VCE General Mathematics 2023 Exam 1

All 40 multiple-choice questions from the 2023 VCE General Mathematics Examination 1, with the correct option and a short worked reason. For the extended-response paper, see the 2023 Examination 2 walkthrough.

How to use this page

• Questions are from the 2023 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 five options.
• Answers match the key in the 2023 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 one approved CAS technology, an optional scientific calculator and one bound reference. That is just over 2 minutes per question. 2023 was the first year of the current study design, and the paper had five options (A to E) 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 of 40 runners' 800 m times from a dot plot. Answer: B - the counts from 134 s are 3, 8, 11, ...; the 20th and 21st times both fall at 136 s (cumulative 3, 11, 22). (90%)

Q2

Shape of that distribution. Answer: A - it is positively skewed. $Q_1 = 135$ and $Q_3 = 138$, so the upper fence is $138 + 1.5 \times 3 = 142.5$, and the times of 143 s and 146 s are possible outliers. (88%)

Q3

Attempts of 8, 11, 5, 6 and 9 over five days. How many on day six for a mean of exactly 8? Answer: E - $6 \times 8 = 48$ and the first five total 39, so $48 - 39 = 9$. (89%)

Q4

Mean 82 minutes, standard deviation 11 minutes, 2380 visitors. How many spend between 60 and 104 minutes? Answer: D - 60 and 104 are two standard deviations either side of the mean, so 95% of 2380 is 2261. (82%)

Q5

Mean 163.56 cm, standard deviation 8.14 cm, $z = -0.85$. Height? Answer: B - $163.56 - 0.85 \times 8.14 \approx 156.6$ cm. (84%)

Q6

How many of $2450, $3175, $4999, $8925, $10 250 and $105 600 lie in the modal class (3.5 to 4.0) of a $\log_{10}$(price) histogram? Answer: C - their logs are 3.39, 3.50, 3.70, 3.95, 4.01 and 5.02, so $3175, $4999 and $8925: three prices. (54%)

Q7

Equation of the least squares line drawn on a scatterplot of test 2 mark against test 1 mark. Answer: A - the line passes through about $(16, 18)$ and $(36, 49)$: slope $\tfrac{31}{20} = 1.55$ and intercept $18 - 1.55 \times 16 \approx -6.8$. (60%)

Q8

How many students scored within two marks of their predicted test 2 mark? Answer: C - checking each point against the line, only $(16, 18)$, $(23, 28)$, $(29, 39)$, $(31, 41)$ and $(36, 49)$ are within two marks: five students. (43%)

Q9

A least squares line gives a birth rate of 32.2 at 8.53 MJ and 9.9 at 14.9 MJ. Slope? Answer: B - $\tfrac{9.9 - 32.2}{14.9 - 8.53} \approx -3.5$. (59%)

Q10

A negative association between test scores and social media time, with $r^2 = 0.72$. What can be concluded? Answer: A - less time on social media is associated with higher scores. Association does not show causation, which rules out B and D. (59%)

Q11

Least squares line for height against $\tfrac{1}{\textit{age}}$ for 11 trees. Answer: D - regression on the transformed data gives height $= 13.04 - 40.22 \times \tfrac{1}{\textit{age}}$. (60%)

Q12

From height $= -3.8 + 12.6 \times \log_{10}$(age), the age of an 8.52 m tree. Answer: D - $\log_{10}(\textit{age}) = \tfrac{8.52 + 3.8}{12.6} = 0.9778$, so age $= 10^{0.9778} \approx 9.5$ years. (76%)

Q13

Seven-median smoothed winning time for 2006. Answer: C - the times for 2003 to 2009 in order are 116.0, 116.0, 116.4, 116.8, 117.2, 117.6, 117.9, with median 116.8 s. (68%)

Q14

Median winning time for all 23 years from 2000 to 2022. Answer: B - the median is the 12th value in order, which is 117.2 s (2006). The common wrong answer, 118.3 s (2011), is the middle of the unordered list. (37%)

Q15

Eight-mean smoothed visitors, with centring, for day 6. Answer: C - the mean of days 2 to 9 is 323 and of days 3 to 10 is 327; their average is 325. (66%)

Q16

Deseasonalising January's visitors increases them by 35%. Seasonal index? Answer: D - deseasonalised $= \tfrac{\text{actual}}{\text{SI}} = 1.35 \times \text{actual}$, so $\text{SI} = \tfrac{1}{1.35} \approx 0.74$. (43%; 39% chose 0.65, from $1 - 0.35$.)

Recursion and financial modelling (Questions 17 to 24)

Q17

$T_0 = 5$, $T_{n+1} = -T_n$. Find $T_2$. Answer: D - $T_1 = -5$ and $T_2 = 5$. (43%)

Q18

$G_0 = 15\,000$, $G_{n+1} = G_n - 1314$. Rule for $G_n$? Answer: C - $G_n = 15\,000 - 1314n$. The 0.04 is per cup, not per year. (60%; 33% chose A.)

Q19

Depreciation is $0.04 per cup. Cups made per year? Answer: E - $\tfrac{1314}{0.04} = 32\,850$. (62%)

Q20

A $3000 computer is worth $600 after four years. Flat rate depreciation rate? Answer: C - it loses $2400 over four years, $600 a year, which is 20% of $3000. (82%)

Q21

Reducing balance rate for the same values? Answer: E - $3000(1 - r)^4 = 600$ gives $1 - r = 0.2^{0.25} \approx 0.669$, so $r \approx 33\%$. (48%)

Q22

$T_0 = 500\,000$, $T_{n+1} = 1.00325T_n - 2611.65$. The final repayment that clears the loan? Answer: D - Finance Solver at 3.9% p.a. gives $N = 300$ payments. After 299 payments $2605.65 is owing; the 300th payment must cover it plus a month's interest: $2605.65 \times 1.00325 \approx 2614.12$. (39%)

Q23

A $20 000 loan with quarterly repayments of $653.65 owes $19 527.56 after one quarter. Effective annual rate? Answer: D - interest in the first quarter is $19\,527.56 - 20\,000 + 653.65 = 181.21$, a quarterly rate of 0.906%. So the nominal rate is 3.6242% and the effective rate is $1.0090605^4 - 1 \approx 3.67\%$. (37%)

Q24

A perpetuity $P_{n+1} = RP_n - d$ with $P_0 = a$. $R$ equals? Answer: B - a perpetuity's balance never changes, so $Ra - d = a$ and $R = \tfrac{a + d}{a}$. (27%; 37% chose D.)

Matrices (Questions 25 to 32)

Q25

What does $m_{21}$ in the temperature matrix show? Answer: A - row 2 (week 2), column 1 (Monday): 29 °C. (89%)

Q26

Which three letters in $Q = \begin{bmatrix} t & e & a & m & s \end{bmatrix}^T$ move when multiplied by the permutation matrix $P$? Answer: B - $PQ = \begin{bmatrix} t & a & m & e & s \end{bmatrix}^T$, so $e$, $a$ and $m$ move. (90%)

Q27

A bird transition matrix whose column $N$ is $\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$. Long term? Answer: A - $N$ is an absorbing state: birds reaching $N$ stay, so in the long term every bird is at $N$ and none remain at $M$ (or $O$). (39%)

Q28

Which dominance matrix matches the ranking U, V, S, T? Answer: D - its row sums are S 1, T 0, U 3, V 2, and every pair has exactly one winner. Option B has U and S each beating the other, and option E has U beating itself. (69%)

Q29

The $3 \times 2$ matrix with $k_{ij} = (i - j)^2$. Answer: E - rows $(0, 1)$, $(1, 0)$ and $(4, 1)$. (62%)

Q30

How many of four statements about inverses are true? Answer: D - three. Not every square matrix has an inverse (determinant 0), but an inverse can equal the transpose (for example a permutation matrix), a zero determinant means no inverse, and the identity matrix has an inverse. (39%; 42% chose C.)

Q31

Meaning of 0.2 in row 2, column 1 of a Leslie matrix. Answer: C - it is the survival rate from the first age group to the second: 20% survive into their second year. (74%)

Q32

From an incomplete lunch-break transition diagram and Monday's numbers (P 150, B 50, O 220, L 40), what percentage of students expected at the oval on Wednesday were also there on Tuesday? Answer: C - complete the loops so each location's proportions sum to 1 (P 0.2, B 0.3, O 0.3, L 0.3). Tuesday's oval total is 140 and Wednesday's is 127. Of those 127, $0.3 \times 140 = 42$ stayed at the oval, and $\tfrac{42}{127} \approx 33\%$. (18%; 54% chose 30%, the proportion who stay at the oval, rather than a share of Wednesday's total.)

Networks and decision mathematics (Questions 33 to 40)

Q33

How many of five statements about a graph with 5 vertices and 4 edges (no cycle) are true? Answer: D - it is a tree, it is connected, it contains a path, and the degree sum is $2 \times 4 = 8$. It has no cycle. Four are true. (56%)

Q34

A bipartite graph is typically used for? Answer: A - allocating tasks (workers on one side, tasks on the other). (77%)

Q35

Weight of the minimum spanning tree. Answer: C - Kruskal's algorithm takes 6, 7, 8, 9 and 10 without forming a cycle, which connects all six vertices: $6 + 7 + 8 + 9 + 10 = 40$. (74%)

Q36

Minimum total time for four employees and four duties. Answer: B - the Hungarian algorithm (or checking allocations) gives Dario duty 1 (7), Anthea duty 2 (7), Cho duty 3 (7) and Bob duty 4 (9): 30 minutes. (75%)

Q37

Faces of the planar graph with the given adjacency matrix. Answer: B - the upper triangle of the matrix sums to 10 edges (including a double edge between $K$ and $L$). With 5 vertices, $5 + f = 10 + 2$, so $f = 7$. (66%)

Q38

Which network matches the precedence table? Answer: D - $G$ needs both $C$ and $F$, but $H$ needs only $F$. So $F$ must end at the node where $H$ starts, with a dummy from there to the node where $C$ ends and $G$ starts. In every other option, $H$ would also wait for $C$. (60%)

Q39

Capacity of the minimum cut in the maze network. Answer: B - cut the edges $G$-$H$ (12), $C$-$E$ (4) and $D$-$E$ (7), which separates $\{A, B, C, D, G\}$ from $\{E, F, H\}$. The edges $F$-$G$ and $F$-$C$ flow back into the entrance side, so they are not counted: $12 + 4 + 7 = 23$. No cut is smaller, and a flow of 23 is achievable. (62%)

Q40

Which single change increases the flow the most? Answer: E - reversing $G$-$F$ (so flow can go from $G$ to $F$) lets the spare capacity into $G$ reach the exit through $F$-$H$: the maximum flow rises from 23 to 30. The other changes give 24, 23, 27 and 29. (40%)

Exam tips from this paper

• The report flagged Questions 8, 14 and 16 (reading the graph, and seasonal indices), 17 and 21 to 24 (recurrence relations and multi-step finance), 27, 30 and 32 (transition matrices) and 40.
• To find a median from a time series, order the values first (Q14).
• Deseasonalising divides by the seasonal index; a 35% increase means $\text{SI} = \tfrac{1}{1.35}$, not 0.65 (Q16).
• For "how many are true" questions, test every statement separately (Q30, Q33).