VCE General Mathematics 2023 Exam 2

Every question from the 2023 VCE General Mathematics Examination 2, the extended-response paper. Each question is summarised, then a full worked solution sits behind a Show worked solution toggle. For the multiple-choice paper, see the 2023 Examination 1 walkthrough.

How to use this page

• Questions are from the 2023 VCE General Mathematics Examination 2, copyright Victorian Curriculum and Assessment Authority (VCAA). Each 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 2023 General Mathematics Examination 2 report (Word document) and every calculation was redone independently. Both files are listed on the VCAA General Mathematics examinations page.
• 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.

Structure and timing

Examination 2 is 14 questions (60 marks) in 90 minutes (plus 15 minutes reading time), with one approved CAS technology, an optional scientific calculator and one bound reference. That is 1.5 minutes per mark.

• Questions 1 to 4: Data analysis (24 marks)
• Questions 5 to 7: Recursion and financial modelling (12 marks)
• Questions 8 to 11: Matrices (12 marks)
• Questions 12 to 14: Networks and decision mathematics (12 marks)

Data analysis

Question 1 (9 marks)

Table 1 records ID, weight (g), volume (cm³), image size (megapixels) and size (small, medium or large) for 15 oysters. The (weight, volume, image size, size) values for IDs 1 to 15 are: (12.9, 13.0, 5.1, L), (11.4, 11.7, 4.8, M), (17.4, 17.4, 6.5, L), (6.8, 7.2, 2.9, S), (9.6, 10.1, 3.7, M), (15.5, 15.6, 5.7, L), (9.7, 9.9, 4.0, S), (7.0, 7.5, 2.7, S), (12.6, 12.7, 5.5, M), (12.5, 12.7, 5.0, M), (10.1, 10.5, 3.9, M), (10.6, 10.8, 4.1, M), (13.0, 13.1, 5.3, L), (8.1, 8.5, 3.5, S), (14.1, 14.2, 5.3, L).

a

How many categorical variables are in Table 1? (1 mark)

b. i

Find the mean weight of all the oysters. (1 mark)

ii

Find the median weight of the large oysters. (1 mark)

c

The least squares line for volume against weight is volume $= 0.780 + 0.953 \times$ weight. i. Name the response variable. (1 mark)

ii

Complete: each 10 g increase in weight is associated with a ___ cm³ increase in volume. (1 mark)

d

Find the least squares line for volume with image size as the explanatory variable, with values to four significant figures. (2 marks)

e

Image sizes are approximately normal: 97.5% are less than 4.6 megapixels and 84% are more than 4.3 megapixels. Use the 68-95-99.7% rule to find the mean and standard deviation. (2 marks)

Show worked solution

a. [1 mark]

2: ID (a label, even though it uses numbers) and size. Weight, volume and image size are numerical.

b. i. [1 mark]

$\bar{x} = \dfrac{171.3}{15} = 11.42$ g. (Do not round: 11.4 was not accepted.)

ii. [1 mark]

The five large oysters weigh 12.9, 13.0, 14.1, 15.5 and 17.4 g in order, so the median is 14.1 g.

c. i. [1 mark]

Volume.

ii. [1 mark]

$10 \times 0.953 = 9.53$ cm³.

d. [2 marks]

Enter image size as $x$ and volume as $y$ and run linear regression: $a = 0.0028571\ldots$ and $b = 2.5714\ldots$ To four significant figures,

$$\text{volume} = 0.002\,857 + 2.571 \times \text{image size}.$$

e. [2 marks]. 97.5% below 4.6 means 2.5% above it, so 4.6 is two standard deviations above the mean: $\mu + 2\sigma = 4.6$. 84% above 4.3 means 16% below, so 4.3 is one standard deviation below: $\mu - \sigma = 4.3$. Subtracting, $3\sigma = 0.3$:

$$\sigma = 0.1, \qquad \mu = 4.4.$$

From the report. Half the students missed part a. In b.ii (39%) many gave 11.4, the median of the whole sample. In c.ii (27%) 0.953 was a frequent response. In d (37% full marks) four significant figures caused trouble, and some placed the coefficients or variables in the wrong order. Part e was answered fully by only 29%.

Question 2 (5 marks)

a

The sizes of 20 oysters are: 7 small, 10 medium and 3 large. i. Complete the frequency table (number and percentage). (1 mark)

ii

Draw a percentage segmented bar chart. (1 mark)

b

Table 3 shows oyster sizes from two farms. Farm A: 42 small, 124 medium, 44 large (total 210). Farm B: 114 small, 160 medium, 46 large (total 320). i. Find the percentage of all these oysters graded large, to the nearest whole number. (1 mark)

ii

The farmer believes farm A has a greater capacity to grow larger oysters. Does Table 3 support this? Compare two percentages, each to the nearest whole number. (2 marks)

Show worked solution

a. i. [1 mark].

Size	Number	Percentage (%)
small	7	35
medium	10	50
large	3	15
Total	20	100

ii. [1 mark]

One bar to 100%: small from 0 to 35, medium from 35 to 85, large from 85 to 100, shaded to match the key.

b. i. [1 mark]

$\dfrac{44 + 46}{210 + 320} \times 100\% = \dfrac{90}{530} \times 100\% \approx 17\%$.

ii. [2 marks]

Yes. At farm A, $\tfrac{44}{210} \approx 21\%$ of oysters are large, compared with $\tfrac{46}{320} \approx 14\%$ at farm B. The larger percentage at farm A supports the belief.

From the report. In b.ii (50% full marks) the answer had to say "yes" explicitly and state that farm A's percentage was higher than farm B's.

Question 3 (6 marks)

A scatterplot shows average monthly ice cream consumption (litres/person) against average monthly temperature (°C), with temperatures from about $-4.5$ °C to 22 °C. The least squares line is consumption $= 0.1404 + 0.0024 \times$ temperature, and the coefficient of determination is 0.7212.

a

Draw the least squares line on the scatterplot (temperature axis from $-10$ to 25). (1 mark)

b

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

c

Describe the association in terms of strength, direction and form. (1 mark)

d

Interpret the intercept in terms of the variables. (1 mark)

e

Predict consumption at $-6$ °C. (1 mark)

f

Is this interpolation or extrapolation? (1 mark)

Show worked solution

a. [1 mark]. Calculate two points at the ends of the axis and rule a line through them:

• at $-10$ °C: $0.1404 + 0.0024 \times (-10) = 0.1164$
• at 25 °C: $0.1404 + 0.0024 \times 25 = 0.2004$

b. [1 mark]

$r = +\sqrt{0.7212} = 0.849$ (positive, because the slope is positive).

c. [1 mark]

Strength: strong. Direction: positive. Form: linear.

d. [1 mark]

On average, when the temperature is 0 °C, ice cream consumption is predicted to be 0.1404 litres per person.

e. [1 mark]

$0.1404 + 0.0024 \times (-6) = 0.126$ litres per person.

f. [1 mark]

Extrapolation: $-6$ °C is outside the range of temperatures in the data (the lowest is about $-4.5$ °C).

From the report. In part a (40%) many students had correct coordinates but plotted them carelessly; use a ruler. In b some squared the coefficient of determination instead of taking its square root. In c "moderate" came from looking at $r^2$ rather than $r$. In d (43%) many interpreted the slope instead of the intercept, and interpreting both was not accepted. In e some rounded to 0.13. Part f (41%) was weak for a two-option question: the value fits on the axes but is outside the data.

Question 4 (4 marks)

A time series plot shows monthly ice cream consumption from January 2010 (month 1) to December 2012 (month 36), with peaks at months 6, 18 and 30. April 2010 (month 4) is about 0.18 litres/person.

a

Identify a feature of the plot consistent with a seasonal component. (1 mark)

b

The long-term seasonal index for April is 1.05. Find the deseasonalised value for April 2010, to two decimal places. (1 mark)

c

The 2011 monthly values are 0.156, 0.150, 0.158, 0.180, 0.200, 0.210, 0.183, 0.172, 0.162, 0.145, 0.134 and 0.154. Show that the seasonal index for July 2011 is 1.10, to two decimal places. (2 marks)

Show worked solution

a. [1 mark]. The peaks (maximum values) occur 12 months apart, at months 6, 18 and 30 (June each year). A pattern that repeats every year signals seasonality.

b. [1 mark].

$$\text{deseasonalised} = \frac{\text{actual}}{\text{seasonal index}} = \frac{0.18}{1.05} = 0.171\ldots \approx 0.17.$$

c. [2 marks]. Show both steps. The 2011 total is 2.004, so the monthly mean is

$$\frac{2.004}{12} = 0.167.$$

The July value is 0.183, so

$$\text{seasonal index} = \frac{0.183}{0.167} = 1.0958\ldots \approx 1.10.$$

From the report. Part a (26%) was poorly answered: many wrote a definition of seasonality instead of pointing to a feature of this plot. Part c needed both calculations for 2 marks; many showed $\tfrac{0.183}{0.167}$ without showing where 0.167 came from.

Recursion and financial modelling

Question 5 (3 marks)

Arthur borrows $30 000 at 6.4% per annum, compounding quarterly, repaid with quarterly repayments over six years: $A_0 = 30\,000$, $A_{n+1} = 1.016A_n - 1515.18$.

a

How many repayments will he make? (1 mark)

b

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

c

The final repayment differs slightly from $1515.18. Find it, to the nearest cent. (1 mark)

Show worked solution

a. [1 mark]. $6 \times 4 =$ 24 repayments.

b. [1 mark].

$$A_1 = 1.016 \times 30\,000 - 1515.18 = 28\,964.82$$

$$A_2 = 1.016 \times 28\,964.82 - 1515.18 = 27\,913.077\ldots \approx 27\,913.08$$

The balance is $27 913.08.

c. [1 mark]. Finance Solver with $N = 24$, $I\% = 6.4$, $PV = 30\,000$, $PMT = -1515.18$, 4 periods per year gives a future value of $0.14 in Arthur's favour: 24 full repayments would overpay by $0.14. So the final repayment is $1515.18 - 0.14 =$ $1515.04.

From the report. In b (51%) the mark needed the recursive steps shown; answers alone, early rounding (28 964.8) and transcription errors lost it. Part c (25%) was not answered well: some added the $0.14 instead of subtracting it.

Question 6 (4 marks)

Arthur invests $600 000 in an annuity paying $3973.00 a month, with interest calculated monthly. The amortisation table shows: payment 0, balance $600 000.00; payment 1, interest $2520.00, principal reduction $1453.00, balance $598 547.00; payment 2, interest $2513.90, principal reduction $1459.10, balance $597 087.90.

a

The rate is 0.42% per month. Find the rate per annum. (1 mark)

b

Using the table values, complete the line for payment 3, to the nearest cent. (1 mark)

c

Write a recurrence relation for the balance $V_n$ after $n$ months. (1 mark)

d

What is an annuity whose balance stays constant called? (1 mark)

Show worked solution

a. [1 mark]. $0.42\% \times 12 = 5.04\%$ per annum.

b. [1 mark].

• Payment: $3973.00
• Interest $= 0.0042 \times 597\,087.90 = 2507.77$
• Principal reduction $= 3973.00 - 2507.77 = 1465.23$
• Balance $= 597\,087.90 - 1465.23 = 595\,622.67$

c. [1 mark]. $V_0 = 600\,000$, $V_{n+1} = 1.0042V_n - 3973$.

d. [1 mark]. A perpetuity.

From the report. In b (40%) values were often rounded too early, and some wrote only the payment. In c the multiplying factor 1.0042 caused trouble for some.

Question 7 (5 marks)

Arthur borrows $60 000 with interest compounding weekly: $V_0 = 60\,000$, $V_{n+1} = 1.0015V_n - d$.

a

Show that the interest rate is 7.8% per annum. (1 mark)

b

Find $d$ if i. he makes interest-only repayments (1 mark)

ii

he fully repays the loan in five years, to the nearest cent. (1 mark)

c

He pays $d = 300$ for the first year. What new value of $d$ repays the loan in exactly three more years, to the nearest cent? (1 mark)

d

For what value of $d$ is the sequence geometric? (1 mark)

Show worked solution

a. [1 mark]. The weekly rate is $1.0015 - 1 = 0.0015$, and there are 52 weeks in a year:

$$(1.0015 - 1) \times 52 \times 100\% = 7.8\%.$$

b. i. [1 mark]

Interest-only repayments equal one week's interest: $0.0015 \times 60\,000 =$ 90.

ii. [1 mark]

Finance Solver: $N = 5 \times 52 = 260$, $I\% = 7.8$, $PV = 60\,000$, $FV = 0$, 52 periods per year gives $PMT \approx -278.86$, so $d = 278.86$.

c. [1 mark]

After one year of $300 payments ($N = 52$, $PMT = -300$), the balance is $FV \approx 48\,651.67$. Then $N = 156$, $PV = 48\,651.67$, $FV = 0$ gives $PMT \approx -350.01$, so $d = 350.01$.

d. [1 mark]

$d = 0$ (no repayments), so the balance is multiplied by 1.0015 each week.

From the report. In a (32%) all working had to be shown; CAS "solve" syntax was not appropriate. Part d was answered correctly by only 11%.

Matrices

Question 8 (3 marks)

Ticket prices are in the column matrix $N = \begin{bmatrix} 36 & 15 & 8 \end{bmatrix}^T$ (family, adult, child), with elements $n_{ij}$.

a

Which element shows the cost of one child ticket? (1 mark)

b

A family ticket admits two adults and two children. Complete $\begin{bmatrix} 0 & 2 & 2 \end{bmatrix} \times N - \square \times N = 10$ to show the $10 saving. (1 mark)

c

The circus sold 204 family, 162 adult and 176 child tickets. Find $K$ such that $KN$ gives the revenue for each ticket type, $\begin{bmatrix} 7344 & 2430 & 1408 \end{bmatrix}^T$. (1 mark)

Show worked solution

a. [1 mark]. Row 3, column 1: $n_{31}$ (lower-case $n$, row then column).

b. [1 mark]. The row matrix $\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$ picks out the family price:

$$\begin{bmatrix} 0 & 2 & 2 \end{bmatrix}N - \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}N = (2 \times 15 + 2 \times 8) - 36 = 46 - 36 = 10.$$

c. [1 mark]. A diagonal matrix multiplies each price by its own number of tickets:

$$K = \begin{bmatrix} 204 & 0 & 0 \\ 0 & 162 & 0 \\ 0 & 0 & 176 \end{bmatrix}.$$

Check: $204 \times 36 = 7344$, $162 \times 15 = 2430$ and $176 \times 8 = 1408$.

From the report. In part a many used a capital $N$ or reversed the subscripts. Parts b (44%) and c (34%) were not answered well; in c a row or column matrix was often given instead of a $3 \times 3$ diagonal matrix.

Question 9 (4 marks)

Total ticket sales from the last 20 performances at locations E, F, G, H and I are $R = \begin{bmatrix} 960\,000 & 990\,500 & 940\,100 & 920\,800 & 901\,300 \end{bmatrix}$.

a

Complete $\square \times R$ to give the average sales per performance at each location. (1 mark)

b

Revenue is to rise by 25% using $tR \times \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \end{bmatrix}^T$. Find $t$. (1 mark)

c

A permutation matrix (columns this month E, F, G, H, I; rows next month) moves the circus each month: E to G, F to I, G to F, H to E and I to H. Starting at I, list the order of towns. (1 mark)

d

A sixth location, J, is added after E and before G. Complete three columns of the new $6 \times 6$ matrix. (1 mark)

Show worked solution

a. [1 mark]. Divide by the 20 performances:

$$\frac{1}{20} \times R = \begin{bmatrix} 48\,000 & 49\,525 & 47\,005 & 46\,040 & 45\,065 \end{bmatrix}.$$

b. [1 mark]

A 25% increase multiplies by $t =$ 1.25.

c. [1 mark]

Follow the 1s column by column from I: I to H, H to E, E to G, G to F, and F back to I. The order is I, H, E, G, F.

d. [1 mark]

VCAA invalidated this part after an error was found in the printed matrix, so it is not solved here. (The intended cycle would be I, H, E, J, G, F.)

From the report. In part a (25%) many could not find the correct scalar, $\tfrac{1}{20}$. In part b many gave 0.25 instead of 1.25.

Question 10 (3 marks)

Communication matrix $G$ (rows sender, columns receiver, order D, M, P, S, C) has rows D: 0, 1, 1, 1, 1; M: 1, 0, 1, 1, 1; P: 0, 1, 0, 0, 0; S: 0, 1, 0, 0, 1; C: 0, 0, 0, 1, 0.

a

What is the shortest communication sequence for a customer's complaint to reach a director? (1 mark)

b. i

Complete $H$, the matrix of two-step links. (1 mark)

ii

What do $g_{21}$ and $h_{21}$ tell you about communication between these employees? (1 mark)

Show worked solution

a. [1 mark]. Customers can only contact sales staff; sales staff can contact managers; managers can contact directors: C, S, M, D.

b. i. [1 mark]. Two-step links are counted by $H = G^2$:

$$H = \begin{bmatrix} 1 & 2 & 1 & 2 & 2 \\ 0 & 3 & 1 & 2 & 2 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 2 & 1 \\ 0 & 1 & 0 & 0 & 1 \end{bmatrix}.$$

ii. [1 mark]. $g_{21} = 1$: managers can communicate directly with directors. $h_{21} = 0$: they cannot do so through exactly one other person.

From the report. Part b.ii (25%) was weak; students who dealt with each element separately wrote more accurate answers.

Question 11 (2 marks)

A circus needs 180 workers per show. From show to show, 95% of workers keep working (W) and the rest leave (L) permanently. $S_0 = \begin{bmatrix} 180 & 0 \end{bmatrix}^T$ and $S_{n+1} = TS_n + B$.

a. Write down the transition matrix $T$ (columns this show W, L; rows next show W, L). (1 mark)

b. Write down $B$ so that there are always 180 workers. (1 mark)

Show worked solution

a. [1 mark].

$$T = \begin{bmatrix} 0.95 & 0 \\ 0.05 & 1 \end{bmatrix}.$$

Workers who leave never return, so the L column is 0 then 1.

b. [1 mark]. Each show, $0.05 \times 180 = 9$ workers leave, so hire 9 new workers:

$$B = \begin{bmatrix} 9 \\ 0 \end{bmatrix}.$$

(The report also accepted $-9$ as the second element.)

From the report. In part a (34%) 0.5 was often written instead of 0.05, and the 1 was left out or put in the first row.

Networks and decision mathematics

Question 12 (4 marks)

A graph of five states has edges A-B, A-C, B-C, B-D, C-D, C-E and D-E (shared borders).

a

Find the sum of the degrees. (1 mark)

b. i

Complete Euler's formula $v + f = e + 2$. (1 mark)

ii

Euler's formula holds because the graph is connected and ___. (1 mark)

c

A map shows state A and four other regions numbered 1 to 4. Region A borders 2 and 3; region 1 borders 2 and 4; region 2 borders A, 1, 3 and 4; region 3 borders A, 2 and 4. Match B, C, D and E to the numbers. (1 mark)

Show worked solution

a. [1 mark]

Degrees A 2, B 3, C 4, D 3, E 2: the sum is 14 (twice the 7 edges).

b. i. [1 mark]

$5 + 4 = 7 + 2$.

ii. [1 mark]

Planar (it can be drawn with no edges crossing).

c. [1 mark]

C borders all four other states, so C is region 2. E borders only C and D, so E is region 1 (bordering 2 and 4), which makes D region 4. B is then region 3, which borders A, C and D as required.

State	B	C	D	E
Number	3	2	4	1

From the report. In b.ii (81%) some wrote "complete" instead of "planar".

Question 13 (3 marks)

Roads between nine landmarks (km): G-H 1.7, G-K 1.5, G-N 1.8, H-I 3.2, H-K 1.1, K-I 1.2, K-L 0.9, I-M 3.2, I-J 2.8, M-J 2.1, L-N 1.8, L-O 1.4, N-O 2.4, O-M 2.6, O-J 3.8.

a

Find the minimum distance from G to M. (1 mark)

b

Write a route that visits every landmark and returns to G with minimum total distance. (1 mark)

c

Shyla travels every road, starting at G and ending elsewhere, repeating two roads. Which two roads? (1 mark)

Show worked solution

a. [1 mark]

G-K-I-M $= 1.5 + 1.2 + 3.2 =$ 5.9 km. (The route G-K-L-O-M is 6.4 km.)

b. [1 mark]

A minimum Hamiltonian cycle: G-H-K-I-J-M-O-L-N-G (or its reverse), which is $1.7 + 1.1 + 1.2 + 2.8 + 2.1 + 2.6 + 1.4 + 1.8 + 1.8 = 16.5$ km.

c. [1 mark]

The odd-degree vertices are G, H, J, L, M and N. A trail using every road needs at most two odd vertices, at its start and end. Starting at G, the other four odd vertices besides the end point must be paired up by repeated roads. The two roads joining odd vertices directly (other than those at G) are L-N and J-M, so she repeats the roads between vertex L and vertex N and between vertex J and vertex M, and finishes at H.

From the report. In a, 6.4 km was a common error, and some gave the route without the distance. In b (44%) landmarks were missed or misplaced, or the route did not end at G. Part c (12%) was not answered well; very few had both pairs.

Question 14 (5 marks)

A renovation has 12 activities. A (6) and B (4) start the project. C (7) follows A. D (5) and E (10) follow B. F (4) follows C. G (3) and H (7) follow D. I (6) follows the end of C (through a dummy) and G. J (6) follows E and H. K (4) follows F and I. L (1) follows J and K.

a

Write down the immediate predecessor(s) of I. (1 mark)

b

Find the earliest start time of J. (1 mark)

c

How many activities have zero float? (1 mark)

d

A and B are each reduced by two days. What is the maximum reduction in the project time? (1 mark)

e

Activities can be shortened by up to two days each at these daily costs: A $1500, B $2000, F $2500, H $1000, I $1500, K $3000. With a $15 000 budget, find the reductions that give the earliest completion. (1 mark)

Show worked solution

a. [1 mark]

C and G (C through the dummy).

b. [1 mark]

J needs E (finishes at $4 + 10 = 14$) and H (finishes at $9 + 7 = 16$), so its EST is 16 days.

c. [1 mark]

The paths are: A-C-I-K-L $= 24$, A-C-F-K-L $= 22$, B-D-G-I-K-L $= 23$, B-D-H-J-L $= 23$ and B-E-J-L $= 21$. The critical path A-C-I-K-L (24 days) contains 5 activities with zero float (the dummy is not an activity).

d. [1 mark]

Shortening A and B by 2 days makes the paths 22, 20, 21, 21 and 19. The project now takes 22 days, a reduction of 2 days.

e. [1 mark]

First find the best possible time. B-D-H-J-L (23 days) can only be shortened through B and H, by at most $2 + 2 = 4$ days, so 19 days is the earliest possible completion.

To reach 19 days as cheaply as possible:

• B-D-H-J-L needs B and H both cut by 2: $2(2000) + 2(1000) = 6000$ dollars.
• A-C-I-K-L (24) needs 5 days from A, I and K (at most 2 each): cut A by 2 and I by 2 (the cheaper ones, $1500 a day), then K by 1: $3000 + 3000 + 3000 = 9000$ dollars.

Activity	A	B	F	H	I	K
Reduction (days)	2	2	0	2	2	1

The paths become A-C-I-K-L 19, A-C-F-K-L 19, B-D-G-I-K-L 18, B-D-H-J-L 19 and B-E-J-L 19, at a total cost of $6000 + 9000 = 15\,000$ dollars, exactly the budget.

From the report. In c (35%) some counted the dummy. In d a few gave the new completion time of 22 days instead of the reduction. Part e was answered correctly by only 9%.

General advice from the 2023 report

• Round only when told to. 11.42 was required in Question 1b.i; 11.4 was not accepted.
• Know significant figures from decimal places (Question 1d) and give nearest-cent answers to two decimal places.
• Show all steps in "show that" questions (Questions 4c and 7a) and show recursive calculations when asked (Question 5b).
• Keep descriptive answers brief. "The intercept" means the vertical-axis intercept read from the equation (Question 3d).
• Answer the question actually asked, such as the distance rather than the route in Question 13a.