Skill Align SACE Stage 2 General Mathematics - Pack 0 - Question Booklet
Questions 1-9 | Total marks: 90 | Total time: 130 minutes
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SACE Stage 2 General Mathematics - examination conditions
- Answer all questions and write your answers in this question booklet.
- Materials are this question booklet and the candidate's SACE registration number label. Attach the registration label in the field below.
- Show appropriate working and steps of logic in this question booklet.
- Use black or blue pen. A sharp dark pencil may be used for diagrams and graphical representations.
- You may bring one unfolded A4 sheet, using both sides, containing your own handwritten notes. No formula sheet is supplied.
- You may use either two approved graphics calculators or one approved graphics calculator and one scientific calculator. Computer algebra system (CAS) calculators are not permitted. Scientific-calculator memory must be cleared; graphics-calculator memory does not need to be cleared. No external storage media may be used.
- Two pages labelled Additional working space follow Question 9. Identify the question and part for any working continued there.
Question booklet
Answer all Questions 1-9 and write answers in this booklet. Show appropriate working and steps of logic. Use black or blue pen; a sharp dark pencil may be used for diagrams and graphical representations. Two labelled additional-working pages follow Question 9.
Question 1
6 marksQuestion 2
8 marksQuestion 3
9 marksQuestion 4
12 marksQuestion 5
7 marksQuestion 6
10 marksQuestion 7
11 marksQuestion 8
14 marksQuestion 9
13 marksWorked Solutions And Marking Guide
Question 1
(a) (0.0912).
Standardising gives z=(15-11.8) / 2.4=1.333ldots. Hence P(W>15)=1-Phi(1.333ldots)=0.091211ldots, so P(W>15)=0.0912.
(b) (14.9text( minutes)).
The target is the 90th percentile: operatorname(invNorm)(0.90,11.8,2.4)=14.8757ldots, which is 14.9 minutes to the nearest 0.1 minute.
Mark allocation
- Part a: uses the lower bound 15 with the upper tail; enters mu=11.8 and sigma=2.4 correctly; reports 0.0912 to three significant figures.
- Part b: identifies the required cumulative probability as 0.90; uses inverse normal with the stated mean and standard deviation; rounds 14.8757ldots to 14.9 minutes.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: uses the lower bound 15 with the upper tail; enters mu=11.8 and sigma=2.4 correctly; reports 0.0912 to three significant figures.
Acceptable alternatives: Accept equivalent normal-CDF or inverse-normal entries that model the same tail, interval or cumulative probability.; Accept z-standardisation followed by standard-normal tables or calculator evaluation when the stated value and rounding are correct.
Do not credit by itself: Do not credit use of the wrong tail, interval bound or cumulative probability.; Do not confuse the stated standard deviation with variance or round an inverse-normal result contrary to the question.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: identifies the required cumulative probability as 0.90; uses inverse normal with the stated mean and standard deviation; rounds 14.8757ldots to 14.9 minutes.
Acceptable alternatives: Accept equivalent normal-CDF or inverse-normal entries that model the same tail, interval or cumulative probability.; Accept z-standardisation followed by standard-normal tables or calculator evaluation when the stated value and rounding are correct.
Do not credit by itself: Do not credit use of the wrong tail, interval bound or cumulative probability.; Do not confuse the stated standard deviation with variance or round an inverse-normal result contrary to the question.
Question 2
(a) PV=24,500 and PMT=-503.40, with N=60 and FV=0.
The borrower receives value of 24,500. Each compulsory monthly outflow is 496.40+7.00=503.40, so the cash-flow signs must be opposite.
(b) Monthly rate =(0.714%); nominal annual comparison rate =(8.56%text( p.a.)).
Solving 24,500=503.40((1-(1+i)⁻⁶⁰) / (i)) gives i=0.007135885ldots. Thus the monthly rate is 0.7135885ldots%=0.714% to three significant figures. Retaining unrounded i, 12i=0.0856306ldots, or 8.56% p.a.
(c) It incorporates the establishment and monthly fees, so it compares the compulsory cost of the credit rather than interest alone.
The monthly fee changes every repayment and the establishment fee reduces the value received. A rate based only on interest omits both costs.
Mark allocation
- Part a: uses the net value received, 24,500, as present value; adds the 7.00 fee to obtain a payment of 503.40; uses opposite cash-flow signs and N=60, FV=0.
- Part b: solves the annuity present-value equation or equivalent TVM entry; reports the monthly rate as 0.714% to three significant figures; multiplies the unrounded monthly rate by 12 and reports 8.56% p.a..
- Part c: states that compulsory fees are included; links inclusion of fees to a fairer comparison of total borrowing cost.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: uses the net value received, 24,500, as present value; adds the 7.00 fee to obtain a payment of 503.40; uses opposite cash-flow signs and N=60, FV=0.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: solves the annuity present-value equation or equivalent TVM entry; reports the monthly rate as 0.714% to three significant figures; multiplies the unrounded monthly rate by 12 and reports 8.56% p.a..
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: states that compulsory fees are included; links inclusion of fees to a fairer comparison of total borrowing cost.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Question 3
(a) (160,015) to the nearest dollar.
With i=0.054 / 12, B=185000(1+i)⁴⁸-1300(((1+i)⁴⁸-1) / (i))=160014.6025ldots.
(b) (1569.80) per month.
Use PV=160014.6025ldots, i=0.061 / 12, N=144, and FV=0. The required payment is 1569.7961ldots, or 1569.80.
(c) No. At the new rate, 1300 is less than the required 1569.80, so a positive balance remains after 144 payments.
The repayment required for a zero future value is 1569.80. A smaller payment cannot amortise the same balance over the same 144 months at the same rate.
Mark allocation
- Part a: uses the monthly rate 0.054 / 12; models 48 end-of-month repayments with correct signs; obtains 160,014.60ldots, hence 160,015.
- Part b: uses the unrounded balance from part (a) as present value; uses 144 months and monthly rate 0.061 / 12; obtains and rounds the repayment to 1569.80.
- Part c: compares 1300 with the calculated required repayment; states that the loan is not cleared in 12 years; justifies the conclusion using the fixed term, rate and present balance.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: uses the monthly rate 0.054 / 12; models 48 end-of-month repayments with correct signs; obtains 160,014.60ldots, hence 160,015.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: uses the unrounded balance from part (a) as present value; uses 144 months and monthly rate 0.061 / 12; obtains and rounds the repayment to 1569.80.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part c (3 marks)
Award the 3 available marks for the following observable evidence: compares 1300 with the calculated required repayment; states that the loan is not cleared in 12 years; justifies the conclusion using the fixed term, rate and present balance.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Question 4
(a) begin(bmatrix)3&14&13&00&8&4&918&2&19&08&1&10&0end(bmatrix).
Subtract row minima 9,16,5,12, respectively, to obtain the stated row-reduced array.
(b) Column reduction gives begin(bmatrix)3&13&9&00&7&0&918&1&15&08&0&6&0end(bmatrix). The smallest uncovered value is 3, giving begin(bmatrix)0&13&6&00&10&0&1215&1&12&05&0&3&0end(bmatrix).
The column minima are 0,1,4,0. After covering the zeros with row B and columns 2,4, subtract 3 from every uncovered entry and add 3 at the two line intersections.
(c) Ato1, Bto3, Cto4, Dto2.
The selected zeros occupy different rows and columns: A1,B3,C4,D2.
(d) (50text( hours)), using 12+20+5+13 from the original array.
Assignment totals must be taken from the original cost array, not the reduced array. The selected entries total 12+20+5+13=50.
Mark allocation
- Part a: identifies row minima 9,16,5,12; subtracts each minimum from every entry in its row; shows the complete row-reduced array correctly.
- Part b: performs the column reduction using minima 0,1,4,0; identifies 3 as the smallest uncovered value under the stated three-line cover; shows the correctly adjusted array.
- Part c: selects one zero in every row; uses each column exactly once; states A1,B3,C4,D2 unambiguously.
- Part d: returns to the original array; adds the four selected times 12,20,5,13; reports the minimum total of 50 hours.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: identifies row minima 9,16,5,12; subtracts each minimum from every entry in its row; shows the complete row-reduced array correctly.
Acceptable alternatives: Accept any valid Hungarian row, column, line-cover and uncovered-minimum sequence that produces independent zeros.; Accept any tied optimal one-to-one assignment when its total is verified using the original cost or benefit array.
Do not credit by itself: Do not credit dependent zeros, an omitted uncovered-minimum adjustment, or an assignment that repeats a row or column.; Do not calculate the final cost or benefit from a reduced or opportunity-loss array.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: performs the column reduction using minima 0,1,4,0; identifies 3 as the smallest uncovered value under the stated three-line cover; shows the correctly adjusted array.
Acceptable alternatives: Accept any valid Hungarian row, column, line-cover and uncovered-minimum sequence that produces independent zeros.; Accept any tied optimal one-to-one assignment when its total is verified using the original cost or benefit array.
Do not credit by itself: Do not credit dependent zeros, an omitted uncovered-minimum adjustment, or an assignment that repeats a row or column.; Do not calculate the final cost or benefit from a reduced or opportunity-loss array.
Part c (3 marks)
Award the 3 available marks for the following observable evidence: selects one zero in every row; uses each column exactly once; states A1,B3,C4,D2 unambiguously.
Acceptable alternatives: Accept any valid Hungarian row, column, line-cover and uncovered-minimum sequence that produces independent zeros.; Accept any tied optimal one-to-one assignment when its total is verified using the original cost or benefit array.
Do not credit by itself: Do not credit dependent zeros, an omitted uncovered-minimum adjustment, or an assignment that repeats a row or column.; Do not calculate the final cost or benefit from a reduced or opportunity-loss array.
Part d (3 marks)
Award the 3 available marks for the following observable evidence: returns to the original array; adds the four selected times 12,20,5,13; reports the minimum total of 50 hours.
Acceptable alternatives: Accept any valid Hungarian row, column, line-cover and uncovered-minimum sequence that produces independent zeros.; Accept any tied optimal one-to-one assignment when its total is verified using the original cost or benefit array.
Do not credit by itself: Do not credit dependent zeros, an omitted uncovered-minimum adjustment, or an assignment that repeats a row or column.; Do not calculate the final cost or benefit from a reduced or opportunity-loss array.
Question 5
(a) ES_E=8 and ES_F=12.
A finishes at 3, C at 8, and B at 4, so E starts at 8 and finishes at 12. Hence F starts at 12.
(b) Duration =(15text( days)); critical path A-C-E-F.
The path A-C-E-F has duration 3+5+4+3=15 days and controls the completion time.
(c) Slack of B=4 days; slack of D=7 days.
A backward scan gives LS_B=4 and LS_D=10. Since ES_B=0 and ES_D=3, their slacks are 4 and 7 days.
Mark allocation
- Part a: uses the maximum predecessor finish to obtain ES_E=8; continues the scan to obtain ES_F=12.
- Part b: obtains the completion time 15 days; identifies A-C-E-F; supports the path using its task durations or zero slack.
- Part c: calculates slack LS-ES=4 days for B; calculates slack LS-ES=7 days for D.
Detailed marking criteria
Part a (2 marks)
Award the 2 available marks for the following observable evidence: uses the maximum predecessor finish to obtain ES_E=8; continues the scan to obtain ES_F=12.
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: obtains the completion time 15 days; identifies A-C-E-F; supports the path using its task durations or zero slack.
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: calculates slack LS-ES=4 days for B; calculates slack LS-ES=7 days for D.
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Question 6
(a) (y=42.03(0.8521)^x).
Exponential regression on the six pairs gives a=42.026859ldots and b=0.852107ldots.
(b) Cooling output is multiplied by about 0.8521 each hour, a decrease of about 14.8% per hour.
Since 1-0.8521=0.1479, the fitted output falls by about 14.8% for each additional hour.
(c) (-0.102text( kW)) to three significant figures.
The model predicts 42.026859(0.8521078)³=26.002225ldots. Thus residual =25.9-26.002225ldots=-0.102225ldotstext( kW).
(d) (16.1text( kW)). It is a short extrapolation beyond the observed range, so it is plausible only if the same decay process continues.
At x=6, the unrounded model gives 16.0877ldotstext( kW), or 16.1text( kW). The value lies just outside 0leq xleq5, so the model assumption must be stated.
Mark allocation
- Part a: enters all six explanatory and response values correctly; selects exponential regression y=ab^x; reports a=42.03 and b=0.8521.
- Part b: interprets b as an hourly multiplicative factor; converts the factor to an approximate 14.8% hourly decrease.
- Part c: calculates the unrounded predicted output near 26.0022; uses observed minus predicted to obtain -0.102text( kW).
- Part d: substitutes x=6 in the unrounded regression model; reports 16.1text( kW); identifies the prediction as extrapolation and states the continuing-process assumption.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: enters all six explanatory and response values correctly; selects exponential regression y=ab^x; reports a=42.03 and b=0.8521.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part b (2 marks)
Award the 2 available marks for the following observable evidence: interprets b as an hourly multiplicative factor; converts the factor to an approximate 14.8% hourly decrease.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: calculates the unrounded predicted output near 26.0022; uses observed minus predicted to obtain -0.102text( kW).
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part d (3 marks)
Award the 3 available marks for the following observable evidence: substitutes x=6 in the unrounded regression model; reports 16.1text( kW); identifies the prediction as extrapolation and states the continuing-process assumption.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Question 7
(a) K:(6,12), L:(12,14), M:(12,15).
G finishes at 2, I at 6, and H at 5, so K runs from 6 to 12. Both L and M then start at 12, finishing at 14 and 15.
(b) H:(1,6), J:(10,13), L:(13,15).
Using finish time 15, L has LF=15,LS=13, M has LS=12, K has LS=6, H has LS=1, and J has LF=13,LS=10.
(c) (15text( days)), with critical path G-I-K-M.
Activities G,I,K,M each have zero slack and total 2+4+6+3=15 days.
(d) (16text( days)), with new critical path H-K-M.
H originally has 1 day of slack. A 2-day delay uses that slack and delays K by 1 day, so H-K-M finishes at 5+2+6+3=16 days.
Mark allocation
- Part a: obtains K:(ES,EF)=(6,12); obtains L:(12,14); obtains M:(12,15).
- Part b: obtains H:(LS,LF)=(1,6); obtains J:(10,13); obtains L:(13,15).
- Part c: states the 15-day completion time; identifies G-I-K-M as the critical path.
- Part d: compares the delay with H's 1-day slack; obtains the new duration 16 days; identifies H-K-M as a new critical path.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: obtains K:(ES,EF)=(6,12); obtains L:(12,14); obtains M:(12,15).
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: obtains H:(LS,LF)=(1,6); obtains J:(10,13); obtains L:(13,15).
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Part c (2 marks)
Award the 2 available marks for the following observable evidence: states the 15-day completion time; identifies G-I-K-M as the critical path.
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Part d (3 marks)
Award the 3 available marks for the following observable evidence: compares the delay with H's 1-day slack; obtains the new duration 16 days; identifies H-K-M as a new critical path.
Acceptable alternatives: Accept an equivalent activity-on-node table or network notation that shows the same precedence relationships and times.; Accept any correct forward / backward-scan layout and any equivalent listing of all zero-slack critical paths.
Do not credit by itself: Do not credit a scan that fails to use the latest predecessor finish at a merge or the earliest successor start in the backward pass.; Do not infer a critical path from path length alone when the required zero-slack evidence or revised schedule is inconsistent.
Question 8
(a) hat y=2.143+2.083x, with r²=0.9446.
Linear regression gives intercept 2.142857ldots, slope 2.082857ldots, and r²=0.944574ldots.
(b) hat y=3.106(1.343)^x, with r²=0.9986.
Exponential regression gives a=3.105703ldots, b=1.342511ldots, and r²=0.998635ldots on the transformed regression.
(c) Linear residual =-0.991text( g); exponential residual =-0.115text( g), to three significant figures.
The linear prediction is 8.391428ldots, giving 7.4-8.391428ldots=-0.991428ldots. The exponential prediction is 7.514736ldots, giving 7.4-7.514736ldots=-0.114736ldots.
(d) The exponential model is better; hat y(7)=(24.4text( g)). This extrapolation assumes the same growth factor continues beyond day 5.
The exponential model has r² much closer to 1 and substantially smaller residuals without the curved pattern produced by a line. It predicts 3.105703(1.342511)⁷=24.41099ldotstext( g).
Mark allocation
- Part a: enters the six data pairs correctly; selects linear regression; reports the model and r²=0.9446.
- Part b: selects exponential regression y=ab^x; reports a=3.106 and b=1.343; reports r²=0.9986.
- Part c: calculates the unrounded linear prediction; obtains linear residual -0.991text( g); calculates the unrounded exponential prediction; obtains exponential residual -0.115text( g).
- Part d: selects the exponential model; uses r² and residual evidence to justify the choice; calculates 24.4text( g) from the unrounded exponential model; identifies extrapolation or the continuing-growth-factor assumption.
Detailed marking criteria
Part a (3 marks)
Award the 3 available marks for the following observable evidence: enters the six data pairs correctly; selects linear regression; reports the model and r²=0.9446.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part b (3 marks)
Award the 3 available marks for the following observable evidence: selects exponential regression y=ab^x; reports a=3.106 and b=1.343; reports r²=0.9986.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part c (4 marks)
Award the 4 available marks for the following observable evidence: calculates the unrounded linear prediction; obtains linear residual -0.991text( g); calculates the unrounded exponential prediction; obtains exponential residual -0.115text( g).
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Part d (4 marks)
Award the 4 available marks for the following observable evidence: selects the exponential model; uses r² and residual evidence to justify the choice; calculates 24.4text( g) from the unrounded exponential model; identifies extrapolation or the continuing-growth-factor assumption.
Acceptable alternatives: Accept direct calculator regression output or equivalent transformed working when the explanatory and response variables are entered in the stated order.; Accept an equivalent residual table or direct observed-minus-predicted calculation using unrounded fitted values.
Do not credit by itself: Do not credit swapped explanatory and response variables, an unsupported model choice, or a residual calculated as predicted minus observed.; Do not use prematurely rounded regression coefficients when they change a residual, forecast or model comparison.
Question 9
(a) (261,542) to the nearest dollar.
With i=0.051 / 12 and N=120, FV=420000(1+i)¹²⁰-2800(((1+i)¹²⁰-1) / (i))=261542.378ldots.
(b) (2975.47) per month.
Use PV=420000, i=0.051 / 12, N=216, and FV=0. The annuity payment is 2975.4681ldots, or 2975.47.
(c) Withdrawal =(2796.37) per month, about 179.09 less each month.
With i=0.043 / 12 and N=216, the payment is 2796.3746ldots. The reduction from the 5.1% result is 2975.4681ldots-2796.3746ldots=179.0935ldots.
Mark allocation
- Part a: uses monthly rate 0.051 / 12; uses 120 end-of-month withdrawals; uses opposite signs for the initial balance and withdrawals; obtains 261,542 to the nearest dollar.
- Part b: uses 216 monthly periods; sets future value to zero; uses correct cash-flow signs in the annuity calculation; rounds the payment to 2975.47.
- Part c: changes the monthly rate to 0.043 / 12; retains PV=420000, N=216, FV=0; obtains 2796.37; calculates a decrease of about 179.09 per month; explains that the lower return supports a smaller sustainable withdrawal.
Detailed marking criteria
Part a (4 marks)
Award the 4 available marks for the following observable evidence: uses monthly rate 0.051 / 12; uses 120 end-of-month withdrawals; uses opposite signs for the initial balance and withdrawals; obtains 261,542 to the nearest dollar.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part b (4 marks)
Award the 4 available marks for the following observable evidence: uses 216 monthly periods; sets future value to zero; uses correct cash-flow signs in the annuity calculation; rounds the payment to 2975.47.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Part c (5 marks)
Award the 5 available marks for the following observable evidence: changes the monthly rate to 0.043 / 12; retains PV=420000, N=216, FV=0; obtains 2796.37; calculates a decrease of about 179.09 per month; explains that the lower return supports a smaller sustainable withdrawal.
Acceptable alternatives: Accept a TVM solver or an explicit compound-interest or annuity equation with the correct periodic rate, number of periods and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite throughout the calculation.
Do not credit by itself: Do not credit use of a nominal annual rate as the periodic rate, an incorrect period count, or fees applied at the wrong time.; Do not accept inconsistent cash-flow signs or premature rounding that changes a balance, repayment or comparison.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Topic 3: Statistical Models | Q1 | 6 | ___ | Rework Q1: select the correct normal-distribution bound, then use inverse normal to determine the requested percentile. |
| Topic 4: Financial Models | Q2 | 8 | ___ | Rework Q2: include every compulsory fee in the payment cash flow and solve for the monthly rate before annualising it. |
| Topic 4: Financial Models | Q3 | 9 | ___ | Rework Q3: use the unrounded outstanding balance as the new present value when the interest rate and repayment conditions change. |
| Topic 5: Discrete Models | Q4 | 12 | ___ | Rework Q4: complete row and column reductions, select independent zeros, and total the chosen entries from the original array. |
| Topic 5: Discrete Models | Q5 | 7 | ___ | Rework Q5: recalculate earliest and latest times from the complete precedence table before identifying the critical path or slack. |
| Topic 3: Statistical Models | Q6 | 10 | ___ | Rework Q6: enter the paired data, select exponential regression, calculate residual as observed minus predicted, and interpret the multiplicative factor. |
| Topic 5: Discrete Models | Q7 | 11 | ___ | Rework Q7: complete both scans, identify every zero-slack activity, and compare a proposed delay with the activity's available slack. |
| Topic 3: Statistical Models | Q8 | 14 | ___ | Rework Q8: fit both models, calculate observed-minus-predicted residuals, and compare residual behaviour and r² before selecting a model. |
| Topic 4: Financial Models | Q9 | 13 | ___ | Rework Q9: use consistent monthly timing and cash-flow signs, retaining the unrounded balance or payment for comparisons under changed rates. |