Secondary Mathematics Repair Check: Matched Practice, Delayed Retest and Transfer

A correction is a useful event. It is not the end of the investigation. A student may understand a worked example while the explanation is visible and still need help when the same relationship appears later. This practice pack separates those observations instead of merging them into a single “topic completed” tick.

The pack contains three sets of original questions with complete solutions. Set A can be used during teaching. Set B changes the numbers while retaining a closely related mathematical structure. Set C changes the representation, direction or decision required. These are author-designed teaching sets, not psychometrically equated test forms. Differences in performance may reflect differences in the tasks as well as changes in the learner.

Use a family only after its prerequisites have been taught. Similarity and probability are optional extensions, not assumptions about what every secondary student should know. Select material using the student’s actual school course and chapter sequence rather than school year alone.

Decide what observation you want

During Set A, explanation and guidance are allowed. Record them. For Set B, the suggested observation is a fresh attempt without the earlier worked solution visible. For Set C, the question is whether the learner can select and adapt the relevant relationship when the task no longer has exactly the same appearance.

A teacher might use Set B at a later lesson and Set C on another occasion. The interval should be recorded, not hidden. This pack does not prescribe a universally optimal delay or claim that a particular number of days guarantees retention. Its purpose is to make the conditions of the observation explicit.

Before beginning, write down the selected families, date, allowed tools, time conditions and support conditions. A calculator may be appropriate for one observation and intentionally unavailable for another; changing that condition changes what the result can support. Do not compare the two results as if nothing else changed.

The question sheets

Keep the answer section out of view during independent attempts. The student should write a governing equation, relationship or explanation, not only a final number.

Set A: Teaching and supported practice

A1 — Brackets. Simplify −2(3x − 4) + 5x.

A2 — Percentage. A quantity increases from 120 units to 138 units. Find the percentage increase.

A3 — Fixed and variable charges. A fictional service charges $12 plus $4 per booking. A bill is $44. Find the number of bookings.

A4 — A straight line. Find the gradient and equation of the line through (1, 5) and (4, 11).

A5 — Similarity. Two similar figures have corresponding lengths in the ratio small:large = 2:3. The smaller area is 20 cm². Find the larger area.

A6 — Two draws. A bag contains three 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 drawing two red counters.

Set B: A fresh attempt on closely related structures

B1 — Brackets. Simplify −4(2x − 3) + 7x.

B2 — Percentage. A quantity increases from 200 units to 230 units. Find the percentage increase.

B3 — Fixed and variable charges. A fictional service charges $15 plus $5 per booking. A bill is $55. Find the number of bookings.

B4 — A straight line. Find the gradient and equation of the line through (2, 8) and (5, 14).

B5 — Similarity. Two similar figures have corresponding lengths in the ratio small:large = 3:5. The smaller area is 27 cm². Find the larger area.

B6 — Two draws. 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 drawing two red counters.

Set C: A change in the task

C1 — From words to an expression. Three strips each have length x + 2 cm. Two further strips each have length x − 1 cm. Write a simplified expression for the combined length. State a condition on x that makes every strip length positive.

C2 — Reverse percentage. A fictional price is $156 after a 20% increase. Find the original price.

C3 — Find the fixed charge. A fictional service costs $5 per booking plus a fixed charge. Eight bookings cost $55. Find the fixed charge and write a formula for the cost of b bookings.

C4 — Build the equation from a context. A container initially holds 7 litres of water. Water is added at a constant 2 litres per minute with no outflow. Write V, the volume in litres, in terms of t, the elapsed time in minutes. Find when V = 19 litres. Assume the container has enough capacity throughout the stated interval.

C5 — Reverse the area scale. Two similar figures have areas 36 cm² and 81 cm². A length on the smaller figure is 8 cm. Find the corresponding length on the larger figure.

C6 — Recognise a changed sampling rule. A bag contains four red counters and two blue counters. Two counters are drawn, but the first counter is replaced and the bag is mixed before the second draw. Every counter is equally likely on each draw. Find the probability of two red counters and compare it with B6.

Solutions: Family 1 — Preserve the bracket before collecting terms

A1 becomes −6x + 8 + 5x = 8 − x. The negative multiplier acts on both terms inside the bracket. In particular, (−2)(−4) gives positive 8. At x = 2, the original expression is −2(6 − 4) + 10 = 6, agreeing with 8 − 2.

B1 becomes −8x + 12 + 7x = 12 − x. A correct answer with no intervening working may be accepted mathematically, but a repair review benefits from seeing whether the learner handled the positive constant deliberately. Ask for an explanation after the attempt, rather than supplying the sign while it is being attempted.

C1 is 3(x + 2) + 2(x − 1) = 3x + 6 + 2x − 2 = 5x + 4 cm. For every strip to have positive length, x + 2 > 0 and x − 1 > 0. Together these require x > 1.

C1 is not simply another negative-bracket question. It adds representation and a contextual restriction. A student who passes B1 but struggles with C1 has not thereby “forgotten brackets”. Inspect whether they constructed the sum and understood the condition. Record which added demand caused the difficulty rather than attributing every transfer error to the original repair.

Solutions: Family 2 — Keep the reference quantity visible

For A2, the increase is 138 − 120 = 18 units. Relative to the original 120, this is 18/120 × 100% = 15%. Equivalently, 138/120 = 1.15, showing the original quantity multiplied by 115%.

For B2, the increase is 30 units and the base is 200. The percentage increase is 30/200 × 100% = 15%. The same percentage with different quantities is intentional. It checks whether the student uses the relationship rather than recalling a particular subtraction.

For C2, let P be the original price. A 20% increase gives 1.2P = 156, so P = 130 dollars. A check is 130 × 1.2 = 156. Subtracting 20% of the final price would give $124.80, which would not return to $156 after a 20% increase.

The direction of the task has changed. A2 and B2 ask for the percentage; C2 supplies a percentage and asks for the starting quantity. If C2 needs repair, return to the equation that states what 100% refers to. Do not invent a special “reverse percentage trick” detached from that meaning.

Solutions: Family 3 — Distinguish one-off and repeated quantities

For A3, let b be the number of bookings. The model is 12 + 4b = 44. Subtracting the fixed charge gives 4b = 32, hence b = 8. The answer is a nonnegative whole number, as expected for a count of bookings in this model.

For B3, the model is 15 + 5b = 55. This gives 5b = 40 and b = 8. Writing 15b + 5 would exchange the fixed and repeated quantities. The correct solution of that different equation would not answer the question.

For C3, eight bookings account for 8 × 5 = 40 dollars. The fixed charge is 55 − 40 = 15 dollars. The formula is C = 15 + 5b. The same mathematical structure has been approached from a different unknown.

A useful explanation connects each term to its unit and role: 15 dollars is paid once; 5 dollars per booking multiplied by b bookings gives the variable charge. This is a mathematical illustration, not a statement about any provider’s actual prices or contract terms.

Solutions: Family 4 — Connect change and starting value

In A4, the gradient is (11 − 5)/(4 − 1) = 2. Substituting (1, 5) into y = 2x + c gives c = 3. The equation is y = 2x + 3, and both supplied points satisfy it.

In B4, the gradient is (14 − 8)/(5 − 2) = 2. Using (2, 8) gives 8 = 4 + c, so c = 4. The equation is y = 2x + 4. The slope is unchanged but the vertical intercept is different. A learner should not carry the intercept from A4 into B4 merely because the gradients match.

In C4, the relationship is V = 7 + 2t. The starting volume is 7 litres; the rate contributes 2t litres after t minutes. For V = 19, solve 7 + 2t = 19 to obtain t = 6 minutes.

The context supplies the rate and starting value directly instead of giving two coordinate pairs. Within the stated model, t is nonnegative, the inflow is constant, and the capacity assumption applies. Those conditions belong to the equation’s use. A formula without its interpretation is not the full model.

Solutions: Family 5 — Separate lengths from areas

For A5, the linear scale factor from small to large is 3/2. Areas scale by the square of that factor, so the larger area is 20 × (3/2)² = 20 × 9/4 = 45 cm².

For B5, the linear scale factor is 5/3. The larger area is 27 × (5/3)² = 27 × 25/9 = 75 cm². Multiplying by 5/3 alone would scale a length, not an area.

For C5, the area ratio large:small is 81:36 = 9:4. The positive linear scale factor is √(9/4) = 3/2. The corresponding larger length is 8 × 3/2 = 12 cm.

The word “similar” is essential. A pair of area measurements alone does not establish corresponding length ratios for arbitrary figures. If similarity were removed from C5, the requested length would not generally be determined. An explanation that names this condition is stronger than one that merely applies a square root whenever two areas appear.

Solutions: Family 6 — Read the replacement condition

For A6, the first-red probability is 3/5. Given that a red was drawn and not replaced, two red counters remain among four counters, so the next-red probability is 2/4. The required probability is 3/5 × 2/4 = 3/10.

For B6, the corresponding calculation is 4/6 × 3/5 = 2/5. The second fraction changes both numerator and denominator because a red counter has been removed.

For C6, replacement restores the original composition before the second random draw. The calculation is 4/6 × 4/6 = 4/9. This exceeds B6’s 2/5 by 4/9 − 2/5 = 2/45.

The different answer is produced by a different sampling rule, not a different convention for multiplication. A student who automatically reuses 3/5 as the second fraction in C6 has failed to incorporate the replacement condition. Ask them to state the contents of the bag before each draw. Do not treat success in this small example as comprehensive mastery of probability.

Record evidence without manufacturing a score

For each selected item, retain five observations: the answer, the mathematical relationship written, the help given, any self-correction, and the conditions of the attempt. A short note can preserve all five: “B3: 8 bookings; wrote 15 + 5b = 55; no hints; checked by substitution; attempted five days after supported A3.”

Contrast that with “B3: 8 bookings after the teacher supplied the equation.” Both may be successful teaching moments, but they are not the same evidence of independent modelling. Do not promote the second observation into the first by recording only a tick.

Use a blank record with fields for date, family, set, attempt, prompt, check and next step. Avoid a combined numerical “mastery index” unless its meaning and validation are separately established. This pack provides no calibrated threshold for pass, grade, subject placement or suitability for A-Math.

What to do with four common result patterns

A correct with help, B incorrect alone. Reopen the relevant relationship and inspect the original B working. The evidence supports a narrower claim than “the student cannot learn”: the supported performance has not yet been demonstrated independently on that B task.

B correct, C incorrect. Identify the additional demand in C. It may be representation, reverse reasoning, a condition, or a different type of unknown. Re-teaching the entire topic may not be necessary, but deciding that no repair is needed would also be premature.

B and C correct with unexplained answers. Ask for one reconstruction or check. The correct answers are real observations; the underlying method remains less visible. Do not retrospectively invent a reasoning chain the student did not supply.

B and C correct with explanation and no help. Record the success within the attempted scope. Move to a suitable next task or reduce unnecessary support on that skill. Do not extend the claim to every question in the chapter, every examination setting or every future date.

Boundaries that protect the review

Do not use the pack to infer a medical, psychological or learning-disability diagnosis. Do not assign blame from one attempt. Do not compare two students as though they received identical teaching and assessment conditions when they did not. A forgotten calculator, an untaught topic, a copied example and a time limit all change what an observation means.

A useful repair record can include uncertainty. “Need another independent word-problem attempt” is a legitimate next state. It is more precise than manufacturing a confident judgement from insufficient evidence.

For correction habits, return to the existing Mistake Ledger. For chapter teaching, use the Secondary Mathematics capability map. For timed-paper decisions, continue to Secondary Mathematics Exam Craft.

Scope and source notes

All eighteen questions and their solutions are original teaching material prepared for this draft. The sets are illustrative, not normed or equated assessments. The suggested recording process is an editorial teaching proposal, not an experimentally validated intervention or a report of actual tuition outcomes.

The larger stage route remains the Mathematics Learning Hub. For examination content and permitted tools, consult SEAB’s SEC information and the official syllabus for the student’s examination year and subject level. Reviewed 6 September 2026. Topic selection in this pack does not certify complete G1, G2 or G3 syllabus coverage.