12. Can its median be determined uniquely from this information?

Worked explanation: No. The mean fixes the total at sixty but not the ordered middle value. For example, 10, 11, 12, 13, 14 has mean and median twelve, while 0, 1, 5, 20, 34 also totals sixty but has median five. A correct response explains why the requested quantity is not determined rather than inventing a median from the mean.

Problem 22 — A proof target needs a general structure

Show that the square of every odd integer is odd.

Worked explanation: Write an odd integer as 2n + 1. Its square is 4n² + 4n + 1 = 2(2n² + 2n) + 1. Since 2n² + 2n is an integer, the square has the form two times an integer plus one and is therefore odd. Several odd-number examples would support the pattern but not cover every integer.

Problem 23 — A rational equation must keep its restrictions

Solve 2/(x − 2) = 3/(x + 1) over the reals.

Worked explanation: The original domain excludes x = 2 and x = −1. Cross multiplication is legitimate on the domain: 2(x + 1) = 3(x − 2). Thus 2x + 2 = 3x − 6 and x = 8. It is allowed. Check: 2/6 = 1/3 and 3/9 = 1/3. Stating the exclusions keeps the transformation honest even though the final candidate does not violate them.

Problem 24 — Decide which method makes the target visible

For f(x) = x² − 8x + 7, find its real zeros and minimum value. Use more than one form of the expression.

Worked explanation: Factorisation gives f(x) = (x − 1)(x − 7), so the zeros are one and seven. Completing the square gives f(x) = (x − 4)² − 9, so the minimum value is −9 at x = 4. Both forms are equivalent. One displays the roots; the other displays the minimum. Method choice is not about declaring one form universally best but about making the requested feature easy to read.

The checkpoint should produce next steps, not a score label. If you translated relationships correctly but lost conditions late in solutions, practise verification and domain tracking. If calculation was accurate after a hint but you could not choose a representation, work on the first modelling step. If one topic was simply untaught, schedule instruction rather than classifying the attempt as a problem-solving failure.

Choose two incorrect or supported questions and create changed versions. Alter one structural feature at a time: the percentage base, replacement condition, integer domain, sign, target quantity or representation. Solve the new versions without the original answers visible. Transfer to a changed problem is stronger evidence than reproducing the correction immediately.

28. Questions students and parents ask about mathematical problem solving

Why can I do exercises but not word problems?

A topical exercise often supplies the method in its heading and uses a familiar representation. A word or mixed problem may require you to identify the relationship, represent it and choose the method before executing it. Compare where your work stops. If you can solve once the equation is supplied, the missing step may be representation rather than the later algebra.

Should I memorise a problem-solving checklist?

A short routine can help you remember to identify the target, represent relationships and check the answer. It should not become a script that replaces judgement. Different problems need different representations, cases and checks. Understand what each stage is trying to accomplish so you can shorten, expand or revisit it appropriately.

How long should I struggle before looking at a solution?

There is no universal number of minutes. Make an informative attempt first: state the target, record the givens, try a representation and identify the exact point where progress stops. If you lack required knowledge, seek instruction. If you have a partial route, a small hint may be enough. After studying help, attempt a fresh problem without it.

Is guessing ever a valid method?

Systematic trial can be mathematically useful when the search space is small or a changing pattern is understood. Record what each trial changes and use information from failed trials. Random guesses that neither narrow the range nor test a hypothesis are less useful. A trial route should become more informed as the work proceeds.

Should I always draw a diagram?

No. Draw when spatial relationships, part-whole structure, cases or dependencies become clearer through a diagram. A forced picture can add clutter to a problem already represented cleanly by an equation. The question is what uncertainty the representation removes.

What if two methods give different answers?

Locate the first claim on which they disagree. Check the representation, domain and transformations before deciding which final number looks more reasonable. If both methods are valid for the same problem and conditions, their conclusions must agree. A disagreement is useful evidence that at least one route contains a modelling, algebraic or arithmetic error.

What if my answer is correct but my method is different from the model solution?

A different valid route is mathematically acceptable unless the task specifically requires a method. Check that every transformation is justified, conditions are preserved and the answer is complete. Model solutions are examples of valid routes, not always unique scripts. Course-specific marking rules may still require particular evidence or working, so follow them when relevant.

How does tuition help problem solving rather than simply give answers?

Useful tutoring observes where the student’s route fails, selects the smallest support that restores progress and then returns decisions to the learner. It can compare representations, repair prerequisites, ask a discriminating question or show a worked example. The long-term test is whether the learner can later start, select, solve and check with less support.

Does more difficult practice automatically improve problem solving?

No. Difficulty can come from untaught knowledge, excessive arithmetic load, unclear wording or genuinely novel structure. Choose problems that test the decision you want to improve. A well-designed near contrast can reveal more than an obscure puzzle whose solution depends on a trick unrelated to the current learning goal.

How do I know whether a problem is impossible rather than whether I am stuck?

Look for contradictions, violated domains, invariants or bounds. A whole-number model producing a noninteger value can indicate inconsistent data. A geometry target may violate triangle inequalities. A parameter condition may have no real solution. Proving impossibility requires a reason that covers all allowed routes, not merely the fact that your attempts have failed so far.

29. A compact system for an unfamiliar secondary mathematics problem

When a question is unfamiliar, begin with the target. What must the finished answer contain? Then read the quantities and conditions: what is known, what can vary, what is fixed, and what is forbidden? Choose a representation that exposes the decisive relationship rather than merely displaying every number.

Next, create one useful subgoal or candidate route. Ask what the route is expected to simplify. Execute it while preserving equality, domain, units and case conditions. Monitor the result against an estimate or structural prediction. If the route stalls, return to the last justified state and change something specific.

Finally, verify against the original problem. Check candidates, boundaries, units and completeness. Interpret the answer in words if the context requires it. A correct final number is strongest when you can explain why it belongs to the problem and why no required alternatives have been omitted.

In compact form:

TARGET → QUANTITIES & CONDITIONS → REPRESENT → SUBGOAL → CHOOSE ROUTE → EXECUTE → MONITOR → RECOVER OR SWITCH → VERIFY → INTERPRET → COMPLETE.

This is not a rigid twelve-step ritual. Short problems may compress most stages into a few lines. Difficult problems may revisit representation or route selection several times. The value of the sequence is diagnostic: it tells you which kind of decision is missing when “I don’t know how to solve it” is too broad to guide the next action.

Return to the thirteen boxes. Clara’s first progress came from a reference case. Ryan’s trial became systematic because each replacement changed the total by four. Aisha’s equations compressed the same comparison. Their routes differed in appearance but shared a mathematical core: preserve the total number of boxes and measure how the contents change when one type replaces the other.

That is what a reusable problem-solving method should do. It should help you see the structure beneath the surface without forcing every question into one template. The aim is not to make unfamiliar problems look familiar by pretending they are identical. It is to find the relationship that makes a justified route possible.

30. Continue from problem solving into study, understanding and specialist mathematics

If the main difficulty is organising learning across weeks and examples, continue to How to Study Maths Effectively. If rules feel arbitrary or conditions keep disappearing, use How to Understand Mathematics Instead of Memorising It.

For the narrower problem-solving leaves, use Represent the Problem Before You Calculate, Solve a Simpler Problem First, Work Backwards When the Route Is Hidden, Split a Difficult Problem into Cases and Check an Answer Through an Independent Route.

For course and school-stage routes, return to the Secondary Mathematics Learning Hub. The Additional Mathematics Directory remains the specialist owner for A-Math. The Mathematics Examination Craft route remains the owner for examination-performance mechanics.

Evidence and scope note: all worked problems, fictional learner scenes and checkpoint tasks in this article are original explanatory material. They are not reported outcomes or a trial of the complete system. Research-informed bridges refer to the What Works Clearinghouse problem-solving guide for Grades 4–8 and algebra guide for middle and high school students; consult the source documents for their populations, methods and evidence ratings. The complete BTT casebook should be adapted to the learner’s taught content and current course rather than treated as a guaranteed formula for examination marks.

A mathematics problem becomes solvable when the learner can turn its conditions into a representation, turn the representation into a route, and keep checking that the route still belongs to the original question.

Which Method—and Why Not the Other One? A Mixed-Problem Decision Casebook

Mathematics becomes harder when the chapter label disappears. A learner who can solve ten substitution questions in a row may still hesitate when a mixed paper presents one equation, one ratio problem, one graph, one geometry question and one modelling task without telling the student which method belongs where.

The missing capability is often not calculation. It is method selection.

A strong problem solver does not ask only, “Which method can work?” The stronger question is: Which method is best justified by the structure here, and why are the obvious alternatives weaker?

Case 1: simultaneous equations—substitution or elimination?

Solve:

[ x+y=10,qquad 3x-2y=5. ]

Both substitution and elimination are valid. The first equation already isolates either variable cheaply, so substitution has a low setup cost:

[ y=10-x. ]

Substitute into the second equation:

[ 3x-2(10-x)=5, ]

so (5x=25), giving (x=5), (y=5).

Why not elimination? Elimination also works. But it requires a scaling step or a deliberate coefficient match. The decisive clue is that one equation is already almost solved for a variable. The method choice is about efficiency, not correctness.

Case 2: when elimination is cleaner

Solve:

[ 2x+3y=19,qquad 3x-2y=9. ]

Neither variable is isolated cheaply. The coefficients invite elimination because multiplying the first equation by (2) and the second by (3) creates opposite (y)-coefficients:

[ 4x+6y=38,qquad 9x-6y=27. ]

Adding gives (13x=65), so (x=5), then (y=3).

Why not substitution? It is valid, but solving for (x) or (y) first introduces fractions earlier. The decisive structural clue is the coefficient pattern.

Case 3: factorisation or quadratic formula?

Solve:

[ x^2-7x+12=0. ]

The integers (3) and (4) multiply to (12) and add to (7), so factorisation gives:

[ (x-3)(x-4)=0. ]

Hence (x=3) or (x=4).

Why not the quadratic formula? It works, but it adds arithmetic and opportunities for error without adding information. The decisive clue is the simple integer factor structure.

Case 4: when the quadratic formula earns its place

Solve:

[ 2x^2+3x-1=0. ]

This does not factorise neatly over the integers. The quadratic formula gives:

[ x=rac{-3pmsqrt{17}}{4}. ]

Why not keep guessing factor pairs? Because the structure does not support a simple integer factorisation. The method decision should respond to the polynomial, not to habit.

Case 5: proportional reasoning or algebra?

A recipe uses (450) g of flour for (6) servings. How much flour is needed for (10) servings?

Direct proportion gives:

[ 450 imesrac{10}{6}=750 ext{ g}. ]

Why not build a full equation? You can. But the ratio structure is already explicit. The shortest defensible method is proportional scaling.

Method choice is not a competition to use the most advanced technique. It is a match between structure and tool.

Case 6: algebra or trial-and-improvement?

Three consecutive integers have a sum of (72). Find them.

Let the integers be (n-1,n,n+1). Then:

[ 3n=72, ]

so (n=24), giving (23,24,25).

Trial-and-improvement could find the same answer, but the consecutive-number structure compresses directly into algebra. The decisive clue is the fixed relationship among the unknowns.

Case 7: diagram or equation?

A rectangle has perimeter (34) cm and length (3) cm more than its width. Find its dimensions.

A labelled diagram helps preserve the relationship, but it does not finish the problem. Let the width be (w), so the length is (w+3). Then:

[ 2w+2(w+3)=34, ]

giving (w=7) and length (10).

Why not choose between diagram and algebra? Because the best route uses both: the diagram stabilises interpretation, and the equation performs the solving.

Case 8: exact algebra or graph?

Solve:

[ x^2-5x+6=0. ]

Factorisation gives exact roots (2) and (3). A graph can confirm the intercepts and show the quadratic’s shape, but it is not the most efficient primary method.

Now compare:

[ x=cos x. ]

Elementary factorisation does not apply. A graph of (y=x) and (y=cos x) reveals an intersection near (0.74), and a numerical method can refine it. The decisive clue is the equation type.

Case 9: Pythagoras or trigonometry?

A right triangle has perpendicular sides (6) cm and (8) cm. Find the hypotenuse.

Pythagoras is direct:

[ c^2=6^2+8^2=100, ]

so (c=10).

Why not trigonometry? It would require introducing an angle that the question does not give or ask for. The decisive clue is that two side lengths in a right triangle already determine the third through Pythagoras.

Case 10: trigonometry when an angle controls the relationship

A ladder (5) m long makes an angle of (68^circ) with the ground. How high up the wall does it reach?

The height is opposite the given angle and the ladder is the hypotenuse:

[ h=5sin68^circ. ]

Pythagoras is not yet usable because only one side length is known. The decisive clue is the angle-side relationship.

Case 11: mean or median?

Five delivery times in minutes are (12,13,13,14,48).

The mean is:

[ rac{12+13+13+14+48}{5}=20. ]

The median is (13).

Which method should summarise the “typical” delivery? That depends on the reader job. The outlier (48) strongly affects the mean. If the goal is to represent a central typical time, the median may be more informative. If the total time burden matters, the mean still carries useful information.

The decisive clue is not a formula keyword. It is what feature the summary must preserve.

Case 12: counting directly or using the complement?

A fair die is rolled twice. Find the probability of getting at least one six.

Direct counting is possible. But the complement is shorter:

[ P( ext{at least one six})=1-P( ext{no six}) ]

[ =1-left(rac56 ight)^2=rac{11}{36}. ]

The decisive clue is the phrase “at least one”, which often creates a clean complement.

Method selection is a classification problem

Before calculating, classify the structure:

• Is there a proportional relationship?
• Is there a factor structure?
• Is a variable already isolated?
• Do coefficients invite elimination?
• Is the question exact or approximate?
• Is the geometry controlled by side lengths, angles or similarity?
• Does “at least one” suggest a complement?
• Does the data contain an outlier that affects the summary?
• Is a diagram needed to expose the relationship?

Do not switch methods merely because the current route feels long

A route can be valid but unfinished. Before abandoning it, ask:

1. Is the current method mathematically invalid?
2. Is it valid but inefficient?
3. Is it valid and efficient but I have not completed the intermediate step?

Only the first case demands repair. The second may justify a switch. The third often needs persistence rather than reinvention.

Fresh mixed-problem transfer

Choose a method before solving each problem. Write one sentence naming the decisive clue.

1. Solve (5x+2y=19) and (3x-2y=5).
2. Solve (x^2-11x+24=0).
3. Solve (3x^2+2x-7=0).
4. A map scale is (1:25,000). A route measures (7.2) cm on the map. Find the real distance.
5. A right triangle has hypotenuse (12) cm and one acute angle (35^circ). Find the side opposite the angle.
6. Find the probability of at least one head in four fair coin tosses.
7. Data values are (4,5,5,6,40). Compare mean and median and decide which better answers “typical”.
8. Solve approximately (x=e^{-x}).

The answer is not complete until the method is explained

A learner who can calculate but cannot explain why the method fits is still dependent on topic labels. A transfer-ready learner can recognise the structure without the chapter heading.

The durable loop is:

Read → classify structure → compare plausible methods → choose the decisive clue → solve → check → explain why the chosen route fits.