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 questions. 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) Yes. The only odd-degree vertices are B and D, so the trail starts at one and ends at the other.
Apply the odd-degree criterion for an Euler trail.
(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 an Euler trail has 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 float is 8-7=1 day.
Subtract the shorter path duration from the critical 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 using the difference between project and path duration; 1 mark for obtaining 1 day of float.
- 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. |