VCE General Mathematics 2024 Exam 1

All 40 multiple-choice questions from the 2024 VCE General Mathematics Examination 1, with the correct option and a short worked reason. For the extended-response paper, see the 2024 Examination 2 walkthrough.

How to use this page

• Questions are from the 2024 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 2024 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

Percentage who chose blue in a percentage segmented bar chart. Answer: B - the blue segment runs from 56% to 74%, so $74 - 56 = 18\%$. (78%)

Q2

Payment preference (cash or electronic) and budget preference (less than $50 or more than $50) are what types of variable? Answer: A - both are categorical; the budget groups are categories, not measured amounts. (79%)

Q3

Frequency of the column containing the median in a histogram of $\log_{10}$(population density) for 27 countries. Answer: D - the median is the 14th value; the cumulative frequencies 2, 4, 6, 12, 21 put it in the 2.0 to 2.2 column, which has frequency 9. (71%)

Q4

The upper outlier lies in the 3.2 to 3.4 column. Which population density could it be? Answer: C - $\log_{10}2030 \approx 3.31$. The others give 2.52, 3.12 and 3.44. (68%)

Q5

Which statement about a boxplot of 24 sibling counts is correct? Answer: C - the data have twelve 1s, seven 2s, three 3s and two 4s. The median is the mean of the 12th and 13th values, $\tfrac{1 + 2}{2} = 1.5$. ($Q_1 = 1$ and $Q_3 = 2$, so the IQR is 1 and the 4s are outliers.) (80%)

Q6

Five-number summary 2, 5, 11, 48, 613 (athletes per country). Smallest number that would be an outlier? Answer: C - upper fence $= 48 + 1.5 \times 43 = 112.5$, so the smallest whole-number outlier is 113. (53%)

Q7

Mean score 55.7; a score of 48 has $z = -1.75$. Standard deviation? Answer: C - $s = \tfrac{48 - 55.7}{-1.75} = 4.4$. (82%)

Q8

Equation of the least squares line drawn on a scatterplot of females against number. Answer: B - females is the response variable, and the line passes close to $(30, 24)$: $-3.39 + 0.91 \times 30 = 23.91$. Option A gives 25.7 at 30. (73%)

Q9

For males $= 67.5 - 1.27 \times$ number, with $s_{\text{number}} = 8.51$ and $s_{\text{males}} = 19.0$, find $r$. Answer: A - $r = b \times \tfrac{s_x}{s_y} = -1.27 \times \tfrac{8.51}{19.0} \approx -0.569$. (52%)

Q10

At which Olympic Games is the predicted males closest to 25.6? Answer: C - $25.6 = 67.5 - 1.27n$ gives $n \approx 32.99$, the 33rd Games. (73%)

Q11

Least squares line after a $\log_{10}$ transformation of the explanatory variable (year) for the parrot-pairs data. Answer: D - regressing pairs on $\log_{10}$(year) gives pairs $= 303 - 151 \times \log_{10}$(year). Option A transforms the response instead. (59%)

Q12

After fitting a line to $\tfrac{1}{\textit{pairs}}$ against year, where is the largest difference between actual and predicted values? Answer: A - year 1 is furthest from the line: the predicted value there is about 264 pairs against an actual 320. (48%)

Q13

Seven-median smoothed value of new staff for 2016. Answer: A - the values for 2013 to 2019 are 7, 6, 11, 13, 12, 6, 10; in order 6, 6, 7, 10, 11, 12, 13, so the median is 10. (72%)

Q14

Adding 2023 makes the 13-year mean 11. New staff in 2023? Answer: B - $13 \times 11 = 143$ and the 12 plotted values sum to 127, so $143 - 127 = 16$. (68%)

Q15

Six-mean smoothed value, with centring, for month 5 of the cans data. Answer: C - the mean of months 2 to 7 is 318 and of months 3 to 8 is 324; their average is 321. (70%)

Q16

Month 3's seasonal index is twice month 6's. Find month 3's. Answer: C - the indices sum to 12 and the other ten sum to 9.93, so months 3 and 6 total 2.07. Then month 6 is 0.69 and month 3 is 1.38. (58%)

Recursion and financial modelling (Questions 17 to 24)

Q17

When does $u_{n+1} = Ru_n + d$ generate a geometric sequence? Answer: D - a geometric sequence multiplies by the same ratio each time, so it needs $d = 0$; only option D has that. (36%)

Q18

$T_{n+1} = 1.00075T_n - 677.55$, weekly, 52 weeks a year. Annual interest rate? Answer: B - $0.00075 \times 52 \times 100\% = 3.9\%$. (74%)

Q19

Reducing balance depreciation of 18% a year, $L_{n+1} = kL_n$. Value of $k$? Answer: C - $1 - 0.18 = 0.82$. (72%)

Q20

$2000 at 4.4% p.a. compounding quarterly for three years. Simple interest rate giving the same interest? Answer: B - $2000 \times 1.011^{12} \approx 2280.57$, so interest is $280.57; then $\tfrac{280.57}{2000 \times 3} \approx 4.68\%$. (48%)

Q21

$121 000 loan repaid with monthly payments of $2228.40 over five years. Total interest? Answer: D - $2228.40 \times 60 = 133\,704$, minus the $121 000 borrowed, is $12 704. (56%)

Q22

Principal reduction for payment 2 of a $240 000 loan (payment $2741.05, first interest $960). Answer: C - the monthly rate is $\tfrac{960}{240\,000} = 0.4\%$; interest on $238 218.95 is $952.88, so principal reduction $= 2741.05 - 952.88 = 1788.17$. (70%)

Q23

Years to repay that loan. Answer: A - the annual rate is $0.4\% \times 12 = 4.8\%$. Finance Solver with $PV = 240\,000$, $PMT = -2741.05$, $FV = 0$, 12 periods per year gives $N \approx 108$ months, which is 9 years. (54%)

Q24

$18 000 with $A_{n+1} = 1.002A_n + 100$ for two years; the new monthly payment for the last three years so the balance reaches $30 000. Answer: A - the rate is $0.2\% \times 12 = 2.4\%$. After 24 months the balance is $21 340.18; then Finance Solver with $N = 36$, $PV = -21\,340.18$ and $FV = 30\,000$ gives $PMT \approx -189.55$, a deposit of $189.55 a month. (46%)

Matrices (Questions 25 to 32)

Q25

$J$ is $2 \times 3$, $K$ is $3 \times 1$ and $L$ is added to $JK$. Order of $L$? Answer: B - $JK$ is $2 \times 1$, and only matrices of the same order can be added. (79%)

Q26

Discount every price in $C$ by 15%. Answer: B - multiply by $0.85$: $0.85C$. (76%)

Q27

When does $\begin{bmatrix} 4 & g \\ 8 & h \end{bmatrix}$ have no inverse? Answer: B - determinant $4h - 8g = 0$ when $g = \tfrac{h}{2}$. (50%)

Q28

Multiplying the sport-by-team matrix $W$ by a column of 1s. What does $x_{31}$ show? Answer: C - it is the row 3 (tennis) total, $63 + 76 + 66 + 75 = 280$. The football row also totals 280, which makes option D a trap, but $x_{31}$ is row 3. (84%)

Q29

Playing order changes by $G_{n+1} = G_nP$. Who has the bye in week 4? Answer: C - the orders are $QRSTU$, then $SQTUR$, then $TSURQ$, then $UTRQS$, so Siobhan has the week-4 bye. (Or compute $G_1P^3$ directly.) (49%)

Q30

Leslie matrix from birth rates 0, 1.8, 1.2 and survival rates 0.7, 0.6, 0. Answer: B - birth rates across row 1, survival rates on the subdiagonal. (78%)

Q31

Two chess games remain (I against J, I against L); ranking uses one-step plus two-step dominances. Which outcome is not possible? Answer: D - checking all four results of the two games: $I$ is first if it wins both and fifth if it loses both, and $J$ is first if it beats $I$ while $I$ beats $L$. $J$ is never fifth. (53%)

Q32

$S_{n+1} = TS_n + A$ with $S_3$ (Sunday) given. Change at the Botanical Gardens from Thursday to Sunday? Answer: D - work backwards with $S_n = T^{-1}(S_{n+1} - A)$ three times to get Thursday's $G = 4924$. Sunday is 5620, an increase of 696. (33%)

Networks and decision mathematics (Questions 33 to 40)

Q33

Sum of degrees of a graph with an isolated vertex and a loop. Answer: C - the degrees are 0, 1, 3, 4 (the loop adds 2), 2 and 2, summing to 12. (75%)

Q34

Which is a Eulerian trail? Answer: C - the odd vertices are $B$ and $F$, so the trail runs from $B$ to $F$. $BACBDCFDEF$ uses all 9 edges exactly once. (85%)

Q35

Number of faces of a graph drawn with crossing edges. Answer: B - the graph can be redrawn as planar, with 7 vertices and 11 edges: $7 + f = 11 + 2$, so $f = 6$. (52%)

Q36

Minimum spanning tree connecting eight houses. Answer: D - Prim's algorithm from $A$ takes $AB$ (15), $AC$ (16), $BF$ (16), $FG$ (14), $FH$ (17), $CD$ (18) and $DE$ (19), total 115 m, which is the tree in option D. (83%)

Q37

The shortest path from carpark to lookout is 34 m. Which values of $x$ and $y$ achieve this? Answer: D - without $x$ or $y$ the best route is $10 + 11 + 8 + 7 = 36$ m. Using $y$ via the top path costs $10 + 11 + 8 + y = 29 + y$, which is 34 when $y = 5$; with $x = 11$ no route through $x$ is shorter. Options A to C give 35, 36 and 35. (40%)

Q38

A connected graph has six vertices and six edges. How many of the four statements must always be true? Answer: A - only "the sum of the degrees is 12" (twice the edges) must hold. One counter-example rules out the other three: a triangle with an extra edge hanging off each corner has six odd vertices (so no Eulerian trail) and three vertices of degree 1 (so no Hamiltonian path). (18%, the hardest question on the paper.)

Q39

Four workers, four sequential tasks; then a fifth worker, Edgar, may take over one task. Reduction in minimum total time? Answer: D - the best original allocation (Carly task 1, Blake task 2, Dexter task 3, Anush task 4) takes $11 + 7 + 16 + 9 = 43$ hours. With Edgar, Dexter task 1, Edgar task 2, Blake task 3 and Anush task 4 take $10 + 5 + 15 + 9 = 39$ hours: 4 hours less. (19%; 45% chose B.)

Q40

Which activity table fits an unlabelled network of 15 activities? Answer: B - the network has four activities leaving the start, two activities that each have four immediate predecessors, one three-step chain into a node, and exactly two activities ending at the finish. Table B matches all of these ($K$ follows $E, G, H, J$; $F$ follows $B, K, M, L$; $C \to A \to E \to K$ is the three-step chain; $F$ and $O$ end the project). Table A has no three-step chain, and tables C and D each leave three activities with no successor. (32%)

Exam tips from this paper

• The report singled out multi-step questions and ones that need several options checked (Q9, Q12, Q20, Q23, Q24, Q29, Q31, Q32 and Q35 to Q40).
• A $\log$ or reciprocal transformation "applied to the explanatory variable" changes the $x$-values only (Q11).
• For a geometric sequence the recurrence must have $d = 0$ (Q17).
• In backward-working matrix recurrences, undo the addition before multiplying by $T^{-1}$ (Q32).
• For "must always be true" graph questions, test each statement against a quick counter-example (Q38).