Skill Align QCE Year 12 Essential Mathematics - Practice Assessment Pack 0
Original CIA-style Question and Response Book - Unit 3 - 50 marks
- Paper
- CIA-style Question and Response Book Showcase
- Perusal time
- 5 minutes
- Working time
- 60 minutes
- Assessment
- 50 marks
Any additional response paper must show your name and the relevant question number, and must be attached as directed by your school.
The school supplies the QCAA Essential Mathematics formula book, planning paper and any additional response paper. Planning paper is collected but is not marked. Candidates must not bring notes or an annotated formula book. Candidates need a black or blue pen or 2B pencil, a ruler and a calculator; mobile phones must not be used as calculators.
Part A: Simple
Questions 1-9 are short-response simple-familiar questions worth 40 marks. Answer all questions and show reasoning or working for every question or part worth more than one mark.
Question 1
3 marksQuestion 2
4 marksQuestion 3
5 marksQuestion 4
5 marksQuestion 5
5 marksQuestion 6
5 marksQuestion 7
4 marksQuestion 8
5 marksQuestion 9
4 marksPart B: Complex
Questions 10-11 are short-response complex questions worth 10 marks. Question 10 is complex familiar and Question 11 is complex unfamiliar. Identify relevant information, show the selected mathematical pathway and justify practical decisions.
Question 10
5 marksQuestion 11
5 marksWorked Solutions And Marking Guide
Question 1
(a) 0.85 and 85%.
136 / 160=0.85=85%.
(b) About 340 seeds.
0.85 × 400=340.
Mark allocation
- Part a: 1 mark - uses successful trials divided by total trials.; 1 mark - states 0.85 and the equivalent 85%.
- Part b: 1 mark - multiplies 400 by 0.85 and states about 340 seeds.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 0.85 and 85%.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the unsuccessful count or the total as the relative-frequency numerator; the required result is 0.85 and 85%.
Part b (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the unsuccessful count or the total as the relative-frequency numerator; the required result is About 340 seeds.
Question 2
(a) 18,750 mL.
18.75 × 1000=18,750 mL.
(b) 25 full bottles.
18,750 / 750=25.
Mark allocation
- Part a: 1 mark - uses the factor 1000 mL per litre.; 1 mark - states 18,750 mL with the correct unit.
- Part b: 1 mark - divides the available millilitres by 750.; 1 mark - states 25 full bottles in context.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 18,750 mL.
Do not credit by itself: Do not award an observable-evidence mark when the response uses 100 mL per litre or rounds the bottle count to the nearest value; the required result is 18,750 mL.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 25 full bottles.
Do not credit by itself: Do not award an observable-evidence mark when the response uses 100 mL per litre or rounds the bottle count to the nearest value; the required result is 25 full bottles.
Question 3
(a) 9.6 m.
6.4 × 150=960 cm=9.6 m.
(b) 6.3 m.
4.2 × 150=630 cm=6.3 m.
(c) 31.8 m.
2(9.6+6.3)=31.8 m.
Mark allocation
- Part a: 1 mark - multiplies 6.4 cm by 150.; 1 mark - converts 960 cm to 9.6 m.
- Part b: 1 mark - multiplies 4.2 cm by 150.; 1 mark - converts 630 cm to 6.3 m.
- Part c: 1 mark - uses both actual dimensions to obtain a perimeter of 31.8 m.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 9.6 m.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the plan dimensions as metres or applies the scale in the wrong direction; the required result is 9.6 m.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 6.3 m.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the plan dimensions as metres or applies the scale in the wrong direction; the required result is 6.3 m.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the plan dimensions as metres or applies the scale in the wrong direction; the required result is 31.8 m.
Question 4
(a) 24 m².
A = bh = 9.6 × 2.5 = 24 m².
(b) 3 m.
24 = 8h, so h = 24 / 8 = 3 m.
(c) The height must be perpendicular to the base.
The sloping side is not the perpendicular distance between the parallel bases.
Mark allocation
- Part a: 1 mark - uses base times perpendicular height.; 1 mark - obtains 24 m².
- Part b: 1 mark - uses 24 = 8h.; 1 mark - obtains perpendicular height 3 m.
- Part c: 1 mark - distinguishes perpendicular height from the sloping side.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 24 m².
Do not credit by itself: Do not award an evidence mark for a contradictory result or an unsupported decision.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 3 m.
Do not credit by itself: Do not award an evidence mark for a contradictory result or an unsupported decision.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an evidence mark for a contradictory result or an unsupported decision.
Question 5
(a) 5.0 m.
√(4.8²+1.4²)=√25=5.0 m.
(b) 12.4 m.
2(5.0+1.2)=12.4 m.
Mark allocation
- Part a: 1 mark - identifies the horizontal and vertical lengths as perpendicular.; 1 mark - calculates 4.8²+1.4²=25.; 1 mark - takes the square root and states 5.0 m.
- Part b: 1 mark - adds the 1.2 m platform to one ramp length.; 1 mark - doubles the one-side length to obtain 12.4 m.
Detailed marking criteria
Part a (3 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 5.0 m.
Do not credit by itself: Do not award an observable-evidence mark when the response adds the two side lengths instead of applying Pythagoras or counts only one handrail; the required result is 5.0 m.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 12.4 m.
Do not credit by itself: Do not award an observable-evidence mark when the response adds the two side lengths instead of applying Pythagoras or counts only one handrail; the required result is 12.4 m.
Question 6
(a) 35.2 m.
2π(5.6)approx35.1858 m, which rounds to 35.2 m.
(b) 8.8 m.
((90) / (360)) × 2π(5.6)approx8.7965 m, which rounds to 8.8 m.
(c) 4 arcs.
360° divided by 90° gives 4 quarter-circle arcs.
Mark allocation
- Part a: 1 mark - substitutes radius 5.6 m into C=2π r.; 1 mark - rounds the result to 35.2 m.
- Part b: 1 mark - uses the fraction 90 / 360 of the circumference.; 1 mark - rounds the arc length to 8.8 m.
- Part c: 1 mark - states 4 arcs.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 35.2 m.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the diameter as the radius, applies 90 / 180 instead of 90 / 360 to circumference or miscounts the quarter-arcs; the required result is 35.2 m.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 8.8 m.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the diameter as the radius, applies 90 / 180 instead of 90 / 360 to circumference or miscounts the quarter-arcs; the required result is 8.8 m.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the diameter as the radius, applies 90 / 180 instead of 90 / 360 to circumference or miscounts the quarter-arcs; the required result is 4 arcs.
Question 7
(a) RR, RB, BR, BB.
There are two possible results on each spin, giving four ordered outcomes.
(b) 0.48.
P(RB)+P(BR)=0.4(0.6)+0.6(0.4)=0.48.
(c) 0.52.
P(RR)+P(BB)=0.4²+0.6²=0.52.
Mark allocation
- Part a: 1 mark - lists RR, RB, BR, BB with no omission.
- Part b: 1 mark - multiplies along the RB and BR branches.; 1 mark - adds the mutually exclusive probabilities to obtain 0.48.
- Part c: 1 mark - adds P(RR) and P(BB) to obtain 0.52.
Detailed marking criteria
Part a (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response omits an ordered outcome or adds probabilities along successive branches; the required result is RR, RB, BR, BB.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 0.48.
Do not credit by itself: Do not award an observable-evidence mark when the response omits an ordered outcome or adds probabilities along successive branches; the required result is 0.48.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response omits an ordered outcome or adds probabilities along successive branches; the required result is 0.52.
Question 8
(a) 6.48 m^2.
3.6 × 1.8=6.48 m^2.
(b) 1.08 m^2.
frac12(1.2)(1.8)=1.08 m^2.
(c) 6 times.
6.48 / 1.08=6, so the rectangular area is six times the triangular area.
Mark allocation
- Part a: 1 mark - uses rectangle area 3.6 × 1.8.; 1 mark - states 6.48 m² with square units.
- Part b: 1 mark - uses one-half base times perpendicular height to obtain 1.08 m^2.
- Part c: 1 mark - divides 6.48 by 1.08.; 1 mark - states that the rectangular area is 6 times the triangular area.
Detailed marking criteria
Part a (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 6.48 m^2.
Do not credit by itself: Do not award an observable-evidence mark when the response omits the one-half factor for the triangle or compares lengths instead of areas; the required result is 6.48 m^2.
Part b (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response omits the one-half factor for the triangle or compares lengths instead of areas; the required result is 1.08 m^2.
Part c (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 6 times.
Do not credit by itself: Do not award an observable-evidence mark when the response omits the one-half factor for the triangle or compares lengths instead of areas; the required result is 6 times.
Question 9
(a) 0.55.
55 / 100=0.55.
(b) 0.483.
(42+48+55) / 300=145 / 300approx0.483.
(c) Chance variation produces different results in repeated finite simulations.
A repeated random experiment is not expected to reproduce the same counts each time.
Mark allocation
- Part a: 1 mark - identifies 55 successes in 100 trials as a relative frequency of 0.55.
- Part b: 1 mark - combines the successes to obtain 145 from 300 trials.; 1 mark - divides and rounds to 0.483.
- Part c: 1 mark - states that random variation can produce different results across repeated finite simulations.
Detailed marking criteria
Part a (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the largest run as the combined estimate or expects repeated runs to match exactly; the required result is 0.55.
Part b (2 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 0.483.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the largest run as the combined estimate or expects repeated runs to match exactly; the required result is 0.483.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses the largest run as the combined estimate or expects repeated runs to match exactly; the required result is Chance variation produces different results in repeated finite simulations.
Question 10
(a) tan32^circ=((h) / (18)).
The opposite side is the vertical rise and the adjacent side is 18 m.
(b) 11.25 m.
h=18tan32^circapprox11.2476 m, so h=11.25 m.
(c) 12.8 m.
11.2476+1.6=12.8476 m, which rounds to 12.8 m.
Mark allocation
- Part a: 1 mark - states a tangent relationship using opposite rise over adjacent 18 m.
- Part b: 1 mark - uses degree mode.; 1 mark - calculates 18tan32^circ without premature rounding.; 1 mark - rounds the rise to 11.25 m.
- Part c: 1 mark - adds the 1.6 m eye height and rounds the total to 12.8 m.
Detailed marking criteria
Part a (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses sine or cosine with the 18 m adjacent side or omits the observer's eye height; the required result is tan32^circ=((h) / (18)).
Part b (3 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives 11.25 m.
Do not credit by itself: Do not award an observable-evidence mark when the response uses sine or cosine with the 18 m adjacent side or omits the observer's eye height; the required result is 11.25 m.
Part c (1 mark)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Do not credit by itself: Do not award an observable-evidence mark when the response uses sine or cosine with the 18 m adjacent side or omits the observer's eye height; the required result is 12.8 m.
Question 11
Indicative answer: One trip is sufficient. The turf requirement is 203.84 m², so 6 whole rolls are needed, exactly the trailer limit. A valid limitation is that the scale dimensions and 12% cutting allowance are treated as exact and adequate for the real site.
The courtyard dimensions are 16 m by 12 m, so its area is 192 m^2. The garden-bed dimensions are 5 m by 4 m, so its area is frac12(5)(4)=10 m^2. Turf before allowance is 192-10=182 m², and with allowance it is 182 × 1.12=203.84 m^2. Since 203.84 / 35approx5.824, 6 whole rolls are required. One trip is sufficient because the trailer limit is 6 rolls. This assumes the plan dimensions are accurate and the stated allowance covers site irregularities.
Mark allocation
- Part task: 1 mark - converts the plan scale to actual dimensions of 16 m by 12 m and 5 m by 4 m.; 1 mark - calculates and subtracts the triangular garden-bed area to obtain 182 m² of turf.; 1 mark - applies the 12% allowance to obtain 203.84 m^2.; 1 mark - rounds 203.84 / 35 up to 6 rolls and concludes that one trip is sufficient.; 1 mark - states a relevant assumption or limitation, such as plan accuracy or adequacy of the cutting allowance.
Detailed marking criteria
Integrated response (5 marks)
Award each mark independently for its corresponding observable evidence. Apply consequential marking when later working is mathematically correct from an earlier result.
Acceptable alternatives: Accept mathematically equivalent working that demonstrates the listed evidence and gives One trip is sufficient. The turf requirement is 203.84 m², so 6 whole rolls are needed, exactly the trailer limit. A valid limitation is that the scale dimensions and 12% cutting allowance are treated as exact and adequate for the real site.
Do not credit by itself: Do not award an observable-evidence mark when the response uses plan areas as actual areas, includes the garden bed, rounds the roll count down or gives no assumption or limitation; the required result is One trip is sufficient. The turf requirement is 203.84 m², so 6 whole rolls are needed, exactly the trailer limit. A valid limitation is that the scale dimensions and 12% cutting allowance are treated as exact and adequate for the real site.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Probability and relative frequencies: Relative frequency from a trial | Q1 | 3 | ___ | Error pattern: uses the unsuccessful count or the total as the relative-frequency numerator. Remediation: divide the 136 successful trials by 160, then use that decimal as an estimated probability. |
| Measurement: Capacity and metric conversion | Q2 | 4 | ___ | Error pattern: uses 100 mL per litre or rounds the bottle count to the nearest value. Remediation: use 1 L = 1000 mL, then keep only complete bottles. |
| Scales, plans and models: Actual dimensions and perimeter from a plan | Q3 | 5 | ___ | Error pattern: uses the plan dimensions as metres or applies the scale in the wrong direction. Remediation: multiply each plan length by 150, convert centimetres to metres, then calculate perimeter. |
| Measurement: Parallelogram area and perpendicular height | Q4 | 5 | ___ | Error pattern: uses the sloping side as perpendicular height or multiplies instead of dividing to recover height. Remediation: use base times perpendicular height, then rearrange the same relationship for the missing height. |
| Scales, plans and models: Pythagoras in a right triangle | Q5 | 5 | ___ | Error pattern: adds the two side lengths instead of applying Pythagoras or counts only one handrail. Remediation: square the perpendicular sides, take the square root, then include both ramp-and-platform handrails. |
| Measurement: Circumference and arc length | Q6 | 5 | ___ | Error pattern: uses the diameter as the radius, applies 90 / 180 instead of 90 / 360 to circumference or miscounts the quarter-arcs. Remediation: calculate circumference using 2πr, divide it by four for the quarter-circle arc length and recall that four quarter-arcs make a full circle. |
| Probability and relative frequencies: Two-stage sample space | Q7 | 4 | ___ | Error pattern: omits an ordered outcome or adds probabilities along successive branches. Remediation: construct all four ordered outcomes, multiply along branches and add mutually exclusive outcomes. |
| Measurement: Area of standard shapes | Q8 | 5 | ___ | Error pattern: omits the one-half factor for the triangle or compares lengths instead of areas. Remediation: find each standard area independently and compare by dividing the areas. |
| Probability and relative frequencies: Repeated-simulation variation | Q9 | 4 | ___ | Error pattern: uses the largest run as the combined estimate or expects repeated runs to match exactly. Remediation: compare run relative frequencies, pool all trials and explain natural chance variation. |
| Scales, plans and models: Angle of elevation | Q10 | 5 | ___ | Error pattern: uses sine or cosine with the 18 m adjacent side or omits the observer's eye height. Remediation: select tangent in degree mode, retain the unrounded vertical rise and then add eye height. |
| Scales, plans and models: Integrated plan, material and delivery decision | Q11 | 5 | ___ | Error pattern: uses plan areas as actual areas, includes the garden bed, rounds the roll count down or gives no assumption or limitation. Remediation: select and combine the scale conversion, composite-area subtraction, allowance, whole-roll and trailer constraints, then evaluate the model. |