Skill Align Mathematics Applications Year 12 Section One Calculator-free for WACE - 2026 Edition
Original Skill Align practice examination content
- Paper
- Section One Calculator-free Question and Response Book Showcase
- Reading
- 5 minutes reading time
- Writing
- 50 minutes
- Assessment
- 51 marks
Standard items: pens, pencils including coloured pencils, sharpener, correction fluid or tape, eraser, ruler and highlighters. Special items: nil. A formula sheet is provided by the supervisor. An examination changeover period of up to 15 minutes applies, during which candidates are not permitted to work.
Section One
Answer all 5 questions. This section contributes 35% of the examination. For any question or part question worth more than two marks, valid working or justification is required to receive full marks.
Question 1
9 marksQuestion 2
10 marksQuestion 3
10 marksQuestion 4
11 marksQuestion 5
11 marksWorked Solutions And Marking Guide
Section One Question 1
(a) hat y=40.
Substitute x=8 into the fitted line.
(b) 43-40=3.
Residual is actual minus predicted.
(c) r=+0.90.
Use the positive square root because the fitted gradient is positive.
(d) The plotted observations end at x=10, so x=20 is extrapolation well outside the observed range and the linear relationship may not continue.
Compare the requested value with the numerical graph scale.
Mark allocation
- Part (a) (2 marks): 1 mark for substituting x=8 into 12+3.5x; 1 mark for obtaining hat y=40.
- Part (b) (2 marks): 1 mark for using actual minus predicted; 1 mark for obtaining the signed residual 3.
- Part (c) (2 marks): 1 mark for calculating √0.81=0.90; 1 mark for selecting the positive sign from the increasing trend.
- Part (d) (3 marks): 1 mark for identifying the observed range 2le xle10; 1 mark for identifying x=20 as extrapolation; 1 mark for explaining that the fitted relationship may not persist outside the observed range.
Section One Question 2
(a) B_1=$23760.
Calculate 1.005(24000)-360.
(b) B_2=$23518.80.
Calculate 1.005(23760)-360.
(c) Interest is calculated from the changing balance, so the interest amount and net reduction change each month.
The recurrence does not have a constant difference.
(d) The month-2 interest is 0.005(23760)=$118.80; the repayment is $360.
Interest is calculated from B_1 before subtracting the fixed repayment.
Mark allocation
- Part (a) (2 marks): 1 mark for applying the interest multiplier before the repayment; 1 mark for obtaining B_1=$23760.
- Part (b) (3 marks): 1 mark for using B_1=23760 as the next initial condition; 1 mark for applying interest before the second repayment; 1 mark for obtaining B_2=$23518.80.
- Part (c) (2 marks): 1 mark for identifying that interest depends on the current balance; 1 mark for linking the changing interest amount to changing balance reductions.
- Part (d) (3 marks): 1 mark for using B_1=23760 as the interest-bearing balance; 1 mark for calculating 0.005(23760)=$118.80; 1 mark for distinguishing the $118.80 interest from the $360 repayment.
Section One Question 3
(a) The series increases overall, with a fall from month 2 to month 3.
Compare the early and late values, then locate the decline.
(b) 138 / 0.92=150.
Divide the observed value by its seasonal index.
(c) 160(1.15)=184 passengers.
Multiply the trend estimate by the summer seasonal index.
(d) Deseasonalising removes a recurring within-year effect so the underlying trend is clearer, but irregular changes and structural changes remain.
Separate the purpose of adjustment from the limitations of forecasting.
Mark allocation
- Part (a) (2 marks): 1 mark for identifying the overall upward trend; 1 mark for identifying the month-2 to month-3 decrease.
- Part (b) (2 marks): 1 mark for dividing by the winter index 0.92; 1 mark for obtaining 150 passengers.
- Part (c) (3 marks): 1 mark for selecting multiplication for reseasonalisation; 1 mark for using the summer index 1.15; 1 mark for obtaining 184 passengers.
- Part (d) (3 marks): 1 mark for explaining that seasonal adjustment removes a recurring seasonal effect; 1 mark for linking adjustment to clearer underlying trend estimation; 1 mark for identifying an irregular or structural source of remaining forecast error.
Section One Question 4
(a) deg(C)=4.
The incident edges are AC,BC,CD,CE.
(b) A-C-E and A-C-D-E are both shortest routes, each taking 11 minutes.
A complete route search gives a shared minimum of 11.
(c) The graph is connected and semi-Eulerian. Its only odd-degree vertices are B and D, so a semi-Eulerian trail starts at one and ends at the other.
A connected graph with exactly two odd-degree vertices has an open trail through every edge once.
(d) One MST uses BC,AC,DE,CD, with weight 2+3+3+5=13.
Select light edges without creating a cycle until all vertices are connected.
Mark allocation
- Part (a) (2 marks): 1 mark for identifying the four edges incident with C; 1 mark for stating deg(C)=4.
- Part (b) (3 marks): 1 mark for showing a valid route-search process; 1 mark for accumulating a minimum weight of 11 minutes; 1 mark for stating either shortest route and acknowledging the equal alternative.
- Part (c) (3 marks): 1 mark for determining the vertex degrees; 1 mark for identifying B and D as the only odd-degree vertices; 1 mark for concluding that the graph is semi-Eulerian with trail endpoints B and D.
- Part (d) (3 marks): 1 mark for selecting four cycle-free edges that connect all five vertices; 1 mark for showing the selected weights 2, 3, 3 and 5; 1 mark for obtaining the MST weight 13.
Section One Question 5
(a) The upper path A-C takes 3+4=7 days; the lower path B-D takes 5+3=8 days.
Add the activity durations on each complete path.
(b) The critical path is B-D, and the project duration is 8 days.
The longest start-to-finish path controls completion.
(c) The difference is 8-7=1 day.
Subtract the shorter path duration from the longer path duration.
(d) The lower path becomes 3+3=6 days, so the upper path A-C is critical and the new duration is 7 days.
Recalculate both paths after changing B.
Mark allocation
- Part (a) (3 marks): 1 mark for identifying the upper complete path; 1 mark for obtaining 7 days for the upper path; 1 mark for obtaining 8 days for the lower path.
- Part (b) (3 marks): 1 mark for comparing the two path durations; 1 mark for identifying B-D as the longer path; 1 mark for stating the project duration of 8 days.
- Part (c) (2 marks): 1 mark for subtracting the shorter path duration from the longer duration; 1 mark for obtaining a difference of 1 day.
- Part (d) (3 marks): 1 mark for recalculating the lower path as 6 days; 1 mark for retaining the upper path duration of 7 days; 1 mark for identifying A-C as critical with a 7-day duration.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Bivariate Data Analysis | Q1 | 9 | ___ | Review the bivariate data analysis hand calculation, method selection, interpretation and justification assessed in the listed questions. |
| Loans, Investments and Annuities | Q2 | 10 | ___ | Review the loans, investments and annuities hand calculation, method selection, interpretation and justification assessed in the listed questions. |
| Time Series Analysis | Q3 | 10 | ___ | Review the time series analysis hand calculation, method selection, interpretation and justification assessed in the listed questions. |
| Graphs and Networks | Q4 | 11 | ___ | Review the graphs and networks hand calculation, method selection, interpretation and justification assessed in the listed questions. |
| Project Planning and Networks | Q5 | 11 | ___ | Review the project planning and networks hand calculation, method selection, interpretation and justification assessed in the listed questions. |