VCE General Mathematics 2022 Exam 1

Every multiple-choice question from the 2022 VCE Further Mathematics Examination 1, with the correct option and a short worked reason. Further Mathematics was the predecessor of General Mathematics: 2022 was its last November examination, General Mathematics replaced it in 2023, and VCAA lists this paper with the General Mathematics examinations. For the extended-response paper, see the 2022 Examination 2 walkthrough.

How to use this page

• Questions are from the 2022 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 and all five options.
• Answers match the key in the 2022 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, which Further Mathematics was examined on from 2016; the current General Mathematics study design began in 2023. Everything in the core (Questions 1 to 24) and in the Matrices and Networks and decision mathematics modules is still on the current Units 3 and 4 course, except Module 1 Question 6: solving simultaneous equations with an inverse matrix is now General Mathematics Unit 1 content, and it is flagged where it appears. 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, time zones, linear inequalities and linear programming are not in the current General Mathematics study design at all. Both modules are flagged question by question, 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) and was worth one mark. That is just over 2 minutes per question.

• Section A, Core (24 marks, 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, each of 8 questions: Matrices, Networks and decision mathematics, Geometry and measurement, and Graphs and relations. All four are worked below.

In General Mathematics there is no choice: Matrices and Networks and decision mathematics are both compulsory, and each has 8 questions on Examination 1.

Data analysis (Questions 1 to 16)

Q1

Shape of a histogram of skull width for 46 female possums. Answer: C - the peak is in the 56 to 57 mm class, the lower tail (down to 49 mm) is longer than the upper tail, and a single possum at 67 to 68 mm sits well clear of the rest. So it is negatively skewed with a possible outlier. (36%; 54% chose E, approximately symmetric with a possible outlier.)

Q2

Percentage of the 46 possums with a skull width less than 55 mm. Answer: B - the columns below 55 mm hold $1 + 1 + 2 + 1 + 2 + 5 = 12$ possums, and $\tfrac{12}{46} \approx 26\%$. (85%)

Q3

A possible value of $Q_3$ for the skull width distribution. Answer: D - $Q_3$ is the median of the top 23 values, the 12th from the top. Counting down from the top: 1 (67 to 68), 1, 3, 5 reaches 10, and the 57 to 58 mm class takes the count past 12, so $Q_3$ lies between 57 and 58 mm: 57.7. (55%)

Q4

Mean and standard deviation of age for 14 possums from a dot plot. Answer: E - the ages are 1, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6, 6, 9, 9, giving $\bar{x} = \tfrac{69}{14} \approx 4.9$ and $s \approx 2.5$ (sample standard deviation). (74%)

Q5

Body lengths are normal with mean 88 cm and standard deviation 4 cm; 2498 possums. How many are between 84 and 96 cm? Answer: A - 84 cm is $z = -1$ and 96 cm is $z = 2$, so the share is $34\% + 47.5\% = 81.5\%$, and $0.815 \times 2498 \approx 2036$. (76%)

Q6

Histogram of $\log_{10}$(spleen weight) for 32 seals. How many have a spleen weight of 1000 g or more? Answer: D - $\log_{10} 1000 = 3$, and the columns from 3.0 up hold $8 + 9 + 7 + 1 = 25$ seals. (75%)

Q7

From $\log_{10}(\textit{spleen weight}) = 2.698 + 0.009434 \times \textit{age}$, the increase in $\log_{10}$(spleen weight) per extra month. Answer: A - that is the slope, 0.009434. (68%)

Q8

Predicted spleen weight of a 30-month-old seal. Answer: D - $2.698 + 0.009434 \times 30 = 2.981$, so the weight is $10^{2.981} \approx 957$ g. (56%; 28% chose A, stopping at the log value.)

Q9

Types of the variables test grade (A to E) and instructional method (individual, group). Answer: E - grades have a natural order, so test grade is ordinal; the methods have no order, so instructional method is nominal. (67%)

Q10

Of the 28 students with an A grade, what percentage were instructed individually? Answer: C - $\tfrac{10}{28} \approx 36\%$. (56%; 34% chose A, dividing by the 115 individual students instead.)

Q11

Column-percentage table of qualified (A or B grade) by instructional method. Answer: B - individual: $\tfrac{10 + 35}{115} \approx 39\%$ yes and 61% no; group: $\tfrac{18 + 30}{126} \approx 38\%$ yes and 62% no. (86%)

Q12

Equation of the least squares line of body length against head length for 17 crocodiles. Answer: B - head length is the explanatory variable, so the equation must give body length in terms of head length, which rules out A and C. The line rises from about $(30, 165)$ to $(85, 550)$, a slope of $\tfrac{385}{55} = 7$, which rules out D and E. That leaves body length $= -40 + 7 \times$ head length. (58%)

Q13

Median head length of the 17 crocodiles. Answer: B - the median is the 9th value in order along the horizontal axis. The head lengths run 32, 35, 36, 38, 38, 40, 40, 49, 51, ..., so the median is about 51 cm. (51%)

Q14

$r = 0.963$. Percentage of variation in body length not explained by head length? Answer: C - $r^2 = 0.963^2 \approx 0.927$, so $100\% - 92.7\% = 7.3\%$ is not explained. (48%; 24% chose B, from $100\% - 96.3\%$.)

Q15

Six-mean smoothed coffee sales, with centring, for Thursday of Week 2. Answer: B - the mean of Monday to Saturday of Week 2 is 128.67 and the mean of Tuesday to Sunday is 146.67; their average is $\tfrac{128.67 + 146.67}{2} \approx 138$. (66%)

Q16

The summer seasonal index for sunscreen sales is 1.25. How should actual summer sales be corrected for seasonality? Answer: A - deseasonalised $= \tfrac{\text{actual}}{1.25} = 0.8 \times \text{actual}$, a 20% reduction. (40%; 24% chose B, reduced by 25%, and 22% chose E.)

Recursion and financial modelling (Questions 17 to 24)

Q17

$R_0 = 2$, $R_{n+1} = 2 - R_n$. Find $R_2$. Answer: D - $R_1 = 2 - 2 = 0$ and $R_2 = 2 - 0 = 2$. (44%; 32% chose B.)

Q18

$V_0 = 400\,000$, $V_{n+1} = 1.003V_n - 2024$. When does the balance first fall below 398 000 dollars? Answer: C - $V_1 = 399\,176$, $V_2 \approx 398\,349.53$ and $V_3 \approx 397\,520.58$, so after 3 months. (78%)

Q19

With a small change to the final payment, how long until the same loan is repaid? Answer: A - the monthly rate is 0.3%, so 3.6% p.a. Finance Solver (or iterating the rule) gives $N = 300$ monthly payments, with only about 5 dollars left after the 300th, which the adjusted final payment covers: $\tfrac{300}{12} = 25$ years. (47%)

Q20

Equipment used 10 hours a day, every day, depreciated by unit cost: $E_0 = 100\,000$, $E_{n+1} = E_n - 5475$. Depreciation rate? Answer: A - $\tfrac{5475}{365} = 15$ per day, and $\tfrac{15}{10} = 1.50$, so $1.50 per hour. (65%)

Q21

How many of four statements about nominal and effective interest rates are true? Answer: D - three are true: the rates are equal when interest compounds annually; the effective rate rises as compounding becomes more frequent (for 12% nominal: 12.55% quarterly, 12.68% monthly); and 12% p.a. nominal is 1% per month. The effective rate is never lower than the nominal rate, so the fourth is false. (33%; 45% chose C.)

Q22

Tim's balance after $n$ years is $T_n = 6000 \times 1.003^{12n}$. Recurrence relation for the balance after $n$ months? Answer: C - the exponent $12n$ counts months, so each month multiplies the balance by 1.003: $R_0 = 6000$, $R_{n+1} = 1.003R_n$. (58%; 23% chose E.)

Q23

Li invests 4000 dollars for five years at 3.88% p.a. compounding annually. Which simple interest recurrence for Joseph ends at the same value? Answer: D - Li finishes with $4000 \times 1.0388^5 \approx 4838.60$, and $3500 + 5 \times 267.72 = 4838.60$. Each other option finishes at 4338.60 or 4588.60. (49%)

Q24

Dion invests 10 500 dollars at 0.52% per quarter and adds a constant amount $C$ at the end of each quarter; after two years the balance is 12 700.95 dollars. Find $C$. Answer: B - $D_1 = C$ and $D_{n+1} = D_n$ mean the addition is the same each quarter. Finance Solver with $N = 8$, $I = 2.08\%$ ($0.52 \times 4$), $PV = -10\,500$, $FV = 12\,700.95$ and 4 periods a year gives $PMT \approx 215.55$. (32%; 31% chose C.)

Module 1: Matrices (Questions 1 to 8)

Q1

Bike rental matrix $B$ (rows child, junior, adult; columns road, mountain). Which element is the adult mountain bike cost? Answer: E - adult is row 3 and mountain is column 2, so the element is $b_{32} = 135$. (83%)

Q2

Sunday prices are 10% higher than those in $B$. Which calculation gives them? Answer: D - a 10% increase multiplies every element by $1 + 0.1 = 1.1$, so calculate $1.1B$ (for example, $1.1 \times 80 = 88$). (68%; 24% chose $0.1B$, which gives only the increase.)

Q3

Which transition matrix matches an incomplete swim-centre diagram (M loop 0.3, M to A 0.2, A loop 0.4, A to N 0.5, N to M 0.6, N loop 0.1)? Answer: C - each "this day" column must sum to 1. Column $M$: $1 - 0.3 - 0.2 = 0.5$ to $N$. Column $A$: $1 - 0.4 - 0.5 = 0.1$ to $M$. Column $N$: $1 - 0.6 - 0.1 = 0.3$ to $A$. That gives columns $(0.3, 0.2, 0.5)$, $(0.1, 0.4, 0.5)$ and $(0.6, 0.3, 0.1)$. (71%; 22% chose E, which puts 0 where the missing arrows belong.)

Q4

Using a five-person communication matrix, which chain lets Ursula reach Steph? Answer: C - U-T-W-S: row $U$ has a 1 under $T$, row $T$ a 1 under $W$, and row $W$ a 1 under $S$. Each other option uses a missing link, for example neither $T$ nor $V$ can send directly to $S$, and $U$ cannot send to $W$. (87%)

Q5

$E$ is $2 \times 2$, $F$ is $2 \times 3$, $G$ is $3 \times 2$, $H$ is $3 \times 3$. Which product could have an inverse? Answer: D - only a square matrix can have an inverse. $EF$ and $FH$ are $2 \times 3$, $GE$ and $HG$ are $3 \times 2$, and $GF$ is $3 \times 3$, the only square product. (Strictly, a $3 \times 3$ product of a $3 \times 2$ and a $2 \times 3$ matrix always has determinant 0, but the question tests the order condition.) (71%)

Q6

Which calculation solves $y + z = 4$, $x - y + z = 1$, $-x + y = 2$? Answer: E - write the system as $AX = K$ with $A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & -1 & 1 \\ -1 & 1 & 0 \end{bmatrix}$ and $K = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$, so $X = A^{-1}K$. Here $\det A = -1$ and $A^{-1} = \begin{bmatrix} 1 & -1 & -2 \\ 1 & -1 & -1 \\ 0 & 1 & 1 \end{bmatrix}$, the matrix in option E, placed before $K$. It gives $x = -1$, $y = 1$, $z = 3$. (31%; 43% chose D, which multiplies by $A$ itself instead of its inverse.)

Study design: using an inverse matrix to solve simultaneous linear equations is now General Mathematics Unit 1 content (Discrete mathematics), not Units 3 and 4. Units 3 and 4 still cover the inverse, the determinant and when an inverse exists.

Q7. Column matrix $M$ is multiplied by the $5 \times 5$ permutation matrix $K$ (row 1 has its 1 in column 3) to give $P = KM$. Where does $m_{31}$ end up? Answer: A - row 1 of $K$ picks out the third element of $M$, so $p_{11} = m_{31}$. In full, $KM = \begin{bmatrix} m_{31} & m_{21} & m_{41} & m_{51} & m_{11} \end{bmatrix}^T$. (42%; 25% chose $p_{31}$.)

Q8. $N$ ($3 \times 2$) gives minutes for Henry, Irvine and Jean to service a laptop and a desktop; $Q$ ($4 \times 2$) gives laptops and desktops in four departments. Which calculation gives each person's total time for all four departments? Answer: A - $Q \times N^T$ is $4 \times 3$ (departments by person), and pre-multiplying by the $1 \times 4$ row of ones adds the four departments, leaving a $1 \times 3$ matrix of totals: $\begin{bmatrix} 706 & 794 & 588 \end{bmatrix}$ minutes (Henry: $25 \times 18 + 32 \times 8 = 706$). Option B totals each department instead, and option E adds everything into one number. (31%; 29% chose B.)

Module 2: Networks and decision mathematics (Questions 1 to 8)

Q1

What can Prim's algorithm be used to find? Answer: E - a minimum spanning tree: it grows the tree from a vertex by repeatedly adding the shortest edge to a new vertex. (80%)

Q2

A map of seven Central American countries becomes a network with an edge for each shared border. How many edges? Answer: C - 7: Belize-Guatemala, Guatemala-Honduras, Guatemala-El Salvador, El Salvador-Honduras, Honduras-Nicaragua, Nicaragua-Costa Rica and Costa Rica-Panama. (49%; 28% chose 6.)

Q3

Allocate four athletes to long jump, high jump, shot put and javelin to maximise total distance. Answer: D - Harsha long jump, Taylor high jump, Shona shot put, Eve javelin: $4.8 + 1.7 + 14.4 + 40.9 = 61.8$ m. The other options total 59.6, 61.6, 60.4 and 60.8, and checking all 24 allocations confirms 61.8 is the maximum. Shona's big throw matters more in shot put than in javelin, where Eve is almost as good. (39%; 27% chose B, 61.6 m.)

Q4

How many edges must be removed to make the drawn graph (7 vertices, 9 edges, drawn with crossing edges) planar? Answer: A - none. The graph is already planar: the degree 1 vertex can be moved so its edge crosses nothing, and the one crossing left disappears when one edge is redrawn around the outside. Crossings in a drawing do not make a graph non-planar. (21%; 37% chose 2.)

Q5

A connected graph has five vertices and four edges. Which statement is not true? Answer: E - a connected graph with $v$ vertices and $v - 1$ edges is a tree, so it cannot contain a cycle. It is a tree, it is planar, any tree is bipartite, and it contains paths. (54%; 21% chose C.)

Q6

Earliest start time of $G$ in a 12-activity landscaping network, where $G$ follows $D$ and a dummy from the end of $B$ and $E$. Answer: C - $G$ must wait for $A$ then $D$ ($6 + 5 = 11$) and for the node after $B$ and $E$, whose earliest time is $\max(10, 5 + 7) = 12$. So the EST of $G$ is $\max(11, 12) = 12$ hours. (48%; 25% chose 11, which ignores the dummy.)

Q7

From a precedence table for activities A to K, where must the dummy go? Answer: B - $F$ needs $D$ and $E$, but $G$ needs only $E$. Let $E$ end at the node where $G$ starts, and draw a dummy from the end of $E$ to the start of $F$ (which begins after $D$). (49%)

Q8

For the same project, what is the sum of all float times when it is completed in minimum time? Answer: D - the forward and backward passes give a minimum time of 28 days with critical path $B$-$E$-$G$-$H$-$I$-$K$ ($7 + 8 + 2 + 3 + 6 + 2 = 28$). The non-critical floats are $A$ 4, $C$ 4, $D$ 5, $F$ 3 and $J$ 4, so the total is $4 + 4 + 5 + 3 + 4 = 20$ days. (34%; 23% chose 18.)

Module 3: Geometry and measurement (Questions 1 to 8)

Study design: Geometry and measurement is not part of the current General Mathematics Units 3 and 4, so this module is optional practice. The notes under each question say where its skills sit now.

Q1. A regular octagon is drawn inside a circle with lines from the centre to each vertex. Find angle $x$ at the centre. Answer: D - the eight central angles are equal and fill a full turn, so $x = \tfrac{360}{8} = 45°$. (93%)

Study design: angle properties of regular polygons are not listed as such in the current study design; the nearest topic is Unit 2 Space and measurement.

Q2. An isosceles triangle in a circle has base 3.0 cm and area 5.25 cm². Find its height $h$. Answer: D - $\tfrac{1}{2} \times 3.0 \times h = 5.25$, so $h = \tfrac{10.5}{3.0} = 3.5$ cm. (81%)

Study design: areas of triangles and circles are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Q3. The triangle's area is 38.5% of the circle's area. Radius of the circle? Answer: B - the circle's area is $\tfrac{5.25}{0.385} \approx 13.64$ cm², so $r = \sqrt{\tfrac{13.64}{\pi}} \approx 2.08$ cm. (67%)

Study design: areas of circles are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Q4. Williamsburg (77° W), Mountain Grove (92° W) and Santa Cruz (122° W) share a latitude, with 15° of longitude per hour. Which time statement is true? Answer: C - further west is earlier. Mountain Grove is $\tfrac{92 - 77}{15} = 1$ hour behind Williamsburg and Santa Cruz is $\tfrac{122 - 77}{15} = 3$ hours behind, so 3 pm in Santa Cruz is 6 pm in Williamsburg. (49%; 32% chose D, but 3 pm in Mountain Grove is 1 pm in Santa Cruz, not 5 pm.)

Study design: latitude and longitude and time zones are not in the current General Mathematics study design.

Q5. A tank is a cylinder 15 m long and 3 m in diameter with a hemisphere on each end. A scale model is 300 mm long overall. Diameter of the model? Answer: C - the two hemispheres add a radius of 1.5 m at each end, so the full tank is $15 + 2 \times 1.5 = 18$ m long. The scale is $\tfrac{300}{18}$, and the model's diameter is $3 \times \tfrac{300}{18} = 50$ mm. (24%; 53% chose D, 60 mm, which uses 15 m as the total length and ignores the hemispheres.)

Study design: similarity and linear scale factors are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Q6. A plane flies 45 km due north from Amberley to Beachwood, 66 km on a bearing of 303° to Chalton, then 98 km back to Amberley. Bearing of Chalton to Amberley? Answer: E - at Beachwood, Amberley is on 180° and Chalton on 303°, so angle $ABC = 303° - 180° = 123°$. The sine rule gives $\sin C = \tfrac{45 \sin 123°}{98}$, so $C \approx 22.65°$. From Chalton, Beachwood is on the back bearing $303° - 180° = 123°$, and Amberley is a further $22.65°$ clockwise: $123° + 22.65° \approx 146°$. (38%)

Study design: the sine rule and three-figure bearings are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Q7. From Fran's window ledge the top of a 40.3 m tower opposite is at an angle of elevation of 20° and its base at an angle of depression of 17°. Height $h$ of the ledge? Answer: B - if $d$ is the horizontal distance between the towers, $d(\tan 20° + \tan 17°) = 40.3$, so $d \approx 60.18$ m and $h = d \tan 17° \approx 18.4$ m. (69%)

Study design: angles of elevation and depression are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Q8. A camera 20 m up a tower looks down at 35° to the bottom of a ramp and at 20° to its top, which is 4 m above the ground. Length of the ramp? Answer: C - the camera is $\tfrac{20}{\sin 35°} \approx 34.87$ m from the bottom of the ramp and $\tfrac{20 - 4}{\sin 20°} \approx 46.78$ m from the top, with $35° - 20° = 15°$ between those lines. The cosine rule gives $\sqrt{34.87^2 + 46.78^2 - 2 \times 34.87 \times 46.78 \cos 15°} \approx 15.9$ m, closest to 16 m. (40%; 29% chose A.)

Study design: angles of depression and the cosine rule are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Module 4: Graphs and relations (Questions 1 to 8)

Study design: Graphs and relations is not part of the current General Mathematics Units 3 and 4, so this module is optional practice. The notes under each question say where its skills sit now.

Q1. A tree's height is $H = 1.5 + 0.12n$ metres after $n$ months. Height after eight months? Answer: E - $1.5 + 0.12 \times 8 = 2.46$ m. (90%)

Study design: linear functions and models are now General Mathematics Unit 1 content (Functions, relations and graphs), not Units 3 and 4.

Q2. A curve shows room temperature over 12 hours, falling from 28 °C to a minimum of 2 °C at 8 hours, then rising to 17 °C. Which statement is not true? Answer: E - in the first two hours the temperature falls from 28 °C to 16 °C, an average of $\tfrac{12}{2} = 6$ °C per hour, not 5. The others hold: the drop over three hours is $28 - 12 = 16$ °C, and from 4 to 12 hours the rise is $\tfrac{17 - 9}{8} = 1$ °C per hour. (62%; 24% chose D.)

Study design: reading average rates of change from a non-linear graph is not listed as such in the current study design; the nearest topic is Unit 1 Functions, relations and graphs.

Q3. Vases cost 40 each to make and sell for 75 each, with a fixed monthly cost of 1600 (all amounts in dollars). Profit on 150 vases? Answer: A - $150 \times (75 - 40) - 1600 = 3650$, a profit of $3650. (78%)

Study design: cost, revenue and profit as linear models built from a worded description are now General Mathematics Unit 1 content (Functions, relations and graphs), not Units 3 and 4.

Q4. Maximum of $Z = 3x - 5y$ over a shaded feasible region. Answer: C - the corner points are $(5, 5)$, $(5, 25)$, $(20, 10)$ and $(10, 5)$, giving $Z = -10$, $-110$, $10$ and $5$. The maximum is 10, at $(20, 10)$. (73%)

Study design: linear programming is not in the current General Mathematics study design.

Q5. 11 sedans and seven station wagons cost 733 000 in total, and each sedan costs 7000 less than each wagon (all in dollars). Cost of a sedan? Answer: B - with wagon price $w$, $11(w - 7000) + 7w = 733\,000$, so $18w = 810\,000$ and $w = 45\,000$. Each sedan cost $45\,000 - 7000 = 38\,000$, so $38 000. (73%)

Study design: simultaneous linear equations from word problems are now General Mathematics Unit 1 content (Functions, relations and graphs), not Units 3 and 4.

Q6. Pink paint uses at least three drops of white ($x$) for every two drops of red ($y$). Which inequality fits? Answer: B - the ratio needs $\tfrac{x}{y} \ge \tfrac{3}{2}$, so $x \ge \tfrac{3}{2}y$. Check: with $y = 2$ it gives $x \ge 3$. (43%; 16% chose A, which has the ratio upside down.)

Study design: linear inequalities are not in the current General Mathematics study design.

Q7. A 1200 L tank drains 400 L at 25 L per minute, stays level for 30 minutes, then empties, finishing at 86 minutes. Equation of the last segment, from $t = b$ to $t = 86$? Answer: A - the first stage takes $\tfrac{400}{25} = 16$ minutes, so $b = 16 + 30 = 46$. The last 800 L drain in $86 - 46 = 40$ minutes, a slope of $-20$, so $V = -20t + c$ with $0 = -20 \times 86 + c$, giving $V = -20t + 1720$. (52%)

Study design: piecewise linear (line segment) graphs are now General Mathematics Unit 1 content (Functions, relations and graphs), not Units 3 and 4.

Q8. An integer feasible region bounded by $y = 24 - 6x$, $y = 18 - 2x$ and $y = 8 - \tfrac{2}{3}x$, with $Z = ax + by$ for $a, b > 0$. Which statement is true? Answer: D - the minimum is always at one of the lower corner points $(2, 12)$, $(3, 6)$ or $(6, 4)$. When $\tfrac{a}{b} > 6$, $Z(3, 6) - Z(2, 12) = a - 6b > 0$ and $Z(6, 4) - Z(2, 12) = 4a - 8b > 0$, so the minimum is at $(2, 12)$. A is false (six points maximise $Z$ when $\tfrac{a}{b} = 2$, not eight), B is false ($(3, 12)$ always beats $(3, 10)$), C is false (the maximum is at $(7, 4)$) and E is false (for small $\tfrac{a}{b}$ the minimum moves to $(6, 4)$). (29%; 25% chose E and 24% chose C.)

Study design: linear programming is not in the current General Mathematics study design.

Exam tips from this paper

• The report flagged core Questions 1 and 16 (distribution shape and seasonal indices) and 17, 21 and 24 (recurrence relations, effective rates and annuity investments) as ones students found challenging.
• A distribution whose lower tail is longer than its upper tail is negatively skewed, even with an outlier at the top (Q1).
• Deseasonalising divides by the seasonal index: an index of 1.25 means a 20% reduction, not 25% (Q16).
• In matrix questions, check orders first: only a square matrix can have an inverse, and a product's order tells you what it totals (Module 1 Q5 and Q8).
• In critical path questions, follow every dummy before taking the earliest start time (Module 2 Q6).