Skill Align Mathematics Applications Year 12 Section Two Calculator-assumed for WACE - 2026 Edition
Original Skill Align practice examination content
- Paper
- Section Two Calculator-assumed Question and Response Book Showcase
- Reading
- 10 minutes reading time
- Writing
- 100 minutes
- Assessment
- 100 marks
Standard items: pens, pencils including coloured pencils, sharpener, correction fluid or tape, eraser, ruler and highlighters. Special items: drawing instruments, templates, notes on two unfolded sheets of A4 paper, and up to three calculators, which may include scientific, graphic and CAS calculators. A formula sheet is provided by the supervisor. Candidates are assumed to have CAS capability. An examination changeover period of up to 15 minutes applies, during which candidates are not permitted to work.
Section Two
Answer all questions. For any question or part question worth more than two marks, valid working or justification is required to receive full marks.
Question 1
12 marksQuestion 2
13 marksQuestion 3
12 marksQuestion 4
13 marksQuestion 5
12 marksQuestion 6
13 marksQuestion 7
12 marksQuestion 8
13 marksWorked Solutions And Marking Guide
Section Two Question 1
(a) hat y=55.118+2.608x.
Technology gives slope 2.608262... and intercept 55.118052....
(b) r^2approx0.9990; about 99.90% of the variation in enquiries is explained by its linear association with advertising contacts in this dataset.
Use the regression output and identify response and explanatory variables.
(c) hat yapprox164.665, so the residual is 161-164.665approx-3.665.
Substitute into the fitted model, then use actual minus predicted.
(d) The residuals show no systematic curve or funnel, which supports linearity over the observed range; the small dataset and extrapolation remain limitations.
Judge the residual pattern rather than the high coefficient alone.
Mark allocation
- Part (a) (3 marks): 1 mark for entering the paired data in corresponding lists; 1 mark for obtaining the gradient 2.608 to three decimal places; 1 mark for obtaining the intercept 55.118 to three decimal places.
- Part (b) (3 marks): 1 mark for reporting r^2approx0.9990; 1 mark for interpreting it as a proportion of variation explained by the linear model; 1 mark for naming enquiries and advertising contacts in the interpretation.
- Part (c) (3 marks): 1 mark for calculating the prediction 164.665; 1 mark for using actual minus predicted for the residual; 1 mark for obtaining the signed residual approximately -3.665.
- Part (d) (3 marks): 1 mark for producing or describing residuals against the explanatory variable; 1 mark for identifying the absence of a systematic pattern; 1 mark for stating a dataset-size, context or extrapolation limitation.
Section Two Question 2
(a) The indices sum to 4 and average 1. Quarter 3 demand is typically 15% above the underlying trend.
Check the mean seasonal index and compare 1.15 with 1.
(b) 282 / 1.15approx245.22.
Divide the observed Quarter 3 value by its index.
(c) hat T=214.562+3.817t, with r^2approx0.564.
Fit regression to time and the deseasonalised values.
(d) At t=13, the trend estimate is 264.19; reseasonalising gives 264.19(0.80)approx211.35, or about 211 units. The moderate r² and extrapolation beyond three years limit reliability.
Forecast the trend first, then restore the Quarter 1 seasonal effect.
Mark allocation
- Part (a) (3 marks): 1 mark for showing that the four indices sum to 4; 1 mark for showing that their mean is 1; 1 mark for interpreting 1.15 as 15% above trend in Quarter 3.
- Part (b) (3 marks): 1 mark for selecting the Year 2 Quarter 3 value 282; 1 mark for dividing by 1.15; 1 mark for obtaining approximately 245.22.
- Part (c) (3 marks): 1 mark for creating all 12 deseasonalised values; 1 mark for obtaining the trend gradient 3.817; 1 mark for obtaining the intercept 214.562 and r^2approx0.564.
- Part (d) (4 marks): 1 mark for substituting t=13 into the trend model; 1 mark for obtaining the trend estimate approximately 264.19; 1 mark for reseasonalising with the Quarter 1 index to obtain about 211; 1 mark for using the moderate r², short series or extrapolation to qualify reliability.
Section Two Question 3
(a) A_(n+1)=1.0055A_n+450, A_0=3000. Interest is applied before the end-of-month deposit.
The multiplier acts on the existing balance only.
(b) A_(36)approx21515.92. Deposits total 19200, so interest earned is 2315.92.
Iterate without premature rounding, then separate contributions from growth.
(c) A_(42)approx24973.18<25000<A_(43)approx25560.53, so the first month is 43.
Compare consecutive recurrence-table entries around the target.
(d) The model is C_(n+1)=1.0055(C_n+450). It gives C_(36)approx21614.15, about 98.24 more because each deposit earns one additional month of interest.
Move the deposit inside the interest multiplication.
Mark allocation
- Part (a) (3 marks): 1 mark for stating the multiplier 1.0055; 1 mark for stating the end-of-month deposit and initial condition; 1 mark for explaining that interest precedes the deposit.
- Part (b) (3 marks): 1 mark for iterating the recurrence to A_(36); 1 mark for obtaining the balance 21,515.92; 1 mark for subtracting total deposits to obtain 2,315.92 interest.
- Part (c) (3 marks): 1 mark for locating month 42 below the target; 1 mark for locating month 43 above the target; 1 mark for concluding that month 43 is the first threshold month.
- Part (d) (3 marks): 1 mark for modelling the beginning-of-month timing correctly; 1 mark for obtaining the 36-month balance 21,614.15; 1 mark for quantifying and explaining the approximately 98.24 advantage.
Section Two Question 4
(a) For example, S-A-C-T has bottleneck min(10,6,8)=6.
Read arc directions and take the smallest residual capacity on the path.
(b) One feasible flow sends 6 on S-A-C-T, 4 on S-A-D-T, 2 on S-B-C-T, and 3 on S-B-D-T.
The stated path flows respect every capacity and conserve flow at intermediate vertices.
(c) The maximum flow is 15. The cut formed by the arcs entering T, Cto T and Dto T, has capacity 8+7=15.
A feasible flow matching a cut capacity is maximum.
(d) The new maximum flow is 18. For example, send 6 from A to C, 4 from A to D, 2 from B to C, and 6 from B to D. This saturates Sto A and Sto B, whose total capacity is 10+8=18.
Construct a feasible 18-unit flow and use the source cut.
Mark allocation
- Part (a) (3 marks): 1 mark for giving a valid directed source-to-sink path; 1 mark for listing the capacities on that path; 1 mark for identifying its bottleneck capacity.
- Part (b) (3 marks): 1 mark for constructing source-to-sink augmentations totalling 15; 1 mark for respecting every arc capacity; 1 mark for showing flow conservation at intermediate vertices.
- Part (c) (3 marks): 1 mark for stating the feasible flow value 15; 1 mark for identifying the cut Cto T,Dto T; 1 mark for using cut capacity 15 to prove optimality.
- Part (d) (4 marks): 1 mark for updating the changed sink-arc capacity; 1 mark for constructing a feasible 18-unit flow; 1 mark for identifying the source cut capacity 18; 1 mark for concluding maximum flow 18 from matching flow and cut values.
Section Two Question 5
(a) EF_A=4, EF_B=5, EF_C=10, EF_D=8, EF_E=14. The project duration is 14 days.
At a merge, use the latest predecessor finish.
(b) Activity E must start by day 10, D by day 7, so B may start as late as day 2.
Work back from day 14 using the earliest required successor start.
(c) The critical activities are A,C,E. Activities B and D each have 2 days of float.
Compare earliest and latest starts.
(d) Both branches into E then finish at day 8, so the project duration is 12 days. The critical paths are A-C-E and B-D-E.
Repeat the forward scan with the changed duration.
Mark allocation
- Part (a) (3 marks): 1 mark for obtaining earliest finishes 4 and 5 for A and B; 1 mark for obtaining earliest finishes 10 and 8 for C and D; 1 mark for obtaining earliest finish 14 for E and the project.
- Part (b) (3 marks): 1 mark for starting the backward scan from project time 14; 1 mark for obtaining latest start 7 for D; 1 mark for obtaining latest start 2 for B.
- Part (c) (3 marks): 1 mark for identifying A,C,E as zero-float activities; 1 mark for calculating 2 days of float for B; 1 mark for calculating 2 days of float for D.
- Part (d) (3 marks): 1 mark for recalculating both branches to finish at day 8; 1 mark for obtaining the new project duration 12 days; 1 mark for identifying both critical paths.
Section Two Question 6
(a) begin(pmatrix)7&0&5&63&1&0&44&7&0&73&2&5&0end(pmatrix).
Subtract the minimum entry from each row.
(b) The first-column minimum is 3. Subtracting it gives begin(pmatrix)4&0&5&60&1&0&41&7&0&70&2&5&0end(pmatrix). Independent zeros may be selected at A2,B1,C3,D4.
The first column requires reduction before independent zeros are selected.
(c) Assign Ato2, Bto1, Cto3, Dto4. The minimum cost is 2+6+1+4=13.
Evaluate the chosen assignment in the original matrix.
(d) An optimal feasible assignment is Ato2, Bto3, Cto1, Dto4, with cost 2+3+5+4=14. The minimum increases by 1.
Remove the prohibited cell and repeat the assignment search.
Mark allocation
- Part (a) (3 marks): 1 mark for identifying row minima 2, 3, 1 and 4; 1 mark for subtracting each row minimum correctly; 1 mark for showing the complete row-reduced matrix.
- Part (b) (3 marks): 1 mark for identifying the first-column minimum as 3; 1 mark for subtracting 3 from the first column correctly; 1 mark for selecting four independent zeros.
- Part (c) (3 marks): 1 mark for stating a one-to-one assignment; 1 mark for reading the four original costs; 1 mark for obtaining the minimum total 13.
- Part (d) (4 marks): 1 mark for excluding the prohibited C-to-3 allocation; 1 mark for finding a feasible one-to-one replacement assignment; 1 mark for obtaining the new minimum cost 14; 1 mark for stating the increase of 1.
Section Two Question 7
(a) A: 6.0411%; B: 5.8000%; C: 5.8439%.
Use the compounding frequency for each nominal rate.
(b) A, then C, then B. Nominal rates use different compounding frequencies and are not directly comparable.
Compare rates on a common effective basis.
(c) A: 22489.42; B: 22387.28; C: 22405.85.
Use each plan's actual period rate and number of periods.
(d) Plan B then grows 19920 to about 22297.73. Plan A remains best at 22489.42.
Apply the fee before compounding and compare final values.
Mark allocation
- Part (a) (3 marks): 1 mark for converting Plan A with 12 compounding periods; 1 mark for retaining Plan B as the stated effective rate; 1 mark for converting Plan C with four compounding periods.
- Part (b) (3 marks): 1 mark for ranking A above C above B; 1 mark for identifying different compounding frequencies; 1 mark for explaining the need for a common effective rate.
- Part (c) (3 marks): 1 mark for obtaining Plan A's value to cents; 1 mark for obtaining Plan B's value to cents; 1 mark for obtaining Plan C's value to cents.
- Part (d) (3 marks): 1 mark for reducing Plan B's invested principal to 19,920; 1 mark for obtaining its fee-adjusted final value about 22,297.73; 1 mark for selecting Plan A using the final-value comparison.
Section Two Question 8
(a) Use L_(mathrm(recent)): it is fitted to the most recent six months and has the smaller absolute residual at each of months 10, 11 and 12.
Compare recency and the stated prediction errors without diagnosing a residual-plot pattern.
(b) L_(mathrm(recent))(18)=32.4+7.45(18)=166.50.
Substitute month 18 into the selected linear trend.
(c) 166.50(1.08)=179.82. Capacity 170 is insufficient by 9.82, so another arrangement is required.
Restore the seasonal effect before comparing with capacity.
(d) The recent line uses only six observations and month 18 is six periods beyond the latest data. Retain both forecasts as a sensitivity range, compare each new actual value with its forecast, and refit if the absolute forecast error exceeds 10 units for three consecutive months.
Evaluate sample size and extrapolation, then give a numerical error rule.
Mark allocation
- Part (a) (3 marks): 1 mark for selecting the recent-window linear trend; 1 mark for using the more recent fitting window; 1 mark for using its smaller supplied absolute residuals.
- Part (b) (3 marks): 1 mark for selecting the recent-window equation; 1 mark for substituting t=18; 1 mark for obtaining the deseasonalised forecast 166.50.
- Part (c) (3 marks): 1 mark for multiplying by the month-18 seasonal index; 1 mark for obtaining demand 179.82; 1 mark for comparing with capacity 170 and quantifying the shortfall.
- Part (d) (4 marks): 1 mark for identifying the six-observation recent window; 1 mark for identifying month 18 as extrapolation; 1 mark for proposing comparison with both forecasts or new actuals; 1 mark for giving the stated numerical error trigger for refitting.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Bivariate Data Analysis | Q1 | 12 | ___ | Review the bivariate data analysis methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Time Series Analysis | Q2 | 13 | ___ | Review the time series analysis methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Loans, Investments and Annuities | Q3 | 12 | ___ | Review the loans, investments and annuities methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Networks and Decision Mathematics | Q4 | 13 | ___ | Review the networks and decision mathematics methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Project Planning and Networks | Q5 | 12 | ___ | Review the project planning and networks methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Assignment Problems | Q6 | 13 | ___ | Review the assignment problems methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Financial Modelling | Q7 | 12 | ___ | Review the financial modelling methods, calculator setup, interpretation and justification assessed in the listed questions. |
| Integrated Modelling | Q8 | 13 | ___ | Review the integrated modelling methods, calculator setup, interpretation and justification assessed in the listed questions. |