Skill Align ACT BSSS Essential Mathematics A Free Online Pack 0 - 2026 Edition
Original Skill Align Essential Mathematics A practice assessment; not an official ACT BSSS examination. The 90-mark total is a Skill Align practice convention. Teachers must adapt content, scaffolding and conditions before any use with Essential Mathematics M students.
- Paper
- Question and Response Booklet Showcase
- Reading
- As set by the supervising teacher
- Writing
- 125 minutes
- Assessment
- 90 marks
Calculator and technology conditions: A scientific calculator may be used. Graphics calculators, CAS, spreadsheet software and internet access are not permitted for this practice assessment. Candidates may use only formula or reference materials supplied by the supervising teacher.
Question and Response Booklet
Answer all questions. Show working, units and an interpretation in context where requested. This paper is worth 90 marks as a Skill Align practice convention and has a suggested working time of 125 minutes. Calculator and technology conditions: A scientific calculator may be used. Graphics calculators, CAS, spreadsheet software and internet access are not permitted for this practice assessment. Candidates may use only formula or reference materials supplied by the supervising teacher. Teachers must adapt this A paper before any use with Essential Mathematics M students.
Question 1
10 marksQuestion 2
10 marksQuestion 3
10 marksQuestion 4
10 marksQuestion 5
10 marksQuestion 6
10 marksQuestion 7
10 marksQuestion 8
10 marksQuestion 9
10 marksWorked Solutions And Marking Guide
Question 1
(a) 9.5 m by 6.5 m.
Multiply by 250 and convert centimetres to metres.
(b) 61.75 m^2.
9.5 × 6.5=61.75.
(c) AUD 2594.
61.75 × 42=2593.50, which rounds to 2594.
Mark allocation
- Part a: uses the scale factor 250 on both plan dimensions; obtains 950 cm and 650 cm, or equivalent intermediate lengths; converts and states 9.5 m by 6.5 m. Accept: accept 9500 mm by 6500 mm before correct metre conversion; accept equivalent working that keeps units explicit.
- Part b: selects rectangle area as length multiplied by width; substitutes the converted dimensions 9.5 and 6.5; states 61.75 m² with square units. Accept: apply consequential marking to correctly multiplied dimensions from part (a); accept 61.8 m² if explicitly rounded to one decimal place.
- Part c: uses the floor area and the rate of AUD 42 per square metre; obtains an unrounded cost of AUD 2593.50, or the consequential equivalent; rounds to and states AUD 2594 in context; identifies the result as the estimated carpet cost. Accept: apply consequential marking to a valid area from part (b); accept $2,594 or AUD 2594.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the scale factor 250 on both plan dimensions; obtains 950 cm and 650 cm, or equivalent intermediate lengths; converts and states 9.5 m by 6.5 m. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept 9500 mm by 6500 mm before correct metre conversion; accept equivalent working that keeps units explicit
Do not credit by itself: using 1:250 as division by 250; converting centimetres to metres by an incorrect factor
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: selects rectangle area as length multiplied by width; substitutes the converted dimensions 9.5 and 6.5; states 61.75 m² with square units. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to correctly multiplied dimensions from part (a); accept 61.8 m² if explicitly rounded to one decimal place
Do not credit by itself: multiplying the plan dimensions without applying the scale; reporting metres rather than square metres
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: uses the floor area and the rate of AUD 42 per square metre; obtains an unrounded cost of AUD 2593.50, or the consequential equivalent; rounds to and states AUD 2594 in context; identifies the result as the estimated carpet cost. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid area from part (b); accept $2,594 or AUD 2594
Do not credit by itself: rounding before multiplying; omitting the currency or reporting AUD 2593.50 when a nearest-dollar answer is requested
Question 2
(a) 2.4 m^3.
2.4 × 1.25 × 0.8=2.4.
(b) 2400 L.
One cubic metre equals 1000 litres.
(c) 100 minutes.
0.75 × 2400=1800 litres and 1800 / 18=100.
Mark allocation
- Part a: selects rectangular-prism volume V=lwh; substitutes 2.4 × 1.25 × 0.8; states 2.4 m^3. Accept: accept any multiplication order; accept a correct diagram-based calculation with cubic-metre units.
- Part b: states or uses 1 m³=1000 L; multiplies 2.4 by 1000; states 2400 L. Accept: apply consequential marking to part (a) using the correct conversion; accept 2.4 kL only when also expressed as 2400 L.
- Part c: calculates 75% of the capacity as 1800 L; divides the filled volume by 18 L / min; obtains 100; states 100 minutes as the draining time. Accept: apply consequential marking to a valid litre capacity from part (b); accept 1 hour 40 minutes.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: selects rectangular-prism volume V=lwh; substitutes 2.4 × 1.25 × 0.8; states 2.4 m^3. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept any multiplication order; accept a correct diagram-based calculation with cubic-metre units
Do not credit by itself: using surface area instead of volume; omitting cubic units
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: states or uses 1 m³=1000 L; multiplies 2.4 by 1000; states 2400 L. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to part (a) using the correct conversion; accept 2.4 kL only when also expressed as 2400 L
Do not credit by itself: dividing by 1000; reporting cubic metres when litres are requested
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: calculates 75% of the capacity as 1800 L; divides the filled volume by 18 L / min; obtains 100; states 100 minutes as the draining time. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid litre capacity from part (b); accept 1 hour 40 minutes
Do not credit by itself: using the full capacity instead of 75%; multiplying by the flow rate or omitting minutes
Question 3
(a) 17(:)40 on Tuesday.
Singapore is two hours behind Canberra.
(b) 19(:)25 on Tuesday.
Add one hour and 45 minutes.
(c) 16(:)15 on Wednesday.
Add ten hours to UTC.
Mark allocation
- Part a: identifies Singapore as two hours behind Canberra for the stated schedule; subtracts two hours from 19(:)40; states 17(:)40 on Tuesday in Singapore. Accept: accept a correct UTC conversion through 09(:)40 UTC; accept 5:40 pm Tuesday.
- Part b: converts 105 minutes to 1 hour 45 minutes; adds the full duration to 17(:)40; states 19(:)25 on Tuesday in Singapore. Accept: accept adding 105 minutes directly; apply consequential marking from a stated part (a) time.
- Part c: uses Canberra as UTC+10 for the stated schedule; adds 10 hours to 06(:)15 UTC; obtains 16(:)15; states Wednesday in Canberra. Accept: accept 4:15 pm Wednesday; accept a correctly labelled 24-hour or 12-hour time.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: identifies Singapore as two hours behind Canberra for the stated schedule; subtracts two hours from 19(:)40; states 17(:)40 on Tuesday in Singapore. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept a correct UTC conversion through 09(:)40 UTC; accept 5:40 pm Tuesday
Do not credit by itself: adding two hours; omitting or changing the day
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: converts 105 minutes to 1 hour 45 minutes; adds the full duration to 17(:)40; states 19(:)25 on Tuesday in Singapore. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept adding 105 minutes directly; apply consequential marking from a stated part (a) time
Do not credit by itself: treating 105 minutes as 1.05 hours; incorrect carry through the hour
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: uses Canberra as UTC+10 for the stated schedule; adds 10 hours to 06(:)15 UTC; obtains 16(:)15; states Wednesday in Canberra. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept 4:15 pm Wednesday; accept a correctly labelled 24-hour or 12-hour time
Do not credit by itself: subtracting the Canberra offset; advancing to Thursday
Question 4
(a) 96 kg.
Add 18+24+15+21+18.
(b) 19.2 kg.
Divide 96 by 5.
(c) 60%.
(24-15) / 15 × 100=60%.
Mark allocation
- Part a: reads the five values 18,24,15,21,18 from the chart; adds all five values; states a total harvest of 96 kg. Accept: accept any correctly grouped addition; accept a table transcribed from the chart before summing.
- Part b: uses the total harvest 96; divides by five garden beds; states a mean of 19.2 kg per bed. Accept: apply consequential marking to a valid total from part (a); accept 19.2 kg / bed.
- Part c: finds the difference 24-15=9 kg; uses Bed C's 15 kg as the comparison denominator; calculates 9 / 15 × 100; states that Bed B exceeds Bed C by 60%. Accept: accept 1.6 as the ratio only when interpreted as a 60% increase; accept equivalent fraction or decimal working.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: reads the five values 18,24,15,21,18 from the chart; adds all five values; states a total harvest of 96 kg. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept any correctly grouped addition; accept a table transcribed from the chart before summing
Do not credit by itself: omitting one repeated 18; reporting 96 without kilograms
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the total harvest 96; divides by five garden beds; states a mean of 19.2 kg per bed. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid total from part (a); accept 19.2 kg / bed
Do not credit by itself: dividing by four intervals; reporting a total rather than a mean
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: finds the difference 24-15=9 kg; uses Bed C's 15 kg as the comparison denominator; calculates 9 / 15 × 100; states that Bed B exceeds Bed C by 60%. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept 1.6 as the ratio only when interpreted as a 60% increase; accept equivalent fraction or decimal working
Do not credit by itself: using Bed B's value as the denominator; stating 160% as the percentage increase
Question 5
(a) 0.70.
126 / 180=0.70.
(b) About 315 riders.
0.70 × 450=315.
(c) Rider age, local rules, route type or observation time may differ.
Relative frequency is sample- and context-dependent.
Mark allocation
- Part a: forms the relative frequency 126 / 180; evaluates the fraction as 0.70; states the relative frequency without attaching a rider count. Accept: accept 0.7, 70% or 7 / 10; accept an equivalent simplified fraction.
- Part b: uses the relative frequency 0.70; multiplies by 450 riders; states an estimate of 315 helmet users. Accept: apply consequential marking to a valid rate from part (a); accept about 315 riders.
- Part c: identifies one changed sampling factor such as rider age, route, rules or observation time; explains how that factor could change helmet-use behaviour; links the limitation to applying this sample rate at another location; states the conclusion in the rider-observation context. Accept: accept another specific, plausible sampling or context difference; accept a concise response when the causal link is explicit.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: forms the relative frequency 126 / 180; evaluates the fraction as 0.70; states the relative frequency without attaching a rider count. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept 0.7, 70% or 7 / 10; accept an equivalent simplified fraction
Do not credit by itself: using 126 / (180-126); reporting 126 as the relative frequency
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the relative frequency 0.70; multiplies by 450 riders; states an estimate of 315 helmet users. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid rate from part (a); accept about 315 riders
Do not credit by itself: dividing 450 by 0.70; omitting that the result is an estimate
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: identifies one changed sampling factor such as rider age, route, rules or observation time; explains how that factor could change helmet-use behaviour; links the limitation to applying this sample rate at another location; states the conclusion in the rider-observation context. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept another specific, plausible sampling or context difference; accept a concise response when the causal link is explicit
Do not credit by itself: giving only 'the sample is different'; claiming relative frequency guarantees the future count
Question 6
(a) B_(n+1)=1.005B_n-280, B_0=5200.
Apply interest first and subtract the payment.
(b) About AUD 4691.
B_1=4946, then B_2=1.005(4946)-280=4690.73.
(c) Monthly interest is added before the fixed payment.
Interest offsets part of each payment.
Mark allocation
- Part a: uses the monthly multiplier 1.005; subtracts the payment 280 after applying interest; states B_(n+1)=1.005B_n-280 with B_0=5200. Accept: accept an equivalent recurrence with clearly defined indexing; accept B_1=1.005(5200)-280 as supporting working, but the general recurrence is still required.
- Part b: calculates B_1=4946; substitutes B_1 into the recurrence for a second iteration; obtains B_2=4690.73 before rounding; states a nearest-dollar balance of about AUD 4691. Accept: accept $4,691 or AUD 4691; apply consequential marking to a correct second iteration of the recurrence stated in part (a).
- Part c: identifies that interest is added each month; identifies that the fixed payment is then subtracted; explains that the interest offsets part of the AUD 280 payment. Accept: accept equivalent wording that correctly relates interest, payment and net decrease; accept a numerical illustration using either monthly balance.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the monthly multiplier 1.005; subtracts the payment 280 after applying interest; states B_(n+1)=1.005B_n-280 with B_0=5200. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept an equivalent recurrence with clearly defined indexing; accept B_1=1.005(5200)-280 as supporting working, but the general recurrence is still required
Do not credit by itself: using 1.05 for 0.5%; subtracting 280 before applying interest
Part b (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: calculates B_1=4946; substitutes B_1 into the recurrence for a second iteration; obtains B_2=4690.73 before rounding; states a nearest-dollar balance of about AUD 4691. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept $4,691 or AUD 4691; apply consequential marking to a correct second iteration of the recurrence stated in part (a)
Do not credit by itself: performing only one payment cycle; rounding the first balance so coarsely that the final dollar changes
Part c (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: identifies that interest is added each month; identifies that the fixed payment is then subtracted; explains that the interest offsets part of the AUD 280 payment. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept equivalent wording that correctly relates interest, payment and net decrease; accept a numerical illustration using either monthly balance
Do not credit by itself: claiming the interest is subtracted; stating only that the balance falls without explaining why it falls by less than AUD 280
Question 7
(a) 5 km.
√(4²+3²)=5.
(b) 11 km.
BC is 6 km, so the total is 11 km.
(c) 2 hours 30 minutes.
11 / 4.4=2.5 hours.
Mark allocation
- Part a: finds coordinate changes of 4 km east and 3 km north; uses Pythagoras √(4²+3²); states AB=5 km. Accept: accept recognition of the 3-4-5 triangle; accept a correctly scaled geometric measurement only with supporting calculation.
- Part b: reads BC=6 km from the horizontal coordinate difference; adds route legs 5+6; states a total route distance of 11 km. Accept: apply consequential marking to a valid AB from part (a); accept explicitly summed segment lengths.
- Part c: uses time = distance divided by speed; calculates 11 / 4.4=2.5 hours; converts 0.5 hour to 30 minutes; states 2 hours 30 minutes. Accept: apply consequential marking to a valid route distance from part (b); accept 150 minutes or 2.5 hours.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: finds coordinate changes of 4 km east and 3 km north; uses Pythagoras √(4²+3²); states AB=5 km. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept recognition of the 3-4-5 triangle; accept a correctly scaled geometric measurement only with supporting calculation
Do not credit by itself: adding 4+3 as the diagonal; omitting kilometres
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: reads BC=6 km from the horizontal coordinate difference; adds route legs 5+6; states a total route distance of 11 km. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid AB from part (a); accept explicitly summed segment lengths
Do not credit by itself: calculating direct displacement from A to C; omitting one route leg
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: uses time = distance divided by speed; calculates 11 / 4.4=2.5 hours; converts 0.5 hour to 30 minutes; states 2 hours 30 minutes. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid route distance from part (b); accept 150 minutes or 2.5 hours
Do not credit by itself: multiplying distance by speed; interpreting 2.5 hours as 2 hours 5 minutes
Question 8
(a) 114 m^2.
14 × 9-4 × 3=126-12=114.
(b) 190 tiles.
114 / 0.6=190.
(c) 204 tiles.
190 × 1.07=203.3, so round up to 204.
Mark allocation
- Part a: calculates the outer area 126 m² and garden area 12 m²; subtracts garden from courtyard; states a paved area of 114 m^2. Accept: accept a correct partition of the L-shaped paved region; accept equivalent square-metre working.
- Part b: uses the paved area 114 m²; divides by 0.6 m² per tile; states a minimum of 190 tiles. Accept: apply consequential marking to a valid paved area from part (a); accept 190.0 before stating 190 whole tiles.
- Part c: uses an extra factor of 1.07 on the base tile count; calculates 190 × 1.07=203.3; recognises tiles must be ordered as whole items; rounds up and states an order of 204 tiles. Accept: apply consequential marking to a valid tile count from part (b); accept adding 7% as 190+13.3.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: calculates the outer area 126 m² and garden area 12 m²; subtracts garden from courtyard; states a paved area of 114 m^2. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept a correct partition of the L-shaped paved region; accept equivalent square-metre working
Do not credit by itself: adding the removed garden; using perimeter or reporting metres rather than square metres
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the paved area 114 m²; divides by 0.6 m² per tile; states a minimum of 190 tiles. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid paved area from part (a); accept 190.0 before stating 190 whole tiles
Do not credit by itself: multiplying by 0.6; rounding down a non-whole consequential result
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: uses an extra factor of 1.07 on the base tile count; calculates 190 × 1.07=203.3; recognises tiles must be ordered as whole items; rounds up and states an order of 204 tiles. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: apply consequential marking to a valid tile count from part (b); accept adding 7% as 190+13.3
Do not credit by itself: rounding 203.3 to the nearest whole item as 203; adding 0.07 tiles instead of 7%
Question 9
(a) An increase of 6 percentage points, from 78% to 84%.
84-78=6. Because two percentages are being subtracted, the absolute change is 6 percentage points.
(b) About 7.7%.
(84-78) / 78 × 100=7.692ldots%, which rounds to 7.7%.
(c) The 75% baseline exaggerates the visual change: the displayed bar heights are 3 and 9 units even though the rate rose by only 6 percentage points. Start the vertical axis at 0%, or show and justify a clear axis break.
A truncated vertical axis makes the second displayed bar three times the first displayed height, which overstates the modest absolute and relative changes.
Mark allocation
- Part a: subtracts 78 from 84; obtains an absolute change of 6; states the change as an increase of 6 percentage points. Accept: accept a sentence stating that satisfaction rose from 78% to 84% by 6 percentage points; accept '6 points' only when percentage points are clear from context.
- Part b: uses the change 84-78=6; uses the original 78% rate as the denominator; calculates and states a relative increase of about 7.7%. Accept: accept 7.69% before one-decimal-place rounding; apply consequential marking to a valid change from part (a).
- Part c: identifies the 75% vertical-axis baseline as truncated; explains that the displayed bar heights are 3 and 9 scale units; explains that this exaggerates a 6-percentage-point or 7.7% increase; proposes a zero baseline or a clearly marked and justified axis break. Accept: accept another improved design that preserves a truthful visual scale and labels the values; accept a precise warning annotation only when the truncated scale remains unmistakable.
Detailed marking criteria
Part a (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: subtracts 78 from 84; obtains an absolute change of 6; states the change as an increase of 6 percentage points. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept a sentence stating that satisfaction rose from 78% to 84% by 6 percentage points; accept '6 points' only when percentage points are clear from context
Do not credit by itself: calling the absolute change 6% without distinguishing units; subtracting the truncated-axis baseline
Part b (3 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 3: uses the change 84-78=6; uses the original 78% rate as the denominator; calculates and states a relative increase of about 7.7%. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept 7.69% before one-decimal-place rounding; apply consequential marking to a valid change from part (a)
Do not credit by itself: using 84 as the denominator; reporting 6 / 78 without converting to a percentage
Part c (4 marks)
Award one mark for each distinct evidence statement demonstrated, to a maximum of 4: identifies the 75% vertical-axis baseline as truncated; explains that the displayed bar heights are 3 and 9 scale units; explains that this exaggerates a 6-percentage-point or 7.7% increase; proposes a zero baseline or a clearly marked and justified axis break. Apply consequential marking when a correct method uses an earlier incorrect value, unless that value makes the task materially easier.
Acceptable alternatives: accept another improved design that preserves a truthful visual scale and labels the values; accept a precise warning annotation only when the truncated scale remains unmistakable
Do not credit by itself: claiming the plotted percentages themselves are false; criticising the graph without linking the concern to the truncated scale
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Unit 3 — Scales, plans and models — Find the actual dimensions in metres | Q1(a) | 3 | ___ | Rework Q1(a): uses the scale factor 250 on both plan dimensions; obtains 950 cm and 650 cm, or equivalent intermediate lengths; converts and states 9.5 m by 6.5 m. |
| Unit 3 — Scales, plans and models — Find the actual floor area | Q1(b) | 3 | ___ | Rework Q1(b): selects rectangle area as length multiplied by width; substitutes the converted dimensions 9.5 and 6.5; states 61.75 m² with square units. |
| Unit 3 — Scales, plans and models — Carpet costs AUD 42 per square metre. Estimate the total cost to the nearest dollar | Q1(c) | 4 | ___ | Rework Q1(c): uses the floor area and the rate of AUD 42 per square metre; obtains an unrounded cost of AUD 2593.50, or the consequential equivalent; rounds to and states AUD 2594 in context; identifies the result as the estimated carpet cost. |
| Unit 3 — Measurement — Calculate its full capacity in cubic metres | Q2(a) | 3 | ___ | Rework Q2(a): selects rectangular-prism volume V=lwh; substitutes 2.4 × 1.25 × 0.8; states 2.4 m^3. |
| Unit 3 — Measurement — Convert the capacity to litres | Q2(b) | 3 | ___ | Rework Q2(b): states or uses 1 m³=1000 L; multiplies 2.4 by 1000; states 2400 L. |
| Unit 3 — Measurement — The trough is filled to 75 percent and drains at 18 litres per minute. Find the draining time | Q2(c) | 4 | ___ | Rework Q2(c): calculates 75% of the capacity as 1800 L; divides the filled volume by 18 L / min; obtains 100; states 100 minutes as the draining time. |
| Unit 4 — Earth geometry and time zones — Find the starting time in Singapore | Q3(a) | 3 | ___ | Rework Q3(a): identifies Singapore as two hours behind Canberra for the stated schedule; subtracts two hours from 19(:)40; states 17(:)40 on Tuesday in Singapore. |
| Unit 4 — Earth geometry and time zones — The webinar lasts 105 minutes. Find its Singapore finishing time | Q3(b) | 3 | ___ | Rework Q3(b): converts 105 minutes to 1 hour 45 minutes; adds the full duration to 17(:)40; states 19(:)25 on Tuesday in Singapore. |
| Unit 4 — Earth geometry and time zones — A replay is released at 06:15 UTC on Wednesday. Find the Canberra release time | Q3(c) | 4 | ___ | Rework Q3(c): uses Canberra as UTC+10 for the stated schedule; adds 10 hours to 06(:)15 UTC; obtains 16(:)15; states Wednesday in Canberra. |
| Unit 2 — Representing and comparing data — Find the total harvest | Q4(a) | 3 | ___ | Rework Q4(a): reads the five values 18,24,15,21,18 from the chart; adds all five values; states a total harvest of 96 kg. |
| Unit 2 — Representing and comparing data — Find the mean harvest per bed | Q4(b) | 3 | ___ | Rework Q4(b): uses the total harvest 96; divides by five garden beds; states a mean of 19.2 kg per bed. |
| Unit 2 — Representing and comparing data — Find the percentage by which Bed B exceeds Bed C | Q4(c) | 4 | ___ | Rework Q4(c): finds the difference 24-15=9 kg; uses Bed C's 15 kg as the comparison denominator; calculates 9 / 15 × 100; states that Bed B exceeds Bed C by 60%. |
| Unit 4 — Probability and relative frequencies — Find the relative frequency of helmet use | Q5(a) | 3 | ___ | Rework Q5(a): forms the relative frequency 126 / 180; evaluates the fraction as 0.70; states the relative frequency without attaching a rider count. |
| Unit 4 — Probability and relative frequencies — Estimate how many helmet users would be expected among 450 similar riders | Q5(b) | 3 | ___ | Rework Q5(b): uses the relative frequency 0.70; multiplies by 450 riders; states an estimate of 315 helmet users. |
| Unit 4 — Probability and relative frequencies — Give one reason the estimate may not apply at another location | Q5(c) | 4 | ___ | Rework Q5(c): identifies one changed sampling factor such as rider age, route, rules or observation time; explains how that factor could change helmet-use behaviour; links the limitation to applying this sample rate at another location; states the conclusion in the rider-observation context. |
| Unit 4 — Loans and compound interest — Write a recurrence for the balance | Q6(a) | 3 | ___ | Rework Q6(a): uses the monthly multiplier 1.005; subtracts the payment 280 after applying interest; states B_(n+1)=1.005B_n-280 with B_0=5200. |
| Unit 4 — Loans and compound interest — Find the balance after two payments, to the nearest dollar | Q6(b) | 4 | ___ | Rework Q6(b): calculates B_1=4946; substitutes B_1 into the recurrence for a second iteration; obtains B_2=4690.73 before rounding; states a nearest-dollar balance of about AUD 4691. |
| Unit 4 — Loans and compound interest — Explain why the balance falls by less than AUD 280 each month | Q6(c) | 3 | ___ | Rework Q6(c): identifies that interest is added each month; identifies that the fixed payment is then subtracted; explains that the interest offsets part of the AUD 280 payment. |
| Unit 3 — Scales, plans and models — Find the distance AB | Q7(a) | 3 | ___ | Rework Q7(a): finds coordinate changes of 4 km east and 3 km north; uses Pythagoras √(4²+3²); states AB=5 km. |
| Unit 3 — Scales, plans and models — Find the total route distance | Q7(b) | 3 | ___ | Rework Q7(b): reads BC=6 km from the horizontal coordinate difference; adds route legs 5+6; states a total route distance of 11 km. |
| Unit 2 — Time and motion — At an average speed of 4.4 kilometres per hour, find the travel time | Q7(c) | 4 | ___ | Rework Q7(c): uses time = distance divided by speed; calculates 11 / 4.4=2.5 hours; converts 0.5 hour to 30 minutes; states 2 hours 30 minutes. |
| Unit 3 — Measurement — Find the paved area | Q8(a) | 3 | ___ | Rework Q8(a): calculates the outer area 126 m² and garden area 12 m²; subtracts garden from courtyard; states a paved area of 114 m^2. |
| Unit 3 — Measurement — Tiles cover 0.6 square metres each. Find the minimum number of tiles | Q8(b) | 3 | ___ | Rework Q8(b): uses the paved area 114 m²; divides by 0.6 m² per tile; states a minimum of 190 tiles. |
| Unit 3 — Measurement — Allow 7 percent extra for cuts. Find the whole number of tiles to order | Q8(c) | 4 | ___ | Rework Q8(c): uses an extra factor of 1.07 on the base tile count; calculates 190 × 1.07=203.3; recognises tiles must be ordered as whole items; rounds up and states an order of 204 tiles. |
| Unit 3 — Graphs — Calculate the increase in percentage points and describe the change accurately | Q9(a) | 3 | ___ | Rework Q9(a): subtracts 78 from 84; obtains an absolute change of 6; states the change as an increase of 6 percentage points. |
| Unit 3 — Graphs — Calculate the percentage increase relative to the Term 1 rate, to one decimal place | Q9(b) | 3 | ___ | Rework Q9(b): uses the change 84-78=6; uses the original 78% rate as the denominator; calculates and states a relative increase of about 7.7%. |
| Unit 3 — Graphs — Explain why the published graph may mislead a reader and describe one improved graph design | Q9(c) | 4 | ___ | Rework Q9(c): identifies the 75% vertical-axis baseline as truncated; explains that the displayed bar heights are 3 and 9 scale units; explains that this exaggerates a 6-percentage-point or 7.7% increase; proposes a zero baseline or a clearly marked and justified axis break. |