VCE General Mathematics 2025 Exam 1

All 40 multiple-choice questions from the 2025 VCE General Mathematics Examination 1, with the correct option and a short worked reason. For the extended-response paper, see the 2025 Examination 2 walkthrough.

How to use this page

• Questions are from the 2025 VCE General Mathematics Examination 1, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised in a line; open the official examination PDF for the graphs, tables and all four options.
• Answers match the key in the 2025 General Mathematics Examination 1 external assessment report (Word document), and every calculation was redone independently. Both files are listed on the VCAA General Mathematics examinations page.
• Percentages in brackets are the share of students who chose the correct option, from the report.

Structure and timing

Examination 1 is 40 multiple-choice questions (40 marks) in 90 minutes (plus 15 minutes reading time), with a CAS calculator, a scientific calculator and one bound reference. That is just over 2 minutes per question.

• Questions 1 to 16: Data analysis
• Questions 17 to 24: Recursion and financial modelling
• Questions 25 to 32: Matrices
• Questions 33 to 40: Networks and decision mathematics

Data analysis (Questions 1 to 16)

Q1

Median bag size from a histogram of 39 customers. Answer: D - the frequencies are 4, 11, 3, 9, 5, 7; the median is the 20th value, and the cumulative counts 4, 15, 18, 27 put it in bag size 4. (75%)

Q2

Total avocados sold. Answer: D - $1(4) + 2(11) + 3(3) + 4(9) + 5(5) + 6(7) = 138$. (76%; 20% gave 39, the number of customers.)

Q3

Which statement about two life-expectancy boxplots is correct? Answer: D - the minimum for Sample T (about 68) is above the median for Sample H (about 67), so every Sample T country exceeds Sample H's median. The other statements fail on the IQRs, medians and quartiles read from the plots. (82%)

Q4

Which column of a $\log_{10}$ histogram contains Singapore (6 028 460 people, 720 km²)? Answer: D - density $\approx 8373$ per km², and $\log_{10}8373 \approx 3.92$, in the 3.5 to 4.0 column. (52%)

Q5

Normal heights: 2.5% above 178.9 cm and 16% below 157.6 cm. Mean and standard deviation? Answer: C - by the 68-95-99.7% rule, $\mu + 2\sigma = 178.9$ and $\mu - \sigma = 157.6$, so $\sigma = 7.1$ and $\mu = 164.7$. (81%)

Q6

$Q_1 = 74.9$, $Q_3 = 78.5$, lowest values 68.5, 68.6, 69.0, 70.1, 74.8. How many low outliers? Answer: C - lower fence $= 74.9 - 1.5(3.6) = 69.5$, and three values are below it. (79%)

Q7

Percentage of female buyers preferring silver (28, 42, 35). Answer: B - $\tfrac{42}{105} = 40\%$. (86%)

Q8

Best display for the car colour by gender table. Answer: D - a segmented bar chart, one bar per gender, suits two categorical variables. (79%)

Q9

Least squares line on the goals-against scatterplot. Answer: A - reading two points on the line, such as $(27, 12)$ and $(44, 9)$, gives a slope of about $-0.18$ and $\textit{games won} = 16.8 - 0.178 \times \textit{goals against}$. (66%)

Q10

Interpreting $r = -0.466$. Answer: D - a negative association: more goals against is associated with fewer wins. Correlation does not show causation, which rules out B and C. (63%)

Q11

Least squares line with $\log_{10}(\textit{life})$ as the response. Answer: C - regressing $\log_{10}(\textit{life})$ on doctors gives $\log_{10}(\textit{life}) = 1.79 + 0.0383 \times \textit{doctors}$. (81%)

Q12

With $\textit{life} = a + b \times \textit{doctors}^2$, predict life for 2 doctors per 1000. Answer: C - the fitted line is about $63.12 + 2.842 \times \textit{doctors}^2$, giving $63.12 + 2.842 \times 4 \approx 74.5$. (61%)

Q13

Five-median smoothed margin in week 8. Answer: C - weeks 6 to 10 are 50, 38, 32, 35, 41; ordered 32, 35, 38, 41, 50, the median is 38. (76%)

Q14

Features of the time series. Answer: D - a decreasing trend with irregular fluctuations. (85%)

Q15

Four-mean smoothed value with centring for day 8. Answer: B - the means of days 6 to 9 and days 7 to 10 are 134.25 and 145.75; their average is 140. (81%)

Q16

A winter seasonal index of 1.75. By what percentage should actual sales be reduced to deseasonalise? Answer: B - deseasonalised $= \tfrac{\text{actual}}{1.75} \approx 0.57 \times \text{actual}$, a reduction of 43%. (50%; 23% chose 75%.)

Recursion and financial modelling (Questions 17 to 24)

Q17

$4000 at 4% p.a. simple interest for three years. Answer: D - $4000 + 3(0.04 \times 4000)$. (53%; 32% chose compound growth, $4000 \times 1.04^3$.)

Q18

$u_{n+1} = Ru_n + d$ with $a > 0$, $R = 0.5$, $d = 0$. Answer: D - with $d = 0$ it is geometric, and $0 < R < 1$ makes it decrease. (52%)

Q19

When is the flat rate value first below the reducing balance value? Answer: B - flat rate is $60\,000 - 4000n$ and reducing balance is $60\,000 \times 0.92^n$. After 5 years they are $40 000 and about $39 545; after 6 years, $36 000 and about $36 382. So 6 years. (49%)

Q20

Equipment worth $12 000 falls to $7680 after two years at 960 hours per year. Depreciation per hour? Answer: A - $\tfrac{12\,000 - 7680}{1920} = 2.25$, that is $2.25 per hour. (55%)

Q21

$250 000 perpetuity at 5% p.a. When do payments first exceed $250 000 in total? Answer: C - $12 500 a year reaches exactly $250 000 after 20 years, so it first exceeds it at the end of year 21. (69%)

Q22

$4000 plus $50 a week for 3 years grows to $14 000. Annual rate (compounding weekly)? Answer: A - Finance Solver with $N = 156$, $PV = -4000$, $PMT = -50$, $FV = 14\,000$, 52 periods per year gives $I \approx 8.4\%$. (68%)

Q23

$5000 at an effective annual rate of 4.51% for five years. Interest earned? Answer: B - $5000 \times 1.0451^5 \approx 6233.89$, so about $1234 interest. (Equivalently, convert to the nominal fortnightly rate and use Finance Solver.) (41%)

Q24

An $800 000 annuity at 4.8% p.a. pays $6000 a month until the balance is $521 118.96, then $4767.66 a month. Total years of payments? Answer: A - Finance Solver gives 84 payments (7 years) to reach $521 118.96, then 144 payments (12 years) to reach zero: 19 years. (52%)

Matrices (Questions 25 to 32)

Q25

Describe $G$, a $3 \times 3$ matrix of 0s and 1s with a zero row. Answer: A - it is a binary matrix. It is not a permutation matrix (row 3 has no 1), identity or diagonal. (83%)

Q26

The calculation for $c_{21}$ in $C = AB$. Answer: C - row 2 of $A$ times column 1 of $B$: $1 \times 3 + 6 \times 5$. (84%)

Q27

For which $m$ and $n$ does $E = \begin{bmatrix} m & -9 \\ 4 & n\end{bmatrix}$ have no inverse? Answer: C - $\det E = mn + 36$, which is 0 when $m = 3$, $n = -12$. (76%)

Q28

Which sequence lets headset $A$ message headset $E$? Answer: C - $A \to B$, $B \to D$ and $D \to E$ are all 1s in the communication matrix; each other option contains a missing link. (93%)

Q29

Leslie matrix for the life-cycle diagram. Answer: A - birth rates $0, 2.1, 4.6, 1.8$ in row 1 (no loop on group 1) and survival rates 0.9, 0.7, 0.2 on the subdiagonal. (84%)

Q30

How many elements of the $4 \times 4$ matrix with $f_{ij} = i^2 - j$ are negative? Answer: B - only row 1 has $j > i^2$, for $j = 2, 3, 4$: three elements. (73%)

Q31

Which computation is defined? Answer: B - $C^T$ is $2 \times 3$, $B$ is $3 \times 1$ and $D$ is $1 \times 3$, so $C^TBD$ is $2 \times 3$. The other sums and products have mismatched orders. (70%)

Q32

Complete the dominance matrix and find $T = D + D^2$. Answer: B - using the clues, Kyle beat Maggie and Neil and lost to Lian and Ophelia; Lian's only win is over Kyle and Neil's only win is over Lian. Completing $D$ and computing $D + D^2$ gives option B. (63%; 21% chose D, which is just $D$.)

Networks and decision mathematics (Questions 33 to 40)

Q33

Number of Hamiltonian cycles starting from $E$. Answer: B - only $EDCBAFE$ and its reverse $EFABCDE$ visit every vertex once. (76%)

Q34

Number of bridges. Answer: C - the three edges on the path joining the rest of the graph to the triangle each disconnect the graph when removed. (85%)

Q35

The minimum spanning tree contains the edge $w$. Its length? Answer: B - Kruskal's algorithm without $w$ gives 50 with three edges of weight 6; including $w$ replaces one of the weight-6 edges on the cycle it creates, giving $44 + w$. (65%)

Q36

Capacity of Cut 1. Answer: A - count only edges flowing from the source side to the sink side: $7 + 3 + 5 + 12 = 27$. (The edges of 4 and 6 flow back across the cut.) (77%)

Q37

Maximum flow. Answer: B - the cut through the three edges into the sink has capacity $7 + 4 + 6 = 17$, the minimum cut. (73%)

Q38

Shortest distance from home to school. Answer: A - $900 + 500 + 400 + 800 + 1100 = 3700$ m. The upper route ($1500 + 1600 + 700 = 3800$) is the common trap. (47%; another 47% chose B.)

Q39

Result of the first two steps of the Hungarian algorithm. Answer: B - subtracting row minima gives table A; then subtracting column minima (4500 for site 1, 13 150 for site 3 and 8700 for site 4) gives table B. (71%)

Q40

Float time of activity $B$. Answer: C - the forward and backward scans give $B$ an earliest start of 4 and a latest start of 12 (it must finish by day 18, when $G$ can start at the latest). Float $= 12 - 4 = 8$ days. The project takes 44 days. (39%)

Exam tips from this paper

• Read the scale on graphs carefully (Q3, Q4) and check the transformation asked for (Q11, Q12).
• In Q16, deseasonalising divides by the seasonal index; the percentage change is $1 - \tfrac{1}{1.75}$.
• For financial questions, set up Finance Solver carefully: signs of $PV$, $PMT$ and $FV$, and the number of compounding periods per year.
• In networks, a cut only counts edges flowing from the source side to the sink side, and in critical path analysis a dummy activity may be needed (Q40).