Skill Align VCE General Mathematics Units 3&4 - Free Online Pack 0
Question and Answer Book | 18 questions | 60 marks | Technology active
- Paper
- Examination 2 Showcase
- Reading
- 15 minutes
- Writing
- 1 hour 30 minutes
- Assessment
- 60 marks
Materials supplied: one General Mathematics formula sheet. Materials permitted: one approved CAS calculator or CAS software, one scientific calculator, one bound reference that may be annotated, and basic stationery. Write every answer in the spaces provided in this Question and Answer Book. Pack 0 remains free online; no public formula-sheet or PDF download is supplied.
Written-Response Questions
Answer all questions in the spaces provided. Show sufficient working and justification to support each answer.
Question 1
5 marksQuestion 2
2 marksQuestion 3
2 marksQuestion 4
8 marksQuestion 5
3 marksQuestion 6
4 marksQuestion 7
4 marksQuestion 8
3 marksQuestion 9
3 marksQuestion 10
2 marksQuestion 11
3 marksQuestion 12
2 marksQuestion 13
4 marksQuestion 14
3 marksQuestion 15
4 marksQuestion 16
2 marksQuestion 17
2 marksQuestion 18
4 marksWorked Solutions And Marking Guide
Question 1
(a) 8 minutes.
The fifth ordered value is 8.
(b) 7 minutes.
Q1 = (5 + 7) / 2 = 6 and Q3 = (12 + 14) / 2 = 13, so the IQR is 7.
(c) It is not an upper outlier.
The upper fence is 13 + 1.5(7) = 23.5, and 22 is below it.
(d) The median.
The median is less affected by an extreme value than the mean.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 8 minutes.
- Part (b) (2 marks): 1 mark for q1 = (5 + 7) / 2 = 6 and Q3 = (12 + 14) / 2 = 13, so the IQR is 7; 1 mark for obtaining or concluding 7 minutes. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
- Part (c) (1 mark): 1 mark for obtaining or identifying It is not an upper outlier.
- Part (d) (1 mark): 1 mark for obtaining or identifying The median.
Question 2
(a) 14 parcels, from day 3 to day 4.
Compare each pair of consecutive days. The largest increase is 118-104=14 parcels, from day 3 to day 4.
(b) 106 parcels.
The mean is (96 + 104 + 118) / 3 = 106.
Mark allocation
- Part (a) (1 mark): 1 mark for identifying 14 parcels, from day 3 to day 4, by comparing consecutive values.
- Part (b) (1 mark): 1 mark for obtaining or identifying 106 parcels.
Question 3
(a) Approximately 81.5%.
This interval is from one standard deviation below to two above the mean.
(b) -2.5.
z = (61.5 - 74) / 5 = -2.5.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying Approximately 81.5%.
- Part (b) (1 mark): 1 mark for obtaining or identifying -2.5.
Question 4
(a) Weekly customer visits, y.
Customer visits is predicted from the number of advertising placements.
(b) y = 127.1 + 12.6x.
Technology gives intercept a = 127.1 and slope b = 12.6, to one decimal place.
(c) 241 visits.
127.1 + 12.6(9) = 240.5, which rounds to 241.
(d) -5.5 visits.
Residual = observed - predicted = 235 - 240.5 = -5.5.
(e) Each additional advertising placement is associated with about 12.6 additional weekly visits.
The slope is the predicted change in visits for one extra placement.
(f) There is a very strong positive linear association, but it does not prove causation.
The positive value close to 1 indicates a very strong positive linear association; correlation alone does not establish cause.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying Weekly customer visits, y.
- Part (b) (2 marks): 1 mark for technology gives intercept a = 127.1 and slope b = 12.6, to one decimal place; 1 mark for obtaining or concluding y = 127.1 + 12.6x. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
- Part (c) (1 mark): 1 mark for obtaining or identifying 241 visits.
- Part (d) (1 mark): 1 mark for obtaining or identifying -5.5 visits.
- Part (e) (1 mark): 1 mark for obtaining or identifying Each additional advertising placement is associated with about 12.6 additional weekly visits.
- Part (f) (2 marks): 1 mark for the positive value close to 1 indicates a very strong positive linear association; correlation alone does not establish cause; 1 mark for obtaining or concluding There is a very strong positive linear association, but it does not prove causation. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
Question 5
(a) 1.30.
The coefficient of log10(x) is the power.
(b) 4.47.
The constant is 10^0.65, approximately 4.47.
(c) y = 4.47x^1.30, approximately.
Undo the logarithmic transformation.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 1.30.
- Part (b) (1 mark): 1 mark for obtaining or identifying 4.47.
- Part (c) (1 mark): 1 mark for obtaining or identifying y = 4.47x^1.30, approximately.
Question 6
(a) 1.12.
Quarterly indices sum to 4, so 4 - (0.78 + 0.94 + 1.16) = 1.12.
(b) 125 thousand.
145 / 1.16 = 125.
(c) 102.96 thousand, or about 103 thousand.
132(0.78) = 102.96.
(d) An upward trend.
Corresponding quarterly values are generally higher in the second year.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 1.12.
- Part (b) (1 mark): 1 mark for obtaining or identifying 125 thousand.
- Part (c) (1 mark): 1 mark for obtaining or identifying 102.96 thousand, or about 103 thousand.
- Part (d) (1 mark): 1 mark for obtaining or identifying An upward trend.
Question 7
(a) L_0 = 28000 and L_(n+1) = 1.0055L_n - 720.
Monthly interest is applied before the $720 repayment.
(b) $20 998.73.
Iterating the recurrence for 12 months gives L_12 = 20998.73.
(c) 43 full repayments and a final repayment of $630.98.
After 43 full repayments the next interest-adjusted balance is $630.98, so the 44th payment is reduced to that amount.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying L_0 = 28000 and L_(n+1) = 1.0055L_n - 720.
- Part (b) (1 mark): 1 mark for obtaining or identifying $20 998.73.
- Part (c) (2 marks): 1 mark for after 43 full repayments the next interest-adjusted balance is $630.98, so the 44th payment is reduced to that amount; 1 mark for obtaining or concluding 43 full repayments and a final repayment of $630.98. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
Question 8
(a) AUD 3 per hour.
The loss is AUD 42,000, and 42000 / 14000 = 3.
(b) AUD 21,000.
48000 - 3(9000) = 21000.
(c) V(h) = 48000 - 3h, for 0 <= h <= 14000.
The initial value is 48000 and the hourly decrease is 3.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying AUD 3 per hour.
- Part (b) (1 mark): 1 mark for obtaining or identifying AUD 21,000.
- Part (c) (1 mark): 1 mark for obtaining or identifying V(h) = 48000 - 3h, for 0 <= h <= 14000.
Question 9
(a) L(0) = 26000 and L(n+1) = 1.0055L(n) - 620.
Apply monthly interest before subtracting the repayment.
(b) AUD 25,523.
1.0055(26000) - 620 = 25523.
(c) AUD 140.38.
0.0055(25523) = 140.3765.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying L(0) = 26000 and L(n+1) = 1.0055L(n) - 620.
- Part (b) (1 mark): 1 mark for obtaining or identifying AUD 25,523.
- Part (c) (1 mark): 1 mark for obtaining or identifying AUD 140.38.
Question 10
(a) 0.55%.
6.6% / 12 = 0.55%.
(b) Approximately 6.80%.
(1.0055)¹² - 1 is approximately 0.06798.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 0.55%.
- Part (b) (1 mark): 1 mark for obtaining or identifying Approximately 6.80%.
Question 11
(a) [0.60, 0.25, 0.15]^T.
The arrows leaving A show the proportions sent to A, B and C.
(b) Each bicycle must be assigned to exactly one station after the transition.
The column records all possible destinations from one starting station.
(c) 30 bicycles.
0.15(200) = 30.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying [0.60, 0.25, 0.15]^T.
- Part (b) (1 mark): 1 mark for obtaining or identifying Each bicycle must be assigned to exactly one station after the transition.
- Part (c) (1 mark): 1 mark for obtaining or identifying 30 bicycles.
Question 12
(a) 0.
2(-1) + 1(2) = 0.
(b) 2.
det(A) = 2(3) - 1(4) = 2.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 0.
- Part (b) (1 mark): 1 mark for obtaining or identifying 2.
Question 13
(a) [710, 870]^T.
Juveniles: 0.20(800) + 1.10(500) = 710. Adults: 0.65(800) + 0.70(500) = 870.
(b) 1580.
710 + 870 = 1580.
(c) An expected 65% of juveniles enter or survive into the adult class during one period.
It is the juvenile-to-adult transition contribution.
Mark allocation
- Part (a) (2 marks): 1 mark for juveniles: 0.20(800) + 1.10(500) = 710. Adults: 0.65(800) + 0.70(500) = 870; 1 mark for obtaining or concluding [710, 870]^T. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
- Part (b) (1 mark): 1 mark for obtaining or identifying 1580.
- Part (c) (1 mark): 1 mark for obtaining or identifying An expected 65% of juveniles enter or survive into the adult class during one period.
Question 14
(a) 60.
The first entry is 1(40) + 0(30) + 1(20) = 60.
(b) [60, 50, 70]^T.
The row products are 60, 50 and 70.
(c) 180.
60 + 50 + 70 = 180.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 60.
- Part (b) (1 mark): 1 mark for obtaining or identifying [60, 50, 70]^T.
- Part (c) (1 mark): 1 mark for obtaining or identifying 180.
Question 15
(a) 15.
The cut crosses S-A with capacity 8 and S-B with capacity 7.
(b) 15.
The maximum flow cannot exceed the capacity of any source-to-sink cut.
(c) For example, 7 units along S-A-T and 6 units along S-B-T.
Each edge capacity is respected and the total reaching T is 13.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 15.
- Part (b) (1 mark): 1 mark for obtaining or identifying 15.
- Part (c) (2 marks): 1 mark for each edge capacity is respected and the total reaching T is 13; 1 mark for obtaining or concluding For example, 7 units along S-A-T and 6 units along S-B-T. Accept an equivalent exact form or the rounding requested in the question. Where this part uses an earlier result, apply consequential marking to consistent subsequent working.
Question 16
(a) 16 hours.
5 + 7 + 4 = 16.
(b) Each designer receives one job and each job receives one designer.
An assignment is one-to-one.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying 16 hours.
- Part (b) (1 mark): 1 mark for obtaining or identifying Each designer receives one job and each job receives one designer.
Question 17
(a) No.
An Euler trail requires exactly zero or two odd-degree vertices.
(b) Two pairings.
Four odd vertices must be paired into two pairs.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying No.
- Part (b) (1 mark): 1 mark for obtaining or identifying Two pairings.
Question 18
(a) B, distance 2.
B has the smallest temporary label from S.
(b) 3.
The route S-B-A has length 2 + 1 = 3, improving the direct label 4.
(c) 13.
The permanent label at T is 13.
(d) S-B-A-C-D-T.
The edge weights sum to 2 + 1 + 5 + 2 + 3 = 13.
Mark allocation
- Part (a) (1 mark): 1 mark for obtaining or identifying B, distance 2.
- Part (b) (1 mark): 1 mark for obtaining or identifying 3.
- Part (c) (1 mark): 1 mark for obtaining or identifying 13.
- Part (d) (1 mark): 1 mark for obtaining or identifying S-B-A-C-D-T.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Data Analysis | Q1(a) | 1 | ___ | Redo the calculation from its defining rule: The fifth ordered value is 8. |
| Data Analysis | Q1(b) | 2 | ___ | Order the data, calculate Q1, median and Q3 using the stated convention, then apply the 1.5 × IQR fences before comparing distributions. |
| Data Analysis | Q1(c) | 1 | ___ | Order the data, calculate Q1, median and Q3 using the stated convention, then apply the 1.5 × IQR fences before comparing distributions. |
| Data Analysis | Q1(d) | 1 | ___ | Redo the calculation from its defining rule: The median is less affected by an extreme value than the mean. |
| Data Analysis | Q2(a) | 1 | ___ | Compare the consecutive changes +8, +8, +14, +4 and -7; the largest one-day increase is 14 parcels from day 3 to day 4. |
| Data Analysis | Q2(b) | 1 | ___ | Redo the calculation from its defining rule: The mean is (96 + 104 + 118) / 3 = 106. |
| Data Analysis | Q3(a) | 1 | ___ | Standardise with z=((x-mu) / (sigma)), then interpret the sign and magnitude in standard-deviation units. |
| Data Analysis | Q3(b) | 1 | ___ | Standardise with z=((x-mu) / (sigma)), then interpret the sign and magnitude in standard-deviation units. |
| Data Analysis | Q4(a) | 1 | ___ | Redo the calculation from its defining rule: Customer visits is predicted from the number of advertising placements. |
| Data Analysis | Q4(b) | 2 | ___ | Redo the calculation from its defining rule: Technology gives intercept a = 127.1 and slope b = 12.6, to one decimal place. |
| Data Analysis | Q4(c) | 1 | ___ | Redo the calculation from its defining rule: 127.1 + 12.6(9) = 240.5, which rounds to 241. |
| Data Analysis | Q4(d) | 1 | ___ | Calculate residual = observed - predicted, then use its sign or the residual-plot pattern to assess the model. |
| Data Analysis | Q4(e) | 1 | ___ | Redo the calculation from its defining rule: The slope is the predicted change in visits for one extra placement. |
| Data Analysis | Q4(f) | 2 | ___ | Redo the calculation from its defining rule: The positive value close to 1 indicates a very strong positive linear association; correlation alone does not establish cause. |
| Data Analysis | Q5(a) | 1 | ___ | Redo the calculation from its defining rule: The coefficient of log10(x) is the power. |
| Data Analysis | Q5(b) | 1 | ___ | Redo the calculation from its defining rule: The constant is 10^0.65, approximately 4.47. |
| Data Analysis | Q5(c) | 1 | ___ | Redo the calculation from its defining rule: Undo the logarithmic transformation. |
| Data Analysis | Q6(a) | 1 | ___ | Divide by the seasonal index to remove seasonality, multiply by it to restore seasonality, and align the forecast with the correct season. |
| Data Analysis | Q6(b) | 1 | ___ | Divide by the seasonal index to remove seasonality, multiply by it to restore seasonality, and align the forecast with the correct season. |
| Data Analysis | Q6(c) | 1 | ___ | Divide by the seasonal index to remove seasonality, multiply by it to restore seasonality, and align the forecast with the correct season. |
| Data Analysis | Q6(d) | 1 | ___ | Divide by the seasonal index to remove seasonality, multiply by it to restore seasonality, and align the forecast with the correct season. |
| Recursion and Financial Modelling | Q7(a) | 1 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q7(b) | 1 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q7(c) | 2 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q8(a) | 1 | ___ | Choose the stated depreciation model, calculate adjacent values around any threshold, and retain the required scrap-value or domain constraint. |
| Recursion and Financial Modelling | Q8(b) | 1 | ___ | Choose the stated depreciation model, calculate adjacent values around any threshold, and retain the required scrap-value or domain constraint. |
| Recursion and Financial Modelling | Q8(c) | 1 | ___ | Choose the stated depreciation model, calculate adjacent values around any threshold, and retain the required scrap-value or domain constraint. |
| Recursion and Financial Modelling | Q9(a) | 1 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q9(b) | 1 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q9(c) | 1 | ___ | Iterate the loan recurrence with interest applied before each repayment; reduce the final repayment when the interest-adjusted balance is below the regular payment. |
| Recursion and Financial Modelling | Q10(a) | 1 | ___ | Redo the calculation from its defining rule: 6.6% / 12 = 0.55%. |
| Recursion and Financial Modelling | Q10(b) | 1 | ___ | Redo the calculation from its defining rule: (1.0055)¹² - 1 is approximately 0.06798. |
| Matrices | Q11(a) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Matrices | Q11(b) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Matrices | Q11(c) | 1 | ___ | Redo the calculation from its defining rule: 0.15(200) = 30. |
| Matrices | Q12(a) | 1 | ___ | Redo the calculation from its defining rule: 2(-1) + 1(2) = 0. |
| Matrices | Q12(b) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Matrices | Q13(a) | 2 | ___ | Multiply the matrix by the state column vector using the printed row-and-column convention, then apply any stated culling, restocking or total constraint. |
| Matrices | Q13(b) | 1 | ___ | Multiply the matrix by the state column vector using the printed row-and-column convention, then apply any stated culling, restocking or total constraint. |
| Matrices | Q13(c) | 1 | ___ | Multiply the matrix by the state column vector using the printed row-and-column convention, then apply any stated culling, restocking or total constraint. |
| Matrices | Q14(a) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Matrices | Q14(b) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Matrices | Q14(c) | 1 | ___ | Use the printed matrix dimensions and row-by-column products, then interpret the requested entry or vector in context. |
| Networks and Decision Mathematics | Q15(a) | 1 | ___ | Construct a feasible source-to-sink flow, calculate a matching cut capacity, and use equality of the flow and cut to prove maximality. |
| Networks and Decision Mathematics | Q15(b) | 1 | ___ | Construct a feasible source-to-sink flow, calculate a matching cut capacity, and use equality of the flow and cut to prove maximality. |
| Networks and Decision Mathematics | Q15(c) | 2 | ___ | Redo the calculation from its defining rule: Each edge capacity is respected and the total reaching T is 13. |
| Networks and Decision Mathematics | Q16(a) | 1 | ___ | Redo the calculation from its defining rule: 5 + 7 + 4 = 16. |
| Networks and Decision Mathematics | Q16(b) | 1 | ___ | Redo the calculation from its defining rule: An assignment is one-to-one. |
| Networks and Decision Mathematics | Q17(a) | 1 | ___ | Redo the calculation from its defining rule: An Euler trail requires exactly zero or two odd-degree vertices. |
| Networks and Decision Mathematics | Q17(b) | 1 | ___ | Identify the odd vertices, compare the shortest required pairings, and add the minimum repeated-path weight to the total edge weight. |
| Networks and Decision Mathematics | Q18(a) | 1 | ___ | Use Dijkstra's algorithm: make the smallest temporary label permanent, update adjacent labels, and trace predecessor labels for the route. |
| Networks and Decision Mathematics | Q18(b) | 1 | ___ | Use Dijkstra's algorithm: make the smallest temporary label permanent, update adjacent labels, and trace predecessor labels for the route. |
| Networks and Decision Mathematics | Q18(c) | 1 | ___ | Use Dijkstra's algorithm: make the smallest temporary label permanent, update adjacent labels, and trace predecessor labels for the route. |
| Networks and Decision Mathematics | Q18(d) | 1 | ___ | Use Dijkstra's algorithm: make the smallest temporary label permanent, update adjacent labels, and trace predecessor labels for the route. |