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2026 VCE Specialist Mathematics exam: What to learn from last year’s mistakes
Hi all, in preparation for the upcoming exam, here's what to learn from last year's mistakes.
These were some of the lowest-scoring written questions in 2025:
Exam 2, Section B, 3g: what it tested Concentration after the outlet closes; average 0.18/1; scored zero 81.95%.
Exam 2, Section B, 3a: what it tested Explaining why the salt quantity increases; average 0.20/1; scored zero 79.57%.
Exam 1, 5c: what it tested Equal acceleration magnitudes; average 0.2/1; scored zero 75%.
Exam 1, 9c: what it tested Rational-function sketch; average 1.0/3; scored zero 50%.
Exam 1, 3c: what it tested Finding a parameter from displacement; average 0.8/2; scored zero 49%.
Exam 2, Section B, 3c: what it tested Euler’s method; average 0.92/2; scored zero 46.55%.
Exam 2 multiple-choice Question 12 was also the weakest multiple-choice question: only 33% chose the correct answer.
Figures and feedback come from VCAA’s Exam 1 report: source and Exam 2 report: source. Averages and percentages are reproduced as published, including rounding.
1. Exam 2, Section B, Question 3g: the tank volume changes
Average: 0.18/1. Full marks: 18.05%.
Where students went wrong
The report identifies 25 minutes as the most common incorrect answer. The correct answer was 50 minutes.
What they should have done
Once the outlet closes, water keeps entering. Both the salt quantity and the water volume increase.
Let t be the minutes since the outlet closed. The tank starts this stage with 100 kg of salt in 3000 litres. Water enters at 20 litres per minute with 0.1 kg of salt per litre.
Salt quantity = 100 + 2t.
Water volume = 3000 + 20t.
Set the concentration equal to the target:
(100 + 2t)/(3000 + 20t) = 1/20
100 + 2t = 150 + t
t = 50 minutes.
Keeping the denominator at 3000 produces the wrong answer of 25 minutes. That ignores the extra water entering the tank.
2. Exam 2, Section B, Question 3a: actually quote the concentrations
Average: 0.20/1. Full marks: 20.43%.
Where students went wrong
Many responses did not quote the concentrations. Saying the salt quantity increases was not enough for the requested explanation.
What they should have done
The initial concentration was 5/3000 = 1/600 kg per litre. The incoming concentration was 0.1 kg per litre.
The incoming solution was more concentrated. With equal inflow and outflow rates, salt initially entered faster than it left, so the salt quantity increased.
Include the comparison, not just the conclusion.
3. Exam 1, Question 5c: check the earlier conditions
Average: 0.2/1. Full marks: 25%.
Where students went wrong
Some students gave a = −5 and b = −6 as well as the correct pair. The report says those responses did not earn the mark.
What they should have done
Use the conditions already established in the question. The particles still had to collide at t = 1 with perpendicular velocities.
Part b gave a = −1 or a = −3/2 from the perpendicular velocities.
Equal acceleration magnitudes also require |6 + 2a| = 4, giving a = −1 or a = −5.
Only a = −1 satisfies both conditions. The collision condition gives b = a − 1, so b = −2.
The valid answer is a = −1 and b = −2. Reject a = −5 because it fails the perpendicular-velocity condition.
4. Exam 1, Question 9c: a hole is not a vertical asymptote
Average: 1.0/3. Full marks: 16%.
Where students went wrong
Students added an incorrect vertical asymptote at x = 1, missed the hole or placed it incorrectly. Some sketches also missed the intercepts or curved away from the asymptotes.
What they should have done
Use the simplified expression from part a:
f(x) = −x − 1 + 1/(x + 1), with x ≠ −1 and x ≠ 1.
The vertical asymptote is x = −1. The oblique asymptote is y = −x − 1.
At x = 1, the simplified expression gives −3/2, but the original function excludes x = 1. Draw an open circle at (1, −3/2).
The graph passes through (−2, 0) and (0, 0). Draw the asymptotes with a ruler and label their equations. The branches must approach them, rather than bend away.
That same missing point was tested in Question 9b, where 54% scored zero: the value needed to make g continuous was k = −3/2.
5. Exam 1, Question 3c: rearrange before squaring
Average: 0.8/2. Full marks: 28%.
Where students went wrong
The report flags unsuccessful rearranging and squaring when students tried to find k.
What they should have done
At t = 3, the second particle is at 3 metres and is 1 metre ahead. The first particle must therefore be at 2 metres.
√(9 + k) − √k = 2
Move one square root before squaring:
√(9 + k) = 2 + √k
9 + k = 4 + 4√k + k
5 = 4√k
k = 25/16.
Keep the 4√k term when expanding (2 + √k)². Substituting back gives a first-particle displacement of 2 metres, as required.
6. Exam 2, Section A, Question 12: halfway along the distance
Correct answer: D. Only 33% selected it; 33% chose A.
Where students went wrong
Option A was (u + v)/2. That averages the endpoint velocities, but the question asked for velocity halfway along the distance under constant acceleration.
What they should have done
Let L be the displacement from A to B and w the velocity at the midpoint.
v² = u² + 2aL
w² = u² + 2a(L/2) = u² + aL
Substitute aL = (v² − u²)/2:
w² = (u² + v²)/2
The required answer was w = √((u² + v²)/2), option D.
Half the distance does not generally mean half the travel time.
7. Exam 2, Section B, Question 3c: show the Euler steps
Average: 0.92/2. Full marks: 38.93%.
Where students went wrong
Several students wrote the final answer without showing Euler’s method. The report says the method needed to be visible to earn both marks.
What they should have done
Start with Q(0) = 5 and use the 15-minute step size:
Q(next) = Q(current) + 15 × (300 − Q(current))/150
Q(15) = 5 + 15 × 295/150 = 34.5
Q(30) = 34.5 + 15 × 265.5/150 = 61.05 kg.
A table showing Q(15) and Q(30) was also accepted. Writing only 61.05 did not show how the approximation was obtained.
Use the VCE Specialist Maths hub and past papers and examiner reports alongside this.
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