From the Human Biology questions graded on AusGrader, these are the 5 topics WACE Human Biology students score lowest on:
Endocrine system: 66% average
Hominid evolutionary trends: 67% average
Homeostasis: 68% average
Gene pools: 69% average
Central and peripheral nervous system: 70% average
From the Physics questions graded on AusGrader, here's every area of study in VCE Physics ranked by average score, hardest first:
How are fields used in electricity generation: 60% average
How do physicists explain motion in two dimensions: 60% average
How do things move without contact: 61% average
How has understanding about the physical world changed: 65% average
How is scientific inquiry used to investigate fields, motion or light: 69% average
From the Maths Methods questions graded on AusGrader, these are the 5 topics QCE Maths Methods students score lowest on:
Further applications of differentiation: 53% average
Further integration: 58% average
Differentiation of trigonometric functions and differentiation rules: 60% average
Discrete random variables: 62% average
Continuous random variables and the normal distribution: 63% average
From the Biology questions graded on AusGrader, here's every topic in WACE Biology ranked by average score, hardest first:
Heredity: 70% average
Continuity of life on earth: 70% average
Infectious disease: 72% average
Homeostasis: 73% average
From the Biology questions graded on AusGrader, here's every topic in QCE Biology ranked by average score, hardest first:
Continuity of life on earth: 68% average
Functioning ecosystems and succession: 70% average
Genetics and heredity: 70% average
Biodiversity and populations: 71% average
From the Chemistry questions graded on AusGrader, these are the 5 topics HSC Chemistry students score lowest on, with the module each one sits in:
Analysis of organic substances (module 8: applying chemical ideas): 54% average
Factors that affect equilibrium (module 5: equilibrium and acid reactions): 60% average
Chemical synthesis and design (module 8: applying chemical ideas): 60% average
Static and dynamic equilibrium (module 5: equilibrium and acid reactions): 60% average
Nomenclature (module 7: organic chemistry): 61% average
From the Psychology questions graded on AusGrader, here's every area of study in VCE Psychology ranked by average score, hardest first:
How is scientific inquiry used to investigate mental processes and psychological functioning: 58% average
How does sleep affect mental processes and behaviour: 62% average
How do people learn and remember: 63% average
How does the nervous system enable psychological functioning: 67% average
What influences mental wellbeing: 70% average
You assign oxidation numbers to every atom in the reactants and products, then find the element whose oxidation number goes up (the one being oxidised).
21.1% chose (C), H in CH₄. In methane, hydrogen is +1, and in hydrogen gas (H₂) it’s 0. Going from +1 to 0 means hydrogen gains electrons, which is reduction, not oxidation. Whoever picked (C) either flipped oxidation and reduction or assigned the oxidation numbers in CH₄ incorrectly.
Assigning oxidation numbers to carbon:
• C in CH₄: with four H at +1 each, carbon is −4.
• C in CO: with O at −2, carbon is +2.
Carbon goes from −4 to +2, an increase of 6, and an increase in oxidation number is oxidation.
To confirm the rest: O in H₂O is −2 and stays −2 in CO (no change); H in H₂O goes from +1 to 0 in H₂ (reduction); H in CH₄ goes from +1 to 0 in H₂ (reduction). The element oxidised is carbon. The answer is (B) C in CH₄.
Question 2b sat in the coffee beans context and asked students to explain why, over time, the price elasticity of supply for coffee beans is likely to change. Three marks. The key phrase was 'over time', this wasn't asking about a snapshot.
Thirty-two percent scored zero.
Three errors dominated.
The first was answering from a PED perspective or conflating PES with a non-price supply factor. The examiners note this shows either a knowledge gap or a failure to read the question. PES measures how responsive quantity supplied is to a price change, it's not the same as a supply factor, and it's not demand-side.
The second error was explaining the law of supply, the direct relationship between price and quantity, instead of explaining the degree of responsiveness. Knowing the direction of change is not the same as knowing how elastic it is.
Third: students who identified a relevant factor (like production time or technology) but didn't explain how that factor would actually change the elasticity over time. Naming the factor is step one. Explaining the mechanism, for example, new growing methods shortening production periods, allowing farmers to respond more quickly to price signals, is what earns the marks.
Q32a (5 marks): List the following milestones in the order in which they first occurred, starting with the first to evolve: marine invertebrates, prokaryotes, land plants, eukaryotes, fishes.
What went wrong:
Confusing timescales of evolution. Many students reversed the order of early life forms or assumed land plants came before aquatic organisms. The marking key requires all five milestones in exact sequence to earn full marks.
What they needed to do: The correct order (1 mark each): First: prokaryotes, Second: eukaryotes, Third: marine invertebrates, Fourth: fishes, Fifth: land plants. This sequence reflects the actual fossil record and represents major transitions from single-celled to multicellular to colonization of terrestrial environments.
Q32 (7 marks): Design an epidemiological study to investigate the risk of developing the disease in response to age at exposure in areas A, B and C over the given time period. Justify your design.
What went wrong:
Students forgot to include a control group. The data showed areas A and B connected by a bridge (populations mix) while area C is isolated. Area C should serve as a control group. A, B, C as treatment groups without recognising C's role as control.
What they needed to do: Include: a clearly defined hypothesis about age at exposure and disease risk. A case-control or cohort design using all three areas. Area C (isolated, unexposed) as the control group. Participant matching or randomisation to reduce confounding. Justify how each feature improves validity and reliability.
Q5(a), Paper 1 (3 marks): Given y²(dy/dx) = x² + 128, use implicit differentiation on both sides with respect to x.
What went wrong:
Product rule mistakes on the left side. Students either differentiated y² and dy/dx separately without applying the product rule correctly, or forgot that d(dy/dx)/dx = d²y/dx².
What they needed to do: Differentiate d/dx[y²(dy/dx)] = d/dx[x² + 128]. Left side requires product rule: y²(d²y/dx²) + 2y(dy/dx)(dy/dx) = 2x. Simplify to y²(d²y/dx²) + 2y(dy/dx)² = 2x. This gives the complete differentiated equation needed for part (b).
Question 11b asked students to figure out how many heat lamps Janine had installed in her shed. A wind turbine supplied 230 V RMS to the shed via 900 m of transmission wire with a total resistance of 6.0 ohms, and the voltage across the shed was 210 V RMS. Three marks. Sounds manageable.
It was not. 85% of the state scored nothing.
The approach required a sequence of steps: use the voltage drop across the line (230 V minus 210 V = 20 V) to find the current in the transmission wire (20 V divided by 6.0 ohms = 3.33 A), which is also the current entering the shed. From there you can find the power consumed by the lamps (210 V times 3.33 A = 700 W), and dividing by 70 W per lamp gives 10 lamps.
The errors in the report tell the story. Many students attempted to use the 5.00 kW figure from the question stem with no justification, the turbine's maximum output, not what was actually being delivered. Others never found the current in the lines at all, so the rest of the working had nowhere to go.
Multi-step application questions like this one require you to understand the physics of the circuit, not just recognise a formula. Practice problems where you have to decide which quantity to find first.
Q27 (6 marks): Two groups of animals are shown. Group A contains: snake, lizard, frog, mammal. Group B contains: sea urchin, oyster, coral. Distinguish between the two groups based on their repro.
What went wrong:
Students identified fertilisation types but didn't compare them thoroughly. marking shows: (2-3 marks) Outlines features of fertilisation without comparison. (4 marks) Correctly identifies types with sound understanding but limited data reference.
What they needed to do: Use the stimulus data effectively to support the key features of internal and external fertilisation. Provide both similarities and differences between internal and external fertilisation.
Q34a (2 marks): Identify two errors in the statement 'Crown gall is a viral, waterborne disease of plants' and justify each one.
What went wrong:
Assuming it's about bacteria vs viruses. Students often identify one error correctly but pick the wrong second error. The marking key requires specific identification of BOTH viral and waterborne claims as incorrect.
Stating 'it's caused by bacteria' without being specific. The marking key requires you to name the BACTERIUM or at least distinguish it from viruses specifically.
Confusing transmission methods. The key error students miss: crown gall is not waterborne, it is SOILBORNE. This is crucial for disease control understanding.
What they needed to do: Error one: Crown gall is a BACTERIAL disease, not viral (caused by a bacterium, not a virus). Error two: Crown gall is a SOILBORNE disease, not waterborne (it is transmitted through contaminated soil or equipment, not water).
You have to nail two things here: what happens to the growth rate as a population approaches K (the carrying capacity), and which type of competition drives it, intraspecific (same species) or interspecific (different species).
The result was a near-even split. 44.1% chose (D). They correctly saw that growth slows near K, but pinned it on interspecific competition. The distinction that matters: as one species grows toward its own carrying capacity, the limiting factor is competition among members of that same species for shared resources like food, territory and mates. That’s intraspecific competition. Interspecific competition is between different species, which is a separate interaction and not what’s tied to a population reaching its own K.
As N approaches K:
• Individuals of the same species compete harder for the same limited resources.
• That’s intraspecific competition, happening within one species.
• Birth rates fall, death rates rise, and the growth rate slows toward zero.
• The logistic model shows it neatly: dN/dt = rN(1 − N/K). As N → K, the (1 − N/K) term goes to 0, so growth goes to 0.
The answer is (C): growth rate decreases due to intraspecific competition.
Q26 (b) (2 marks): From the graph, describe the trend between acid strength and the identity of the halogen X.
What went wrong:
Students read the graph but didn't link it to periodic properties. Many wrote 'the pKa values decrease from left to right' without connecting this to the actual question about acid STRENGTH. They described the data without interpreting it.
What they needed to do: From the graph, as the halogen changes going down the group, the strength of the haloethanoic acid decreases.
point (p, f(p)) has a unique x-intercept equal to the x-intercept of f itself. The setup had already been given in earlier parts of the question.
Two thirds of the cohort scored zero.
Including p = 4.71 when it shouldn't be there. The examiner report specifically flags p = 4.71... as a common incorrect inclusion. Students found it as a solution to an intermediate equation but failed to check whether it actually satisfied the original condition. It did not, and including it cost the mark.
Missing the correct answers entirely. The valid values were p = 2.38 and p = 7.04, correct to two decimal places. Students who set up the condition incorrectly arrived at wrong values like p = 4.49 and p = 7.73 instead.
The x-intercept of f occurs at x = 3pi/2. You needed to substitute that into the tangent line expression from part f(i) and solve t(3pi/2) = 0 for p, then verify each solution.
Only 10% of students scored full marks.
Q1(c) (6 marks): 'Contrast the duration, capacity and encoding between parts A and C of the multi-store model in Figure 1' (sensory register vs long-term store).
What went wrong:
Vague duration statements. Students wrote 'sensory register is very short' and 'long-term memory lasts a long time' without specifics. Marking key required: sensory register is several hundred milliseconds, long-term store is virtually unlimited. Saying 'seconds' for sensory register or 'years' for long-term were marked as describing the wrong construct.
Imprecise capacity descriptions. The sensory register has very large capacity capturing all sensory experiences, but students wrote 'small' or 'limited'. Long-term is not limited to just sensory experience. Exact phrasing mattered because it shows understanding of how each system works.
Wrong encoding terminology. Sensory register encodes 'in the form of sense' (visual, auditory, etc.) but many students wrote 'visual only'. Long-term encodes semantically, which means by meaning, not just sensory. Students who wrote 'words' or 'concepts' without saying 'semantic' lost marks for precision.
What they needed to do: Create a clear contrast table stating: Duration, Part A (sensory register): several hundred milliseconds. Part C (long-term): virtually unlimited. Capacity, Part A: very large, capturing all sensory experiences. Part C: not limited to just sensory experience. Encoding, Part A: in the form of sense (visual, auditory). Part C: semantic (by meaning). Each difference had to use the exact technical language the marking key expected.
This one’s two steps: find the gradient of the tangent at x = 0.8, then apply the perpendicular gradient rule to finish.
The common error: 37.0% chose (D) 1.25, which is the tangent gradient itself. They differentiated correctly to get dy/dx = 1/x, found 1/0.8 = 1.25 at x = 0.8, and stopped. That’s the tangent gradient, not the perpendicular one. A few others chose (A) −1.25, negating but forgetting to take the reciprocal. You need both moves: flip and negate.
The full method:
Step 1: Differentiate. d/dx[ln(4x)] = 1/x (chain rule: 1/(4x) × 4 = 1/x).
Step 2: Tangent gradient at x = 0.8 is m₁ = 1/0.8 = 1.25.
Step 3: Perpendicular gradient is m₂ = −1/m₁ = −1/1.25 = −0.80.
So the answer is (B) −0.80. The perpendicular gradient is always the negative reciprocal: flip the fraction, change the sign.
