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Practice examples · Free questions with worked answers.

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New South Wales · Mathematics Standard 2 · 12 marks

Compound interest with monthly compounding

Question

A school invests $8,000 in a reserve account earning 4.8% per annum, compounded monthly. The monthly interest rate is 0.4%.

(a) Calculate the balance after 30 months, correct to the nearest cent.

(b) Immediately after the 30-month balance is calculated, the school withdraws $1,500 and leaves the remainder in the account for a further 18 months at the same rate. Calculate the final balance, correct to the nearest cent.

(c) Explain why using an annual growth factor once per year would not give the same result as the monthly-compounding model.

Reveal worked answer

Worked answer

(a) Monthly multiplier: 1.004. Balance: 8000 x 1.004³⁰ = $9017.82.

(b) Carry the unrounded first balance: (8000 x 1.004³⁰ - 1500) x 1.004¹⁸ = $8077.90.

(c) Monthly interest is added before later monthly interest is calculated. An annual factor changes the compounding timing and does not model the withdrawal at the stated monthly stage.

Why this earns credit

The response keeps the monthly rate and number of monthly periods together in each growth factor. It carries the first balance unrounded before applying the withdrawal and second growth period, so the displayed final amount is reproducible. It also explains why the timing of compounding affects the result.

Common errors

Using 4.8% as the monthly rate or using 30 years instead of 30 months.
Rounding the first balance to cents and then reporting a final amount that cannot be reproduced from the displayed rounded working.
Applying 1.048 for each monthly period or applying one annual factor without modelling the withdrawal at the stated monthly stage.

Could you explain why the withdrawal must be applied before the further 18 months of growth?

Victoria · General Mathematics · 10 marks

Conditional relative frequency and association

Question

A class compares whether students used an online revision module or printed notes before a test. The results are: online module, 80 students with 56 passing and 24 not passing; printed notes, 100 students with 60 passing and 40 not passing.

(a) Find the conditional pass rate for each revision method.

(b) State the difference in percentage points.

(c) Calculate the percentage increase in the online-module pass rate compared with the printed-notes pass rate, correct to one decimal place.

(d) Explain why these results alone do not establish that using the online module caused the higher pass rate.

Reveal worked answer

Worked answer

(a) Online module: $\frac{56}{80} = 70 %$ . Printed notes: $\frac{60}{100} = 60 %$ .

(b) Difference: $70 % - 60 % = 10$ percentage points.

(d) The groups may not have been formed randomly. Prior achievement or available study time could differ, so the results show an association in this class rather than proving that the module caused the difference.

Why this earns credit

The denominators are the totals for each revision group, so the two conditional pass rates are comparable. The answer distinguishes percentage points from percentage increase, then separates the observed association from a causal claim by identifying a plausible difference between the groups.

Common errors

Using the combined 180 students as the denominator for both conditional pass rates.
Calling the 10 percentage-point difference a 10% increase without using the printed-notes rate as the reference value.
Claiming that the online module caused the difference without considering how students selected their revision method.

What other difference between the two groups could affect the comparison?

Victoria · Mathematical Methods · 12 marks

Exponential functions, differentiation and definite integrals

Question

The output rate of a device is modelled by g(x) = xe^(-x/2), for x greater than or equal to 0.

(a) Find g'(x).

(b) Show that the only stationary point occurs at x = 2.

(c) Determine whether this stationary point is a maximum or minimum.

(d) Find its value.

(e) Find the exact area under the graph of g between x = 0 and x = 2.

(f) Interpret this area in the context of the device.

Reveal worked answer

Worked answer

(a) g'(x) = e^-x/2(1 - x/2).

(b) Since e^-x/2 is never zero, g'(x) = 0 only when 1 - x/2 = 0, so x = 2.

(d) g(2) = 2/e.

(e) integral from 0 to 2 of xe^-x/2 dx = [-(2x + 4)e^-x/2] from 0 to 2 = 4 - 8/e.

(f) The area represents the total output accumulated while x increases from 0 to 2.

Why this earns credit

The response combines product-rule differentiation, stationary-point analysis and integration. Exact form is retained throughout, and the definite integral is interpreted as accumulated output because g is an output rate.

Common errors

Forgetting the derivative of the exponential factor when applying the product rule.
Assuming the exponential factor can equal zero.
Giving the stationary point without checking whether the derivative changes sign.
Using an incorrect antiderivative for xe^(-x/2).
Reporting a decimal approximation when an exact value is requested.

Could you justify the maximum using the second derivative instead of a sign chart?

New South Wales · Mathematics Advanced · 12 marks

Calculus: stationary points, tangents and area

Question

The curve C has equation y = f(x) = x^3 - 6x^2 + 9x + 2.

(a) Find the coordinates of the stationary points of C and determine the nature of each stationary point. (4 marks)

(b) Find the equation of the tangent to C at x = 2. (3 marks)

(c) Find the total area between C and this tangent for 1 <= x <= 3. Give an exact answer. (5 marks)

Reveal worked answer

Worked answer

f'(x) = 3(x - 1)(x - 3), so the stationary points are (1, 6) and (3, 2). Since f''(1) < 0 and f''(3) > 0, these are a local maximum and local minimum respectively.

At x = 2, f(2) = 4 and f'(2) = -3. The tangent is y - 4 = -3(x - 2), or y = -3x + 10.

The difference is f(x) - (-3x + 10) = (x - 2)³. Splitting at x = 2 gives areas 1/4 and 1/4, so the total area is 1/2 square units.

Why this earns credit

The item combines core Advanced calculus skills without requiring Extension-only methods. The tangent is used as a comparison function, so the sign change at x = 2 must be handled before integrating.

Common errors

Uses f(x) = 0 instead of f'(x) = 0 when finding stationary points.
Uses the curve's y-value but not its derivative as the tangent gradient.
Integrates the signed difference from 1 to 3 without splitting where the curve and tangent meet.
Classifies the stationary points without checking local behaviour.

Which step made the area calculation require a split at x = 2?

New South Wales · Mathematics Standard 2 · 14 marks

Annuities, depreciation and financial comparison

Question

At the end of each year, Mia deposits $1000 into a savings account earning 5% interest per year. Let $B_{n}$ be the balance immediately after the nth deposit, with $B_{0} = 0$ .

A laptop costs $12 000 and depreciates by 15% per year on a reducing-balance basis.

(a) Write a recurrence relation for $B_{n}$ and calculate $B_{3}$ .

(b) Write a model for the laptop value $V_{n}$ after n years and calculate $V_{3}$ .

(c) After three years, compare Mia's savings with the laptop's value. Decide whether Mia can buy an equivalent $12 000 laptop using only the savings account, and justify your decision.

Reveal worked answer

Worked answer

(a) $B_{n} = 1.05 B_{n - 1} + 1000$ , with $B_{0} = 0$ . Therefore $B_{1} = 1000$ , $B_{2} = 2050$ and $B_{3} = 3152.50$ .

(b) $V_{n} = 12000 {0.85}^{n}$ . Thus $V_{3} = 12000 {0.85}^{3} = 7369.50$ .

(c) Mia has $3152.50 in savings and the laptop is worth $7369.50 after three years. She cannot buy an equivalent new $12 000 laptop because the savings are $12000 - 3152.50 = 8847.50$ short.

Why this earns credit

The financial models use end-of-year deposits and reducing-balance depreciation consistently.

Common errors

Applying interest after the deposit despite the deposits being made at the end of each year.
Using straight-line depreciation instead of a 15% reducing-balance model.
Comparing the savings only with the depreciated value and concluding that a new laptop can be bought.
Rounding the balance or value before completing the calculation.

How would the savings balance change if deposits were made at the start of each year instead?

Victoria · General Mathematics · 12 marks

Matrices, transition models and long-term behaviour

Question

A city transport survey classifies commuters as either bus users or train users. Each month, the transition matrix $P$ is applied to the column vector $v_{n}$ , where the first entry is the number of bus users and the second entry is the number of train users.

$P = [0.80 0.30; 0.20 0.70]$ and $v_{0} = [700, 300] T$ .

(a) Calculate $v_{1} = P v_{0}$ . (4 marks)

(b) Calculate $v_{2} = P v_{1}$ . (3 marks)

(c) Determine the long-term numbers of bus users and train users, assuming the total remains one thousand commuters. (3 marks)

(d) Explain what the long-term result means in this context. (2 marks)

Reveal worked answer

Worked answer

(a) $v_{1} = [0.80 (700) + 0.30 (300), 0.20 (700) + 0.70 (300)] T = [650, 350] T$ .

(b) $v_{2} = [0.80 (650) + 0.30 (350), 0.20 (650) + 0.70 (350)] T = [625, 375] T$ .

(c) At equilibrium, $0.20 B = 0.30 T$ , so $B : T = 3 : 2$ . With $B + T = 1000$ , the long-term numbers are $B = 600$ and $T = 400$ .

(d) Over many months, the commuter numbers approach six hundred bus users and four hundred train users, regardless of these initial numbers.

Why this earns credit

This transition model uses column vectors and a stable long-term distribution.

Common errors

Multiplying the initial vector in the wrong order or treating it as a row vector.
Adding transition probabilities across a row or column without applying them to commuter numbers.
Assuming the second-month vector is already the long-term distribution.
Finding the 3:2 ratio but not converting it to numbers out of 1000.

What change to the transition probabilities might increase the long-term proportion of train users?

New South Wales · Mathematics Standard 2 · 12 marks

Time series, moving averages and seasonal interpretation

Question

A café recorded monthly cold-drink sales: January 80, February 86, March 92, April 98, May 104, June 110.

(a) Calculate the centred 3-month moving averages for February, March, April and May. (4 marks)

(b) Describe the trend shown by the data. Explain why a seasonal pattern cannot be reliably identified from this data set. (3 marks)

(c) Use the pattern to forecast sales for July. State one limitation of this forecast. (5 marks)

Reveal worked answer

Worked answer

(a) February: $\frac{80 + 86 + 92}{3} = 86$ . March: $\frac{86 + 92 + 98}{3} = 92$ . April: $\frac{92 + 98 + 104}{3} = 98$ . May: $\frac{98 + 104 + 110}{3} = 104$ .

(b) Sales increase by 6 drinks each month, so there is a clear upward trend. Six months is not enough to compare the same months across different years, so a seasonal effect cannot be identified reliably.

(c) The forecast is $110 + 6 = 116$ drinks for July. It assumes the current linear trend continues; weather, promotions or other factors may change actual sales.

Why this earns credit

This question uses a short time series to distinguish an observed trend from an unsupported seasonal conclusion.

Common errors

Using an uncentred or incorrectly grouped 3-month average.
Claiming that increasing sales alone proves a seasonal pattern.
Forecasting July without extending the observed increase of 6 drinks.
Giving a limitation unrelated to extrapolating the time series.

What local event or weather change could make the July forecast less reliable?

Victoria · General Mathematics · 12 marks

Networks, flows and scheduling decisions

Question

A warehouse sends parcels from source S to destination T through two sorting areas, A and B, and two dispatch areas, C and D. The directed network has these daily capacities: S to A: 9 parcels; S to B: 7 parcels; A to C: 5 parcels; A to D: 4 parcels; B to C: 4 parcels; B to D: 5 parcels; C to T: 8 parcels; D to T: 7 parcels.

(a) Determine the maximum number of parcels that can reach T each day. Show one feasible allocation of flows. (5 marks)

(b) Identify the bottleneck in the network and justify your answer. (3 marks)

(c) The warehouse can increase one arc capacity by one parcel per day. State one arc that should be increased and show that the maximum flow would then increase. (4 marks)

Reveal worked answer

Worked answer

(a) The maximum flow is $F = 15$ parcels per day.

One feasible allocation is $S \to A = 9, S \to B = 6, A \to C = 5, A \to D = 4, B \to C = 3, B \to D = 3$ .

The flows into T are $C \to T = 8, D \to T = 7$ .

(b) C-T and D-T form the bottleneck because $8 + 7 = 15$ . No greater flow can enter T.

(c) If C-T is increased by one parcel per day, a feasible flow is $S \to A = 9, S \to B = 7, A \to C = 5, A \to D = 4, B \to C = 4, B \to D = 3$ .

The resulting flow into T is $C \to T = 9, D \to T = 7, F = 16$ . Increasing D-T by one instead is also valid and permits a flow of 16.

Why this earns credit

The destination arcs form a combined bottleneck, so increasing either arc entering T by one parcel per day raises the achievable flow.

Common errors

Adding all arc capacities instead of identifying a feasible source-to-destination flow.
Claiming a flow greater than 15 without accounting for the capacities of the arcs entering T.
Increasing an upstream arc even though the original restriction is at the destination.
Claiming that only C-T can be increased; increasing either C-T or D-T by one permits a flow of 16.
Giving flows that do not conserve parcels at A, B, C or D.

Why can improving either arc in a downstream bottleneck increase the maximum flow?

New South Wales · Mathematics Standard 2 · 12 marks

Bivariate data, regression and residual reasoning

Question

The table shows weekly independent study time, $x$ hours, and an assessment score, $y$ out of 100, for six students.

Study time: 2, 3, 4, 5, 6, 7 hours.

Score: 58, 61, 65, 69, 70, 75.

(a) Describe the association between study time and assessment score. State why this association does not prove that increased study time causes higher scores.

(b) A least-squares linear model for the data is $y = 3.4 x + 51.5$ . Use the model to predict the score of a student who studies independently for 5.5 hours per week.

(c) For the student who studied for 6 hours, calculate the residual. Comment on what this residual means in context and give one limitation of using this model to make predictions.

Reveal worked answer

Worked answer

(a) There is a strong positive linear association: students who reported more independent study time generally had higher scores. This does not establish causation because other factors, such as prior achievement or study quality, may affect both study time and score.

(b) $y = 3.4 \times 5.5 + 51.5 = 70.2$ . The predicted score is 70.2, or about 70 out of 100.

$residual = actual - predicted = 70 - 71.9 = - 1.9$ . The student's score was 1.9 marks below the model prediction. A limitation is that the model is based on only six students, so it may not predict reliably for other students or study-time values outside this range.

Why this earns credit

This draft tests interpretation of association, use of a fitted model, prediction and residual reasoning without making a causal claim.

Common errors

Stating that extra study time definitely causes higher scores.
Using 6 hours instead of 5.5 hours for the prediction in part (b).
Calculating the residual as predicted minus actual and giving a positive value.
Describing the negative residual as an above-prediction result.
Treating the fitted line as reliable for all possible study times.

What other student variables could be collected to investigate the relationship more carefully?

Victoria · General Mathematics · 12 marks

Networks, shortest paths and critical activities

Question

A project network has start node A and finish node F. Activity durations are:

$A \to B : 3 days$

$A \to C : 4 days$

$B \to D : 5 days$

$C \to D : 2 days$

$C \to E : 6 days$

$D \to F : 4 days$

$E \to F : 2 days$

(a) Find the shortest route from A to F and its total duration. (4 marks)

(b) Find the critical path or paths and the earliest possible project completion time. (5 marks)

Reveal worked answer

Worked answer

(a) The route

A \to C \to D \to F

has duration

4 + 2 + 4 = 10 days

It is the shortest route.

(b) The route

A \to B \to D \to F

has duration

3 + 5 + 4 = 12 days

The route

A \to C \to E \to F

has duration

4 + 6 + 2 = 12 days

Both are critical paths, so the earliest completion time is 12 days.

12 + 1 = 13 days

The project completion time therefore becomes 13 days.

Why this earns credit

The network contains two critical paths, showing that more than one sequence can control a project's completion time.

Common errors

Choosing the shortest route as the critical path.
Omitting one of the two critical paths.
Adding the delay to the shortest route rather than to the path containing C-E.
Treating the 1-day delay as having no effect despite C-E being on a critical path.

Which non-critical activity could be delayed without necessarily changing the completion date?

New South Wales · Mathematics Standard 2 · 12 marks

Networks, routes and allocation constraints

Question

A delivery depot D is connected to suburbs A, B and C by the following road distances: D-A 4 km, D-B 6 km, A-B 2 km, A-C 7 km and B-C 3 km. Each day, 18 parcels must be delivered to C. A van can carry at most 7 parcels. The B-C road can carry at most 14 parcels per day.

(a) Determine all shortest routes from D to C and state the distance of each. (4 marks)

(b) Explain why all 18 parcels cannot use a shortest route. Determine the greatest number of parcels that can use shortest routes. (3 marks)

(c) Recommend an allocation of the 18 parcels between routes from D to C. Include the number of van trips required and justify your recommendation. (5 marks)

Reveal worked answer

Worked answer

(a) There are two shortest routes.

D-A-B-C: $4 + 2 + 3 = 9 km$

D-B-C: $6 + 3 = 9 km$

(b) Both shortest routes use B-C. Its daily capacity is $14 parcels$ , so at most 14 parcels can use shortest routes.

(c) Send 14 parcels by D-A-B-C and 4 parcels by D-A-C. The trip calculation is $\frac{18}{7} = 2.571...$ , so 3 trips are required. Two full trips carry 14 parcels through B-C; the remaining 4 parcels avoid that road. D-B-C is an equally short alternative for the parcels using B-C.

Why this earns credit

This network has two equally short routes, but both depend on the same capacity-constrained final road segment.

Common errors

Identifying only one shortest route without recognising that D-A-B-C and D-B-C are both 9 km.
Treating the 14-parcel B-C limit as a limit for each van trip rather than for the day.
Sending all 18 parcels through B-C.
Calculating two van trips by rounding 18 divided by 7 down.

How would the recommended allocation change if the B-C daily capacity increased to 21 parcels?

Victoria · General Mathematics · 12 marks

Bivariate data, correlation and regression limits

Question

Six students recorded the number of hours, $x$ , spent completing a revision program and their score, $y$ , on a later quiz.

The paired data are $(1, 53), (2, 57), (3, 62), (4, 66), (5, 70), (6, 68)$ .

(a) Use a calculator to find the Pearson correlation coefficient, $r$ , correct to three decimal places. Describe the direction and strength of the association. (3 marks)

(b) Find the least-squares regression line of $y$ on $x$ , correct to two decimal places. Use it to estimate the score for a student who completes 7 hours. State a limitation of this estimate. (4 marks)

(c) Find the residual for the student who completed 6 hours. Comment on whether this point may warrant further investigation. Explain why this data cannot establish that completing more revision hours causes a higher quiz score. (5 marks)

Reveal worked answer

Worked answer

(a) $r = 0.952$ . There is a strong positive linear association between recorded revision hours and quiz score.

(b) The least-squares regression line is $\hat{y} = 50.87 + 3.37 x$ .

For seven hours, $\hat{y} = 50.87 + 3.37 (7) = 74.46$ . The estimated score is about 74.5. Seven hours is outside the observed range of 1 to 6 hours, so this extrapolated estimate should be used cautiously.

The residual is $68 - 71.09 = - 3.09$ . Using unrounded regression values, the residual is $- 3.10$ . The point is below the regression line and may warrant investigation, but it is not automatically invalid. The data are observational, so confounding factors may explain the association; correlation does not establish causation.

Why this earns credit

This question combines regression calculations with appropriate cautions about extrapolation, unusual residuals and causal interpretation.

Common errors

Describing a positive correlation as proof that revision causes higher scores.
Using the actual score of 68 instead of the regression estimate when finding the residual.
Calculating the residual as predicted minus actual without stating the convention.
Treating the point at 6 hours as automatically invalid rather than noting that it may need investigation.
Presenting the seven-hour estimate as equally reliable to an estimate within the observed range.

Why should the seven-hour regression estimate be treated more cautiously than an estimate within the observed range?

New South Wales · Mathematics Standard 2 · 14 marks

Loan repayments and amortisation

Question

A borrower takes out a $20,000 fixed-rate loan. Interest is charged monthly at 0.5% per month. The loan is to be repaid by 36 equal monthly repayments, with the first repayment one month after the loan is taken out.

(a) Use the loan repayment formula to calculate the monthly repayment, correct to the nearest cent.

A = \frac{P r {(1 + r)}^{n}}{{(1 + r)}^{n} - 1}

(b) Using the calculator value before rounding from part (a), calculate the total interest paid under the 36-payment plan.

(c) Instead, the borrower pays $650 at the end of each month. Calculate the balance immediately after the 33rd payment, without rounding intermediate values. Hence calculate the final repayment one month later and the total interest saved compared with the original plan.

(d) Explain why rounding should be delayed until the final stage of a loan calculation.

Reveal worked answer

Worked answer

(a) Substitute the given principal, monthly rate and number of repayments.

A = \frac{20000 \times 0.005 \times {(1.005)}^{36}}{{(1.005)}^{36} - 1} = 608.438748 \dots

The monthly repayment is $608.44.

(b)

36 \times 608.438748 \dots = 21903.7949 \dots

21903.7949 \dots - 20000 = 1903.7949 \dots

The total interest is $1,903.79.

(c)

B_{33} = 20000 {(1.005)}^{33} - 650 \frac{{(1.005)}^{33} - 1}{0.005} = 320.0833 \dots

320.0833 \dots \times 1.005 = 321.6837 \dots

The final repayment is $321.68.

33 \times 650 + 321.6837 \dots = 21771.6837 \dots

21903.7949 \dots - 21771.6837 \dots = 132.1112 \dots

The interest saved is $132.11.

(d) Early rounding changes later balances and interest calculations. Small differences can accumulate, affecting a final repayment or total interest.

Why this earns credit

This question compares a standard reducing-balance repayment plan with a higher monthly repayment plan, using a stated no-intermediate-rounding convention.

Common errors

Using 0.5 rather than 0.005 as the monthly interest rate.
Using the rounded monthly repayment when part (b) specifies the calculator value.
Calculating the balance after 33 payments without applying monthly interest before each repayment.
Forgetting to apply one further month of interest before calculating the final repayment.
Calling total repayments the total interest.

How would increasing a repayment by a small amount each month change the cost of a long-term loan?

Victoria · General Mathematics · 12 marks

Recurrence relations and financial growth

Question

Mia opens a savings account with $1000. At the end of each year, the balance earns 4% interest and then Mia deposits $500. Let the following sequence represent the balance immediately after the deposit at the end of each year.

S_{n}

S_{0} = 1000

(a) Write a recurrence relation for the next balance in terms of the current balance.

S_{n + 1}

S_{n}

(b) Calculate the first two balances after the initial balance.

S_{1}

S_{2}

(c) Describe the long-term behaviour of this model and state one limitation of using a constant interest rate for future savings.

Reveal worked answer

Worked answer

(a)

S_{n + 1} = 1.04 S_{n} + 500

S_{0} = 1000

(b)

S_{1} = 1.04 (1000) + 500 = 1540

S_{2} = 1.04 (1540) + 500 = 2101.60

(c) The balance increases without bound as the number of years increases because interest is positive and a deposit is made every year. One limitation is that actual interest rates can change, so a constant rate may not represent future account conditions.

Why this earns credit

The recurrence applies interest before the annual deposit, so the deposit made at the end of a year does not earn interest until the next year.

Common errors

Adding the $500 deposit before applying interest, despite the deposit occurring at the end of each year.
Using 0.04 as the multiplier instead of 1.04.
Calculating the second balance from the initial balance instead of from the first balance.
Claiming the balance approaches a fixed limit even though annual deposits and positive growth continue.

How would the recurrence change if Mia made each deposit at the beginning of the year instead?

New South Wales · Mathematics Standard 2 · 12 marks

Compound interest with monthly compounding

Question

A school invests $$ 8000$ in a reserve account earning 4.8% per annum, compounded monthly. The monthly interest rate is $\frac{0.048}{12}$ .

(a) Calculate the balance after 30 months, correct to the nearest cent.

(b) Immediately after the 30-month balance is calculated, the school withdraws $$ 1500$ and leaves the remainder in the account for a further 18 months at the same rate. Calculate the final balance, correct to the nearest cent.

(c) Explain why using an annual growth factor once per year would not give the same result as the monthly-compounding model.

Reveal worked answer

Worked answer

(a) The monthly multiplier is $1 + \frac{0.048}{12}$ .

$A = 8000 {(1 + \frac{0.048}{12})}^{30} = $ 9017.82$

(b) Carry the unrounded first balance: $(8000 {(1 + \frac{0.048}{12})}^{30} - 1500) {(1 + \frac{0.048}{12})}^{18} = $ 8077.90$

(c) Monthly interest is added before later monthly interest is calculated. An annual factor changes the compounding timing and does not model the withdrawal at the stated monthly stage.

Why this earns credit

Common errors

Using 4.8% as the monthly rate or using 30 years instead of 30 months.
Rounding the first balance to cents and then reporting a final amount that cannot be reproduced from the displayed rounded working.
Applying 1.048 for each monthly period or applying one annual factor without modelling the withdrawal at the stated monthly stage.

Could you explain why the withdrawal must be applied before the further 18 months of growth?

Victoria · General Mathematics · 10 marks

Conditional relative frequency and association

Question

A class compares whether students used an online revision module or printed notes before a test. The results are shown below.

Online module: 80 students, with 56 passing and 24 not passing. Printed notes: 100 students, with 60 passing and 40 not passing. The pass-rate calculations use $\frac{56}{80}$ and $\frac{60}{100}$ .

(a) Find the conditional pass rate for each revision method.

(b) State the difference in percentage points.

(c) Calculate the percentage increase in the online-module pass rate compared with the printed-notes pass rate, correct to one decimal place.

(d) Explain why these results alone do not establish that using the online module caused the higher pass rate.

Reveal worked answer

Worked answer

(a) Online module: $\frac{56}{80} = 70 %$ . Printed notes: $\frac{60}{100} = 60 %$ .

(b) Difference: $70 % - 60 % = 10$ percentage points.

Why this earns credit

Common errors

Using the combined 180 students as the denominator for both conditional pass rates.
Calling the 10 percentage-point difference a 10% increase without using the printed-notes rate as the reference value.
Claiming that the online module caused the difference without considering how students selected their revision method.

What other difference between the two groups could affect the comparison?

Victoria · Mathematical Methods · 14 marks

Probability density, quantiles and expectation

Question

A continuous random variable $X$ has probability density function $f (x) = k x$ for $0 \leq x \leq 2$ , and $f (x) = 0$ otherwise.

(a) Find the value of $k$ .

(b) Find $P (0.5 \leq X \leq 1.5)$ .

(c) Find $E (X)$ .

(d) Find the 75th percentile of $X$ and interpret its meaning.

Reveal worked answer

Worked answer

(a) $\int_{0}^{2} k x d x = 1$ , so $2 k = 1$ and $k = \frac{1}{2}$ .

(b) $P (0.5 \leq X \leq 1.5) = \int_{0.5}^{1.5} \frac{x}{2} d x = \frac{1}{2}$ .

(c) $E (X) = \int_{0}^{2} x \frac{x}{2} d x = \frac{4}{3}$ .

(d) $F (x) = \frac{x^{2}}{4}$ . Setting $\frac{x^{2}}{4} = 0.75$ gives $x = \sqrt{3} \approx 1.73$ . About 75% of values of $X$ are at or below 1.73.

Why this earns credit

The density increases linearly over a bounded interval, allowing probabilities, expectation and a percentile to be found by integration.

Common errors

Using k = 1/4 by treating the density as a rectangle rather than integrating.
Calculating the probability from the interval width without accounting for the changing density.
Using the density function directly as the cumulative distribution function.
Choosing the negative solution when finding the percentile, despite X being between 0 and 2.

Why is the 75th percentile closer to 2 than to 1 for this increasing density?

New South Wales · Mathematics Advanced · 12 marks

Trigonometric functions, derivatives and optimisation

Question

A sensor is modelled by the height function below, where height is measured in metres and time is measured in seconds.

h (t) = 2 + 3 \sin (\frac{π}{6} t)

0 \leq t \leq 12

(a) Differentiate the height function with respect to time. (3 marks)

(b) Find all stationary times in the stated domain. Use a second-derivative test to classify each stationary point. (4 marks)

(d) Determine the interval of time for which the sensor is at least 4 metres high. Give exact endpoints and sensible decimal approximations, then interpret the result in the context of the model. (3 marks)

Reveal worked answer

Worked answer

(a) Differentiating gives

h^{'} (t) = \frac{π}{2} \cos (\frac{π}{6} t)

(b) At a stationary point, the first derivative is zero.

\cos (\frac{π}{6} t) = 0

Within the stated domain, the stationary times are

t = 3, 9

The second derivative is

h^{″} (t) = - \frac{π^{2}}{12} \sin (\frac{π}{6} t)

At the first stationary time, the second derivative is negative, so the height has a local maximum. At the second stationary time, the second derivative is positive, so the height has a local minimum.

t = 3

and is

h (3) = 5

Thus the sensor reaches a greatest height of 5 metres after 3 seconds.

(d) The required condition is

h (t) \geq 4

which is equivalent to

\sin (\frac{π}{6} t) \geq \frac{2}{3}

Let the principal angle be

α = \arcsin (\frac{2}{3})

Therefore, in the stated domain,

\frac{6}{π} \arcsin (\frac{2}{3}) \leq t \leq 6 - \frac{6}{π} \arcsin (\frac{2}{3})

So the sensor is at least 4 metres high from approximately 1.39 seconds to 4.61 seconds.

Why this earns credit

This question assesses differentiation of trigonometric functions, stationary-point classification using the second derivative, optimisation on a bounded domain and interpretation of a trigonometric inequality in context.

Common errors

Omitting the chain-rule factor when differentiating the trigonometric function.
Finding only one stationary time in the bounded domain.
Using the first derivative alone without a valid stationary-point test.
Selecting a local minimum rather than the maximum height.
Solving the trigonometric inequality in the wrong interval or outside the stated domain.

Why must the stated time domain be checked before interpreting a trigonometric model's maximum or solution interval?

Victoria · Mathematical Methods · 12 marks

Implicit differentiation and curve analysis

Question

A curved arch is modelled by $x^{2} + 4 y^{2} = 36$ , where distances are measured in metres. Only the upper part of the arch is used, so $y \geq 0$ .

(a) Use implicit differentiation to find $\frac{d y}{d x}$ in terms of $x$ and $y$ . (3 marks)

(b) Find the equation of the tangent to the arch at the point $(3, \frac{3 \sqrt{3}}{2})$ . (3 marks)

(d) State the practical horizontal domain of the arch and explain what its endpoints represent. (2 marks)

Reveal worked answer

Worked answer

(a) $2 x + 8 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = \frac{- x}{4 y}$ .

(b) At the stated point, the gradient is $- \frac{\sqrt{3}}{6}$ . The tangent is $y - \frac{3 \sqrt{3}}{2} = - \frac{\sqrt{3}}{6} (x - 3)$ .

(c) For a stationary point, $\frac{d y}{d x} = 0$ . As $y$ is positive on the upper arch, $x = 0$ . Substitution gives $y = 3$ , so the point is $(0, 3)$ . The upper branch rises then falls, so this is a maximum.

(d) The practical domain is $- 6 \leq x \leq 6$ . Its endpoints are where the arch meets ground level.

Why this earns credit

This question links implicit differentiation to the geometry of a physical arch.

Common errors

Differentiating 4y^2 as 8y without multiplying by dy/dx.
Using x = 0 without checking that the upper-arch value of y is positive.
Classifying the stationary point as a minimum because it is the lowest point on the full ellipse.
Giving the domain for y instead of the requested horizontal domain.

How would the maximum height change if the arch were widened while its equation kept the same general form?

New South Wales · Mathematics Advanced · 12 marks

Exponential models, logarithms and doubling time

Question

A colony of laboratory yeast has a mass of $80 g$ at 8:00 am. After $3 hours$ , its mass is $138.24 g$ . Its mass is modelled by $M (t) = A b^{t}$ , where $t$ is the number of hours after 8:00 am and $M$ is measured in grams.

(a) Determine the values of $A$ and $b$ , and state the model. (4 marks)

(b) Calculate the predicted mass at 2:00 pm. (2 marks)

(d) State the doubling time predicted by the model. Give one limitation of using this model over many days. (3 marks)

Reveal worked answer

Worked answer

(a) At $t = 0$ , $M (0) = A = 80$ .

Using the second observation, $80 b^{3} = 138.24$ . Therefore $b^{3} = 1.728$ and $b = 1.20$ . The model is $M (t) = 80 {(1.20)}^{t}$ .

(b) At 2:00 pm, $t = 6$ . Thus $M (6) = 80 {(1.20)}^{6} = 238.87872 g$ , or approximately $239 g$ .

(c) $80 {(1.20)}^{t} = 200$ , so ${(1.20)}^{t} = 2.5$ . Hence $t = \frac{\ln (2.5)}{\ln (1.20)} = 5.03$ hours. This is about 1:02 pm.

(d) For doubling, $80 {(1.20)}^{t} = 160$ . Therefore $t = \frac{\ln (2)}{\ln (1.20)} = 3.80$ hours, approximately 3 hours 48 minutes. Over many days, changing nutrients, space and waste products make a constant growth rate unrealistic.

Why this earns credit

This modelling question requires students to determine an exponential model from data, make a prediction, solve a logarithmic equation and evaluate the model's limitations.

Common errors

Using the mass after three hours as the initial value of A.
Treating 1.20 as a 1.20% hourly growth rate rather than a growth factor.
Using t = 14 for 2:00 pm instead of measuring time from 8:00 am.
Taking logarithms before isolating the exponential term.

How would the predicted mass change if the hourly growth factor began to decrease after the first six hours?

Victoria · Mathematical Methods · 14 marks

Related rates and optimisation

Question

A school fundraiser models its revenue, in dollars, from selling a number of hundred raffle tickets.

R (x) = 120 x - 2 x^{2}

0 \leq x \leq 50

(a) Find the derivative and explain the meaning of its value when the number of hundred tickets is 20.

R^{'} (x)

R^{'} (20)

(b) Find the stationary point of the revenue function and use a second-derivative test to classify it.

(c) Determine the number of raffle tickets that maximises revenue and state the maximum revenue. Justify that your answer is optimal on the stated domain.

Reveal worked answer

Worked answer

(a)

R^{'} (x) = 120 - 4 x

R^{'} (20) = 40

When 2000 tickets have been sold, revenue is increasing at $40 for each additional hundred tickets sold.

(b)

R^{'} (x) = 0

120 - 4 x = 0

x = 30

R^{″} (x) = - 4

R^{″} (30) < 0

The stationary point is a maximum.

R (30) = 120 (30) - 2 30^{2} = 1800

The maximum revenue is $1800. The stationary point is within the stated domain and the negative second derivative confirms it is optimal.

Why this earns credit

This question uses a quadratic revenue model and connects derivative values to a practical rate of change.

Common errors

Differentiating the quadratic term as negative 2x instead of negative 4x.
Interpreting x = 30 as 30 tickets rather than 30 hundred tickets.
Stating that the stationary point is a maximum without testing the second derivative.
Giving the maximum number of tickets but not evaluating the corresponding revenue.

How would the maximum change if the fundraiser could sell no more than 2500 tickets?

New South Wales · Mathematics Advanced · 13 marks

Integration, average value and area

Question

Water enters a tank at the positive rate $r (t) = 2 + t^{2}$ litres per minute, where $0 \leq t \leq 3$ and $t$ is measured in minutes.

(a) Find an antiderivative $R (t)$ of $r (t)$ . (3 marks)

(b) Find the total volume of water entering the tank during the first 3 minutes. (3 marks)

(d) An endpoint estimate uses $3 r (3)$ litres. Compare this estimate with the exact volume. (5 marks)

Reveal worked answer

Worked answer

(a) An antiderivative is $R (t) = 2 t + \frac{t^{3}}{3} + C$ .

(b) $\int_{0}^{3} (2 + t^{2}) d t = {[2 t + \frac{t^{3}}{3}]}_{0}^{3} = 15$ litres.

(d) $r (3) = 11$ litres per minute, so the endpoint estimate is $3 r (3) = 33$ litres. It is an overestimate by $33 - 15 = 18$ litres because the rate is increasing over the interval.

Why this earns credit

This question links antiderivatives, accumulated change, average value and a right-endpoint estimate for an increasing rate function.

Common errors

Using t²/2 as an antiderivative of t².
Evaluating the definite integral as 18 litres rather than 15 litres.
Reporting the total volume rather than the average rate.
Using 3r(0) rather than 3r(3) for the stated endpoint estimate.
Calling the endpoint estimate an underestimate despite the increasing rate.

Why does using the rate at the final time give an overestimate in this situation?

Victoria · Mathematical Methods · 12 marks

Normal distribution and inverse probability

Question

The mass $X$ , in grams, of packets produced by a machine is normally distributed with mean 50 g and standard deviation 4 g.

(a) Standardise a packet mass of 55 g by finding its $z$ -score.

(b) Find the probability that a randomly selected packet has mass greater than 55 g. Give your answer correct to four decimal places.

(d) State one assumption of this normal model and explain why checking this assumption matters for the probability calculated in part (b).

Reveal worked answer

Worked answer

(a) $z = \frac{55 - 50}{4} = 1.25$ .

(b) $P (X > 55) = P (Z > 1.25) = 0.1056$ .

$m = 50 + 1.2816 \times 4 = 55.1264$ . Therefore, $m = 55.13$ g.

(d) One assumption is that packet masses are approximately normally distributed with a stable mean and standard deviation. This matters because the normal-distribution probability of 0.1056 is only a suitable estimate if that model reasonably represents the packet masses.

Why this earns credit

This draft assesses standardisation, upper-tail probability, inverse normal reasoning and critical interpretation of a normal model.

Common errors

Using 55/4 instead of subtracting the mean before dividing by the standard deviation.
Reporting the lower-tail probability 0.8944 instead of the required upper-tail probability.
Using the 10th-percentile z-score when finding the 90th percentile.
Subtracting the percentile z-score from the mean rather than adding it.
Stating that the normal model is exact rather than describing it as an assumption or approximation.

What production data would you inspect to decide whether the normal model is reasonable for this machine?

New South Wales · Mathematics Advanced · 12 marks

Trigonometric modelling and exact identities

Question

The height $H (t)$ metres of a rotating platform above ground, $t$ hours after 6:00 am, is modelled by $H (t) = 4 + 3 \sin (\frac{π t}{6})$ , where $0 \leq t \leq 12$ .

(a) State the midline, amplitude and period of the model.

(b) Find the exact height of the platform at 8:00 am.

(d) Interpret your solution to part (c) in the context of the platform.

Reveal worked answer

Worked answer

(a) The midline is $H = 4$ , the amplitude is $3$ metres, and the period is $12$ hours.

(b) At 8:00 am, $t = 2$ . Therefore $H (2) = 4 + 3 \sin (\frac{π}{3}) = 4 + \frac{3 3^{\frac{1}{2}}}{2}$ metres.

(d) The platform is higher than $\frac{11}{2}$ metres from just after 7:00 am until just before 11:00 am.

Why this earns credit

This draft assesses interpretation of sinusoidal parameters, an exact special-angle value and solving a trigonometric inequality over one full cycle.

Common errors

Using 3 metres as the midline rather than as the amplitude.
Giving a period of 6 hours by confusing the denominator with the period.
Using sin(pi/3) = 1/2.
Including t = 1 or t = 5 despite the strict inequality.
Selecting the complementary interval where the platform is below the threshold.

How would the interval change if the threshold height were increased?

Victoria · Mathematical Methods · 12 marks

Exponential functions, differentiation and definite integrals

Question

The output rate of a device is modelled by $g (x) = x e^{- \frac{x}{2}}$ , for $x \geq 0$ .

(a) Find $g' (x)$ .

(b) Show that the only stationary point occurs at $x = 2$ .

(c) Determine whether this stationary point is a maximum or minimum.

(d) Find its value.

(e) Find the exact area under the graph of $g$ between $x = 0$ and $x = 2$ .

(f) Interpret this area in the context of the device.

Reveal worked answer

Worked answer

(a) $g' (x) = e^{- \frac{x}{2}} (1 - \frac{x}{2})$

(b) Since $e^{- \frac{x}{2}}$ is never zero, the derivative is zero only when $1 - \frac{x}{2} = 0$ , so $x = 2$ .

(c) The derivative is positive before 2 and negative after 2, so the stationary point is a maximum.

(d) $g (2) = \frac{2}{e}$ .

(e) $\int_{0}^{2} x e^{- \frac{x}{2}} dx = [- (2 x + 4) e^{- \frac{x}{2}}] 02 = 4 - \frac{8}{e}$ .

(f) The area represents the total output accumulated while x increases from 0 to 2.

Why this earns credit

Common errors

Forgetting the derivative of the exponential factor when applying the product rule.
Assuming the exponential factor can equal zero.
Giving the stationary point without checking whether the derivative changes sign.
Using an incorrect antiderivative for xe^(-x/2).
Reporting a decimal approximation when an exact value is requested.

Could you justify the maximum using the second derivative instead of a sign chart?

New South Wales · Mathematics Advanced · 12 marks

Calculus: stationary points, tangents and area

Question

The curve C has equation $y = f (x) = x^{3} - 6 x^{2} + 9 x + 2$ .

(a) Find the coordinates of the stationary points of C and determine the nature of each stationary point. (4 marks)

(b) Find the equation of the tangent to C at $x = 2$ . (3 marks)

(c) Find the total area between C and this tangent for $1 \leq x \leq 3$ . Give an exact answer. (5 marks)

Reveal worked answer

Worked answer

(a) $f' (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ , so the stationary points occur at x = 1 and x = 3. Since f(1) = 6 and f(3) = 2, they are (1, 6) and (3, 2). Also, $f'' (x) = 6 x - 12$ , so f''(1) < 0 gives a local maximum at (1, 6), while f''(3) > 0 gives a local minimum at (3, 2).

(b) $f (2) = 4, f' (2) = - 3$ . Therefore the tangent is $y - 4 = - 3 (x - 2), y = - 3 x + 10$ .

(c) The vertical difference is $f (x) - (- 3 x + 10) = {(x - 2)}^{3}$ . The curve is below the tangent on [1, 2] and above it on [2, 3]. Thus the total area is $- \int_{1}^{2} {(x - 2)}^{3} dx + \int_{2}^{3} {(x - 2)}^{3} dx = 1 / 4 + 1 / 4 = 1 / 2$ square units.

Why this earns credit

The item combines core Advanced calculus skills without requiring Extension-only methods. The tangent is used as a comparison function, so the sign change at x = 2 must be handled before integrating.

Common errors

Uses f(x) = 0 instead of f'(x) = 0 when finding stationary points.
Uses the curve's y-value but not its derivative as the tangent gradient.
Integrates the signed difference from 1 to 3 without splitting where the curve and tangent meet.
Classifies the stationary points without checking local behaviour.

Which step made the area calculation require a split at x = 2?

New South Wales · Business Studies · 12 marks

Human resources, training and workplace change

Question

Coastal Kitchens Pty Ltd is an Australian manufacturer of fitted kitchen cabinets. It is introducing computer-controlled cutting equipment at its Newcastle factory. The new equipment is expected to reduce material waste and shorten production time. However, 24 production employees have not used digital design or cutting software before. Management is considering a two-day paid training program, followed by mentoring from experienced machine operators for six weeks.

(a) Analyse how the proposed training and mentoring strategy could support employees during this workplace change. (4 marks)

(b) Explain how this strategy could contribute to business outcomes for Coastal Kitchens. (4 marks)

(c) Recommend whether Coastal Kitchens should proceed with the proposed strategy. Justify your recommendation using the stimulus and both employee and business outcomes. (4 marks)

Reveal worked answer

Worked answer

(a) Paid training gives employees the digital and equipment skills needed before they use unfamiliar machinery. Because employees are paid, the business recognises training as part of their work rather than shifting the cost to them. Six weeks of mentoring gives practical support while employees apply the new skills, ask questions and correct mistakes. This can reduce uncertainty and resistance to the workplace change.

(b) Better skills can reduce incorrect cuts and rework, supporting the expected reduction in material waste. Employees who can operate the equipment confidently are more likely to use it efficiently, helping to shorten production time. Mentoring can identify operating problems early and support more consistent cabinet quality.

(c) Coastal Kitchens should proceed with the strategy. The two-day paid program directly addresses the skills gap affecting the 24 employees, while mentoring supports them as they transfer learning into everyday work. Employees are more likely to feel capable and supported, reducing resistance and disruption. The business is therefore more likely to achieve lower waste, faster production and consistent product quality.

Why this earns credit

This case-study response links training and mentoring to employee support during workplace change and to operational business outcomes.

Common errors

Describing the technology itself without analysing the training and mentoring response.
Treating mentoring as a one-off induction rather than ongoing workplace support.
Listing employee benefits without linking them to waste, time, quality or rework outcomes.
Giving an unsupported recommendation that does not use details from the case.

Which performance measures could show whether mentoring is improving both employee confidence and production quality?

Victoria · Specialist Mathematics · 14 marks

Differential equations, slope fields and modelling

Question

The temperature T degrees C of a cooling liquid is modelled by

\frac{d T}{d t} = - 0.20 (T - 20)

where t is measured in minutes. Initially,

T = 80

(a) Show that

T = 20 + 60 e^{- 0.20 t}

satisfies both the differential equation and the initial condition.

(b) State the equilibrium temperature and explain its meaning in this model.

(d) State one limitation of this model.

Reveal worked answer

Worked answer

(a)

\frac{d T}{d t} = - 12 e^{- 0.20 t}

and

- 0.20 (T - 20) = - 0.20 [60 e^{- 0.20 t}] = - 12 e^{- 0.20 t}

so the differential equation is satisfied. At

t = 0, T = 20 + 60 = 80

so the initial condition is satisfied.

(b) The equilibrium temperature is 20 degrees C. The liquid approaches the surrounding temperature as time increases.

(c)

30 = 20 + 60 e^{- 0.20 t}

e^{- 0.20 t} = \frac{1}{6}

t = \frac{\ln (6)}{0.20} = 9.0 \min

(d) One limitation is that the surrounding temperature is assumed to remain constant at 20 degrees C.

Why this earns credit

This draft uses a first-order cooling model to assess verification, equilibrium, logarithmic solving and modelling assumptions.

Common errors

Differentiates the exponential term without multiplying by -0.20.
Treats 20 degrees C as a temperature reached in finite time.
Uses log(1/6) but omits the negative sign when solving for time.
Gives a limitation unrelated to the assumptions of the cooling model.

How would the model change if the surrounding temperature increased during the cooling process?

New South Wales · Business Studies · 12 marks

Globalisation, operations and ethical sourcing

Question

Harbour Trail Gear is an Australian business that sells reusable hiking bottles and lunch containers. Its current Australian supplier charges a higher unit price but provides recycled stainless steel and audited records showing safe working conditions. A proposed overseas supplier would reduce the purchase cost by 18% and could deliver larger orders. However, it has not yet provided evidence about worker safety, wage practices or the source of its steel. Shipping from the proposed supplier would also increase the business's transport distance.

(a) Analyse how the proposed supplier's lower purchase cost could affect Harbour Trail Gear's cost efficiency. (4 marks)

(b) Explain one ethical issue Harbour Trail Gear should investigate before changing suppliers. (4 marks)

(c) Recommend whether Harbour Trail Gear should change suppliers at this stage. Justify your decision using the stimulus evidence. (4 marks)

Reveal worked answer

Worked answer

(a) The lower purchase cost could improve cost efficiency by reducing Harbour Trail Gear's input cost for each product. This may increase its profit margin if prices remain unchanged, or allow the business to lower prices and compete more strongly. The benefit should be assessed against total supply costs, including longer-distance transport and any additional inventory needed for longer lead times.

(b) Harbour Trail Gear should investigate worker safety and wage practices. The proposed supplier has provided no evidence that workers are safe or paid fairly. If poor labour practices are found, continuing with that supplier could breach the business's ethical responsibilities and damage customer trust and reputation.

(c) Harbour Trail Gear should not change suppliers at this stage. The lower purchase cost and capacity for larger orders are potential benefits, but the supplier has not supplied evidence about worker conditions or steel sourcing. The business should first obtain credible audit evidence, check delivery reliability and compare the environmental effect of additional transport with the current supplier's recycled steel.

Why this earns credit

The response focuses each part on its stated requirement: cost efficiency, one ethical issue, then an evidence-based supplier decision.

Common errors

Assuming lower purchase cost guarantees higher profit without considering total supply costs.
Calling the overseas supplier unethical solely because it is overseas rather than referring to the missing labour-practice evidence.
Ignoring the current supplier's audited working-condition records and recycled steel.
Making a recommendation without balancing the stated benefits against the missing evidence.

What audit and delivery evidence should Harbour Trail Gear obtain before it makes a final supplier decision?

Victoria · Specialist Mathematics · 12 marks

Complex numbers and loci

Question

On an Argand plane, let $z = x + y i$ satisfy $| z - (2 + i) | = 3$ .

(a) Write the locus as an equation in $x$ and $y$ . (3 marks)

(b) Describe the locus geometrically. (3 marks)

(c) Find the exact points where the locus intersects the real axis. (6 marks)

Reveal worked answer

Worked answer

(a) $| z - (2 + i) | = \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}}$ , so ${(x - 2)}^{2} + {(y - 1)}^{2} = 9$ .

(b) The locus is a circle with centre $2 + i$ and radius $3$ .

(c) On the real axis, $y = 0$ . Therefore ${(x - 2)}^{2} + 1 = 9$ , so ${(x - 2)}^{2} = 8$ . Hence $x = 2 \pm 2 \sqrt{2}$ . The intersection points are $z = 2 + 2 \sqrt{2}$ and $z = 2 - 2 \sqrt{2}$ .

Why this earns credit

The modulus condition gives the distance from the fixed complex number 2 + i, which identifies a circle directly.

Common errors

Using 3 rather than 9 on the right-hand side after squaring the modulus.
Giving the centre as 2 - i instead of 2 + i.
Setting x = 0 rather than y = 0 when finding real-axis intersections.
Leaving the intersections as decimal approximations rather than exact values.

How would the equation change if the centre moved to -1 + 4i while the radius remained 3?

New South Wales · Business Studies · 12 marks

Marketing strategy, segmentation and digital channels

Question

HarbourMove is a Sydney-based service that collects, stores and returns sporting equipment for apartment residents. Its current customers are professionals aged 25 to 40 who live in inner-city apartments, play social sport and value app-based booking. HarbourMove has grown through local social-media advertisements and referrals. The business is considering two strategies: Strategy 1: use customer-location and sport-preference data to send tailored app offers for collection times and storage packages. Strategy 2: partner with three inner-city gyms, offering gym members a first-month storage discount through each gym's digital newsletter.

(a) Analyse how HarbourMove's current target market reflects market segmentation.

(b) Analyse how each proposed strategy could support HarbourMove's marketing objectives.

(c) Recommend one strategy for HarbourMove. Justify your recommendation using the case evidence.

Reveal worked answer

Worked answer

(a) HarbourMove uses demographic segmentation by targeting professionals aged 25 to 40. It uses geographic segmentation by concentrating on inner-city apartment residents. It also uses behavioural and psychographic characteristics because customers play social sport and prefer convenient app-based booking. These segments identify people likely to have equipment-storage needs and value a convenient service.

(b) Strategy 1 can improve customer retention and sales conversion because tailored offers match a customer's location, sport and likely booking needs. It also uses HarbourMove's existing app-based service model. Strategy 2 can increase market reach by placing the service before gym members who are likely to own sporting equipment. The gym newsletters provide a relevant digital channel and the introductory discount may encourage trial.

(c) HarbourMove should prioritise Strategy 2. Its gym partners offer access to a concentrated group with a clear connection to sporting equipment, while the newsletters provide a credible digital route to new customers. This suits the business's existing local growth pattern through social media and referrals. Strategy 1 could be introduced later after the business has stronger customer-data systems and clear consent processes.

Why this earns credit

This case asks students to connect segmentation variables, digital marketing channels and a justified strategy choice to evidence from a service business.

Common errors

Listing segmentation categories without linking them to HarbourMove's customers or service need.
Describing the two strategies without analysing how they could meet a marketing objective.
Recommending both strategies equally rather than making a supported priority choice.
Claiming that tailored app offers can be sent without recognising the need for appropriate customer data and consent.

What customer information would HarbourMove need before it could make tailored offers useful rather than intrusive?

Victoria · Specialist Mathematics · 10 marks

Vectors and geometric reasoning in three dimensions

Question

Let A = (1, 2, -1), B = (4, -1, 2) and C = (2, 4, 1).

(a) Find the vectors AB and AC. (2 marks)

(b) Find the angle BAC, correct to one decimal place. (3 marks)

(c) The foot H of the perpendicular from A to line BC has the form H = B + lambda(C - B). Find lambda and the coordinates of H. (3 marks)

(d) Find the exact length AH. (2 marks)

Reveal worked answer

Worked answer

AB = (3, -3, 3) and AC = (1, 2, 2).

AB · AC = 3, |AB| = 3 × √3 and |AC| = 3. Therefore $\cos (BAC) = \frac{1}{3 \times √3}$ and angle BAC = 78.9 degrees.

Let d = C - B = (-2, 5, -1). The perpendicular condition gives (A - H) · d = 0. Since (A - B) · d = 24 and d · d = 30, $λ = \frac{4}{5}$ and H = (12/5, 3, 6/5).

AH = (7/5, 1, 11/5), so |AH| = √(39/5).

Why this earns credit

The response uses dot products for both angle and perpendicularity. The parameter on line BC is found by imposing the geometric condition AH perpendicular to BC, then the exact distance is calculated from the resulting coordinates.

Common errors

Subtracts coordinates in the wrong order when forming AB or AC.
Uses a two-dimensional slope instead of the three-dimensional dot-product angle formula.
Uses H = B + lambda(C - B) without applying the perpendicular condition.
Reports a decimal distance when an exact form is requested.

Could you verify the perpendicular foot by checking both H lies on BC and AH dot BC equals zero?

New South Wales · Business Studies · 12 marks

Finance, cash flow and short-term decisions

Question

Coastline Catering prepares weekday lunches for offices. At the beginning of September, it has $$ 18,000$ cash in its bank account. During September, it expects the following cash movements:

Cash collected from earlier invoices on 5 September: $$ 10,000$
Cash received from current customers on 30 September: $$ 35,000$
Wages paid by 15 September: $$ 16,000$
Payments to food suppliers due by 15 September: $$ 28,000$
Loan repayment due by 15 September: $$ 5,000$
Online advertising payment due by 15 September: $$ 4,000$

Its September profit forecast is $$ 14,000$ because some supplier costs will be paid in October. Assume no other cash is available during September.

(a) Calculate Coastline Catering's forecast closing cash balance for September. Then calculate its cash position immediately after all payments due by 15 September have been made but before the 30 September customer receipts are received. Identify the cash-flow problem. (5 marks)

(b) Explain one suitable short-term finance option that could assist the business during September. (3 marks)

(c) Recommend a short-term response to the cash-flow problem. Justify your recommendation using the timing and amount of the cash shortfall, and distinguish clearly between profit and liquidity. (4 marks)

Reveal worked answer

Worked answer

(a) Total cash inflows: $$ 35,000 + $ 10,000 = $ 45,000$ . Total cash outflows: $$ 16,000 + $ 28,000 + $ 5,000 + $ 4,000 = $ 53,000$ .

Forecast closing cash: $$ 18,000 + $ 45,000 - $ 53,000 = $ 10,000$ .

Cash position before the 30 September customer receipts: $$ 18,000 + $ 10,000 - $ 53,000 = - $ 25,000$ . This is a temporary liquidity shortfall, despite the positive month-end cash balance.

(b) A bank overdraft is suitable short-term finance. It allows Coastline Catering to draw up to an agreed limit when its account lacks sufficient funds, then repay the amount when the customer receipts arrive. Interest is generally charged on the amount used.

(c) Coastline Catering should arrange an overdraft limit of at least $$ 25,000$ before the payments fall due. It matches the temporary shortfall and can be repaid when the customer receipt arrives on 30 September. Profit is the accounting surplus of revenue over expenses for a period; the forecast profit does not mean cash is available when bills are due. Liquidity is the ability to meet obligations on time, which is the immediate problem.

Why this earns credit

The month-end cash balance alone does not show whether a business can pay obligations as they fall due. Cash-flow timing creates a temporary liquidity shortfall that requires a short-term response.

Common errors

Treating the $14,000 profit forecast as cash available before bills are due.
Omitting the opening cash balance when calculating closing cash.
Ignoring the dates of cash receipts and payments when assessing liquidity.
Assuming a positive month-end closing balance means the business can meet every September obligation on time.
Recommending long-term finance without explaining why it suits a temporary timing shortfall.

How would delaying the $35,000 customer receipt until October affect Coastline Catering's short-term finance requirement?

South Australia · Mathematical Methods · 12 marks

Optimisation with derivatives and constraints

Question

A closed rectangular box has a square base with side length $x$ centimetres and height $h$ centimetres. Its volume is

500 {cm}^{3}

The base and lid material costs three cents per square centimetre. The material for the four side faces costs one cent per square centimetre. The feasible domain is

2 \leq x \leq 20

(a) Show that the material cost, in cents, can be modelled by

C (x) = 6 x^{2} + \frac{2000}{x}

(3 marks)

(b) Find the derivative of the cost function and solve the stationary-point equation. (4 marks)

(c) Use a stationary-point test and the stated domain to determine the dimensions that minimise the cost. Give the minimum cost to the nearest cent. (5 marks)

Reveal worked answer

Worked answer

(a) The volume constraint gives

h = \frac{500}{x^{2}}

The base and lid cost

3 (2 x^{2}) = 6 x^{2}

cents. The sides cost

4 x h = \frac{2000}{x}

cents. Hence the stated model follows.

(b) The derivative is

C^{'} (x) = 12 x - \frac{2000}{x^{2}}

Setting this equal to zero gives

x = {\frac{500}{3}}^{\frac{1}{3}} \approx 5.503

C^{″} (x) = 12 + \frac{4000}{x^{3}} > 0

Thus the stationary point is a minimum. It lies in the domain, and its cost is less than the endpoint costs. The dimensions are

x \approx 5.503 cm, h \approx 16.510 cm

The minimum cost is

C (5.503) \approx 545.14

cents, so the minimum cost to the nearest cent is 545 cents.

Why this earns credit

This question assesses construction of a constrained cost model, differentiation and stationary-point solving, and justification of an absolute minimum on a bounded feasible domain.

Common errors

Omitting either the lid or base from the higher-cost material area.
Using the base side length rather than its square in the volume constraint.
Differentiating the reciprocal cost term with the wrong sign.
Stopping at the stationary point without checking the stationary-point type and feasible domain.
Rounding the cost before completing the calculation.

How would increasing the cost of the side material affect the optimal proportions of the box?

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