Choose your next learning step
Diagnose G3 Maths transfer with five paired tasks, worked answers, first-failure decisions and a delayed independent retest.
1. Understand · 2. Percentages · 3. Graphs · 4. Geometry · 5. Probability · 6. Averages · 7. Next step
This original G3 K310 diagnostic is a teaching workshop, not an official paper, grade predictor or validated psychometric test. Attempt the tasks independently before reading their checks.
Chapter index
Chapters 1–3
Chapters 4–6
CHAPTER 1 OF 7 · Understand
1. Separate knowing a method from recognising when to use it
A student may solve a labelled percentage exercise yet fail when the same relationship appears in an invoice. Another may calculate a triangle height correctly but multiply it by the wrong dimension in a volume problem. The lost mark does not always identify the missing capability.
This workshop uses paired tasks. The first makes the structure easier to see. The second changes wording, representation or context while preserving a central relationship. Compare the two solution routes, not only the final scores.
If both tasks fail at the same mathematical step, inspect the prerequisite or method. If the first succeeds but the second never reaches a useful equation, inspect recognition and translation. If both models are correct but execution breaks, inspect algebra, units or arithmetic. These are working hypotheses: verify them with a fresh task rather than labelling the learner from one attempt.
Allow a first untimed attempt with no topic labels beyond the task itself. Record the first equation and the first uncertain decision. Timing can be introduced later; it should not hide a conceptual problem during the initial diagnosis.
Task A: An amount after a 20% increase is 96. Find the original amount. Task B: A fictional repair invoice shows a $144 total after a 20% service surcharge. Find the charge before that surcharge. Assume this is the only addition.
Attempt both, then check. A: 1.20x = 96, so x = 80. B: 1.20c = 144, so c = 120. The final total is 120% of the original charge. Subtracting 20% of the final total incorrectly uses the new amount as the base.
Now change the wording: a $144 total includes a fixed $24 service fee. The pre-fee charge is 144 − 24 = 120. The answer happens to match Task B, but the model is different. A correct answer alone would conceal whether the student understood the relationship.
Diagnosis: if the student writes 0.80 × 144 for Task B, repair the reference-whole decision. If the student writes the right multiplier but divides incorrectly, repair execution. Retest with a discount rather than a surcharge so the routine cannot be copied mechanically.
Task A: A straight line passes through (2, 11) and (6, 23). Find its equation. Task B: A fictional equipment hire costs $17 for 4 hours and $29 for 8 hours. Assume cost is a fixed fee plus a constant hourly charge. Find the model and interpret its intercept.
A: gradient = (23 − 11)/(6 − 2) = 3. Substituting (2, 11) into y = 3x + b gives b = 5, so y = 3x + 5. B: hourly rate = (29 − 17)/(8 − 4) = 3 dollars per hour. Thus C = 3h + 5 and the fixed fee is $5.
The two models have the same algebraic form. In Task B, the gradient and intercept have contextual meanings and units. Checking h = 8 gives C = 29, matching the data.
Boundary check: the model’s intercept is a mathematical fixed-fee parameter, but the data alone do not prove a real company sells zero-hour rentals. State the domain allowed by the problem. If the fixed fee is missed, investigate whether the student believes every rate model passes through the origin.
Task A: A right triangle has hypotenuse 10 cm and an acute angle of 30°. Find the opposite side. Task B: A triangular prism has length 12 cm. Its right-triangular cross-section has hypotenuse 10 cm and a 30° angle between that hypotenuse and one leg. Find the volume.
A: opposite side = 10 sin 30° = 5 cm. For B, the other perpendicular leg is 10 cos 30° = 5√3 cm. Cross-sectional area = ½ × 5 × 5√3 = 25√3/2 cm². Volume = 12 × 25√3/2 = 150√3 cm³, approximately 260 cm³ to three significant figures.
The hypotenuse is not perpendicular to either leg. Using ½ × 10 × 5 for the triangle area ignores the base-height relationship. The formula may be remembered while the geometry is misrepresented.
Diagnosis: if Task A succeeds but B fails at triangle area, the first repair is not necessarily sine. Ask the student to identify the perpendicular pair. If that succeeds but prism length is confused with triangle height, repair the separation of cross-section and extrusion length.
CHAPTER 5 OF 7 · Probability
5. Pair 4: combined probability with and without replacement
Task A: A bag has 3 red and 2 blue counters. Two counters are drawn without replacement. Find the probability both are red. Task B: A second task uses the same bag, but the first counter is returned and the bag remixed before the second draw. Find the probability both are red.
A: P(RR) = 3/5 × 2/4 = 3/10. B: P(RR) = 3/5 × 3/5 = 9/25. The changed sentence changes the second-stage sample space. The original total cannot be reused automatically when an item has been removed.
Transfer question: without replacement, find the probability of one red and one blue in either order. RB contributes 3/5 × 2/4 = 3/10; BR contributes 2/5 × 3/4 = 3/10. These orders are mutually exclusive, so add them to obtain 3/5.
A learner who multiplies correctly but chooses the wrong branch probability needs a sample-space repair. A learner who finds RB correctly but omits BR needs an event-definition repair. Drawing more elaborate trees is useful only when the tree reflects the stated event and changing bag contents.
Task A: 10 readings have mean 12; 20 readings have mean 18. Find the combined mean. Task B: A cyclist travels 6 km at 12 km/h and another 6 km at 24 km/h. Find the average speed for the whole journey, with no stop.
A: total = 10 × 12 + 20 × 18 = 480; count = 30; mean = 16. B: times are 6/12 = 0.5 h and 6/24 = 0.25 h. Average speed = total distance/total time = 12/0.75 = 16 km/h.
The answers happen to share the number 16. Their denominators count different things. In B, averaging 12 and 24 to obtain 18 km/h incorrectly gives equal weight to unequal time intervals. Equal distances do not imply equal travel times.
Change B to one hour at each speed. Then total distance = 12 + 24 = 36 km, total time = 2 h, and average speed = 18 km/h. Ask the student to explain why the answer changes. A reliable method reconstructs the relevant totals rather than treating every average as an arithmetic average of displayed numbers.
Keep a short record for each pair: model chosen, first wrong decision, whether a hint was needed, and independent checking method. Distinguish a missing concept from a translation failure and an execution failure. Do not infer all three from one wrong answer.
Choose the earliest consequential weakness. For reverse percentages, label the original whole and the multiplier. For the prism, redraw and mark the perpendicular legs. For probability, update the bag contents after the first draw. Each repair should end with one changed problem where the student must choose the structure again.
Retest after a gap using new numbers and a different surface story. Remove the chapter label, then add a modest time constraint once the reasoning is stable. Do not claim secure transfer from an immediate repetition of the same question after seeing its solution.
Frequently asked question: Is passing all five pairs evidence of an A1? No. The sample is small and does not cover the entire syllabus or examination conditions. It provides specific evidence about these relationships. Use a complete syllabus map and broader independent work for examination readiness.
Return to the existing subject guides for targeted teaching, then revisit the diagnostic. The purpose is a useful next action: repair the actual broken decision and verify that the repaired method survives a fresh representation.
Continue learning and official sources
Official K310 Mathematics syllabus · Probability Event Logic · Dimensional Reasoning · Mathematics Hub
