VCE General Mathematics 2021 Exam 2

Every question from the 2021 VCE Further Mathematics Examination 2, the extended-response paper. Further Mathematics was the predecessor of General Mathematics: 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 2021 Examination 1 walkthrough.

How to use this page

• Questions are from the 2021 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 2021 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 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 9) and in the Matrices and Networks and decision mathematics modules is still on the current Units 3 and 4 course. 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 (with its inequalities) 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 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 9, 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 (44.8%), followed by Networks and decision mathematics (28%), Graphs and relations (13.7%) and Geometry and measurement (12%). In General Mathematics there is no choice: Matrices and Networks and decision mathematics are both compulsory.

Data analysis

Question 1 (7 marks)

Table 1 gives the results, in metres, of 15 heptathletes (athlete numbers 1 to 15) in three events. High jump: 1.76, 1.79, 1.83, 1.82, 1.87, 1.73, 1.68, 1.82, 1.83, 1.87, 1.87, 1.80, 1.83, 1.87, 1.78. Shot-put: 15.34, 16.96, 13.87, 14.23, 13.78, 14.50, 15.08, 13.13, 14.22, 13.62, 12.01, 12.88, 12.68, 12.45, 11.31. Javelin: 41.22, 42.41, 46.53, 40.62, 45.64, 42.33, 40.88, 39.22, 42.51, 42.75, 38.12, 42.65, 45.68, 41.32, 42.88.

a

Write down the number of numerical variables in Table 1. (1 mark)

b

Table 2 gives the high jump standard deviation (0.06 m) and the shot-put mean (13.74 m) and standard deviation (1.43 m). Find the mean high jump. (1 mark)

c

Athlete 6, Jamilia, threw the shot 14.50 m. Find her standardised score, to one decimal place. (1 mark)

d

High jump heights are expected to be approximately normal. Chara's jump gives $z = -1.0$. Use the 68-95-99.7% rule to find the percentage of athletes expected to jump higher than Chara. (1 mark)

e

A boxplot of the 15 high jump heights has no whisker at its upper end. Explain why. (1 mark)

f

On a boxplot of the javelin results, an athlete shown as an outlier at the upper end is a potential medal winner. What is the minimum distance an athlete needs to throw? (2 marks)

Show worked solution

a. [1 mark]. 3: high jump, shot-put and javelin. The athlete number is only a label.

b. [1 mark]. The 15 heights add to 27.15, so the mean is

$$\bar{x} = \frac{27.15}{15} = 1.81 \text{ m}.$$

c. [1 mark].

$$z = \frac{14.50 - 13.74}{1.43} = 0.531\ldots \approx 0.5.$$

d. [1 mark]

68% of heights lie within one standard deviation of the mean, so 16% lie below $z = -1$. Everyone else jumps higher: $100\% - 16\% = 84\%$ (that is, $34\% + 50\%$).

e. [1 mark]

In order, the heights are 1.68, 1.73, 1.76, 1.78, 1.79, 1.80, 1.82, 1.82, 1.83, 1.83, 1.83, 1.87, 1.87, 1.87, 1.87. With 15 values, $Q_3$ is the 12th value, 1.87, which is also the largest value. The maximum equals $Q_3$, so the upper whisker has zero length.

f. [2 marks]

In order, the javelin results are 38.12, 39.22, 40.62, 40.88, 41.22, 41.32, 42.33, 42.41, 42.51, 42.65, 42.75, 42.88, 45.64, 45.68, 46.53. $Q_1$ is the 4th value, 40.88, and $Q_3$ is the 12th, 42.88, so $\text{IQR} = 2.00$.

$$\text{upper fence} = Q_3 + 1.5 \times \text{IQR} = 42.88 + 1.5 \times 2.00 = 45.88.$$

An outlier must be greater than 45.88 m. The results are recorded to two decimal places, so the minimum throw is 45.89 m.

From the report. Part a (52%) was not answered well: a common error was 45, the number of values rather than variables. In d (60%) the most common wrong answer was 16%. In e (47%) many said the maximum was 1.87 without saying it was also $Q_3$, and some confused the upper fence with the end of the box. Part f was answered fully by only 9%: most students who found the right quartiles gave 45.88 m, missing that a potential medal winner must throw further than the fence.

Question 2 (3 marks)

The times, in seconds, for the 200 m and 800 m runs, time200 and time800, are linearly related. The least squares line is time800 $= 0.03931 + 5.2756 \times$ time200.

a. Round the intercept and the slope to three significant figures. (1 mark)

b. The means are 24.6492 s (time200) and 136.054 s (time800), and the standard deviations are 0.96956 s and 8.2910 s. Find the coefficient of determination as a percentage, to the nearest percentage. (2 marks)

Show worked solution

a. [1 mark]. time800 $= 0.0393 + 5.28 \times$ time200. (Three significant figures of 0.03931 is 0.0393, not 0.039.)

b. [2 marks]. Rearrange $b = r\dfrac{s_y}{s_x}$, with time200 as $x$ and time800 as $y$:

$$r = b \times \frac{s_x}{s_y} = 5.2756 \times \frac{0.96956}{8.2910} = 0.6169\ldots$$

$$r^2 = 0.3806\ldots \approx 38\%.$$

The means are not needed.

From the report. In part a (52%) many students rounded to three decimal places instead of three significant figures. Part b was answered fully by only 29%: many struggled to find the correlation coefficient from the summary statistics, and some did not attempt it.

Question 3 (6 marks)

A time series plot shows the winning time, in seconds, of the women's 100 m freestyle at each Olympic Games from 1956 to 2016, falling from about 62 s to about 53 s. The least squares line is winning time $= 357.1 - 0.1515 \times$ year, and the coefficient of determination is 0.8794.

a

Name the explanatory variable. (1 mark)

b

Find the correlation coefficient $r$, to three decimal places. (1 mark)

c

Write down the average decrease in winning time, in seconds per year. (1 mark)

d

The predicted winning time for 2000 was 54.10 s and the actual time was 53.83 s. Find the residual. (1 mark)

e. i

Show that the predicted winning time in 2032 is 49.252 s. (1 mark)

ii

What assumption is made when the equation is used to predict the winning time in 2032? (1 mark)

Show worked solution

a. [1 mark]. year. In a time series plot, time is the explanatory variable.

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

$$r = -\sqrt{0.8794} = -0.9377\ldots \approx -0.938.$$

c. [1 mark]

0.1515 seconds per year, the slope of the line. No calculation is needed.

d. [1 mark]

Residual $=$ actual $-$ predicted $= 53.83 - 54.10 =$ $-0.27$ s.

e. i. [1 mark]

Show the substitution:

$$\text{winning time} = 357.1 - 0.1515 \times 2032 = 357.1 - 307.848 = 49.252 \text{ s}.$$

ii. [1 mark]. That the same decreasing trend continues into the future, up to 2032.

From the report. Part b was answered correctly by only 19%: many found 0.938 but left off the negative sign. In c (32%) "write down" meant no calculation was needed; estimating the slope from two points on the plot gave the less accurate 0.15. In d (68%) some left out the negative sign. Part e.ii (18%) was poorly answered: many mentioned extrapolation or linearity without linking it to the trend in the equation continuing.

Question 4 (5 marks)

A time series plot shows the men's and women's 100 m freestyle Olympic winning times from 1912 to 2016, both decreasing, with the year axis running from 1908 to 2020. The men's least squares line, winning time men $= 356.9 - 0.1544 \times$ year, is already drawn. For women, winning time women $= 538.9 - 0.2430 \times$ year.

a

Draw the least squares line for the women on the plot. (1 mark)

b

Using difference $=$ winning time women $-$ winning time men, find the predicted difference for the 2024 Games, to one decimal place. (2 marks)

c

The Games are held every four years: 2024, 2028, 2032 and so on. In which Olympic year do the two lines first predict that the women's winning time will be faster than the men's? (2 marks)

Show worked solution

a. [1 mark]. Calculate two points far apart and join them with a ruler:

• at 1908: $538.9 - 0.2430 \times 1908 = 75.256$
• at 2020: $538.9 - 0.2430 \times 2020 = 48.04$

Plot $(1908, 75.256)$ and $(2020, 48.04)$ and rule the line across the whole plot.

b. [2 marks].

$$\text{women: } 538.9 - 0.2430 \times 2024 = 47.068, \qquad \text{men: } 356.9 - 0.1544 \times 2024 = 44.3944$$

$$\text{difference} = 47.068 - 44.3944 = 2.6736 \approx 2.7 \text{ s}.$$

c. [2 marks]. The women are faster when their predicted time is smaller:

$$538.9 - 0.2430 \times \text{year} < 356.9 - 0.1544 \times \text{year} \implies 182 < 0.0886 \times \text{year} \implies \text{year} > 2054.176\ldots$$

The first Olympic year after 2054.176 is 2056. Check: in 2052 the predictions are 40.264 s (women) and 40.071 s (men), and in 2056 they are 39.292 s and 39.454 s.

From the report. In part a (46%) the line had to be positioned correctly; the best way is to calculate the endpoints (1908, 75.256) and (2020, 48.04) and join them with a ruler, and many students left the question out. Part b (67%) was generally well answered, but rounding too early was common. In c (40%) only a small proportion solved the two equations, and many substituted Olympic years one by one. A method mark was available for finding about 2054.176 even without the final step to 2056.

Question 5 (3 marks)

The table gives difference (seconds) at each Games: 1912 18.8, 1920 12.2, 1924 13.4, 1928 12.4, 1932 8.6, 1936 8.3, 1948 9.0, 1952 9.4, 1956 6.6, 1960 6.0, 1964 6.1, 1968 7.8, 1972 7.4, 1976 5.7, 1980 4.4, 1984 6.1, 1988 6.3, 1992 5.6, 1996 5.8, 2000 5.5, 2004 5.7, 2008 5.9, 2012 5.5, 2016 5.1. (No Games were held in 1916, 1940 and 1944.) Its time series plot is clearly non-linear.

a. Apply a reciprocal transformation to difference, fit a least squares line to the transformed data and write its equation, with the intercept and slope to four significant figures. (2 marks)

b. Use the equation to predict the difference in 2032, to one decimal place. (1 mark)

Show worked solution

a. [2 marks]. Enter the 24 years as $x$ and $\dfrac{1}{\text{difference}}$ as $y$, then run linear regression: $a = -2.2342\ldots$ and $b = 0.0012088\ldots$ To four significant figures,

$$\frac{1}{\text{difference}} = -2.234 + 0.001209 \times \text{year}.$$

b. [1 mark].

$$\frac{1}{\text{difference}} = -2.234 + 0.001209 \times 2032 = 0.222688, \qquad \text{difference} = \frac{1}{0.222688} = 4.49\ldots \approx 4.5 \text{ s}.$$

The unrounded coefficients also give 4.5 s.

From the report. Only 16% scored both marks in part a: many students who used the correct transformation rounded incorrectly or wrote the variables incorrectly. Part b was answered correctly by only 14%; a common answer was 0.2, from using the right coefficients but forgetting to take the reciprocal at the end.

Recursion and financial modelling

Question 6 (3 marks)

Sienna invests $420 000 in a perpetuity earning 5.4% per annum.

The perpetuity pays her a regular monthly payment of $1890.

a

Find the total she receives from one year of monthly payments. (1 mark)

b

Write down the value of the perpetuity after one year of payments. (1 mark)

c

With $S_n$ the value after $n$ months, complete $S_0 = \square$, $S_{n+1} = \square \times S_n - 1890$. (1 mark)

Show worked solution

a. [1 mark]

$12 \times 1890 = 22\,680$, so $22 680.

b. [1 mark]

$420 000. A perpetuity pays out exactly the interest earned each month ($0.0045 \times 420\,000 = 1890$), so its value never changes.

c. [1 mark]

The monthly rate is $5.4\% \div 12 = 0.45\%$, so

$$S_0 = 420\,000, \qquad S_{n+1} = 1.0045 \times S_n - 1890.$$

From the report. Part a (59%) was often answered incorrectly. Part b (47%) showed that the definition of a perpetuity, a value that does not change over time, is still misunderstood. In c (38%) 1.054 was a common wrong factor.

Question 7 (3 marks)

Sienna's coffee machine cost $12 000 and is depreciated by the unit cost method.

The rate of depreciation is $0.05 per cup of coffee made.

The year-to-year value, in dollars, is modelled by $M_0 = 12\,000$, $M_{n+1} = M_n - 1440$.

a

How many cups of coffee does the machine make per year? (1 mark)

b

The same recurrence relation could represent flat rate depreciation. What annual flat rate percentage does it represent? (1 mark)

c

Complete the rule $M_n = \square + \square \times n$ for the value of the machine after $n$ cups have been made. (1 mark)

Show worked solution

a. [1 mark].

$$\frac{1440}{0.05} = 28\,800 \text{ cups}.$$

b. [1 mark].

$$\frac{1440}{12\,000} \times 100\% = 12\%.$$

c. [1 mark]. Here $n$ counts cups, not years, and each cup takes 0.05 dollars off the value:

$$M_n = 12\,000 + (-0.05) \times n, \quad \text{that is} \quad M_n = 12\,000 - 0.05n.$$

From the report. Parts a (54%) and b (49%) were each answered correctly by about half the students. Part c was answered correctly by only 21%: many did not recognise that the rule gives the value in terms of the number of cups made, not the number of years.

Question 8 (3 marks)

Sienna takes out a reducing balance loan of $570 000, with interest calculated fortnightly.

The balance after $n$ fortnights is modelled by $S_0 = 570\,000$, $S_{n+1} = 1.001S_n - 1193$.

a

Find the balance of the loan after the first repayment. (1 mark)

b

Show that the compound interest rate is 2.6% per annum. (1 mark)

c

For the loan to be fully repaid, to the nearest cent, the final repayment will be a larger amount. Find this final repayment, to the nearest cent. (1 mark)

Show worked solution

a. [1 mark]. $1.001 \times 570\,000 - 1193 = 570\,570 - 1193 = 569\,377$, so $569 377.

b. [1 mark]. The fortnightly rate is $1.001 - 1 = 0.001$, and there are 26 fortnights in a year:

$$(1.001 - 1) \times 26 \times 100\% = 2.6\%.$$

c. [1 mark]. Finance Solver: $I\% = 2.6$, $PV = 570\,000$, $PMT = -1193$, $FV = 0$, 26 payments per year gives $N = 650.004\ldots$ So there are 650 repayments, the last one slightly larger.

Now set $N = 650$ and solve for the future value: $FV = -5.59\ldots$ This means 650 regular repayments would leave $5.59 still owing, so the final repayment is

$$1193 + 5.59 = 1198.59.$$

The final repayment is $1198.59. (Equivalently, the balance before the last repayment is 1197.39, and one more fortnight of interest makes it $1197.39 \times 1.001 = 1198.59$.)

From the report. In b (27%) the working had to lead to 2.6%: students who worked back from 2.6% to the factor 1.001 could not get the mark, and a response written in CAS syntax (an equation in $r$ followed by "Solve") cannot earn full marks. Part c was answered correctly by only 15%; a common wrong answer was $1197.39, the balance before the last repayment without the interest added for that final fortnight.

Question 9 (3 marks)

Sienna invests $152 431 in an annuity earning 5.1% per annum, compounding monthly.

The annuity pays her $900 a month for 25 years. Let $V_n$ be its balance after $n$ monthly payments.

a. Showing recursive calculations, find the value of the annuity after two months, to the nearest cent. (2 marks)

b. After two years the interest rate will fall to 4.6% per annum. To keep receiving the same number of monthly payments, Sienna will add a one-off amount to the annuity at that time. Find this amount, to the nearest cent. (1 mark)

Show worked solution

a. [2 marks]. The monthly rate is $5.1\% \div 12 = 0.425\%$, so $V_0 = 152\,431$ and $V_{n+1} = 1.00425V_n - 900$:

$$V_1 = 1.00425 \times 152\,431 - 900 = 152\,178.83$$

$$V_2 = 1.00425 \times 152\,178.83 - 900 = 151\,925.59$$

The value after two months is $151 925.59.

b. [1 mark]. First find the balance after two years (24 payments) at 5.1%. Finance Solver: $N = 24$, $I\% = 5.1$, $PV = -152\,431$, $PMT = 900$, 12 payments per year gives $FV = 146\,073.74$.

The payments run for $25 \times 12 = 300$ months, so 276 remain. Find the balance needed to fund them at 4.6%: $N = 276$, $I\% = 4.6$, $PMT = 900$, $FV = 0$ gives $PV = -153\,112.94$, so $153 112.94 is needed.

$$153\,112.94 - 146\,073.74 = 7039.20$$

Sienna must add $7039.20.

From the report. Only 21% scored both marks in part a: many found the factor 1.00425 but then added the $900, or did not show the recursive calculations in full. Part b was answered correctly by only 8%, and a nearest-cent answer needs two decimal places: 7039.2 could not be awarded the mark.

Module 1: Matrices

Question 1 (2 marks)

Elena imports three brands of olive oil: Carmani (C), Linelli (L) and Ohana (O). January 2021 sales of 1 litre bottles are $J = \begin{bmatrix} 2800 & 1700 & 2400 \end{bmatrix}^T$ (C, L, O).

a. What is the order of matrix $J$? (1 mark)

b. Elena expects February sales of all three brands to rise by 5%, and multiplies $J$ by a scalar $k$. Write down $k$. (1 mark)

Show worked solution

a. [1 mark]. Three rows and one column: $3 \times 1$.

b. [1 mark]. A 5% increase multiplies each element by $k =$ 1.05.

From the report. Part a was answered well (93%). Part b (48%) was not: 0.05 was a common answer.

Question 2 (3 marks)

Matrix $M$ shows which of Alex (A), Brie (B), Chai (C), Dex (D) and Elena (E) can send information to whom (rows sender, columns receiver, in the order A, B, C, D, E): row A 0, 1, 0, 0, 1; row B 0, 0, 1, 1, 0; row C 1, 0, 0, 1, 0; row D 0, 1, 0, 0, 0; row E 0, 0, 0, 1, 0.

a

Which two staff members can send information directly to each other? (1 mark)

b

Elena needs to send documents to Chai. What sequence of communication links will get them there? (1 mark)

c

The printed $M^2$, the number of two-step links, has rows A 0, 0, 1, 2, 0; B 0, 1, 0, 1, 0; C 0, 1, 0, 0, 0; D 0, 0, 1, 1, 0; E 0, 1, 0, 0, 0. Only one pair of individuals has two different two-step links. List each two-step link for this pair. (1 mark)

Show worked solution

a. [1 mark]

Brie and Dex: row B, column D and row D, column B are both 1. No other pair has a 1 in both directions.

b. [1 mark]

Elena can send only to Dex, Dex only to Brie, and Brie can send to Chai: Elena, Dex, Brie, Chai.

c. [1 mark]

The only 2 in $M^2$ is in row A, column D. Alex sends to Brie and to Elena, and both of them send to Dex: Alex-Brie-Dex and Alex-Elena-Dex.

(If you square the printed $M$ on CAS, you will not get the printed $M^2$: the 1 in row C, column A (Chai can send to Alex) adds the two-step links Brie-Chai-Alex, Chai-Alex-Brie and Chai-Alex-Elena, so Chai to Brie would also have two links (Chai-Alex-Brie and Chai-Dex-Brie). The printed $M^2$ is the square of $M$ without that 1. The question gives $M^2$ and says only one pair has two links, so answer from the printed $M^2$, as the report does. Parts a and b are not affected.)

From the report. Parts a (92%) and b (83%) were answered well. Part c (21%) was not: most students appeared not to interpret the question correctly.

Question 3 (5 marks)

Each month shoppers change brand according to the transition matrix $T$ (columns this month C, L, O; rows next month C, L, O):

$$T = \begin{bmatrix} 0.85 & 0.10 & 0.05 \\ 0.05 & 0.80 & 0.05 \\ 0.10 & 0.10 & 0.90 \end{bmatrix}.$$

In July 2021, $S_0 = \begin{bmatrix} 3200 & 2000 & 2800 \end{bmatrix}^T$ shoppers bought C, L and O, and $S_n$ gives the numbers $n$ months after July, with $S_{n+1} = TS_n$.

a

How many of these 8000 shoppers bought a different brand in August 2021 from the brand they bought in July? (1 mark)

b

Complete $S_1$ (its first element is 3060). (1 mark)

c

Of the shoppers expected to buy Carmani in August, what percentage also bought Carmani in July, to the nearest percentage? (1 mark)

d

Write a calculation showing that Ohana is the brand bought by 50% of these shoppers in the long run. (1 mark)

e

A new model is $R_{n+1} = TR_n + B$, with $R_0 = S_0$ and $B = \begin{bmatrix} 200 & 100 & k \end{bmatrix}^T$, where $k$ is the extra number of shoppers buying Ohana each month. If $R_2 = \begin{bmatrix} 3333 & 2025 & 3642 \end{bmatrix}^T$, find $k$. (1 mark)

Show worked solution

a. [1 mark]. The leading diagonal gives the shoppers who stay with their brand, so the rest switch:

$$0.15 \times 3200 + 0.20 \times 2000 + 0.10 \times 2800 = 480 + 400 + 280 = 1160.$$

b. [1 mark].

$$S_1 = TS_0 = \begin{bmatrix} 3060 \\ 1900 \\ 3040 \end{bmatrix}.$$

For example, the L element is $0.05 \times 3200 + 0.80 \times 2000 + 0.05 \times 2800 = 1900$.

c. [1 mark]. Of the 3060 August Carmani buyers, $0.85 \times 3200 = 2720$ had also bought Carmani in July:

$$\frac{2720}{3060} \times 100\% = 88.9\ldots\% \approx 89\%.$$

d. [1 mark]. Raise $T$ to a large power, such as $n = 50$ (any $n \ge 35$ works):

$$T^{n} \times S_0 \approx \begin{bmatrix} 2400 \\ 1600 \\ 4000 \end{bmatrix}, \qquad \frac{4000}{8000} = 50\%.$$

e. [1 mark]. First, $R_1 = TR_0 + B$, which has elements 3260, 2000 and $3040 + k$. The Ohana element of $R_2 = TR_1 + B$ is

$$0.10 \times 3260 + 0.10 \times 2000 + 0.90(3040 + k) + k = 3262 + 1.9k.$$

Setting $3262 + 1.9k = 3642$ gives $1.9k = 380$, so $k =$ 200. (Check with the Carmani element: $3323 + 0.05k = 3333$ also gives $k = 200$.)

From the report. Part a (28%) was not answered well: many students made little progress. Part b (78%) was answered well. In c (17%) 85%, taken straight from the transition matrix, was a very common wrong answer. In d (22%) many found a long-term matrix but were unsure how to use it to answer the question. Part e was answered correctly by only 15%.

Question 4 (2 marks)

Five staff, Ike (I), Joelene (J), Katie (K), Leslie (L) and Mikki (M), played a round-robin video game tournament: each played every other player once, and every game had a winner. Their one-step and two-step dominances are I 3 and 5, J 3 and 4, K 1 and 1, L 1 and 2, and M 2 and 4. In the results matrix a 1 means the row player beat the column player. Katie's row is 0, 0, 0, 1, 0 (she beat only Leslie). Use all of the information to complete the results matrix. (2 marks)

Show worked solution

[2 marks]. Work from the players with the fewest wins.

• Katie beat only Leslie, so Ike, Joelene and Mikki all beat Katie.
• Leslie won one game and has two-step dominance 2, so the player she beat has exactly 2 wins. Only Mikki has 2 wins: Leslie beat Mikki, and lost to Ike and Joelene.
• So far Ike has beaten Katie and Leslie, Joelene has beaten Katie and Leslie, and Mikki has beaten Katie. Each of them needs one more win from the games Ike v Joelene, Ike v Mikki and Joelene v Mikki.
• If Mikki had beaten Joelene, then Joelene beat Ike and Ike beat Mikki, and Ike's two-step dominance would be $1 + 1 + 2 = 4$, not 5. So Mikki beat Ike, Ike beat Joelene and Joelene beat Mikki.

Winner \ loser	I	J	K	L	M
I	0	1	1	1	0
J	0	0	1	1	1
K	0	0	0	1	0
L	0	0	0	0	1
M	1	0	1	0	0

Check the two-step dominances: Ike beat J, K and L, who have $3 + 1 + 1 = 5$ wins; Joelene beat K, L and M ($1 + 1 + 2 = 4$); Mikki beat I and K ($3 + 1 = 4$); Leslie beat M (2); Katie beat L (1).

From the report. Only 10% completed the whole matrix correctly (20% scored one mark).

Module 2: Networks and decision mathematics

Question 1 (4 marks)

Maggie's house has five rooms, A to E, and eight doors; F is the outside area. On the floor plan, room A runs across the top, with two doors to the outside and one door down into B. B and E form the middle row, and C and D the bottom row. The other doors join B to C, B to E, C to D, D to E, and C to the outside. A graph of the plan (vertices for the rooms and F, edges for the doors) shows A-F twice, C-F, B-C, B-E, C-D and D-E: one edge is missing.

a

Draw the missing edge on the graph. (1 mark)

b

What is the degree of vertex E? (1 mark)

c

A cleaner enters from F, walks through each room only once, cleaning as he goes, and finishes outside at F. i. Complete one possible route F-?-?-?-?-?-F. (1 mark)

ii

What is the mathematical term for such a journey? (1 mark)

Show worked solution

a. [1 mark]

The door between rooms A and B has no edge, so draw an edge joining A and B.

b. [1 mark]

E has doors to B and D only: degree 2.

c. i. [1 mark]

F-A-B-E-D-C-F, or the reverse, F-C-D-E-B-A-F.

ii. [1 mark]

A Hamiltonian cycle: every vertex is visited once, and the route returns to its start.

From the report. Parts a (93%), b (88%) and c.i (92%) were answered well. In c.ii (60%) some students wrote just "cycle", which was not accepted.

Question 2 (2 marks)

Main roads join the towns G to O, with distances in kilometres: G-H 26, H-I 20, I-K 35, K-L 25, L-M 30, G-J 30, J-K 15, K-N 44, J-N 50, J-O 32, G-O 28, O-N 42 and N-M 16. George lives in town G and Maggie in town M.

a. What is the shortest distance between town G and town M? (1 mark)

b. George will pass through all the towns, taking the shortest route possible. Which town will he pass through twice? (1 mark)

Show worked solution

a. [1 mark]. 86 km, along G-O-N-M ($28 + 42 + 16$). Other routes are longer, such as G-J-N-M (96 km) and G-J-K-L-M (100 km).

b. [1 mark]. K. No route can visit every town exactly once: H and I have only two roads each, which forces the route to start G-H-I-K, and L has only roads to K and M, which forces it to end K-L-M. That leaves no way to reach J, N and O, so a town must be repeated. The shortest route is

$$\text{G-H-I-K-L-K-J-O-N-M} = 26 + 20 + 35 + 25 + 25 + 15 + 32 + 42 + 16 = 236 \text{ km},$$

which passes through K twice.

From the report. Part a (85%) was answered well, although a small number of students wrote the route without giving the distance asked for. Part b (65%) was answered reasonably well.

Question 3 (3 marks)

A directed network shows the roads of Town M and their capacities, in vehicles per hour: entrance to A 1500; A to B 700; B to A 720; A to D 650; D to A 600; B to C 680; C to B 710; D to C 300; C to E 320; C to F 620; D to E 740; E to D 710; E to F 840; F to exit 1500.

a

Find the maximum number of vehicles per hour that can travel from the entrance to the exit. (1 mark)

b

The council will increase the capacity of one road only, to increase this flow. i. Which road should be increased (from which vertex to which vertex)? (1 mark)

ii

What is the minimum capacity of this road that maximises the flow? (1 mark)

Show worked solution

a. [1 mark]. Look for the minimum cut. The cut separating the entrance, A and B from everything else crosses A to D (650) and B to C (680). The road from B to A flows back towards the entrance side, so it is not counted:

$$650 + 680 = 1330.$$

A flow of 1330 is possible (680 along A-B-C and 650 along A-D, then on through E and F), so the maximum flow is 1330 vehicles per hour.

b. i. [1 mark]. From A to D. Raising B to C alone gains at most 20 vehicles, because only 700 can reach B from A. Raising A to D lifts the flow until another cut takes over.

ii. [1 mark]. The smallest cut that does not use A to D is the one into F: $620 + 840 = 1460$, so the flow cannot pass 1460. With A to D at capacity $x$, the original cut carries $680 + x$, so

$$680 + x = 1460 \implies x = 780.$$

The minimum capacity is 780 vehicles per hour.

From the report. Part a (24%) was not answered well: many students did not show that they had tried to find the minimum cut. Part b.i was answered correctly by 38%, and b.ii by only 6%.

Question 4 (3 marks)

A roadworks project has 13 activities (durations in weeks). A (6) and B (3) start the project. C (2) and F (5) follow B. D (3) and E (5) follow A and C. G (4) follows D. H (2), I (2) and J (3) follow E. K (3) follows F and J. L (5) follows G and H. M (1) follows I and L. K and M finish the project.

a

What is the earliest start time of activity K? (1 mark)

b

How many activities have zero float time? (1 mark)

c

Activities A, E, F, L and K can each be shortened by up to two weeks, at weekly costs (in dollars) of A 140 000, E 100 000, F 100 000, L 120 000 and K 80 000. The completion time can be reduced to 16 weeks. What is the minimum cost of this change? (1 mark)

Show worked solution

a. [1 mark]

Forward scan: C and F can start at 3; D and E at $\max(6, 3 + 2) = 6$; H, I and J at $6 + 5 = 11$. K needs F (finishes at $3 + 5 = 8$) and J (finishes at $11 + 3 = 14$), so its earliest start time is 14 weeks.

b. [1 mark]

Continue the scan: G starts at 9, L at $\max(9 + 4, 11 + 2) = 13$ and M at $\max(13 + 5, 11 + 2) = 18$, so the project takes $\max(18 + 1, 14 + 3) = 19$ weeks. There are two critical paths, A-D-G-L-M and A-E-H-L-M, and the zero-float activities are A, D, E, G, H, L and M: 7.

c. [1 mark]

List the paths through the network:

Path	Weeks
A-D-G-L-M	19
A-E-H-L-M	19
B-C-D-G-L-M	18
B-C-E-H-L-M	18
A-E-J-K	17
B-C-E-J-K	16
A-E-I-M	14
B-C-E-I-M	13
B-F-K	11

Every path must be 16 weeks or less.

• B-C-D-G-L-M (18) can be shortened only through L, so L must be cut by 2 weeks: $2 \times 120\,000 = 240\,000$.
• A-D-G-L-M is then 17 weeks and, with L at its limit, can be shortened only through A: cut A by 1 week, costing $140\,000$.

Now A-D-G-L-M, A-E-H-L-M, B-C-D-G-L-M, B-C-E-H-L-M, A-E-J-K and B-C-E-J-K are all 16 weeks, and the rest are shorter. Both reductions were forced, so this is the cheapest way: $240\,000 + 140\,000 = 380\,000$, a minimum cost of $380 000.

From the report. Part a (46%) was answered reasonably well. In b (22%) many students did not realise there were two critical paths. Part c was answered correctly by only 9%: few recognised the need to reduce A by one week and L by two weeks.

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 (4 marks)

A squash ball is a sphere of radius 2 cm.

a. Show that the volume of one ball, to two decimal places, is 33.51 cm³. (1 mark)

Each ball is sold in a cube-shaped box of side 4.1 cm.

b

Find the empty space around the ball in the box, in cubic centimetres, to two decimal places. (1 mark)

c

Find the total surface area of one box, in square centimetres. (1 mark)

d

Shops store the boxes in a display space 17.0 cm long, 12.5 cm wide and 8.5 cm high (a shelf sits above it). What is the maximum number of boxes that fit in the space? (1 mark)

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

Show worked solution

a. [1 mark].

$$V = \frac43\pi r^3 = \frac43 \times \pi \times 2^3 = 33.510\ldots \approx 33.51 \text{ cm}^3.$$

b. [1 mark].

$$4.1^3 - 33.51 = 68.921 - 33.51 = 35.41 \text{ cm}^3.$$

(The unrounded volume gives 35.4107, which is also 35.41.)

c. [1 mark]. Six square faces:

$$6 \times 4.1^2 = 6 \times 16.81 = 100.86 \text{ cm}^2.$$

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

d. [1 mark]. Count whole boxes along each edge: $17.0 \div 4.1 = 4.1\ldots$ gives 4, $12.5 \div 4.1 = 3.0\ldots$ gives 3, and $8.5 \div 4.1 = 2.07\ldots$ gives 2. The maximum is $4 \times 3 \times 2 =$ 24 boxes.

From the report. Parts a (87%), b (74%) and c (81%) were answered well, but in c an answer such as 101 could not be accepted because no rounding was asked for. In d (43%) 26 was a very common wrong answer, from dividing the volume of the space by the volume of one box.

Question 2 (3 marks)

Each side wall of a squash court is a 9.75 m wide rectangle. The shaded playing area is the part of the wall below a straight line that slopes from 4.57 m high at one end to 2.13 m high at the other.

a

Find the area of the shaded region, in square metres, to two decimal places. (1 mark)

b

Find the perimeter of the shaded region, in metres, to one decimal place. (1 mark)

c

On the court floor, Wei-Yi serves from point A and Bao stands at point B. A is 2.7 m from point C, B is 3.1 m from C, and angle ACB is 119°. Show that the distance between the players is 5 m, to the nearest metre. (1 mark)

Study design: areas, perimeters, Pythagoras' theorem and the cosine rule are now General Mathematics Unit 2 content, not Units 3 and 4.

Show worked solution

a. [1 mark]. The shaded region is a trapezium with parallel sides 4.57 m and 2.13 m, 9.75 m apart:

$$A = \frac{4.57 + 2.13}{2} \times 9.75 = 32.6625 \approx 32.66 \text{ m}^2.$$

b. [1 mark]. The sloping edge is the hypotenuse of a right-angled triangle with sides 9.75 m and $4.57 - 2.13 = 2.44$ m:

$$\sqrt{9.75^2 + 2.44^2} = 10.0506\ldots$$

$$\text{perimeter} = 4.57 + 9.75 + 2.13 + 10.0506\ldots = 26.5006\ldots \approx 26.5 \text{ m}.$$

c. [1 mark]. Use the cosine rule, showing the substitution and the square root:

$$AB = \sqrt{2.7^2 + 3.1^2 - 2 \times 2.7 \times 3.1 \times \cos(119^\circ)} = \sqrt{25.0157\ldots} = 5.0015\ldots \approx 5 \text{ m}.$$

From the report. In a (51%) some students did not use efficient methods. Part b was answered correctly by 48%. In c (42%) the correct substitution and the square root sign had to be shown.

Question 3 (5 marks)

Players travel to a squash competition in New York City (41° N, 74° W). Wei-Yi is from Shanghai (31° N, 121° E), Camilla from Durban (30° S, 31° E) and Ozlem from Istanbul (41° N, 29° E).

a

Which players are from the Northern Hemisphere? (1 mark)

b

A diagram of the globe shows the equator, the Greenwich meridian and the parallels through New York City, Shanghai and Durban. Mark the location of Istanbul with an X. (1 mark)

c

The flight from Istanbul to New York City travels along a small circle. Take the radius of Earth as 6400 km. i. Show that the radius of this small circle is 4830 km, to the nearest kilometre. (1 mark)

ii

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

d

Wei-Yi and Camilla both arrive in New York City at 10.00 pm local time on Monday 11 January. Wei-Yi left Shanghai at 8.00 pm local time that Monday, and Camilla left Durban at 4.00 am local time that Monday. Take the time difference as 13 hours between Shanghai and New York City, and 7 hours between Durban and New York City. How many hours longer was Camilla's flight than Wei-Yi's? (1 mark)

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

Show worked solution

a. [1 mark]

Wei-Yi and Ozlem: Shanghai (31° N) and Istanbul (41° N) are north of the equator, while Durban is at 30° S.

b. [1 mark]

Istanbul is on the same parallel as New York City (41° N), and its longitude, 29° E, is close to Durban's (31° E). Mark the X on the front (solid) part of that parallel, east of the Greenwich meridian and a little to the left of the point directly above Durban. In the report's answer the X sits just above, and slightly left of, the Shanghai dot.

c. i. [1 mark]

The radius of the 41° N small circle is

$$r = 6400\cos(41^\circ) = 4830.14\ldots \approx 4830 \text{ km}.$$

ii. [1 mark]. One city is 74° west and the other 29° east, so the angle between their meridians is $74^\circ + 29^\circ = 103^\circ$:

$$\text{distance} = \frac{103}{360} \times 2\pi \times 4830.14\ldots = 8683.09\ldots \approx 8683 \text{ km}.$$

(Using $r = 4830$ also gives 8683 km.)

d. [1 mark]. Convert both departure times to New York time. Both cities are east of New York, so their clocks are ahead of it.

• Shanghai is 13 hours ahead, so 8.00 pm Monday in Shanghai is 7.00 am Monday in New York. Wei-Yi flew from 7.00 am to 10.00 pm: 15 hours.
• Durban is 7 hours ahead, so 4.00 am Monday in Durban is 9.00 pm Sunday in New York. Camilla flew from 9.00 pm Sunday to 10.00 pm Monday: 25 hours.

Camilla's flight was $25 - 15 =$ 10 hours longer.

From the report. In a (70%) some students gave only one name. In b (41%) many students did not choose the correct parallel, and others chose it but placed the cross on the dotted side. Part c.i (41%) was not answered well, and many students did not attempt c.ii (30%). Part d was answered correctly by only 16%.

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 (2 marks)

A graph shows the height (m) of a drone over six minutes. It rises from 0 m at time 0, passing 50 m at 1 minute and 200 m at 2 minutes, levels out near 250 m, peaks at about 310 m after 3.5 minutes, then falls back through 50 m at 5 minutes to 0 m at 6 minutes.

a. For how long, in minutes, was the drone at least 50 m high? (1 mark)

b. What was the average rate of change of its height, in metres per minute, over the first two minutes? (1 mark)

Study design: outside the current Units 3 and 4; reading a graph like this is still a useful general skill.

Show worked solution

a. [1 mark]. The height is at least 50 m from 1 minute to 5 minutes: 4 minutes.

b. [1 mark].

$$\frac{200 - 0}{2 - 0} = 100 \text{ metres per minute}.$$

From the report. Parts a (73%) and b (67%) were answered correctly by most students.

Question 2 (3 marks)

Christy sells blocks of land, at most 20 a week. A step graph gives her weekly pay in dollars: 0 for no blocks, 3000 for 1 to 5 blocks, 6000 for 6 to 9, 11 000 for 10 to 14, 15 000 for 15 to 19 and 20 000 for 20 blocks.

a. What is the minimum number of blocks Christy must sell to receive $6000 in one week? (1 mark)

John also sells blocks of land. He is paid $1000 for each block he sells.

b. Write down all the numbers of blocks sold for which Christy and John receive the same weekly pay. (1 mark)

c. John sells no more than three blocks for every five blocks that Christy sells. With $x$ the number Christy sells and $y$ the number John sells, complete an inequality for $y$ in terms of $x$. (1 mark)

Study design: step graphs are now General Mathematics Unit 1 content (piecewise linear graphs). Inequalities like part c are not in the current General Mathematics study design.

Show worked solution

a. [1 mark]

6: the pay steps up to 6000 dollars at 6 blocks.

b. [1 mark]

John earns $1000n$ dollars for $n$ blocks. This matches one of Christy's steps at 3 blocks (3000), 6 blocks (6000), 11 blocks (11 000), 15 blocks (15 000) and 20 blocks (20 000): 3, 6, 11, 15 and 20. (Both earn nothing if no blocks are sold; the report's answer does not list 0.)

c. [1 mark]

The ratio of John's blocks to Christy's is at most $3 : 5$, so $\dfrac{y}{x} \le \dfrac35$, which gives

$$y \le \frac35x.$$

From the report. Part a (83%) was answered reasonably well. In b (37%) many students found some but not all of the values, and all of them were needed for the mark. Part c (21%) was not answered well.

Question 3 (3 marks)

The cost, in dollars, for Lam to work onsite for $n$ weeks is $C = 10\,000 + k \times n$.

a. The cost of 15 weeks of onsite work is $92 500. Show that $k = 5500$. (1 mark)

Lam's revenue for this job is $6500 per week.

b. How many weeks must Lam work onsite to break even? (1 mark)

c. Complete the profit equation $P = \square \times n + \square$. (1 mark)

Study design: linear models and simultaneous linear equations, such as this break-even problem, are now General Mathematics Unit 1 content.

Show worked solution

a. [1 mark]. Substitute $n = 15$ and $C = 92\,500$:

$$92\,500 = 10\,000 + 15k \implies k = \frac{92\,500 - 10\,000}{15} = \frac{82\,500}{15} = 5500.$$

b. [1 mark]. Break even when revenue equals cost:

$$6500n = 10\,000 + 5500n \implies 1000n = 10\,000 \implies n = 10.$$

Lam breaks even after 10 weeks.

c. [1 mark]. Profit is revenue minus cost:

$$P = 6500n - (10\,000 + 5500n) = 1000 \times n + (-10\,000).$$

The boxes hold 1000 and $-10\,000$, so $P = 1000n - 10\,000$.

From the report. Part a was answered correctly by only 29%, part b by 46% and part c by only 17%.

Question 4 (4 marks)

Let $x$ be the number of health and training sessions for children each day, and $y$ the number for adults. There must be at least 10 sessions a day, adult sessions must not be fewer than children's sessions, and sessions take 30 minutes (children) and 40 minutes (adults). The constraints are $x \ge 0$, $y \ge 0$, $x + y \ge 10$, $y \ge x$ and $30x + 40y \le 600$ (Inequalities 1 to 5), and a graph shows the feasible region.

a

Explain the meaning of Inequality 5 in this context. (1 mark)

b

What is the maximum number of sessions for children each day? (1 mark)

c

Each session for children makes a profit of $45.

Each session for adults makes a profit of $60.

i. Find the maximum profit per day from all sessions. (1 mark)

ii. List all the points in the feasible region that give this maximum profit. (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]

Children's sessions take $30x$ minutes and adults' sessions take $40y$ minutes, so the total time for all sessions each day cannot be more than 600 minutes (10 hours).

b. [1 mark]

The feasible region reaches furthest right where $y = x$ meets $30x + 40y = 600$: $70x = 600$, so $x = 8.57\ldots$ Sessions must be whole numbers, so the maximum is 8 (for example 8 sessions each: $240 + 320 = 560 \le 600$ minutes).

c. i. [1 mark]

The profit is $P = 45x + 60y = 1.5(30x + 40y)$, so the profit line has the same slope ($-\tfrac34$) as the boundary $30x + 40y = 600$, and the maximum is reached all along that boundary:

$$P = 1.5 \times 600 = 900.$$

The maximum profit is $900 per day. (The corner points confirm it: $(0, 10)$ gives 600, $(5, 5)$ gives 525, and $(0, 15)$ and $\left(\tfrac{60}{7}, \tfrac{60}{7}\right)$ both give 900.)

ii. [1 mark]. Whole-number points on $30x + 40y = 600$ (that is, $3x + 4y = 60$) with $y \ge x$: (0, 15), (4, 12) and (8, 9). The next such point, (12, 6), breaks $y \ge x$.

From the report. In a (43%) some students confused time with the number of sessions, and others said only that the time must be less than 600 minutes. In b (43%) 8.5 was quite often given by students who did not recognise that the number of sessions must be a whole number. Part c.i was answered correctly by 43%. In c.ii (20%) many students found the two corner points with the maximum profit but missed the third point on the line segment.

General advice from the 2021 report

• Reread the question after answering it. In core Question 7c the rule was for the value in terms of cups made, not years, and many students missed the change.
• Round only when told to, and exactly as told. With no rounding instruction an exact answer is needed (100.86 cm² in Geometry Question 1c, not 101). Know significant figures from decimal places (Question 2a), and give nearest-cent answers to two decimal places (7039.20, not 7039.2, in Question 9b).
• Show working on any question worth more than one mark, so a method mark is possible: finding about 2054.176 in Question 4c earned a mark even without 2056.
• In "show that" questions, work towards the given result with every step shown. Starting from the answer (Question 8b) did not earn the mark, and a response in CAS syntax cannot earn full marks.
• Know your definitions, such as a perpetuity keeping a constant value (Question 6b), and keep descriptive answers brief; point form is fine.
• Copy calculator results carefully, since transcription errors were common, and bring a ruler for drawing lines in Data analysis and Graphs and relations.