Skill Align QCE General Mathematics Paper 2 Question and Response Book - Free Online Pack 0
Original Skill Align Units 3&4 independent practice question and response book.
- Paper
- Paper 2 Question and Response Book - Free Online Pack 0
- Reading
- 5 minutes perusal
- Writing
- 90 minutes
- Assessment
- 38 marks
Any additional response paper must show your name and the relevant question number, and must be attached as directed by your school.
Independent practice conditions
- Answer all questions. Show relevant working, mathematical reasoning and conclusions in the spaces provided.
- You may use writing equipment, a QCAA-approved handheld scientific calculator and the QCAA formula book.
- During perusal time, read the questions and make notes only on planning paper supplied by your school.
- If you use additional response paper, show your name and the relevant question and part number, then attach it as directed by your school.
- This independent practice resource is not a secure QCAA examination paper. Complete the candidate details above so printed work can be identified.
Section 1
Attempt all questions. Select and apply mathematical models, show relevant working, justify decisions and evaluate results in context.
Question 1
6 marksQuestion 2
5 marksQuestion 3
5 marksQuestion 4
6 marksQuestion 5
5 marksQuestion 6
6 marksQuestion 7
5 marksWorked Solutions And Marking Guide
General marking principles
- Accept mathematically equivalent methods and forms. Apply follow-through where a later result correctly uses an earlier student value, unless that removes the mathematical demand being assessed.
Section 1 Question 1
(a) hat y=10.280+3.707x, r=0.998, R²=0.996.
Evidence: enter all six ordered pairs; report the intercept and gradient, then both technology statistics with the positive sign of r.
(b) Residual approx-0.640 thousand dollars; the actual cost is about $640 below the prediction.
Evidence: residual =43-(10.28+3.706666ldots(9))=-0.64; a negative residual places the observation below the fitted line.
(c) Observed domain 2le xle9 years. A 15-year prediction is a long extrapolation and is not reliable for the replacement decision without older-vehicle evidence.
Evidence: identify the minimum and maximum observed ages, classify x=15 as outside that interval, and link the limitation to possible changes in maintenance behaviour.
Mark allocation
- Part a (2 marks): Evidence: enter all six ordered pairs; report the intercept and gradient, then both technology statistics with the positive sign of r. Required result: hat y=10.280+3.707x, r=0.998, R²=0.996.
- Part b (2 marks): Evidence: residual =43-(10.28+3.706666ldots(9))=-0.64; a negative residual places the observation below the fitted line. Required result: Residual approx-0.640 thousand dollars; the actual cost is about $640 below the prediction.
- Part c (2 marks): Evidence: identify the minimum and maximum observed ages, classify x=15 as outside that interval, and link the limitation to possible changes in maintenance behaviour. Required result: Observed domain 2le xle9 years. A 15-year prediction is a long extrapolation and is not reliable for the replacement decision without older-vehicle evidence.
Section 1 Question 2
(a) The ratios are 1.12, 1.08, 1.10, so the average-percentage seasonal index is 1.10.
Evidence: form all three actual-to-trend ratios before averaging; retain the dimensionless index 1.10, equivalent to 110%.
(b) The model gives 240(1.10)=264 thousand visits. The proposal is 6 thousand higher and is not supported by the seasonal calculation alone.
Evidence: identify that the trend forecast must be reseasonalised, calculate 264, compare with 270, and state the evidence-bound decision.
(c) The historical seasonal pattern may no longer represent visitor behaviour, so the index may not remain stable.
Evidence: identify the stationarity assumption and connect the transport change to a possible change in seasonal demand.
Mark allocation
- Part a (2 marks): Evidence: form all three actual-to-trend ratios before averaging; retain the dimensionless index 1.10, equivalent to 110%. Required result: The ratios are 1.12, 1.08, 1.10, so the average-percentage seasonal index is 1.10.
- Part b (2 marks): Evidence: identify that the trend forecast must be reseasonalised, calculate 264, compare with 270, and state the evidence-bound decision. Required result: The model gives 240(1.10)=264 thousand visits. The proposal is 6 thousand higher and is not supported by the seasonal calculation alone.
- Part c (1 mark): Evidence: identify the stationarity assumption and connect the transport change to a possible change in seasonal demand. Required result: The historical seasonal pattern may no longer represent visitor behaviour, so the index may not remain stable.
Section 1 Question 3
(a) The cut immediately after S crosses SA and SB, with capacity 13+10=23, so no daily flow can exceed 23.
Evidence: select a valid directed cut, name its crossing edges, calculate its capacity and state the upper-bound consequence.
(b) Use SA=13, SB=10, AC=9, AD=4, BC=4, BD=6, CT=13, DT=10, giving sink inflow 23.
Evidence: derive the allocation rather than assuming the target; every flow is within capacity and conservation holds because inflow equals outflow at A,B,C,D.
(c) Maximum flow =23 units per day because a feasible flow reaches the capacity of a cut.
Evidence: connect the attainable lower bound to the equal cut upper bound.
Mark allocation
- Part a (2 marks): Evidence: select a valid directed cut, name its crossing edges, calculate its capacity and state the upper-bound consequence. Required result: The cut immediately after S crosses SA and SB, with capacity 13+10=23, so no daily flow can exceed 23.
- Part b (2 marks): Evidence: derive the allocation rather than assuming the target; every flow is within capacity and conservation holds because inflow equals outflow at A,B,C,D. Required result: Use SA=13, SB=10, AC=9, AD=4, BC=4, BD=6, CT=13, DT=10, giving sink inflow 23.
- Part c (1 mark): Evidence: connect the attainable lower bound to the equal cut upper bound. Required result: Maximum flow =23 units per day because a feasible flow reaches the capacity of a cut.
Section 1 Question 4
(a) Earliest C=max(4+5,6+4)=10; earliest D=9; earliest T=max(10+6,9+5)=16 days.
Evidence: use the later predecessor completion at each merge and identify the earliest terminal time.
(b) It does not meet the requirement: completion is 16 days. The controlling path is S-B-C-T, with duration 6+4+6=16 days.
Evidence: compare the derived completion time with the requirement and identify the zero-float path, not merely the longest visible arc.
(c) The activity has float 10-4-5=1 day, so a two-day delay would delay the project by one day and should be rejected unless the schedule is changed.
Evidence: derive total float from the event times, compare it with the proposed delay, and state the schedule consequence.
Mark allocation
- Part a (2 marks): Evidence: use the later predecessor completion at each merge and identify the earliest terminal time. Required result: Earliest C=max(4+5,6+4)=10; earliest D=9; earliest T=max(10+6,9+5)=16 days.
- Part b (2 marks): Evidence: compare the derived completion time with the requirement and identify the zero-float path, not merely the longest visible arc. Required result: It does not meet the requirement: completion is 16 days. The controlling path is S-B-C-T, with duration 6+4+6=16 days.
- Part c (2 marks): Evidence: derive total float from the event times, compare it with the proposed delay, and state the schedule consequence. Required result: The activity has float 10-4-5=1 day, so a two-day delay would delay the project by one day and should be rejected unless the schedule is changed.
Section 1 Question 5
(a) deg(P)=3,deg(A)=2,deg(B)=3,deg(C)=2,deg(T)=2. An Euler trail exists from P to B, so the proposed start is feasible.
Evidence: determine all degrees, identify exactly two odd vertices and connect them to the required trail endpoints.
(b) Duplicate edge PB. This changes the parity of the two odd vertices P and B, producing an Euler circuit.
Evidence: identify the direct odd-vertex connection and explain its parity effect.
(c) The inspection-time weights for the alternative P-to-B paths are required so their total repeated weights can be compared.
Evidence: recognise that edge counts alone do not determine the minimum when inspection times are unequal.
Mark allocation
- Part a (2 marks): Evidence: determine all degrees, identify exactly two odd vertices and connect them to the required trail endpoints. Required result: deg(P)=3,deg(A)=2,deg(B)=3,deg(C)=2,deg(T)=2. An Euler trail exists from P to B, so the proposed start is feasible.
- Part b (2 marks): Evidence: identify the direct odd-vertex connection and explain its parity effect. Required result: Duplicate edge PB. This changes the parity of the two odd vertices P and B, producing an Euler circuit.
- Part c (1 mark): Evidence: recognise that edge counts alone do not determine the minimum when inspection times are unequal. Required result: The inspection-time weights for the alternative P-to-B paths are required so their total repeated weights can be compared.
Section 1 Question 6
(a) FV=750((1.012²⁰-1) / (0.012))=$16,839.65; the deposits are end-of-quarter payments.
Evidence: select the future-value model from the timing, use n=20 and quarterly rate 0.012, and round only the final result.
(b) Required capital =((16500) / (0.055))=$300,000.
Evidence: recognise a perpetuity and match the annual payment with the annual rate.
(c) Combined resources are $290,000+$16,839.65=$306,839.65. The plan is supportable with a margin of $6,839.65.
Evidence: connect the two time-consistent capital amounts, compare with the perpetuity requirement, and state the decision with its margin.
Mark allocation
- Part a (2 marks): Evidence: select the future-value model from the timing, use n=20 and quarterly rate 0.012, and round only the final result. Required result: FV=750((1.012²⁰-1) / (0.012))=$16,839.65; the deposits are end-of-quarter payments.
- Part b (2 marks): Evidence: recognise a perpetuity and match the annual payment with the annual rate. Required result: Required capital =((16500) / (0.055))=$300,000.
- Part c (2 marks): Evidence: connect the two time-consistent capital amounts, compare with the perpetuity requirement, and state the decision with its margin. Required result: Combined resources are $290,000+$16,839.65=$306,839.65. The plan is supportable with a margin of $6,839.65.
Section 1 Question 7
(a) Departure is 12:30 pm Monday UTC; arrival is 9:30 pm Monday UTC, which is 11:30 am Monday in Honolulu.
Evidence: convert through UTC, add the elapsed time, then apply the destination offset while retaining the date.
(b) Distance =((72) / (360))(40000)=8000 km, so average speed =8000 / 9approx889 km / h. The below-900 claim is supported.
Evidence: identify the great-circle circumference, calculate the arc distance, divide by elapsed hours and compare with the claim.
(c) The eastward trip crosses the International Date Line, so the calendar moves back one day relative to the westward side.
Evidence: identify the direction / date-line effect and reconcile it with the UTC calculation.
Mark allocation
- Part a (2 marks): Evidence: convert through UTC, add the elapsed time, then apply the destination offset while retaining the date. Required result: Departure is 12:30 pm Monday UTC; arrival is 9:30 pm Monday UTC, which is 11:30 am Monday in Honolulu.
- Part b (2 marks): Evidence: identify the great-circle circumference, calculate the arc distance, divide by elapsed hours and compare with the claim. Required result: Distance =((72) / (360))(40000)=8000 km, so average speed =8000 / 9approx889 km / h. The below-900 claim is supported.
- Part c (1 mark): Evidence: identify the direction / date-line effect and reconcile it with the UTC calculation. Required result: The eastward trip crosses the International Date Line, so the calendar moves back one day relative to the westward side.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Bivariate Data Analysis | Q1 | 6 | ___ | Enter every ordered pair, report the least-squares equation, r and R², calculate residual as observed minus predicted, and check domain before interpreting a prediction. |
| Time Series | Q2 | 5 | ___ | Use an odd centred moving mean or median, derive actual-to-trend ratios before averaging a seasonal index, and match the index to the forecast period when reseasonalising. |
| Networks and Flow | Q3 | 5 | ___ | Identify a cut, calculate its capacity, construct a flow of the same value without exceeding capacities, and verify conservation at every intermediate vertex. |
| Project Networks | Q4 | 6 | ___ | Complete forward and backward event-time scans, identify zero-float activities and the critical path, then calculate and interpret float from earliest and latest times. |
| Graphs and Networks | Q5 | 5 | ___ | Audit vertex degrees and connectivity, apply the relevant Euler, planar, shortest-path or spanning-tree condition, and justify the operational route consequence. |
| Loans, Investments and Annuities | Q6 | 6 | ___ | Match interest and payment periods before selecting a present-value, future-value, perpetuity or financial recurrence model; retain unrounded values until the final financial decision. |
| Earth Geometry and Time Zones | Q7 | 5 | ___ | Convert departure and arrival times through UTC, track the calendar date before applying the destination offset, and explain any International Date Line adjustment. |