VCE General Mathematics 2022 Exam 2

Every question from the 2022 VCE Further Mathematics Examination 2, the extended-response paper. Further Mathematics was the predecessor of General Mathematics: 2022 was the last year it was examined, General Mathematics replaced it in 2023, and VCAA lists this paper with the General Mathematics examinations. Each question is summarised, then a full worked solution sits behind a Show worked solution toggle. For the multiple-choice paper, see the 2022 Examination 1 walkthrough.

How to use this page

• Questions are from the 2022 VCE Further Mathematics Examination 2, 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 paraphrased with the data you need; open the official examination PDF for the exact wording, graphs and answer spaces.
• Solutions are our own. Final answers were checked against the 2022 Further Mathematics Examination 2 external assessment report (Word document) and every calculation was redone independently.
• From the report notes summarise what the examiners said students got wrong. Percentages are the share of students who scored full marks on that part.
• Study design. This paper was set on the previous Further Mathematics study design; the current General Mathematics study design began in 2023. Everything in the core (Questions 1 to 8) and in the Networks and decision mathematics module is still on the current Units 3 and 4 course, and so is most of the Matrices module: only the simultaneous equations parts of Matrices Question 2 are now Units 1 and 2 content. Geometry and measurement (Module 3) and Graphs and relations (Module 4) are not part of the current Units 3 and 4: some of their skills are now General Mathematics Units 1 and 2 content, while latitude and longitude, time zones and linear programming are not in the current General Mathematics study design at all. These are flagged where they appear, so skip them or use them only as extra practice.

Structure and timing

Examination 2 was 60 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. That is 1.5 minutes per mark.

• Section A, Core (36 marks, compulsory): Questions 1 to 5, Data analysis (24 marks), and Questions 6 to 8, Recursion and financial modelling (12 marks).
• Section B, Modules (24 marks): students chose two of four 12-mark modules: Matrices, Networks and decision mathematics, Geometry and measurement, and Graphs and relations. All four are worked below.

In the report's module table, Matrices was the most popular module (46.4%), followed by Networks and decision mathematics (31.1%), Graphs and relations (11.75%) and Geometry and measurement (10.4%). In General Mathematics there is no choice: Matrices and Networks and decision mathematics are both compulsory.

Data analysis

Question 1 (6 marks)

A back-to-back stem plot (key: 4 | 2 = 42) shows the daily maximum wind speed, in km/h, at a weather station on the 30 days of April 2021 and the 30 days of November 2021. The April values are 11, 11, 11, 13, 13, 13, 13, 13, 15, 15, 17, 19, 20, 20, 20, 20, 20, 22, 28, 31, 31, 31, 31, 33, 33, 33, 39, 39, 48, 54. The November values are 15, 17, 19, 19, 20, 20, 22, 22, 24, 24, 24, 26, 28, 28, 28, 28, 30, 30, 30, 31, 33, 35, 35, 39, 43, 46, 48, 54, 59, 70.

a

For April 2021, find i. the median wind speed (1 mark) and ii. the percentage of days with a wind speed less than 25 km/h. (1 mark)

b

The five-number summary for November is 15, 22, 28, 35, 70. Show that 59 km/h and 70 km/h would appear as outliers on a boxplot. (2 marks)

c

Construct a histogram of the November wind speeds, using class intervals of width 5 starting at 15 km/h. (2 marks)

Show worked solution

a. i. [1 mark]. With 30 values the median is the mean of the 15th and 16th values. In order, the April values start 11, 11, 11, 13, 13, 13, 13, 13, 15, 15, 17, 19, 20, 20, 20, 20, ..., so the 15th and 16th values are both 20. The median is 20 km/h.

ii. [1 mark]. The values below 25 are the 18 values from 11 to 22:

$$\frac{18}{30} \times 100\% = 60\%.$$

b. [2 marks]. Find the upper fence:

$$\text{IQR} = 35 - 22 = 13, \qquad \text{upper fence} = Q_3 + 1.5 \times \text{IQR} = 35 + 1.5 \times 13 = 54.5.$$

Both 59 and 70 are greater than the upper fence of 54.5, so both are outliers. (The lower fence is $22 - 19.5 = 2.5$, so there are no low outliers.)

c. [2 marks]. Count the November values in each class:

Class (km/h)	15 to under 20	20 to under 25	25 to under 30	30 to under 35	35 to under 40	40 to under 45	45 to under 50	50 to under 55	55 to under 60	60 to under 65	65 to under 70	70 to under 75
Frequency	4	7	5	5	3	1	2	1	1	0	0	1

Draw touching columns of these heights, leaving the 60 to 65 and 65 to 70 classes empty, and put the single 70 km/h value in the 70 to 75 column.

From the report. Parts a.i (85%) and a.ii (79%) were well answered. In b (64%) most students found the upper fence, but some stopped there without explaining what made 59 and 70 outliers, and some drew a boxplot without the key calculations. In c (61%) some students missed a column, and the outlier was often put in the wrong place.

Question 2 (5 marks)

Parallel boxplots show the relative humidity (%) at 9 am and at 3 pm on the 30 days of November 2021. The 9 am boxplot has a whisker from 40 to 65, a box from 65 to 100 with the median at 75, and no upper whisker. The 3 pm boxplot runs 27, 51, 63, 80, 100.

a. i

Is the relative humidity more variable at 9 am or 3 pm? Give the approximate values of both interquartile ranges. (1 mark)

ii

Write down the percentage of days on which the 3 pm relative humidity was less than 80%. (1 mark)

iii

Complete the five-number summaries (9 am: $Q_1 = 65$, median 75; 3 pm: minimum 27, $Q_1 = 51$, median 63). (1 mark)

b

Do the boxplots support the contention that relative humidity is associated with the time of day? Refer to the values of an appropriate statistic. (2 marks)

Show worked solution

a. i. [1 mark]

$\text{IQR}_{9\text{ am}} = 100 - 65 = 35$ and $\text{IQR}_{3\text{ pm}} = 80 - 51 = 29$. The relative humidity is more variable at 9 am, because its IQR (35) is larger than the 3 pm IQR (29).

ii. [1 mark]

80% is $Q_3$ for 3 pm, so 75% of days were below it.

iii. [1 mark]

Time	Minimum	$Q_1$	Median	$Q_3$	Maximum
9 am	40	65	75	100	100
3 pm	27	51	63	80	100

At 9 am the box reaches 100 and there is no upper whisker, so $Q_3$ and the maximum are both 100.

b. [2 marks]. Yes. The median relative humidity is 75% at 9 am but only 63% at 3 pm. The median changes with the time of day, so there is an association between relative humidity and time of day.

From the report. In a.i (59%) some students gave both IQR values correctly but never said which time was more variable. Part a.ii was answered correctly by 70% and a.iii by 90%, although some students could not find the 9 am minimum. In b (41%) many gave both medians but did not say that they differ.

Question 3 (4 marks)

Table 1 gives seven weather variables for the first eight days of December 2021: day number; minimum temperature (°C) 19.4, 17.6, 7.6, 7.5, 5.7, 9.9, 11.0, 6.5; maximum temperature (°C) 28.3, 29.7, 16.5, 15.9, 19.0, 23.8, 11.9, 14.2; rainfall (mm) 0, 1.0, 11.6, 0, 0.2, 0, 0, 0; maximum wind speed (km/h) 35, 35, 26, 30, 24, 39, 22, 28; direction of maximum wind speed ENE, WSW, WSW, WSW, ESE, NE, SSW, ESE; and temperature 9 am (°C) 22.9, 24.2, 12.7, 10.9, 10.4, 17.8, 11.7, 9.5.

a

Write down i. the number of numerical variables in Table 1 (1 mark) and ii. the median rainfall, in mm, for the eight days. (1 mark)

b

With unrounded coefficients, the least squares line is maximum temperature $= 9.235946\ldots + 1.002493\ldots \times$ minimum temperature. Round the intercept and slope to four significant figures. (1 mark)

c

Find the coefficient of determination for this association as a percentage, to one decimal place. (1 mark)

Show worked solution

a. i. [1 mark]

5: minimum temperature, maximum temperature, rainfall, maximum wind speed and temperature at 9 am. The day number is only a label, and the wind direction is categorical.

ii. [1 mark]

In order the rainfalls are 0, 0, 0, 0, 0, 0.2, 1.0, 11.6. The 4th and 5th values are both 0, so the median is 0 mm.

b. [1 mark]

$$\text{maximum temperature} = 9.236 + 1.002 \times \text{minimum temperature}.$$

c. [1 mark]. Run linear regression with minimum temperature as $x$ and maximum temperature as $y$: $r = 0.78178\ldots$, so

$$r^2 = 0.61118\ldots \approx 61.1\%.$$

From the report. In a.i (58%) 6 was a common error from counting day number as a numerical variable. In a.ii (70%) some students appear to have found the mean instead. In b (62%) the slope was the usual problem, often rounded to four decimal places. In c (48%) 0.6 was a common wrong answer.

Question 4 (5 marks)

A scatterplot shows relative humidity (%) at 9 am against temperature (°C) at 9 am for the 30 days of November 2021. The least squares line is relative humidity $= 120.1 - 3.417 \times$ temperature, and the coefficient of determination is 0.6073.

a

The line is used to predict relative humidity from temperature. Name the explanatory variable. (1 mark)

b

Describe the association between relative humidity and temperature in terms of strength and direction. (1 mark)

c

Interpret the slope of the least squares line in terms of the two variables. (1 mark)

d

On the day when the 9 am temperature was 16.3 °C, the relative humidity was 45.0%. Find the residual, to one decimal place. (2 marks)

Show worked solution

a. [1 mark]. Temperature, the variable used to make the prediction.

b. [1 mark]. Strong, negative. The slope is negative, so $r$ is negative:

$$r = -\sqrt{0.6073} = -0.7793\ldots,$$

and by the usual VCE guide $0.75 \le |r| < 0.9$ is a strong association.

c. [1 mark]. On average, relative humidity decreases by 3.417% for each 1 °C increase in temperature.

d. [2 marks].

$$\text{predicted} = 120.1 - 3.417 \times 16.3 = 64.4029$$

$$\text{residual} = \text{actual} - \text{predicted} = 45.0 - 64.4029 = -19.4029 \approx -19.4\%.$$

From the report. Part a was well answered (93%), though some chose relative humidity. Part b was answered correctly by only 28%: many called the association moderate, and others added comments on form or outliers when only strength and direction were asked for. In c (44%) many answers left out the 1 degree increase in temperature. In d (54%) some students stopped at the predicted value, and arithmetic or transcription errors cost marks.

Question 5 (4 marks)

A time series plot shows the monthly rainfall (mm) at a weather station over 36 months. The rainfall rises and falls in a repeated pattern, with peaks at months 5, 17 and 29, and the peaks and troughs fall over time, from about 206 mm at month 5 to about 20 mm at month 35. From the plot, months 18 to 22 have rainfalls of about 155, 112, 78, 61 and 100 mm.

a

The plot contains irregular fluctuations. Give two other descriptions of the pattern. (1 mark)

b

Write down the five-median smoothed rainfall for month 20. (1 mark)

c

The rainfalls for months 1 to 12 are 109.0, 147.2, 129.6, 179.6, 206.2, 155.2, 127.6, 83.8, 61.8, 134.6, 122.8 and 113.6 mm. i. Find the nine-mean smoothed rainfall for month 7, to one decimal place. (1 mark)

ii

How many points would the smoothed plot have if nine-mean smoothing were applied to the full 36 months? (1 mark)

Show worked solution

a. [1 mark]

A decreasing trend and seasonality. Both are needed for the mark.

b. [1 mark]

Months 18 to 22 have rainfalls of 155, 112, 78, 61 and 100 mm. In order: 61, 78, 100, 112, 155. The five-median smoothed value is 100 mm.

c. i. [1 mark]

The nine months centred on month 7 are months 3 to 11:

$$\frac{129.6 + 179.6 + 206.2 + 155.2 + 127.6 + 83.8 + 61.8 + 134.6 + 122.8}{9} = \frac{1201.2}{9} = 133.466\ldots \approx 133.5 \text{ mm}.$$

ii. [1 mark]. A nine-mean needs four months on each side, so the first four and the last four months get no smoothed value:

$$36 - 4 - 4 = 28 \text{ points}.$$

From the report. Part a (40%) was not answered well: many mentioned a structural change, and both terms were needed. In b (54%) answers varied, and some students found the mean instead of the median. Part c.i (54%) was handled fairly well by students who tried it, and the answer had to be rounded to one decimal place. In c.ii (34%) 4 was a common wrong answer.

Recursion and financial modelling

Question 6 (4 marks)

Pina depreciates her workplace equipment using flat rate depreciation. Its value, in dollars, after $n$ years is $V_n = 200\,000 - 12\,500n$.

a

Find $V_1$, the value after one year. (1 mark)

b

After how many years will the equipment first have a value of zero? (1 mark)

c

Write a recurrence relation for the value in terms of $V_0$, $V_{n+1}$ and $V_n$. (1 mark)

d

Pina's model takes a fixed percentage of the original value each year. What is the name of the depreciation that instead takes a fixed percentage of the current value each year? (1 mark)

Show worked solution

a. [1 mark]

$V_1 = 200\,000 - 12\,500 \times 1 = 187\,500$, so $187 500.

b. [1 mark]

$200\,000 - 12\,500n = 0$ gives $n = \dfrac{200\,000}{12\,500} =$ 16 years.

c. [1 mark]

$$V_0 = 200\,000, \qquad V_{n+1} = V_n - 12\,500.$$

d. [1 mark]. Reducing balance depreciation.

From the report. Parts a (94%) and b (88%) were well answered. In c (62%) notation was a problem, and a common error was to end the relation with $12\,500n$. In d (48%) many answered flat rate or unit cost.

Question 7 (4 marks)

Pina invests $540 000 in an annuity paying 3% per annum, compounding monthly.

The annuity pays her $5214.28 a month for 10 years.

An amortisation table shows: payment 0, balance 540 000.00; payment 1, payment 5214.28, interest 1350.00, principal reduction 3864.28, balance 536 135.72; payment 2, payment 5214.28, interest 1340.34, principal reduction 3873.94, balance 532 261.78. The line for payment 3 is missing.

a

What is the value of payment number 3? (1 mark)

b

Find the interest for payment number 3, to the nearest cent. (1 mark)

c

Let $P_n$ be the balance after $n$ months. Write a recurrence relation in terms of $P_0$, $P_{n+1}$ and $P_n$. (1 mark)

d

If Pina had invested the $540 000 as a simple perpetuity instead, what monthly payment would she have drawn? (1 mark)

Show worked solution

a. [1 mark]. Every payment is the same: $5214.28.

b. [1 mark]. The monthly rate is $3\% \div 12 = 0.25\%$. Interest is earned on the balance after payment 2:

$$0.0025 \times 532\,261.78 = 1330.654\ldots \approx 1330.65.$$

The interest is $1330.65.

c. [1 mark].

$$P_0 = 540\,000, \qquad P_{n+1} = 1.0025P_n - 5214.28.$$

d. [1 mark]. A perpetuity pays out exactly the interest each month, so the balance never changes: $0.0025 \times 540\,000 = 1350$. The payment is $1350.

From the report. Part a was well answered (86%). In b (51%) rounding errors came from students who worked out their own interest rate instead of using the one given, and nearest-cent rounding was needed. Part c (35%) was not answered well: the factor 1.0025 was often wrong, 5214.28 was often added, and some mixed $P$ and $V$ notation. In d (49%) many could not define a perpetuity, and 5214.28 was the most common mistake.

Question 8 (4 marks)

Pina takes out a reducing balance loan of $580 000, with interest calculated monthly.

The balance after $n$ months is modelled by $L_0 = 580\,000$, $L_{n+1} = 1.002L_n - 3045.26$.

a

Showing recursive calculations, find the balance of the loan after two months, to the nearest cent. (1 mark)

b

Find the annual compound interest rate for this loan. (1 mark)

c

The final repayment needed to fully repay the loan is smaller than the other repayments by less than one dollar. Find this small amount, to the nearest cent. (1 mark)

d

The recurrence relation models the loan being fully repaid in a whole number of years. With a different multiplication factor (instead of 1.002), the loan would be fully repaid one year sooner. Find this factor, to four decimal places. (1 mark)

Show worked solution

a. [1 mark].

$$L_1 = 1.002 \times 580\,000 - 3045.26 = 578\,114.74$$

$$L_2 = 1.002 \times 578\,114.74 - 3045.26 = 576\,225.709\ldots \approx 576\,225.71$$

The balance after two months is $576 225.71.

b. [1 mark]. The monthly rate is $1.002 - 1 = 0.002 = 0.2\%$, so the annual rate is

$$0.2\% \times 12 = 2.4\%.$$

c. [1 mark]. Finance Solver: $I\% = 2.4$, $PV = 580\,000$, $PMT = -3045.26$, $FV = 0$, 12 payments per year gives $N = 239.99\ldots$, so the loan takes 240 repayments (20 years).

Now set $N = 240$ and solve for the future value: $FV = 0.1493\ldots$ A positive future value means 240 full repayments would overpay by about 15 cents, so the final repayment is smaller by $0.15.

d. [1 mark]. One year sooner is $240 - 12 = 228$ repayments. Finance Solver: $N = 228$, $PV = 580\,000$, $PMT = -3045.26$, $FV = 0$, 12 payments per year gives $I\% = 1.9466\ldots$ The monthly rate is $1.9466\ldots\% \div 12 = 0.16222\ldots\%$, so

$$\text{factor} = 1 + \frac{1.9466\ldots}{1200} = 1.0016222\ldots \approx 1.0016.$$

(A lower interest rate means the same repayments clear the loan sooner.)

From the report. In a (44%) some students gave the right final answer without showing the recursive calculations, and many lost the mark through copying or rounding slips. In b (51%) 0.2% was a common wrong answer. Part c (23%) was not answered well, and some students did not clearly give the small amount of less than one dollar. Part d was answered correctly by only 19%.

Module 1: Matrices

Question 1 (2 marks)

Matrix $C = \begin{bmatrix} 80 & 140 & 270 \end{bmatrix}$ gives the nightly cost, in dollars, of a hostel (H), a motel (M) and an apartment (A) at a ski resort. The Dwyer family plans to stay five nights.

a. Complete the matrix equation $[\;\;] \times \begin{bmatrix} 80 & 140 & 270 \end{bmatrix} = [\;\;]$ to show the cost of five nights at each type of accommodation. (1 mark)

b. The family will stay two nights at the motel and three nights in an apartment. Write down the column matrix $A$ for which $CA$ gives the total cost of the five nights. (1 mark)

Show worked solution

a. [1 mark].

$$\begin{bmatrix} 5 \end{bmatrix} \times \begin{bmatrix} 80 & 140 & 270 \end{bmatrix} = \begin{bmatrix} 400 & 700 & 1350 \end{bmatrix}.$$

b. [1 mark]. The rows of $A$ are the numbers of nights in H, M and A order:

$$A = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}.$$

(Check: $CA = 0 + 280 + 810 = 1090$ dollars.)

From the report. Part a was well answered (94%). In b (60%) many wrong answers were seen, and some students swapped the three numbers around.

Question 2 (4 marks)

Two school groups hire skis and snowboards. With $x$ the daily cost, in dollars, of one set of skis and $y$ the daily cost of one snowboard:

$$20x + 40y = 3700, \qquad 30x + 50y = 4900.$$

a

Write the equations in matrix form. (1 mark)

b. i

Show that the determinant of $\begin{bmatrix} 20 & 40 \\ 30 & 50 \end{bmatrix}$ is $-200$. (1 mark)

ii

Interpret this value in relation to the solution of the equations. (1 mark)

c

Complete the matrix $\begin{bmatrix} x \\ y \end{bmatrix}$ showing the daily cost of hiring one set of skis and one snowboard. (1 mark)

Study design: the determinant (part b.i) and the condition for a matrix to have an inverse are still Units 3 and 4 content, but using matrices to solve simultaneous linear equations (parts a and c, and the unique-solution half of b.ii) is now General Mathematics Unit 1 content.

Show worked solution

a. [1 mark].

$$\begin{bmatrix} 20 & 40 \\ 30 & 50 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3700 \\ 4900 \end{bmatrix}.$$

b. i. [1 mark]. Show the calculation:

$$\det = 20 \times 50 - 40 \times 30 = 1000 - 1200 = -200.$$

ii. [1 mark]. The determinant is not zero, so the matrix has an inverse and the equations have a unique solution.

c. [1 mark]. Multiply both sides by the inverse:

$$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-200}\begin{bmatrix} 50 & -40 \\ -30 & 20 \end{bmatrix}\begin{bmatrix} 3700 \\ 4900 \end{bmatrix} = \begin{bmatrix} 55 \\ 65 \end{bmatrix}.$$

Skis cost 55 dollars a day and a snowboard costs 65 dollars a day. (Check: $20 \times 55 + 40 \times 65 = 3700$.)

From the report. Part a was well answered (86%). Part b.i (76%) was a "show that" question, so the calculation had to be set out. Part b.ii (21%) was not answered well: few could interpret the determinant, and both points (non-zero, so a unique solution) were needed. In c (59%) some students got the answer even after mistakes in earlier parts.

Question 3 (3 marks)

Students spend a school term skiing, from 16 July to 22 September. Each day they use a beginner (B), intermediate (I) or advanced (A) run. The transition matrix (columns this day B, I, A; rows next day B, I, A) and the numbers on 16 July are

$$T = \begin{bmatrix} 0.7 & 0.1 & 0 \\ 0.3 & 0.7 & 0.2 \\ 0 & 0.2 & 0.8 \end{bmatrix}, \qquad S_0 = \begin{bmatrix} 210 \\ 190 \\ 80 \end{bmatrix},$$

where $S_n$ gives the numbers $n$ days after 16 July.

a

How many skiers are expected to choose the same run on 17 July as on 16 July? (1 mark)

b

Of the skiers expected on the advanced run on 17 July, what percentage also chose it on 16 July, to the nearest whole number? (1 mark)

c

What is the maximum number of students expected on the intermediate run on any one day, to the nearest whole number? (1 mark)

Show worked solution

a. [1 mark]. The leading diagonal gives the skiers who stay:

$$0.7 \times 210 + 0.7 \times 190 + 0.8 \times 80 = 147 + 133 + 64 = 344.$$

b. [1 mark]. On 17 July the advanced run has $0.2 \times 190 + 0.8 \times 80 = 38 + 64 = 102$ skiers, and 64 of them were on it the day before:

$$\frac{64}{102} \times 100\% = 62.7\ldots\% \approx 63\%.$$

c. [1 mark]. Generate the state matrices with $S_{n+1} = TS_n$ and watch the I element:

Day	16 July ($S_0$)	17 July ($S_1$)	18 July ($S_2$)	19 July ($S_3$)	20 July ($S_4$)	21 July ($S_5$)
Intermediate	190	212	218.6	219.04	217.32	215.12

It peaks at 219.04 on 19 July, three days after the start, then falls towards the long-run value of about 206. The maximum is 219 students.

From the report. Part a (56%) drew many incorrect methods. Part b was answered correctly by only 25%. In c (19%) many students assumed the long-run number was needed, when the maximum came three days after the start.

Question 4 (3 marks)

Ali, Lee and Max are nominated for ski team captain. Seven days before the end of term, 160 students plan to vote for Ali, 140 for Lee and 180 for Max. Each day: 40% of Ali's supporters stay with Ali and 30% move to Lee; 50% of Lee's supporters stay with Lee and 30% move to Max; 50% of Max's supporters stay with Max and 20% move to Lee. (The remaining percentages are not given.)

a. How many students plan to vote for Ali after one day? (1 mark)

b. Max withdraws after one day. For the remaining six days,

$$T = \begin{bmatrix} 0.5 & 0.7 & 0.5 \\ 0.5 & 0.3 & 0.5 \\ 0 & 0 & 0 \end{bmatrix}$$

(columns this day A, L, M; rows next day A, L, M). At the end of the seven days, who is expected to become captain, and with how many votes, to the nearest whole number? (2 marks)

Show worked solution

a. [1 mark]. Each column must add to 1, so the missing percentages are: 30% of Ali's supporters move to Max, 20% of Lee's move to Ali, and 30% of Max's move to Ali. The full first-day matrix is

$$\begin{bmatrix} 0.4 & 0.2 & 0.3 \\ 0.3 & 0.5 & 0.2 \\ 0.3 & 0.3 & 0.5 \end{bmatrix}.$$

Ali after one day: $0.4 \times 160 + 0.2 \times 140 + 0.3 \times 180 = 64 + 28 + 54 =$ 146.

b. [2 marks]. After one day the state is Ali 146, Lee $0.3 \times 160 + 0.5 \times 140 + 0.2 \times 180 = 154$ and Max $0.3 \times 160 + 0.3 \times 140 + 0.5 \times 180 = 180$. Apply $T$ for the remaining six days:

$$T^6 \begin{bmatrix} 146 \\ 154 \\ 180 \end{bmatrix} = \begin{bmatrix} 280.00\ldots \\ 199.99\ldots \\ 0 \end{bmatrix}.$$

Ali is expected to become captain with 280 votes (Lee 200).

From the report. Part a was answered correctly by only 32%. In b (39%) a good number of students named the winner and gave the vote count, but many showed no working; a method mark was available for the correct state matrix even without a clear conclusion.

Module 2: Networks and decision mathematics

Question 1 (3 marks)

A holiday park has an office and seven cabins, A to G. The roads join office-A, A-B, B-C, C-D, D-E, E-F, F-G, G-office, office-F, B-F and B-E.

a

In the morning Joe leaves his office, visits each cabin once only and returns to the office. Write down a route he could follow. (1 mark)

b

Later, Joe leaves his office and travels along each road once only. i. At which vertex will he finish? (1 mark)

ii

What is the mathematical term for this route? (1 mark)

Show worked solution

a. [1 mark]

A Hamiltonian cycle: office-A-B-C-D-E-F-G-office (or the same route in reverse, office-G-F-E-D-C-B-A-office).

b. i. [1 mark]

The vertex degrees are office 3, A 2, B 4, C 2, D 2, E 3, F 4 and G 2. Only the office and E have odd degree, so a route using every edge once must start at one and finish at the other. Joe finishes at E.

ii. [1 mark]

An Eulerian trail (it uses every edge once but starts and ends at different vertices, so it is not a circuit).

From the report. In a (76%) some students left out part of the route, which had to begin and end at the office. Part b.i was answered correctly by 82%. In b.ii (70%) some wrote only "trail", and others called it a circuit.

Question 2 (5 marks)

A project to add a new cabin has 10 activities, with durations in weeks. From the activity network: A (10) starts the project; B (3), C (7) and D (12) follow A; E (7) follows B; F (5) follows C; G (2) follows E; H (3) and I (8) both follow D and F; J (4) follows G and H; I and J end the project.

a

How many activities have two immediate predecessors? (1 mark)

b

Find the minimum completion time, in weeks. (1 mark)

c

One activity can be reduced by two weeks and another by one week, reducing the minimum completion time by three weeks. Which two activities, and which reduction for each? (1 mark)

d

A new cabin, H, is added, and an adjacency matrix (in the order office, A, B, C, D, E, F, G, H) gives the new road network. Its rows are: office 0 1 0 0 0 0 1 1 0; A 1 0 1 0 0 0 0 0 0; B 0 1 0 1 0 1 1 0 0; C 0 0 1 0 1 0 0 0 0; D 0 0 0 1 0 1 0 0 1; E 0 0 1 0 1 0 1 0 1; F 1 0 1 0 0 1 0 1 0; G 1 0 0 0 0 0 1 0 0; H 0 0 0 0 1 1 0 0 1. Add H and any new roads to the network diagram. (2 marks)

Show worked solution

a. [1 mark]. 3: H (after D and F), I (after D and F) and J (after G and H).

b. [1 mark]. The paths through the network are:

• A-B-E-G-J: $10 + 3 + 7 + 2 + 4 = 26$
• A-C-F-H-J: $10 + 7 + 5 + 3 + 4 = 29$
• A-C-F-I: $10 + 7 + 5 + 8 = 30$
• A-D-H-J: $10 + 12 + 3 + 4 = 29$
• A-D-I: $10 + 12 + 8 = 30$

The minimum completion time is the longest path: 30 weeks. There are two critical paths, A-C-F-I and A-D-I.

c. [1 mark].

Activity	Reduction in completion time
A	2 weeks
I	1 week

A and I are on both critical paths. Reducing A by 2 and I by 1 makes A-C-F-I and A-D-I 27 weeks; the other paths become 24, 27 and 27 weeks, so the new minimum completion time is 27 weeks, three weeks less. (Reducing A by 1 and I by 2 would leave A-C-F-H-J and A-D-H-J at 28 weeks, a reduction of only two weeks, so it matters which activity gets the 2 weeks.)

d. [2 marks]. Compare the matrix with the existing network. The rows for the office and cabins A to G match the old roads, apart from the new 1s in the H column: D and E. So add vertex H with an edge H-D, an edge H-E, and a loop at H (the 1 in row H, column H).

From the report. In a (49%) the most common error was 2. Part b was answered correctly by 62%. Part c (27%) was answered incorrectly by many students. Part d was well done (80%).

Question 3 (4 marks)

Stormwater pipes form a directed network from the source to the sink (R), with capacities in litres per minute (each pipe written from its start vertex to its end vertex): source-J 20, source-K 35, J-L 18, L-K 12, K-N 19, K-O 20, M-L 5, L-R 15, N-M 15, M-Q 10, N-P 14, Q-P 12, Q-R 16, O-N 8, O-P 11 and O-R 19. Cut 1 separates source, J, K, L, N and O from M, P, Q and R.

a

Find the capacity of Cut 1. (1 mark)

b

Find the maximum flow from the source to the sink. (1 mark)

c

The direction of flow is reversed in one pipe. Which pipe should be reversed to give the largest increase in flow? (1 mark)

d

Instead, the capacity of one pipe is increased. Which pipe should be increased to give the largest increase in flow, and what is the least new capacity that achieves it? (1 mark)

Show worked solution

a. [1 mark]. Count only the pipes that flow across the cut from the source side to the sink side: L-R 15, N-M 15, N-P 14, O-P 11 and O-R 19. Pipe M-L flows back towards the source side, so it is not counted.

$$15 + 15 + 14 + 11 + 19 = 74.$$

b. [1 mark]. Look for the minimum cut. Vertex P has no pipe leading out of it, so nothing that reaches P can get to the sink. A cut through L-R, M-Q and O-R (with Q and the sink on one side) has capacity

$$15 + 10 + 19 = 44,$$

(Q-P flows back across this cut, so it is not counted). A flow of 44 is achievable, for example 15 through L-R, 10 through N-M-Q-R and 19 through O-R, so the maximum flow is 44 litres per minute.

c. [1 mark]. From vertex Q to vertex P (reverse it so it runs from P to Q). Flow that reaches P from N and O can then go on through Q to the sink, and Q-R has 6 L/min of spare capacity, so the maximum flow rises to 50. Reversing any other pipe gives no more than 44.

d. [1 mark]. From vertex L to vertex R, new capacity at least 23. At most $18 + 5 = 23$ L/min can reach L (from J and from M), and L-R currently carries only 15. Raising L-R to 23 lifts the maximum flow to 52, the largest increase available from any one pipe.

From the report. Part a was answered correctly by 62%. In b (37%) many different incorrect approaches were seen. Part c was answered correctly by 44%. Part d was answered correctly by only 2%: some students named the right vertices but not the new capacity.

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. The notes under each question say where its skills sit now.

Question 1 (6 marks)

An airship's cabin floor is a 9 m by 2.5 m rectangle. The cockpit takes up a 1.5 m by 2.5 m area at the front; the rest is for passengers and cargo.

a

Find the area available for passengers and cargo, in m². (1 mark)

b

The cabin is a rectangular prism 9 m by 3 m by 2.5 m. Find its volume, in m³. (1 mark)

c

The cabin has eight identical rectangular windows, each 0.85 m high, with a total area of 14.62 m². Find the length of each window, in metres. (1 mark)

d

A 60 m cable, pulled tight, joins the airship nose to a concrete block on level ground. The nose is 55 m horizontally from the block. Find the height of the nose above the ground, to the nearest metre. (1 mark)

e

The airship body is modelled as a cylinder 60 m long and 15 m in diameter, with a hemisphere at each end. i. Find the total length of the body. (1 mark)

ii

The body is covered in fabric, except where the cabin roof is attached, which saves 22.6 m² of fabric. Find the total area of fabric needed, to the nearest m². (1 mark)

Study design: areas, volumes, surface areas and Pythagoras' theorem are now General Mathematics Unit 2 content (Space and measurement), not Units 3 and 4.

Show worked solution

a. [1 mark].

$$9 \times 2.5 - 1.5 \times 2.5 = 22.5 - 3.75 = 18.75 \text{ m}^2.$$

There is no rounding instruction, so give the exact value.

b. [1 mark]. $9 \times 3 \times 2.5 = 67.5$ m³.

c. [1 mark]. Each window has area $14.62 \div 8 = 1.8275$ m², so its length is

$$\frac{1.8275}{0.85} = 2.15 \text{ m}.$$

d. [1 mark]. The cable is the hypotenuse:

$$h = \sqrt{60^2 - 55^2} = \sqrt{575} = 23.97\ldots \approx 24 \text{ m}.$$

e. i. [1 mark]. Each hemisphere adds one radius, 7.5 m, to the length: $60 + 7.5 + 7.5 =$ 75 m.

ii. [1 mark]. The two hemispheres make one sphere of radius 7.5 m:

$$\text{surface area} = 2\pi \times 7.5 \times 60 + 4\pi \times 7.5^2 = 900\pi + 225\pi = 1125\pi = 3534.29\ldots \text{ m}^2$$

$$\text{fabric} = 3534.29\ldots - 22.6 = 3511.69\ldots \approx 3512 \text{ m}^2.$$

From the report. Parts a (87%), b (90%) and d (76%) were well answered, and in part a, a rounded value like 18.8 scored no mark because the question gave no rounding instruction. Part c was answered correctly by 56%. In e.i (58%) 67.5 was a common wrong answer. Part e.ii was answered correctly by only 26%.

Question 2 (4 marks)

a

Survey equipment flew from Frankfurt (50° N, 9° E) to Brisbane (27° S, 153° E). The flight took 23 hours, and that month the time difference between Frankfurt and Brisbane was nine hours. The flight arrived in Brisbane at 6.30 am on Thursday. At what time and on what day did it leave Frankfurt? (1 mark)

b

The airship flew from Brisbane (27° S, 153° E) to a survey site (27° S, 141° E) along a small circle. Take the radius of Earth as 6400 km. i. Show that the radius of the small circle is 5702 km, to the nearest kilometre. (1 mark)

ii

Find the distance flown, to the nearest kilometre. (1 mark)

c

Two spotlights 8 m apart on the cabin base are set at the same angle $\theta$ to the base, so that their beams meet in a single spot on the ground when the cabin base is 15 m above the ground. Find $\theta$, to the nearest degree. (1 mark)

Study design: time zones, latitude and longitude and small circles (parts a and b) are not in the current General Mathematics study design. Right-angled trigonometry (part c) is now General Mathematics Unit 2 content.

Show worked solution

a. [1 mark]. Arrival is 6.30 am Thursday, Brisbane time. The flight took 23 hours, so it left at 7.30 am Wednesday, Brisbane time. Frankfurt is nine hours behind Brisbane, so in Frankfurt it was 10.30 pm on Tuesday.

b. i. [1 mark]. The radius of the 27° S small circle is

$$r = 6400 \cos 27^\circ = 5702.44\ldots \approx 5702 \text{ km}.$$

(Equivalently, $6400 \sin 63^\circ$.)

ii. [1 mark]. The two places are $153^\circ - 141^\circ = 12^\circ$ of longitude apart:

$$\text{distance} = \frac{12}{360} \times 2\pi \times 5702.44\ldots = 1194.3\ldots \approx 1194 \text{ km}.$$

c. [1 mark]. The beams meet halfway between the spotlights, so each beam forms a right-angled triangle with a horizontal side of 4 m and a vertical side of 15 m:

$$\tan\theta = \frac{15}{4}, \qquad \theta = 75.06\ldots^\circ \approx 75^\circ.$$

From the report. Part a (27%) showed that time zone questions were not handled well. Part b.i (45%) was a "show that" question: writing an equation in $r$ to be solved was not enough, and $6400 \sin 63^\circ$ was also accepted. Part b.ii was answered correctly by 39% and part c by 41%.

Question 3 (2 marks)

a. The airship flew south along the 141° E meridian from the survey site (27° S, 141° E) to Mount Gambier (38° S, 141° E). Take the radius of Earth as 6400 km. Find the distance flown, to the nearest kilometre. (1 mark)

b. With the cabin base at ground level, a person's eye is 1.6 m above the ground, 2.0 m from a 2 m tall vertical post. The top of the post lies on their line of sight to the airship nose, which is 10.5 m above the ground. Find the horizontal distance between the person and the airship nose. (1 mark)

Study design: distances along a meridian (part a) are not in the current General Mathematics study design. Similar triangles (part b) are now General Mathematics Unit 2 content.

Show worked solution

a. [1 mark]. A meridian is a great circle of radius 6400 km, and the two places are $38^\circ - 27^\circ = 11^\circ$ apart:

$$\text{distance} = \frac{11}{360} \times 2\pi \times 6400 = 1228.7\ldots \approx 1229 \text{ km}.$$

b. [1 mark]. Measure heights from eye level. The line of sight rises $2 - 1.6 = 0.4$ m over the 2.0 m to the post, and must rise $10.5 - 1.6 = 8.9$ m to reach the nose. By similar triangles,

$$\frac{d}{2.0} = \frac{8.9}{0.4}, \qquad d = 2.0 \times 22.25 = 44.5 \text{ m}.$$

From the report. In a (42%) some students wrote 1129 km. Part b was answered correctly by only 9%: it was often not attempted or answered incorrectly.

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. The notes under each question say where its skills sit now.

Question 1 (4 marks)

A company's weekly profit, in dollars, from selling $n$ single-size bed frames is $P = 130n - 1560$.

a

Find the weekly profit if 25 bed frames are sold. (1 mark)

b

How many bed frames must be sold each week to break even? (1 mark)

c

Sketch $P$ for $0 \le n \le 40$ on the grid provided (profit axis from $-2500$ to 4500). (1 mark)

d

The bed frames sell for 350 dollars each. Complete the equation $C = \square \times n + \square$ for the weekly cost of producing $n$ bed frames. (1 mark)

Study design: linear models and break-even analysis are now General Mathematics Unit 1 content (linear graphs and modelling), not Units 3 and 4.

Show worked solution

a. [1 mark]

$P = 130 \times 25 - 1560 = 1690$, so $1690.

b. [1 mark]

$130n - 1560 = 0$ gives $n = \dfrac{1560}{130} =$ 12 bed frames.

c. [1 mark]

Work out the endpoints and join them with a ruler:

• at $n = 0$: $P = -1560$
• at $n = 40$: $P = 130 \times 40 - 1560 = 3640$

Draw a straight line from $(0, -1560)$ to $(40, 3640)$. It crosses the $n$-axis at $n = 12$.

d. [1 mark]. Profit is revenue minus cost, so $C = 350n - (130n - 1560)$:

$$C = 220 \times n + 1560.$$

From the report. Part a was well answered (95%). In b (66%) a number of students gave 24. In c (40%) the most accurate graphs came from students who first wrote down the endpoint coordinates. Part d was answered correctly by only 22%.

Question 2 (4 marks)

The company buys single-size mattresses for 480 dollars each and queen-size mattresses for 1200 dollars each.

a

Find the cost of 10 single-size and seven queen-size mattresses. (1 mark)

b

A recent order of 25 mattresses cost 18 480 dollars. How many queen-size mattresses were ordered? (1 mark)

c

Under a deal with the supplier, the total cost of $s$ single-size mattresses is $T = 480s$ for $0 < s \le 20$ and $T = 420s + p$ for $s \ge 20$. Find $p$. (1 mark)

d

The springs compress 32 mm under a force of 500 N. Using $F = kx$, with $F$ in newtons and $x$ in mm, find the compression under a force of 800 N. (1 mark)

Study design: simultaneous linear equations and piecewise (step and line segment) graphs are now General Mathematics Unit 1 content, and direct variation (part d) is Unit 2 content, not Units 3 and 4.

Show worked solution

a. [1 mark]. $10 \times 480 + 7 \times 1200 = 4800 + 8400 = 13\,200$, so $13 200.

b. [1 mark]. Let $q$ be the number of queen-size mattresses, so $25 - q$ are single-size:

$$480(25 - q) + 1200q = 18\,480 \implies 12\,000 + 720q = 18\,480 \implies q = 9.$$

9 queen-size mattresses were ordered. (Check: $16 \times 480 + 9 \times 1200 = 7680 + 10\,800 = 18\,480$.)

c. [1 mark]. The two rules must agree at $s = 20$:

$$420 \times 20 + p = 480 \times 20 \implies 8400 + p = 9600 \implies p = 1200.$$

d. [1 mark]. $k = \dfrac{500}{32} = 15.625$, so

$$x = \frac{800}{15.625} = 51.2 \text{ mm}.$$

(Or by proportion: $32 \times \dfrac{800}{500} = 51.2$.)

From the report. Part a was well answered (87%). Part b was answered correctly by 56% and part c by only 24%. In d (44%) some students found $k$ but could not go further.

Question 3 (4 marks)

The company makes $x$ queen-size and $y$ king-size bed frames each week. A queen-size frame takes three hours to make and one hour to paint; a king-size frame takes four hours to make and two hours to paint. The constraints are

$$x \ge 20, \qquad y \ge 10, \qquad 3x + 4y \le 190 \text{ (framing)}, \qquad x + 2y \le 76 \text{ (painting)},$$

and a graph shows the shaded feasible region, whose corners (read from the graph, not labelled on it) are at (20, 10), (20, 28), (38, 19) and (50, 10).

a

What is the minimum number of king-size bed frames produced each week? (1 mark)

b

In a week when 15 king-size frames are produced, what is the maximum number of queen-size frames? (1 mark)

c

The profit is 210 dollars on each queen-size frame and 350 dollars on each king-size frame. Find the maximum weekly profit. (1 mark)

d

The profit on a queen-size frame is reduced. Below what value must it fall so that the maximum weekly profit occurs only at the point (20, 28)? (1 mark)

Study design: linear programming (feasible regions and objective functions) is not in the current General Mathematics study design.

Show worked solution

a. [1 mark]. Inequality 2 says $y \ge 10$: 10 king-size frames.

b. [1 mark]. With $y = 15$:

• framing: $3x + 60 \le 190$ gives $x \le 43.3\ldots$
• painting: $x + 30 \le 76$ gives $x \le 46$

Framing is the tighter limit, and $x$ must be a whole number, so at most 43 queen-size frames.

c. [1 mark]. The framing and painting lines meet where $3x + 4y = 190$ and $x + 2y = 76$: subtracting twice the second equation from the first gives $x = 38$, then $y = 19$. Test the corners with $\text{profit} = 210x + 350y$:

Corner	Profit (dollars)
(20, 10)	7700
(20, 28)	14 000
(38, 19)	14 630
(50, 10)	14 000

The maximum weekly profit is $14 630, from 38 queen-size and 19 king-size frames.

d. [1 mark]. Let the queen-size profit be $a$ dollars, so $\text{profit} = ax + 350y$. The point (20, 28) is where $x = 20$ meets the painting line $x + 2y = 76$, which has gradient $-\tfrac{1}{2}$. The maximum is only at (20, 28) when the profit line is less steep than the painting line:

$$-\frac{a}{350} > -\frac{1}{2} \implies a < 175.$$

The profit must fall below $175. (At exactly 175 dollars, every point on the painting edge from (20, 28) to (38, 19) gives the same profit.)

From the report. Part a was answered correctly by 66% and part b by 52%. In c (37%) a reasonable number of students scored the mark. Part d was answered correctly by only 12%.

General advice from the 2022 report

• Answer every part of the question. In core Question 2a.i many students gave both IQRs but never said which time was more variable, and in Question 4b only strength and direction were wanted, with no extra comments.
• Round only when told to. With no rounding instruction an exact answer is needed (18.75 m² in Geometry Question 1a, not 18.8), and in financial questions round to the nearest dollar or cent only when asked.
• Show working on questions worth more than one mark, so a method mark is possible: in Matrices Question 4b the correct state matrix earned a mark even without a clear conclusion.
• In "show that" questions, set out every step from the given information; you cannot use the value you are trying to show. Show recursive calculations step by step when asked (core Question 8a).
• Copy calculator results carefully. Transcription errors, including an extra zero in financial answers, were common.
• Bring a ruler for drawing straight lines in Data analysis and Graphs and relations.