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SACE SACE Stage 2 Essential Mathematics Free Online Pack 0 — Question Booklet Showcase

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SACE Stage 2 Examination 2026 Edition - Pack 0 v1.0
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Question Booklet Showcase

9 questions

90 marks

Estimated duration: 130 minutes

Reading: No separate reading time · Writing: 130 minutes total

Read Question Booklet Showcase online

Skill Align

Skill Align SACE Stage 2 Essential Mathematics - Pack 0 - Question Booklet

Topic 2: Measurement | Topic 4: Statistics | Topic 5: Investments and loans | 90 marks

Paper
Question Booklet Showcase
Reading
No separate reading time
Writing
130 minutes total
Assessment
90 marks

Candidates may bring one unfolded A4 sheet, using both sides, containing their own handwritten notes. Candidates 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, and no external storage media may be used.

Question booklet

Answer all Questions 1-9 and write answers in this booklet. Questions 1-3 assess Topic 2: Measurement; Questions 4-6 assess Topic 4: Statistics; Questions 7-9 assess Topic 5: Investments and loans. 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

7 marks
A triangular shade area has sides AB=8.4text( m) and AC=6.7text( m), with included angle angle BAC=58^circ.
Diagram PreviewABChorizontal distance (m)vertical distance (m)
(a) 2 marks
Calculate the area of the shade region, correct to one decimal place.
(b) 3 marks
Calculate BC, correct to two decimal places.
(c) 2 marks
Edging is installed around the complete shade region. Find the required length, allowing an extra 5%, to the nearest 0.1text( m).

Question 2

10 marks
A revegetation strip beside a creek is measured at equal 5text( m) intervals. The perpendicular widths, in metres, are begin(array)(c|rrrrrrr) text(distance along baseline (m))&0&5&10&15&20&25&30 hline text(width (m))&0&5.4&7.8&8.6&7.2&4.1&0 end(array) Use Simpson's Rule to model the irregular area.
Graph Preview
0510152025308.66.454.32.150distance along baseline (m)perpendicular width (m)
(a) 1 mark
State the Simpson's Rule interval width h.
(b) 4 marks
Estimate the area of the strip, correct to the nearest square metre.
(c) 2 marks
Express the unrounded estimate in hectares.
(d) 3 marks
Seed is sold in bags that cover 120text( m)^2. The order must include an 8% contingency. Find the minimum number of whole bags.

Question 3

13 marks
A water-treatment vessel consists of a vertical cylinder of internal radius 1.2text( m) and height 2.4text( m), topped by a hemisphere of the same radius. Use the internal dimensions.
Diagram Preview 2.4 m 1.2 m
(a) 3 marks
Calculate the cylinder volume, correct to three decimal places.
(b) 3 marks
Calculate the total vessel volume, including the hemisphere.
(c) 3 marks
Only 82% of the total volume is usable. Find the usable capacity in litres.
(d) 2 marks
A pump delivers 110text( L / min). Estimate the time to reach usable capacity, rounded up to the next minute.
(e) 2 marks
Explain why using external rather than internal dimensions would not be appropriate for this capacity calculation.

Question 4

8 marks
A training provider has 180 day-shift learners, 120 evening-shift learners and 60 weekend learners. It wants a proportional stratified sample of 48 learners to investigate weekly travel time.
(a) 2 marks
State the population and the variable being measured.
(b) 3 marks
Calculate the number selected from each shift.
(c) 2 marks
A coordinator instead emails all learners and uses the first 48 replies. Explain one likely bias.
(d) 1 mark
State one improvement.

Question 5

9 marks
The dispatch times, in minutes, for 12 service calls are 18, 21, 22, 24, 24, 25, 27, 28, 30, 31, 35, 48. Use the median-of-halves convention for quartiles.
(a) 2 marks
Write the data as an ordered stem-and-leaf plot using tens as stems. Include a key.
(b) 3 marks
Find the five-number summary.
(c) 2 marks
Calculate the IQR and determine whether 48 is an outlier.
(d) 2 marks
Describe the distribution using its centre, spread and shape.

Question 6

13 marks
A workplace trial records supervised training time x, in hours, and the number of correctly completed tasks y: begin(array)(c|rrrrrrrr) x&2&3&4&5&6&7&8&9 hline y&14&18&21&27&29&34&37&41 end(array) The scatterplot shows the same paired data.
Graph Preview 23.755.57.2591420.7527.534.2541supervised training time (hours)correctly completed tasks
(a) 4 marks
Find the least-squares regression line hat y=a+bx, giving a and b to four significant figures.
(b) 3 marks
Find r and r², then interpret r² in context.
(c) 2 marks
Calculate the residual for the learner with x=6. Use observed minus predicted.
(d) 4 marks
Use the model to predict the result after 12 hours and evaluate the reliability of the prediction.

Question 7

7 marks
A household currently spends 420 each week on a fixed basket of essentials. The cost is expected to rise by 3.2% each year for four years.
(a) 2 marks
Write a calculation for the expected weekly cost after four years.
(b) 3 marks
Calculate the expected weekly cost to the nearest cent.
(c) 2 marks
The household plans a future weekly budget of 450. Evaluate whether it is sufficient.

Question 8

10 marks
A business deposits 350 at the end of every month into a training fund earning 5.4% p.a. compounded monthly for six years. No initial lump sum is invested.
(a) 2 marks
State the periodic interest rate and number of deposits.
(b) 4 marks
Find the fund balance immediately after the final deposit.
(c) 2 marks
If every deposit were instead made at the beginning of the month, find the balance and the increase.
(d) 2 marks
Explain why the beginning-of-month balance is higher although the total deposits are unchanged.

Question 9

13 marks
A home loan of 280000 is charged at 6.0% p.a. compounded monthly. Repayments of 1850 are made at the end of each month. After 60 repayments, the rate changes to 6.6% p.a. compounded monthly.
(a) 2 marks
Find the interest charged in the first month and the balance immediately after the first repayment.
(b) 4 marks
Find the outstanding balance immediately after the 60th repayment.
(c) 4 marks
Using the unrounded balance from part (b), find the monthly repayment required to clear the loan in a further 15 years at the new rate.
(d) 3 marks
Evaluate what happens if the borrower keeps paying only 1850 for those 180 months.

SACE Stage 2 subjects and examinations are administered by the SACE Board of South Australia. Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by, or endorsed by the SACE Board of South Australia or the South Australian Government.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank.

Worked Solutions And Marking Guide

Question 1

(a) boxed(23.9text( m)²).

Area =frac12(8.4)(6.7)sin58^circ=23.8640ldotstext( m)², hence 23.9text( m)^2.

(b) boxed(7.47text( m)).

By the cosine rule, BC²=8.4²+6.7²-2(8.4)(6.7)cos58^circ, so BC=7.47009ldotstext( m).

(c) boxed(23.7text( m)).

The perimeter is 8.4+6.7+7.47009ldots=22.57009ldotstext( m). Including 5% gives 23.69859ldotstext( m), or 23.7text( m).

Mark allocation

  • Part a: substitutes the two adjacent sides and included angle into frac12absin C; reports 23.9text( m)² with correct units.
  • Part b: forms the cosine-rule expression with the included 58^circ angle; evaluates the positive square root 7.47009ldots; reports 7.47text( m).
  • Part c: uses the unrounded third side to obtain the complete perimeter; applies the 1.05 allowance and reports 23.7text( m).

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: substitutes the two adjacent sides and included angle into frac12absin C; reports 23.9text( m)² with correct units.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: forms the cosine-rule expression with the included 58^circ angle; evaluates the positive square root 7.47009ldots; reports 7.47text( m).

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: uses the unrounded third side to obtain the complete perimeter; applies the 1.05 allowance and reports 23.7text( m).

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Question 2

(a) boxed(h=5text( m)).

Consecutive baseline positions differ by 5text( m).

(b) boxed(171text( m)²).

Aapprox((5) / (3))[0+0+4(5.4+8.6+4.1)+2(7.8+7.2)] =170.666ldotstext( m)^2.

(c) boxed(0.0171text( ha)) to three significant figures.

Since 1text( ha)=10000text( m)², 170.666ldots / 10000=0.0170666ldotstext( ha).

(d) boxed(2text( bags)).

Required coverage =170.666ldots(1.08)=184.32text( m)^2. Then 184.32 / 120=1.536, so two whole bags are required.

Mark allocation

  • Part a: states h=5text( m).
  • Part b: uses the factor h / 3=5 / 3; assigns weight 4 to ordinates 5.4,8.6,4.1; assigns weight 2 to ordinates 7.8,7.2; obtains 170.666ldotstext( m)², hence 171text( m)^2.
  • Part c: divides the unrounded square-metre area by 10000; reports 0.0171text( ha).
  • Part d: applies the 1.08 contingency to the unrounded area; divides by 120text( m)² per bag; rounds 1.536 upwards to 2 bags.

Detailed marking criteria

Part a (1 mark)

Award the 1 available mark for the following observable evidence: states h=5text( m).

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part b (4 marks)

Award the 4 available marks for the following observable evidence: uses the factor h / 3=5 / 3; assigns weight 4 to ordinates 5.4,8.6,4.1; assigns weight 2 to ordinates 7.8,7.2; obtains 170.666ldotstext( m)², hence 171text( m)^2.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: divides the unrounded square-metre area by 10000; reports 0.0171text( ha).

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part d (3 marks)

Award the 3 available marks for the following observable evidence: applies the 1.08 contingency to the unrounded area; divides by 120text( m)² per bag; rounds 1.536 upwards to 2 bags.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Question 3

(a) boxed(10.857text( m)³).

V_(rm cyl)=pi(1.2)²(2.4)=10.857344ldotstext( m)^3.

(b) boxed(14.476text( m)³) to three decimal places.

The hemisphere volume is frac23pi(1.2)³=3.619114ldotstext( m)^3. Total =10.857344ldots+3.619114ldots=14.476458ldotstext( m)^3.

(c) boxed(11871text( L)) to the nearest litre.

14.476458ldots × 1000 × 0.82=11870.696ldotstext( L).

(d) boxed(108text( minutes)).

11870.696ldots / 110=107.915ldots, which must be rounded up to 108 minutes.

(e) External dimensions include wall thickness and would overstate the space available for water.

Capacity depends on the internal space occupied by the liquid, not the outside envelope.

Mark allocation

  • Part a: selects V=pi r^2h; substitutes r=1.2 and h=2.4; reports 10.857text( m)^3.
  • Part b: uses frac23pi r³ for the hemisphere; adds the hemisphere to the cylinder; reports 14.476text( m)^3.
  • Part c: converts the unrounded total from cubic metres to litres; applies the 0.82 usable fraction; reports 11871text( L).
  • Part d: divides usable capacity by 110text( L / min); rounds upwards to 108 minutes.
  • Part e: identifies wall thickness as the excluded material; states that external dimensions would overestimate usable capacity.

Detailed marking criteria

Part a (3 marks)

Award the 3 available marks for the following observable evidence: selects V=pi r^2h; substitutes r=1.2 and h=2.4; reports 10.857text( m)^3.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: uses frac23pi r³ for the hemisphere; adds the hemisphere to the cylinder; reports 14.476text( m)^3.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part c (3 marks)

Award the 3 available marks for the following observable evidence: converts the unrounded total from cubic metres to litres; applies the 0.82 usable fraction; reports 11871text( L).

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part d (2 marks)

Award the 2 available marks for the following observable evidence: divides usable capacity by 110text( L / min); rounds upwards to 108 minutes.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Part e (2 marks)

Award the 2 available marks for the following observable evidence: identifies wall thickness as the excluded material; states that external dimensions would overestimate usable capacity.

Acceptable alternatives: Accept an equivalent measurement formula or decomposition that uses the same dimensions and produces the same physical quantity.; Accept consequential values from a clearly shown earlier result when later units, rounding and practical interpretation are consistent.

Do not credit by itself: Do not credit a formula using incompatible dimensions, an incorrect trigonometric angle, or a conversion between unlike units.; Do not accept premature rounding that changes a whole-item decision, coverage allowance, capacity or stated tolerance.

Question 4

(a) Population: all 360 learners. Variable: each learner's weekly travel time.

The population is the full group about which conclusions are sought; travel time is the measured quantitative variable.

(b) Day 24, evening 16, weekend 8.

The allocations are 48(180 / 360)=24, 48(120 / 360)=16, and 48(60 / 360)=8.

(c) It is a voluntary-response sample; learners who reply quickly may have different travel experiences from those who do not reply.

The selection probability depends on response behaviour rather than random selection within each shift.

(d) Randomly select the required number from each shift's complete learner list.

This preserves the proportional strata while giving members of each stratum a random chance of selection.

Mark allocation

  • Part a: identifies all 360 learners as the population; identifies weekly travel time as the variable.
  • Part b: calculates the day-shift allocation 24; calculates the evening-shift allocation 16; calculates the weekend allocation 8.
  • Part c: identifies voluntary-response or response-time bias; explains how respondents may differ in travel experience.
  • Part d: proposes random selection within every stratum.

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: identifies all 360 learners as the population; identifies weekly travel time as the variable.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: calculates the day-shift allocation 24; calculates the evening-shift allocation 16; calculates the weekend allocation 8.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: identifies voluntary-response or response-time bias; explains how respondents may differ in travel experience.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part d (1 mark)

Award the 1 available mark for the following observable evidence: proposes random selection within every stratum.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Question 5

(a) 1|8; 2|1,2,4,4,5,7,8; 3|0,1,5; 4|8, with key 2|4=24 minutes.

The values are already ordered. Group each units digit beside its tens stem and state the numerical meaning of one leaf.

(b) boxed(18, 23, 26, 30.5, 48).

The lower six are 18,21,22,24,24,25, so Q_1=(22+24) / 2=23. The median is (25+27) / 2=26. The upper six are 27,28,30,31,35,48, so Q_3=(30+31) / 2=30.5. Thus the summary is 18,23,26,30.5,48.

(c) IQR=7.5. The upper fence is 41.75, so 48 is an outlier.

IQR=30.5-23=7.5. Upper fence =30.5+1.5(7.5)=41.75; 48>41.75.

(d) The median is 26 minutes and the middle 50% spans 7.5 minutes. The high outlier at 48 minutes gives a right-skewed tail.

A complete description uses the median and IQR, then links the isolated high value to right skew.

Mark allocation

  • Part a: places every leaf beside the correct tens stem in ascending order; includes a correct key such as 2|4=24 minutes.
  • Part b: identifies minimum 18 and maximum 48; calculates median 26; calculates Q_1=23 and Q_3=30.5.
  • Part c: calculates IQR=7.5 and upper fence 41.75; correctly classifies 48 as an outlier.
  • Part d: states the median 26 and IQR 7.5 in context; describes the right skew or high outlier.

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: places every leaf beside the correct tens stem in ascending order; includes a correct key such as 2|4=24 minutes.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: identifies minimum 18 and maximum 48; calculates median 26; calculates Q_1=23 and Q_3=30.5.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: calculates IQR=7.5 and upper fence 41.75; correctly classifies 48 as an outlier.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part d (2 marks)

Award the 2 available marks for the following observable evidence: states the median 26 and IQR 7.5 in context; describes the right skew or high outlier.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Question 6

(a) boxed(hat y=6.345+3.869x).

Linear regression gives a=6.345238ldots and b=3.869047ldots.

(b) r=0.9975, r²=0.9950. About 99.5% of the variation in correctly completed tasks is explained by its linear association with supervised training time.

Calculator output gives r=0.997500ldots and r²=0.995007ldots.

(c) boxed(-0.560text( task)) to three significant figures.

The unrounded prediction is 6.345238ldots+3.869047ldots(6)=29.559523ldots. Residual =29-29.559523ldots=-0.559523ldots.

(d) boxed(52.8text( tasks)). This is an extrapolation beyond the observed 2-to-9-hour range and assumes the linear trend continues; fatigue or task limits may make it unreliable.

hat y(12)=6.345238ldots+3.869047ldots(12)=52.773809ldots. The value is outside the data range.

Mark allocation

  • Part a: enters all eight x-values as the explanatory variable; enters the paired y-values as the response variable; selects linear regression; reports a=6.345 and b=3.869.
  • Part b: reports r=0.9975; reports r²=0.9950; interprets r² using task variation and training time.
  • Part c: calculates the unrounded prediction 29.5595ldots; uses observed minus predicted to obtain -0.560.
  • Part d: substitutes x=12 in the unrounded model; reports 52.8 tasks; identifies extrapolation beyond 9 hours; states a contextual limitation such as fatigue, saturation or a finite task pool.

Detailed marking criteria

Part a (4 marks)

Award the 4 available marks for the following observable evidence: enters all eight x-values as the explanatory variable; enters the paired y-values as the response variable; selects linear regression; reports a=6.345 and b=3.869.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: reports r=0.9975; reports r²=0.9950; interprets r² using task variation and training time.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: calculates the unrounded prediction 29.5595ldots; uses observed minus predicted to obtain -0.560.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Part d (4 marks)

Award the 4 available marks for the following observable evidence: substitutes x=12 in the unrounded model; reports 52.8 tasks; identifies extrapolation beyond 9 hours; states a contextual limitation such as fatigue, saturation or a finite task pool.

Acceptable alternatives: Accept direct approved-calculator output when the entered data, variable order and requested statistic or regression model are identified.; Accept an equivalent statistically valid explanation that refers explicitly to sampling, distribution, residual, correlation or model range as required.

Do not credit by itself: Do not credit a biased or non-random selection method as representative without a valid limitation.; Do not swap explanatory and response variables, confuse correlation with causation, reverse the residual sign, or extrapolate without evaluating reliability.

Question 7

(a) boxed(420(1.032)⁴).

A 3.2% annual increase has multiplier 1.032, applied four times.

(b) boxed(476.40).

420(1.032)⁴=476.395970ldots, which rounds to 476.40.

(c) No. It is short by boxed(26.40) per week.

476.395970ldots-450=26.395970ldots, so the budget does not maintain the same purchasing power.

Mark allocation

  • Part a: identifies the multiplier 1.032; uses exponent 4 with the current 420 cost.
  • Part b: evaluates the compound-inflation expression; retains sufficient precision; reports 476.40.
  • Part c: compares 450 with the projected unrounded cost; states the 26.40 shortfall and insufficiency.

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: identifies the multiplier 1.032; uses exponent 4 with the current 420 cost.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part b (3 marks)

Award the 3 available marks for the following observable evidence: evaluates the compound-inflation expression; retains sufficient precision; reports 476.40.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: compares 450 with the projected unrounded cost; states the 26.40 shortfall and insufficiency.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Question 8

(a) i=0.054 / 12=0.0045 per month and N=72.

Six years contains 72 months and the nominal annual rate is divided by 12.

(b) boxed(29683.32).

FV=350(((1.0045)⁷²-1) / (0.0045)) =29683.3239ldots .

(c) Balance boxed(29816.90); increase boxed(133.57).

An annuity due gains one additional month's interest: 29683.3239ldots(1.0045)=29816.8989ldots. The increase is 133.57496ldots.

(d) Each deposit earns interest for one extra month.

The deposit amount and count are identical; only the time invested changes.

Mark allocation

  • Part a: states monthly rate 0.0045; states 72 deposits.
  • Part b: uses an ordinary-annuity future-value model; uses payment 350 and rate 0.0045; uses 72 periods with zero initial value; reports 29683.32.
  • Part c: multiplies the ordinary-annuity value by 1.0045 to obtain 29816.90; finds the increase 133.57.
  • Part d: states that each deposit is invested one month longer; distinguishes timing from total contributions.

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: states monthly rate 0.0045; states 72 deposits.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part b (4 marks)

Award the 4 available marks for the following observable evidence: uses an ordinary-annuity future-value model; uses payment 350 and rate 0.0045; uses 72 periods with zero initial value; reports 29683.32.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part c (2 marks)

Award the 2 available marks for the following observable evidence: multiplies the ordinary-annuity value by 1.0045 to obtain 29816.90; finds the increase 133.57.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part d (2 marks)

Award the 2 available marks for the following observable evidence: states that each deposit is invested one month longer; distinguishes timing from total contributions.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Question 9

(a) Interest boxed(1400.00); balance boxed(279550.00).

First-month interest is 280000(0.06 / 12)=1400. The repayment then gives 280000+1400-1850=279550.

(b) boxed(248603.49).

With i=0.06 / 12=0.005, B_(60)=280000(1.005)⁶⁰-1850(((1.005)⁶⁰-1) / (0.005)) =248603.4863ldots .

(c) boxed(2179.29) per month.

Use PV=248603.4863ldots, i=0.066 / 12=0.0055, N=180, and FV=0. The required payment is 2179.2933ldots.

(d) The loan is not cleared; approximately boxed(100820.29) remains.

At the new rate, FV=248603.4863ldots(1.0055)¹⁸⁰-1850(((1.0055)¹⁸⁰-1) / (0.0055)) =100820.2876ldots .

Mark allocation

  • Part a: calculates first-month interest 1400; applies the repayment after interest to obtain 279550.
  • Part b: uses monthly rate 0.005; uses 60 end-of-month repayments; uses opposite signs for the loan and repayments; reports 248603.49.
  • Part c: uses the unrounded balance as present value; uses monthly rate 0.0055; uses 180 remaining months and zero future value; reports 2179.29.
  • Part d: models 180 payments of 1850 at monthly rate 0.0055; obtains a positive balance near 100820.29; concludes that the loan is not cleared.

Detailed marking criteria

Part a (2 marks)

Award the 2 available marks for the following observable evidence: calculates first-month interest 1400; applies the repayment after interest to obtain 279550.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part b (4 marks)

Award the 4 available marks for the following observable evidence: uses monthly rate 0.005; uses 60 end-of-month repayments; uses opposite signs for the loan and repayments; reports 248603.49.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part c (4 marks)

Award the 4 available marks for the following observable evidence: uses the unrounded balance as present value; uses monthly rate 0.0055; uses 180 remaining months and zero future value; reports 2179.29.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Part d (3 marks)

Award the 3 available marks for the following observable evidence: models 180 payments of 1850 at monthly rate 0.0055; obtains a positive balance near 100820.29; concludes that the loan is not cleared.

Acceptable alternatives: Accept a financial solver, recurrence, table or explicit annuity equation using the correct periodic rate, period count and payment timing.; Accept an alternative cash-flow sign convention when inflows and outflows remain consistently opposite and the contextual conclusion is unchanged.

Do not credit by itself: Do not use a nominal annual rate as the periodic rate, an incorrect payment count, or beginning-of-period timing when end-of-period timing is stated.; Do not accept inconsistent financial-solver signs or a comparison that ignores fees, inflation, remaining balance or the stated time horizon.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Topic 2: Measurement Q1 7 ___ Rework Q1: select the included angle for the area formula and cosine rule, retain unrounded lengths, and attach metre units.
Topic 2: Measurement Q2 10 ___ Rework Q2: apply Simpson's Rule using the stated equal interval width, preserve the 1-4-2-4-2-4-1 weighting, and round whole-item orders upwards.
Topic 2: Measurement Q3 13 ___ Rework Q3: separate the cylinder and hemisphere, convert cubic metres to litres, and base filling time on usable rather than total capacity.
Topic 4: Statistics Q4 8 ___ Rework Q4: calculate each stratum from its population proportion and distinguish random selection from voluntary response.
Topic 4: Statistics Q5 9 ___ Rework Q5: construct the ordered display before finding quartiles, then use the 1.5 × IQR fences to identify an outlier.
Topic 4: Statistics Q6 13 ___ Rework Q6: enter the explanatory and response variables in the correct calculator lists, use observed minus predicted for residuals, and identify extrapolation.
Topic 5: Investments and loans Q7 7 ___ Rework Q7: use the annual inflation multiplier for the full period and compare budgets using amounts measured at the same future date.
Topic 5: Investments and loans Q8 10 ___ Rework Q8: use the monthly interest rate, 72 end-of-month deposits, and distinguish an ordinary annuity from beginning-of-month deposits.
Topic 5: Investments and loans Q9 13 ___ Rework Q9: retain the unrounded outstanding balance, use correct cash-flow signs, and calculate the changed repayment over the remaining monthly term.

What is included

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