Secondary Mathematics Mixed-Question Clinic: Choose and Justify a Method

In a chapter exercise, the heading often tells the student which method to use. A mixed problem removes that signal. The first decision becomes part of the mathematics: which quantity is unknown, which relationship constrains it, and which conditions make the proposed method valid?

This clinic contains nine original problems with contrasting approaches, full solutions and checks. It is not a mock examination or a substitute for the existing chapter guides. Its job is to make the first decision visible. Several cases connect topics; others show why an apparently convenient calculation answers the wrong question.

Select cases that match the student’s taught curriculum. Quadratic equations, similarity, coordinate geometry and combined probability require their own prerequisites. Do not infer weakness from an untaught method, and do not assume that every item belongs in every G1, G2 or G3 course at the same stage.

Use a three-line plan before calculating

Ask the learner to write three sentences: “I need to find…”, “The relationship I can use is…”, and “This works because…”. The third sentence should include the condition that licenses the method. Similar figures permit a scale-factor argument; arbitrary figures do not. A constant rate permits a linear time relationship; an unspecified changing rate does not.

After the plan, the learner completes the calculation and checks it against the original problem. A correct method with a calculation slip needs a different response from a perfectly executed calculation based on the wrong model. Keep both possibilities visible.

These planning sentences are a suggested teaching routine, not a prescribed examination format. In an assessment, students should follow the paper’s instructions and show appropriate working without adding unnecessary prose.

Clinic 1: Compare two plans, then respect the whole-number domain

Problem. Two fictional activity plans cover between zero and twenty-four sessions. Plan A costs $18 plus $3 per session. Plan B costs $42 plus $1 per session. When do the costs match, and for which permitted session counts is B cheaper?

A first useful move is to define n as the number of sessions. The cost models are A = 18 + 3n and B = 42 + n, with n a whole number from 0 to 24. Comparing only the per-session prices ignores the fixed charges. Comparing only $18 with $42 ignores repeated use.

Set the costs equal: 18 + 3n = 42 + n. Then 2n = 24 and n = 12. Both plans cost $54 at twelve sessions.

For B to be cheaper, 42 + n < 18 + 3n. This gives 24 < 2n, or n > 12. Within the permitted whole-number domain, B is cheaper for 13 through 24 sessions. The costs match at 12; A is cheaper for 0 through 11.

Check. At eleven sessions, A costs $51 and B costs $53. At thirteen, A costs $57 and B costs $55. The comparison changes at the predicted boundary. These are invented mathematical prices, not consumer advice about a real service.

Method lesson. Equality finds the boundary; an inequality determines the side of the boundary; the context decides which values count. A numerical crossing point is not automatically the complete answer.

Clinic 2: Two roots can describe the same rectangle

Problem. A rectangle has perimeter 36 cm and area 72 cm². Find its side lengths.

Let one side be x cm. Half the perimeter is 18 cm, so the other side is 18 − x cm. Positive lengths require 0 < x < 18. The area condition gives x(18 − x) = 72.

Rearrange to x² − 18x + 72 = 0 and factorise: (x − 6)(x − 12) = 0. Thus x = 6 or x = 12. If x is 6, the other side is 12; if x is 12, the other is 6. The dimensions are therefore 6 cm by 12 cm. The two algebraic roots label the same unordered pair of side lengths, not two differently sized rectangles.

Check. The perimeter is 2(6 + 12) = 36 cm and the area is 6 × 12 = 72 cm². Both original constraints are satisfied.

A common false start is to divide 36 by 4 and assume a square. That would produce sides of 9 cm and area 81 cm², which fails the stated area condition. A rectangle is not necessarily a square.

Method lesson. The shape supplies a relationship between two unknown lengths. Using one variable reduces the problem, but interpreting the solutions is still necessary after factorisation.

Clinic 3: A similar triangle changes area and perimeter differently

Problem. A right triangle has perpendicular sides 6 cm and 8 cm and hypotenuse 10 cm. A larger triangle is similar to it, with corresponding hypotenuse 15 cm. Find the larger triangle’s other sides, perimeter and area.

The linear scale factor is 15/10 = 3/2. Corresponding side lengths are 6 × 3/2 = 9 cm and 8 × 3/2 = 12 cm. The perimeter is 9 + 12 + 15 = 36 cm.

The smaller area is one half of 6 × 8, or 24 cm². The larger area is 24 × (3/2)² = 54 cm². Alternatively, calculate directly from its perpendicular sides: one half of 9 × 12 = 54 cm².

A mistaken solution might multiply the smaller area by 3/2 and obtain 36 cm². That applies a length scale factor to a two-dimensional measure. Showing two valid methods for the area makes the mismatch inspectable.

Check. The scaled sides still satisfy the right-triangle relationship: 9² + 12² = 15². The perimeter is multiplied by 3/2, from 24 cm to 36 cm. The area is multiplied by 9/4, from 24 cm² to 54 cm².

Method lesson. First identify what type of quantity is being scaled. Length, area and volume do not respond to similarity with the same power. The similarity condition must be given or established, not inferred merely from a diagram’s appearance.

Clinic 4: A tank problem needs both units and capacity

Problem. A tank is a rectangular cuboid with internal base dimensions 80 cm by 50 cm and internal height 20 cm. It initially contains water to a depth of 12 cm. A further 24 litres is added, with no leakage or other outflow. Find the final depth and state whether the tank overflows.

The base area is 80 × 50 = 4,000 cm². The added volume is 24,000 cm³ because one litre is 1,000 cm³. The increase in depth is 24,000/4,000 = 6 cm. The final depth is 12 + 6 = 18 cm, below the internal height of 20 cm, so the tank does not overflow.

There is room for another 2 cm of depth. The remaining capacity is 4,000 × 2 = 8,000 cm³, or 8 litres. This second calculation checks the capacity conclusion rather than merely repeating the first answer.

A student who obtains 0.006 cm may have divided 24 by 4,000 without converting litres. A student who gives 6 cm as the final depth may have calculated the increase correctly but omitted the initial water. These are different errors, despite arising in the same question.

Method lesson. Label each calculated quantity: base area, added volume, depth increase, final depth. Units and initial conditions are part of the model. The constant-base assumption is justified here by the tank’s rectangular-cuboid shape; it would not hold in the same way for every vessel.

Clinic 5: Journey sections do not necessarily have equal weight

Problem. A vehicle covers 45 km at 30 km/h and then 90 km at 45 km/h. There are no stops or other elapsed times in the model. Find the average speed for the complete journey.

The first section takes 45/30 = 1.5 hours. The second takes 90/45 = 2 hours. The total distance is 135 km and total time is 3.5 hours. Average speed is 135/3.5 = 270/7 km/h, approximately 38.6 km/h to three significant figures.

Averaging 30 and 45 gives 37.5 km/h, but the vehicle spends different lengths of time at those speeds. Averaging by the number of sections is not the required total-distance-over-total-time calculation.

Check. The result lies between 30 and 45 km/h. That range check is useful but not sufficient: the incorrect 37.5 also lies between them. A stronger check multiplies the exact average by the exact total time: (270/7) × (7/2) = 135 km.

If a stop were added, it would affect the elapsed-time denominator. If the question excluded stopping time, that would need to be stated. Reading which time interval is being measured is therefore part of the method, not a minor afterthought.

Method lesson. A plausible numerical answer can survive a weak check. Choose a check that tests the relationship responsible for the answer, not just whether the number looks reasonable.

Clinic 6: “One of each colour” includes two orders

Problem. A bag contains four red counters and two blue counters. Two counters are drawn without replacement, with each remaining counter equally likely at each draw. Find the probability of obtaining one counter of each colour.

There are two relevant orders. Red then blue has probability 4/6 × 2/5 = 4/15. Blue then red has probability 2/6 × 4/5 = 4/15. These ordered outcomes cannot both happen in the same two-draw experiment, so add their probabilities: 4/15 + 4/15 = 8/15.

A student who obtains 4/15 may have calculated one valid route and stopped before covering the full event. That is different from using an incorrect denominator. Ask the learner to describe what the requested event includes before offering a calculation prompt.

Check by the complement. The probability of two red counters is 4/6 × 3/5 = 2/5. The probability of two blue is 2/6 × 1/5 = 1/15. Together the same-colour probability is 7/15, leaving 1 − 7/15 = 8/15 for different colours.

Method lesson. Multiplication describes a specified successive route; addition combines the disjoint routes that make up the event. The words of the event determine which routes belong in the answer.

Clinic 7: Coordinates can supply geometry without a scale drawing

Problem. A triangle has vertices A(1, 2), B(7, 2) and C(7, 10). Find its area and perimeter.

A and B share a vertical coordinate, so AB is horizontal with length 7 − 1 = 6 units. B and C share a horizontal coordinate, so BC is vertical with length 10 − 2 = 8 units. The sides are perpendicular. The area is one half of 6 × 8 = 24 square units.

The remaining side AC has length √(6² + 8²) = 10 units. The perimeter is 6 + 8 + 10 = 24 units. Area and perimeter happen to have the same numerical value, but they measure different things and have different units.

A quick sketch may help organise the data, but the proof of perpendicularity comes from the coordinate relationships, not the drawing looking like a right angle. An inaccurately scaled sketch should not change those relationships.

Check. The enclosing rectangle has area 6 × 8 = 48 square units, and the diagonal divides it into two equal triangles. That gives a second way to see the area of 24. For the perimeter, ensure that the diagonal is included instead of an unused rectangle side.

Method lesson. A coordinate question need not require the most elaborate coordinate formula available. Use the simplest justified structure in the data.

Clinic 8: The same mean does not establish the same variation

Problem. Two fictional sets of five observed completion times, in minutes, are A: 12, 12, 12, 12, 12 and B: 8, 10, 12, 14, 16. Compare the mean, median and range of the listed observations. Does the same mean show that the two sets have the same spread?

Each set sums to 60 minutes, so each mean is 60/5 = 12 minutes. Both ordered sets have median 12 minutes. Set A has range 12 − 12 = 0 minutes. Set B has range 16 − 8 = 8 minutes. The means and medians agree, but the ranges do not.

The data therefore refute the claim that matching means alone establish matching spread. They do not establish which underlying process will be more reliable in the future. The problem gives only these small fictional observation sets, not a representative study of an actual service.

A student may correctly calculate both means and then stop because the numbers match. Ask which part of “compare the observations” remains unanswered. Calculation of one summary measure is not the same as interpreting the whole requested comparison.

Method lesson. Use a measure that addresses the claim. A statement about typical value calls for a different inspection from a statement about variation. Do not infer a causal explanation or a future guarantee from a descriptive calculation alone.

Clinic 9: Sometimes the correct first decision is “not enough information”

Problem. A rectangle has perimeter 28 cm. Find its area.

The perimeter implies that adjacent side lengths add to 14 cm. It does not determine both lengths. A rectangle measuring 6 cm by 8 cm has perimeter 28 cm and area 48 cm². A rectangle measuring 4 cm by 10 cm has the same perimeter and area 40 cm².

Because two permitted rectangles give different areas, the information does not determine a unique area. Giving 49 cm² assumes a square of side 7 cm, an additional condition not stated in the question. That area is possible, but possibility is not uniqueness.

A useful follow-up asks what extra information would resolve the problem. One side length would suffice. A statement that the shape is a square would also suffice. The area itself, naturally, would make the request redundant rather than create an interesting solution.

Method lesson. Recognising insufficient information is mathematical work. Do not reward an invented assumption merely because it produces a neat answer. A counterexample can show exactly why the stated information is inadequate.

Review the decision, not just the final line

After each attempted clinic, ask the learner to identify the controlling relationship and one tempting alternative that would not answer the question. For the tank, the controlling relationship is volume equals constant base area times depth. For the journey, it is total distance divided by total elapsed time. For the insufficient-information case, it is the fact that the perimeter leaves one degree of freedom.

Record an actual error at the point it occurs: model choice, condition, operation, unit, event coverage or interpretation. These labels organise this review; they are not a standardised diagnostic classification. More than one issue can appear in a single attempt.

A strong next question changes one important feature. Replace “without replacement” with “with replacement”. Supply area rather than perimeter. Ask for the fixed charge rather than the number of sessions. The purpose is to inspect adaptation, not to make the question difficult through irrelevant wording.

Routes after the clinic

For the wider exam-working discussion, use the existing Secondary Mathematics Exam Craft. For chapter explanations and worked stage guides, use the Secondary Mathematics capability map. For diagnosis, return to the existing Secondary Mathematics gaps guide.

The Mathematics Learning Hub remains the broad learner route. Additional Mathematics is a separate branch; these selected general mathematics tasks do not establish readiness for all A-Math content.

Scope and source notes

All nine clinic problems, contexts and solutions are original material prepared for this draft. Fictional plans, journeys and observations are not reports about actual people or providers. The pack is not a past paper, official specimen, grade predictor or complete syllabus checklist.

For the current assessment and content boundaries, consult SEAB’s SEC information and the student’s exact examination-year syllabus. SEC begins in 2027, while a 2026 candidate should use the applicable 2026 examination documents. Reviewed 6 September 2026. The mathematical examples above are illustrative teaching choices rather than a claim of universal placement across subject levels.