VCE General Mathematics 2021 Exam 1

Every multiple-choice question from the 2021 VCE Further Mathematics Examination 1, with the correct option and a worked reason. Further Mathematics was the predecessor of General Mathematics: General Mathematics replaced it in 2023, and VCAA lists this paper with the General Mathematics examinations. All 24 core questions and all 32 module questions (four modules of 8) are answered below. For the extended-response paper, see the 2021 Examination 2 walkthrough.

How to use this page

• Questions are from the 2021 VCE Further Mathematics Examination 1, copyright Victorian Curriculum and Assessment Authority (VCAA). Further Mathematics was the predecessor of General Mathematics, so this paper is listed on the VCAA General Mathematics examinations page. Each question is summarised in a line; open the official examination PDF for the graphs, tables, diagrams and all five options.
• Answers match the key in the 2021 Further Mathematics Examination 1 external assessment report (Word document), and every calculation was redone independently.
• Percentages in brackets are the share of students who chose the correct option, from the report.
• Study design. This paper was set on the previous study design; the current General Mathematics study design began in 2023. The core (Data analysis, and Recursion and financial modelling) and the Matrices and Networks and decision mathematics modules are still on the current Units 3 and 4 course, except that Matrices Question 3 (using matrices to decide whether simultaneous equations have a unique solution) now leans on Unit 1 content. Geometry and measurement (Module 3) and Graphs and relations (Module 4) are not part of the current Units 3 and 4: most of their skills are now General Mathematics Units 1 and 2 content, while latitude and longitude and linear programming are not in the current General Mathematics study design at all. Both modules are flagged where they appear, so skip them or use them only as extra practice.

Structure and timing

Examination 1 was 40 marks in 90 minutes (plus 15 minutes reading time), with one approved technology (calculator or software), an optional scientific calculator, one bound reference and a supplied formula sheet. Every question had five options (A to E).

• Section A, Core (24 questions, compulsory): Questions 1 to 16, Data analysis, and Questions 17 to 24, Recursion and financial modelling.
• Section B, Modules (16 marks): students chose two of four modules of 8 questions each: Matrices, Networks and decision mathematics, Geometry and measurement, and Graphs and relations. All four are worked below.

That is just over 2 minutes per question. In General Mathematics there is no choice: Matrices and Networks and decision mathematics are both compulsory, and the paper is 40 questions in the same 90 minutes.

Data analysis (Questions 1 to 16)

Q1

A percentaged segmented bar chart of age (under 55 years, 55 years and over) against preferred travel destination (domestic, international). What type are the two variables? Answer: A - both are categorical: age is recorded as one of two groups, not as a number. (70%; 21% chose C.)

Q2

Why does the chart support an association between destination and age? Answer: D - the percentage preferring domestic travel changes with age group: 65% of visitors under 55 but only 45% of those 55 and over. A single percentage (B or C) cannot show an association. (82%)

Q3

Which two-way frequency table could match the chart? Answer: A - under 55, $\tfrac{91}{140} = 65\%$ domestic and $\tfrac{49}{140} = 35\%$ international; 55 and over, $\tfrac{90}{200} = 45\%$ domestic and $\tfrac{110}{200} = 55\%$ international. (50%; 28% chose C, which has the percentages in the wrong rows.)

Q4

Two boxplots of fish length, Pond A and Pond B. Answer: B - Pond B's median lies to the right of Pond A's maximum, so at least 50% of Pond B fish are longer than every Pond A fish. Pond B's $Q_1$ sits at Pond A's $Q_3$, below Pond A's maximum, so E fails, and Pond B's minimum is below Pond A's $Q_3$, so D fails too. (60%)

Q5

A stem plot of 20 heights; why is 179 cm an outlier? Answer: E - $Q_1 = \tfrac{148 + 148}{2} = 148$ and $Q_3 = \tfrac{158 + 160}{2} = 159$, so $\text{IQR} = 11$ and the upper fence is $159 + 1.5 \times 11 = 175.5$ cm. (63%; 20% chose A, the value 162.)

Q6

Best display for resting pulse rate (numerical) against age group (four categories)? Answer: C - parallel boxplots, one for each age group. A back-to-back stem plot handles only two groups. (51%; 28% chose A.)

Q7

Scores are approximately normal, mean 69.5, standard deviation 6.5, with 800 participants. Scoring at least 76.0 is successful. How many were unsuccessful? Answer: D - 76.0 is one standard deviation above the mean, so 84% score below it: $0.84 \times 800 = 672$. (54%; 32% chose B, the 128 who were successful.)

Q8

A leading role needs a standardised score of at least 1.80. Amy 81.5, Brian 80.5, Cherie 82.0. Answer: C - $z = \tfrac{81.5 - 69.5}{6.5} \approx 1.85$ for Amy, $\tfrac{80.5 - 69.5}{6.5} \approx 1.69$ for Brian and $\tfrac{82.0 - 69.5}{6.5} \approx 1.92$ for Cherie, so only Brian missed out. (67%)

Q9

Normal heights: 16% are above 160 cm and 2.5% are below 115 cm. Mean and standard deviation? Answer: C - 16% above means 160 is one standard deviation above the mean, and 2.5% below means 115 is two below. So $3s = 45$, $s = 15$ and $\bar{x} = 145$. (71%)

Q10

Oscar's walking times for days 1 to 9 (46, 40, 45, 34, 36, 38, 39, 40, 33 minutes). The least squares line predicts day 10 equals which day? Answer: B - regression gives time $= 44 - 1 \times$ day, so day 10 predicts 34 minutes, the time on day 4. (71%)

Q11

A least squares line predicts weight from height for 10 adults. How many predicted weights exceed the actual weight? Answer: D - the line is weight $\approx -83.48 + 0.8424 \times$ height; the prediction is above the actual weight for six adults (negative residuals). Note that height is the explanatory variable here. (40%; 29% chose B.)

Q12

Quarterly sales from 2010 to 2020. Best description? Answer: D - the peaks and troughs drift downward (a decreasing trend), but they do not repeat in a regular pattern within each year, so there is no seasonality: a decreasing trend with irregular fluctuations. (34%; 41% chose E, which adds seasonality.)

Q13

Nine-median smoothed points for game 10, from a time series plot. Answer: C - games 6 to 14 read about 109.5, 108.5, 96.5, 110, 118, 102, 118, 133.5 and 112. In order, the middle (5th) value is 110. (53%; 24% chose D.)

Q14

Soil sold: Monday 234, Tuesday 186, Saturday 346, Sunday 346 cubic metres; Wednesday to Friday unknown. The five-mean smoothed value for Thursday is 206. Three-mean smoothed value for Thursday? Answer: B - Tuesday to Saturday total $5 \times 206 = 1030$, so Wednesday + Thursday + Friday $= 1030 - 186 - 346 = 498$, and $\tfrac{498}{3} = 166$. (48%; 25% chose D.)

Q15

Art gallery visitors by quarter for 2017 to 2019, with yearly quarterly averages. Seasonal index for summer? Answer: C - $\tfrac{29\,685}{27\,194.0} \approx 1.0916$, $\tfrac{25\,420}{23\,183.5} \approx 1.0965$ and $\tfrac{31\,496}{29\,243.0} \approx 1.0770$, with mean $\approx 1.088$. (59%)

Q16

Deseasonalised visitors $= 2349 - 198.5 \times$ month number, with January 2020 as month 1; February's seasonal index is 1.25. Predicted actual visitors for February 2020? Answer: E - deseasonalised $2349 - 198.5 \times 2 = 1952$, then re-seasonalise: $1952 \times 1.25 = 2440$. (52%; 29% chose C, forgetting to re-seasonalise.)

Recursion and financial modelling (Questions 17 to 24)

Q17

$L_0 = 37$, $L_{n+1} = L_n + C$ and $L_2 = 25$. Find $C$. Answer: A - $L_2 = 37 + 2C = 25$, so $C = -6$. (83%)

Q18

An annuity amortisation table; the principal reduction for payment 3 is blank. Answer: C - the principal reduction is the payment less the interest: 44 970.55 minus 17 962.40 gives $27 008.15. Check: 449 060.08 minus 27 008.15 is the listed balance, 422 051.93. (93%)

Q19

The same annuity (500 000 dollars paying 44 970.55 dollars a year). For how many years is it paid? Answer: B - the first year's interest is 20 000 on 500 000, so the rate is 4% p.a. Finance Solver with $I\% = 4$, $PV = -500\,000$, $PMT = 44\,970.55$, $FV = 0$ gives $N = 15$ years. (44%; 24% chose A.)

Q20

A boat bought for 72 000 dollars depreciates by 10% a year (reducing balance). In the third year it loses 10% of what value? Answer: C - the value at the start of year 3 is $72\,000 \times 0.9^2 = 58\,320$ dollars. (44%; 40% chose B, 52 488, the value at the end of year 3.)

Q21

An investment of 3000 dollars grows to 3728.92 dollars in four years, compounding monthly. Effective annual rate to two decimal places? Answer: D - Finance Solver (48 months) gives a nominal rate of about 5.450% p.a., so $r_{\text{effective}} = \left(1 + \tfrac{0.0545}{12}\right)^{12} - 1 \approx 5.59\%$. (40%; 22% chose A, the nominal rate.)

Q22

12 000 dollars invested at 2.8% p.a. compounding monthly, with an extra monthly payment, must reach 25 000 dollars in five years. Minimum payment? Answer: B - Finance Solver with $N = 60$, $I\% = 2.8$, $PV = -12\,000$, $FV = 25\,000$ gives a payment of about 174.11 dollars. (56%; 23% chose D.)

Q23

$B_0 = 450\,000$, $B_{n+1} = RB_n - 2633$ (monthly), repaid in 20 years. $R$ is closest to? Answer: A - Finance Solver with $N = 240$, $PV = 450\,000$, $PMT = -2633$, $FV = 0$ gives $I\% \approx 3.6$ p.a., a monthly rate of 0.3%, so $R = 1 + \tfrac{3.6}{1200} = 1.003$. (31%; 33% chose D, 1.036, from the annual rate.)

Q24

A loan of 400 000 dollars at 3.14% p.a. monthly; the scheduled 20-year repayment is set, but the first two years are interest-only. After a rate change, the scheduled repayment still clears the loan in the remaining 18 years. New rate? Answer: B - the scheduled repayment is about 2246.53 dollars a month ($N = 240$). The balance after two interest-only years is still 400 000. Solving $N = 216$, $PV = 400\,000$, $PMT = -2246.53$, $FV = 0$ gives $I\% \approx 2.21$. (33%; 25% chose C.)

Module 1: Matrices

Q1

Transpose of $M = \begin{bmatrix} 3 & 2 \\ 8 & 9 \\ 13 & 7 \end{bmatrix}$. Answer: D - rows become columns: $M^T = \begin{bmatrix} 3 & 8 & 13 \\ 2 & 9 & 7 \end{bmatrix}$. (82%)

Q2

65% of deli customers return to the deli, 55% of cafe customers return to the cafe. Transition matrix (columns this Friday, rows next Friday, order D, C)? Answer: D - $\begin{bmatrix} 0.65 & 0.45 \\ 0.35 & 0.55 \end{bmatrix}$: each column sums to 1. (73%)

Q3

$ax + 4y = 10$ and $18x + by = 6$ have no unique solution when? Answer: A - there is no unique solution when the coefficient matrix has determinant zero: $ab - 72 = 0$. Only $a = 2$, $b = 36$ gives $ab = 72$. (56%)

Study design: the determinant and the condition for an inverse are still Units 3 and 4 content, but using matrices to solve simultaneous linear equations is now General Mathematics Unit 1 content.

Q4

Which permutation matrix changes $\begin{bmatrix} R & A & M & O & N \end{bmatrix}^T$ into $\begin{bmatrix} N & O & R & M & A \end{bmatrix}^T$? Answer: E - each row picks one letter: row 1 takes the 5th (N), row 2 the 4th (O), row 3 the 1st (R), row 4 the 3rd (M) and row 5 the 2nd (A). So the 1s sit in columns 5, 4, 1, 3, 2. (73%)

Q5

$A$ is $7 \times 7$ and $B$ is $10 \times 7$. Which expression is defined? Answer: E - $B^T$ is $7 \times 10$, so $AB^T$ is $(7 \times 7)(7 \times 10)$, a $7 \times 10$ matrix. $AB$ and $B^2$ are undefined, $BA$ is $10 \times 7$ (not square, so no inverse), and $A^2 - BA$ subtracts matrices of different orders. (45%)

Q6

A $4 \times 4$ class transition matrix and a partly labelled transition diagram with 20%, 25% and $w$. Find $w$. Answer: B - the nonzero off-diagonal entries join A with C and D, and B with C and D, so the diagram is the cycle A, C, B, D. The only 25% transition is B to D and the 20% on the opposite side is A to C, which places A top right, C top left, B bottom left and D bottom right. Then $w$ is the transition from A to D: 0.15, so $w = 15\%$. (66%; 17% chose A.)

Q7

$S_{n+1} = TS_n - C$ with $T$, $S_0 = \begin{bmatrix} 21 & 51 & 31 \end{bmatrix}^T$ and $S_1 = \begin{bmatrix} 24.0 & 54.3 & 20.7 \end{bmatrix}^T$. Find $S_2$. Answer: A - $TS_0 = \begin{bmatrix} 27 & 53.3 & 22.7 \end{bmatrix}^T$, so $C = TS_0 - S_1 = \begin{bmatrix} 3 & -1 & 2 \end{bmatrix}^T$. Then $S_2 = TS_1 - C = \begin{bmatrix} 23.04 & 55.78 & 16.18 \end{bmatrix}^T$. (40%; 27% chose C.)

Q8

On Monday 50% of marsupials feed at A, 50% at B and none at C; $S_{n+1} = TS_n$ with diagonal entries 0.4, 0.5 and 0.6. Percentage not expected to change station from Tuesday to Wednesday? Answer: D - Tuesday is $S_1 = TS_0 = \begin{bmatrix} 25 & 35 & 40 \end{bmatrix}^T$. Staying: $0.4 \times 25 + 0.5 \times 35 + 0.6 \times 40 = 51.5\%$. (29%; 27% chose C. Using Monday's state instead gives 45%.)

Module 2: Networks and decision mathematics

Q1

Number of vertices of degree 3 in a seven-vertex graph. Answer: E - the degrees are 5 (the central vertex), 2 (the right-hand vertex) and 3 for each of the other five. (91%)

Q2

A bipartite graph of five friends and five fruits. Which statement is not true? Answer: C - Van ate orange as well as strawberry. Orange (Kai, Quinn and Van) is the most eaten fruit, so E is true. (95%)

Q3

Number of faces of a four-vertex graph drawn with two crossing edges. Answer: C - it is the complete graph on 4 vertices, which is planar. Redrawn without the crossing, $v = 4$ and $e = 6$, so Euler's formula $v + f = e + 2$ gives $f = 4$. (34%; 61% chose D, counting faces on the drawing with the crossing.)

Q4

In a directed network, how many vertices cannot be reached from $X$? Answer: A - from $X$ you reach $W$ and $Z$, then $V$, $T$ and $Y$ from $W$, then $U$ from $T$. No edge leads into $S$, so only $S$ is unreachable. (62%; 20% chose E.)

Q5

How many of five statements (planar, a cycle, a bridge, an Eulerian trail, a Hamiltonian path) are true of a six-vertex graph? Answer: D - four. It is planar (drawn with no crossings), it has a cycle (the triangle), the edge to the degree-1 vertex is a bridge, and a Hamiltonian path starts at that vertex. Four vertices have odd degree, so there is no Eulerian trail. (17%; 42% chose C.)

Q6

A project network with minimum completion time 18 hours; activity $E$ takes $x$ hours. Maximum $x$? Answer: B - two paths use $E$: B, E, J takes $5 + x + 7$ hours and C, G, F, E, J takes $4 + 3 + 1 + x + 7 = 15 + x$ hours. For the project to stay at 18 hours, $15 + x \le 18$, so $x \le 3$. (46%; 31% chose E, from the first path only.)

Q7

Number of zeros in the adjacency matrix of a network of pathways between J, K, L, M and N. Answer: C - reading the diagram as the report does: J has no edge to M or to itself, K none to N, L none to itself, M none to J, N or itself, and N none to K, M or itself. That is $2 + 1 + 1 + 3 + 3 = 10$ zeros (K has a loop, so its diagonal entry is not zero). (33%; 24% chose E.)

Q8

A weighted road network with unknown weights $x$ and $y$; the minimum length of road to clear so every town is connected. Answer: B - this is a minimum spanning tree. The fixed edges chosen are 22, 29, 30, 36, 40 and 42 (a total of 199). With $x = 50$ and $y = 60$ the tree adds 50, 52 and 60, giving $199 + 162 = 361$ km, as option B says. The other pairs give 356, 358, 363 and 363, not the totals their options state. (48%)

Module 3: Geometry and measurement

Study design: Geometry and measurement is not part of the current General Mathematics Units 3 and 4, so this module is optional practice. Most of its skills are now General Mathematics Unit 2 content (Space and measurement); latitude and longitude (Q1) is not in the current study design.

Q1

Which city is closest to the Greenwich meridian? Answer: D - Lomé (6° N, 1° E) has the smallest longitude. Distance from the meridian depends on longitude, not latitude. (65%)

Q2

Angle between the clock hands at one o'clock. Answer: C - each hour mark is $\tfrac{360°}{12} = 30°$. (86%)

Q3

A 12 cm long photograph is enlarged by an area scale factor of 9. New length? Answer: D - the length scale factor is $\sqrt{9} = 3$, so $12 \times 3 = 36$ cm. (38%; 46% chose E, multiplying by 9.)

Q4

An equilateral triangle of side 4 cm. Which is not a correct area calculation? Answer: C - $\tfrac12 \times 4 \times 4 = 8$ uses the side as the height. The others all equal $4\sqrt3 \approx 6.93$ cm²: $\tfrac{\sqrt3 \times 4^2}{4}$, $2\sqrt{12}$, $\tfrac{4^2}{2}\sin 60°$ and Heron's $\sqrt{6(6 - 4)^3}$. (68%)

Q5

A cone and a cylinder have the same radius and volume; the cone is 12 cm high. Cylinder height? Answer: A - $\pi r^2 h = \tfrac13 \pi r^2 \times 12$ gives $h = 4$ cm. (46%)

Q6

A 24 cm by 6 cm toy with four circles of diameter 6 cm; shaded area? Answer: B - the shaded rectangle starts at the centre of the first circle, so it is $21 \times 6 = 126$ cm², and three and a half circles lie inside it: $126 - 3.5 \times \pi \times 3^2 \approx 27$ cm². (45%; 25% chose C, about 31, from the full $24 \times 6$ rectangle less four circles.)

Q7

Two triangles share a side; the left triangle is isosceles (two equal sides marked) with apex angle $a°$, and $b°$ is the angle beside its base on the right. Which statement is always true? Answer: D - each base angle of the isosceles triangle is $180° - b°$ (supplementary to $b°$ on the straight line), so $a° + 2(180° - b°) = 180°$, giving $a° = 2b° - 180°$. (32%; 27% chose B.)

Q8

A lookout is 1400 m east of a car park. The swimming hole is on a bearing of 290° from the lookout and 950 m from the car park (closer to the lookout); the cafe is on 240° and 700 m from the car park. How much further did Rod (via the cafe) walk than Lucia (via the swimming hole)? Answer: B - the angle at the lookout is 20° to the swimming hole and 30° to the cafe. The cosine rule gives 495 m (the other root, 2136 m, is further from the lookout than from the car park) and 1212 m. Lucia walked $495 + 950 = 1445$ m and Rod $1212 + 700 = 1912$ m: 467 m further. (30%; 27% chose D.)

Study design: the cosine rule and three-figure bearings are now General Mathematics Unit 2 content.

Module 4: Graphs and relations

Study design: Graphs and relations is not part of the current General Mathematics Units 3 and 4, so this module is optional practice. Linear graphs, break-even models and step graphs (Q2 to Q5) are now General Mathematics Unit 1 content, and the reciprocal relationship in Q6 is Unit 2 content (Variation). Linear programming (Q7 and Q8) is not in the current study design.

Q1

Average daily sunlight hours for each month; how many months are below 4 hours? Answer: B - months 4, 5 and 6 (3.5, 3 and 3 hours). (92%)

Q2

Which line has the same slope as $y = 2x + 3$? Answer: A - $y - 2x = 0$ is $y = 2x$, gradient 2. (57%; 19% chose D.)

Q3

A line through $(8, 0)$, $(0, 6)$ and $(5, m)$. Answer: D - the gradient is $-\tfrac{6}{8} = -0.75$, so $y = 6 - 0.75x$ and $m = 6 - 3.75 = 2.25$. (62%)

Q4

Eastpark charges a step fee (up to 3 hours, 3 to 6 hours, 6 to 10 hours: 6, 10 or 14 dollars); Northpark charges 2.30 dollars an hour. Minimum total for 7 hours and then 4 hours? Answer: C - 7 hours: Eastpark's 14 dollars beats Northpark's 16.10. 4 hours: Northpark's $2.30 \times 4 = 9.20$ beats Eastpark's 10. The total is $23.20. (44%; 31% chose D.)

Q5

Fixed costs 16 000 dollars a month, cost 52 dollars and price 280 dollars per heater. Minimum sales for a profit? Answer: E - each heater contributes $280 - 52 = 228$, and $\tfrac{16\,000}{228} \approx 70.2$, so 71 heaters. (58%)

Q6

A straight line through the origin and $(5, 2)$ on a graph of $y$ against $\tfrac1x$. Answer: D - the gradient is $\tfrac25$, so $y = \tfrac25 \times \tfrac1x = \tfrac{2}{5x}$. (42%; 27% chose C.)

Q7

A feasible region with corner $K(40, 20)$, where the boundaries $2x + y = 100$ and $x + 2y = 80$ meet. Which objective function is not maximised at $K$? Answer: B - $Z = x + 3y$ is 120 at $(0, 40)$ but only 100 at $K$. The other four functions are largest at $K$, because their gradients lie between $-2$ and $-\tfrac12$. (39%)

Q8

Constraints $x, y \ge 0$, $x + 2y \le 100$, $x + y \ge 20$, $x \le 60$; profit $P = ax + by$ is maximised at $(20, 40)$ and $a = 15$. Maximum profit? Answer: E - $(20, 40)$ is not a corner: it lies partway along the edge $x + 2y = 100$. A maximum there means $P$ is parallel to that edge, so $b = 2a = 30$ and $P = 15 \times 20 + 30 \times 40 = 1500$. (29%; 22% chose D.)

Exam tips from this paper

• The report singled out core Questions 12, 23 and 24, Matrices Question 8, Networks Questions 3, 5 and 7, Geometry Questions 3, 7 and 8, and Graphs and relations Questions 7 and 8.
• Seasonality needs a regular, calendar-linked pattern; irregular ups and downs around a falling trend are not seasonal (Q12).
• With a recurrence $B_{n+1} = RB_n - d$ for a monthly loan, $R = 1 + \tfrac{r}{1200}$, using the annual percentage rate $r$ (Q23).
• Several finance questions take two Finance Solver steps: first find the missing quantity (the interest rate from the amortisation table in Q19, the scheduled repayment in Q24), then solve for what is asked.
• In a matrix recurrence, count carefully which state is "Tuesday": $S_0$ is Monday (Matrices Q8).
• Redraw a graph in planar form before counting faces (Networks Q3).