Skill Align VCE Foundation Mathematics Units 3&4 - Free Online Pack 0
Full-length question and answer book showcase
- Paper
- Question and Answer Book Showcase
- Reading
- 15 minutes
- Writing
- 2 hours
- Assessment
- 80 marks
One scientific calculator and one annotated bound reference may be used. Use the current official VCAA Foundation Mathematics Formula Sheet and sample Multiple-Choice Answer Sheet linked with this pack.
Section A - Multiple-Choice Questions
Answer all questions on the separate Multiple-Choice Answer Sheet using pencil. Mark only one response for each question and erase an incorrect mark completely before changing it. Each question is worth 1 mark.
Question 1
1 mark- $5.58
- $6.23
- $6.54
- $18.96
Question 2
1 mark- $33
- $187
- $205
- $253
Question 3
1 mark- $75.00
- $150.00
- $152.25
- $2,652.25
Question 4
1 mark- 45
- 57
- 77
- 89
Question 5
1 mark- 4
- 6
- 8
- 12
Question 6
1 mark- 18
- 19
- 20
- 21
Question 7
1 mark- 2
- 2.5
- 3
- 3.5
Question 8
1 mark- 1 / 3
- 2 / 5
- 2 / 3
- 4 / 5
Question 9
1 mark- 3 kWh
- 4 kWh
- 6 kWh
- 11 kWh
Question 10
1 mark- the sample may under-represent members who attend later
- the data cannot contain categories
- the sample is necessarily too large
- a mean cannot be calculated
Question 11
1 mark- 45 m²
- 51 m²
- 57 m²
- 63 m²
Question 12
1 mark- 288 L
- 1920 L
- 2880 L
- 4700 L
Question 13
1 mark- 0.36 km
- 1.44 km
- 3.6 km
- 36 km
Question 14
1 mark- 3.3 m
- 6.0 m
- 7.5 m
- 8.3 m
Question 15
1 mark- 2.51 m
- 5.02 m
- 8.04 m
- 10.05 m
Question 16
1 mark- $36
- $360
- $396
- $400
Question 17
1 mark- 1 hour 35 minutes
- 1 hour 45 minutes
- 2 hours 5 minutes
- 2 hours 45 minutes
Question 18
1 mark- 5 kits per worker-hour
- 6 kits per worker-hour
- 30 kits per worker-hour
- 36 kits per worker-hour
Question 19
1 mark- 48 km / h
- 54 km / h
- 60 km / h
- 72 km / h
Question 20
1 mark- 24
- 25
- 27
- 29
Section B - Written-Response Questions
Answer all questions in the spaces provided. Show appropriate working. Round only when specifically instructed. Diagrams are not necessarily drawn to scale.
Question 1
6 marksQuestion 2
5 marksQuestion 3
4 marksQuestion 4
5 marksQuestion 5
6 marksQuestion 6
4 marksQuestion 7
7 marksQuestion 8
5 marksQuestion 9
4 marksQuestion 10
5 marksQuestion 11
5 marksQuestion 12
4 marksWorked Solutions And Marking Guide
Section A Question 1
Answer: $6.54
Usage costs 18 x $0.31 = $5.58. Adding the $0.96 supply charge gives $6.54.
Section A Question 2
Answer: $187
The discount is 0.15 x 220 = $33, so the price is 220 - 33 = $187.
Section A Question 3
Answer: $152.25
The balance is 2500(1.03)² = $2,652.25, so the interest is $152.25.
Section A Question 4
Answer: 77
Substitute C = 25: F = 1.8(25) + 32 = 45 + 32 = 77.
Section A Question 5
Answer: 8
Solve 48 + 6h = 72 + 3h. Then 3h = 24, so h = 8.
Section A Question 6
Answer: 20
The total is 100 and 100 / 5 = 20.
Section A Question 7
Answer: 3
The cumulative frequency is 8 by rating 2 and 17 by rating 3, so the 10th and 11th values are both 3.
Section A Question 8
Answer: 2 / 3
There are 15 counters and 10 are not black, so the probability is 10 / 15 = 2 / 3.
Section A Question 9
Answer: 6 kWh
The values are 42, 38, 35, 29 and 31. The greatest decrease is 35 - 29 = 6 kWh.
Section A Question 10
Answer: the sample may under-represent members who attend later
A convenience sample taken at one time may not represent members with different attendance patterns.
Section A Question 11
Answer: 57 m²
The area is 9 x 7 - 3 x 2 = 63 - 6 = 57 square metres.
Section A Question 12
Answer: 2880 L
The volume is 2.4 x 1.5 x 0.8 = 2.88 cubic metres, which is 2880 litres.
Section A Question 13
Answer: 3.6 km
7.2 x 50 000 = 360 000 cm = 3.6 km.
Section A Question 14
Answer: 7.5 m
The scale factor is 5 / 0.8 = 6.25, so the height is 1.2 x 6.25 = 7.5 m.
Section A Question 15
Answer: 5.02 m
Circumference = π x diameter = 3.14 x 1.6 = 5.024 m.
Section A Question 16
Answer: $396
GST is 0.10 x 360 = $36. The invoice total is $360 + $36 = $396.
Section A Question 17
Answer: 1 hour 45 minutes
From 1:25 pm to 2:25 pm is one hour, then to 3:10 pm is 45 minutes.
Section A Question 18
Answer: 6 kits per worker-hour
There are 5 x 6 = 30 worker-hours, so the rate is 180 / 30 = 6 kits per worker-hour.
Section A Question 19
Answer: 60 km / h
Average speed = 135 / 2.25 = 60 km / h.
Section A Question 20
Answer: 27
The common difference is 5, so the next term is 22 + 5 = 27.
Section B Question 1
(a) Expenses $1,250; remaining income $430.
The expenses total 720 + 310 + 128 + 92 = $1,250. The amount remaining is $1,680 - $1,250 = $430.
(b) 25.6%; $2,408 saved.
The remaining-income percentage is 430 / 1680 x 100 = 25.595..., or 25.6%. Monthly savings are 0.70 x $430 = $301, so eight months gives $2,408.
Detailed marking criteria
Part (a) (3 marks)
Part a.1 (1 mark)
Adds all four listed expenses and obtains a total of $1,250.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Subtracts the expenses from the $1,680 income.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.3 (1 mark)
States the remaining income as $430.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Uses $1,680 as the denominator and calculates the remaining-income percentage.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Rounds the percentage to 25.6% and calculates monthly savings as $301.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Multiplies by eight months and states a total saving of $2,408.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 2
(a) 1.5 years; $808.50 interest.
The time is 18 / 12 = 1.5 years. Then I = Prt = 9800 x 0.055 x 1.5 = $808.50.
(b) Repayment total $10,608.50; monthly payment $353.62.
The repayment total is $9,800 + $808.50 = $10,608.50. Dividing by 30 gives $353.616..., which rounds to $353.62.
Detailed marking criteria
Part (a) (2 marks)
Part a.1 (1 mark)
Converts 18 months to 1.5 years.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Substitutes the principal, rate and time into I = Prt and obtains $808.50.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Adds the principal and interest to obtain a repayment total of $10,608.50.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Divides the repayment total by all 30 payments.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Rounds to the nearest cent and states a monthly payment of $353.62.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 3
(a) C = 65 + 18h.
The fixed fee is 65 and each hour adds 18.
(b) $191; 10 hours.
For seven hours, C = 65 + 18(7) = $191. For the second booking, solve 245 = 65 + 18h to obtain h = 10.
Detailed marking criteria
Part (a) (1 mark)
Part a.1 (1 mark)
Writes C = 65 + 18h with the fixed fee as the constant and hourly charge as the coefficient.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Substitutes h = 7 into the cost rule and obtains $191.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Forms 245 = 65 + 18h and isolates 18h = 180.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Solves and states a duration of 10 hours.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 4
(a) 23 minutes.
The fifth of nine ordered values is 23.
(b) Range 19 minutes; mean 24.4 minutes.
The range is 35 - 16 = 19. The total is 220, and 220 / 9 = 24.44..., or 24.4.
(c) Mean 28.8 minutes; median 23.5 minutes.
The new total is 288, so the mean is 28.8. The middle values are 23 and 24, giving median 23.5.
Detailed marking criteria
Part (a) (1 mark)
Part a.1 (1 mark)
Identifies the fifth ordered value and states a median of 23 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (2 marks)
Part b.1 (1 mark)
Subtracts the minimum from the maximum and obtains a range of 19 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Totals the nine values, divides by nine and rounds the mean to 24.4 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (c) (2 marks)
Part c.1 (1 mark)
Includes the tenth value and calculates the new mean as 28.8 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part c.2 (1 mark)
Averages the fifth and sixth ordered values to obtain a median of 23.5 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 5
(a) 108 people; increase 42 people.
The plotted week-4 value is 108. The increase from week 1 to week 6 is 126 - 84 = 42.
(b) 50%; moving mean 98 people, which is 8 more than week 3.
The percentage increase is 42 / 84 x 100 = 50%. The moving mean is (96 + 90 + 108) / 3 = 98, which is 8 more than the week-3 value of 90.
Detailed marking criteria
Part (a) (2 marks)
Part a.1 (1 mark)
Reads the week-4 value as 108 people from the graph.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Subtracts the week-1 value from the week-6 value and obtains an increase of 42 people.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (4 marks)
Part b.1 (1 mark)
Uses the week-1 attendance of 84 as the percentage-change denominator.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Calculates the increase as 50%.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Averages the week 2, 3 and 4 values and obtains a moving mean of 98 people.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.4 (1 mark)
Compares 98 with week 3 and states that it is 8 people higher.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 6
(a) 4 / 15.
There are 4 green tokens among 15 tokens.
(b) Both blue 1 / 7; one yellow and one green 4 / 21.
Both blue has probability 6 / 15 x 5 / 14 = 1 / 7. One yellow and one green has probability 5 / 15 x 4 / 14 + 4 / 15 x 5 / 14 = 4 / 21.
Detailed marking criteria
Part (a) (1 mark)
Part a.1 (1 mark)
States the probability as 4 / 15 from four green tokens out of 15.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Updates the denominator after the first selection and calculates the both-blue probability as 1 / 7.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Calculates the yellow-then-green and green-then-yellow probabilities.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Adds both valid orders and simplifies the result to 4 / 21.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 7
(a) Total 80 m²; planted area 68 m².
The garden area is 10 x 8 = 80 m² and the pond area is 3 x 4 = 12 m², leaving 68 m².
(b) 71.4 m².
The ordered area is 68 x 1.05 = 71.4 m².
(c) $1,570.80; the budget is sufficient with $29.20 remaining.
The cost is 71.4 x $22 = $1,570.80. Since $1,600 - $1,570.80 = $29.20, the budget is sufficient.
Detailed marking criteria
Part (a) (2 marks)
Part a.1 (1 mark)
Calculates the full garden area as 80 m² and the pond area as 12 m².
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Subtracts the pond area and states a planted area of 68 m².
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (2 marks)
Part b.1 (1 mark)
Applies the 5% extra-soil multiplier to the 68 m² planted area.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
States the ordered area as 71.4 m².
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (c) (3 marks)
Part c.1 (1 mark)
Multiplies the ordered area by $22 per square metre and obtains $1,570.80.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part c.2 (1 mark)
Compares the soil cost with the $1,600 budget and concludes that it is sufficient.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part c.3 (1 mark)
Calculates and states $29.20 remaining.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 8
(a) Base area 2.54 m²; capacity 6104 L; about 4270 L of water.
The base area is 3.14 x 0.9² = 2.5434 m². The volume is 2.5434 x 2.4 = 6.10416 m³ = 6104.16 L. Seventy per cent is 4272.912 L, which rounds to 4270 L.
Detailed marking criteria
Part (a) (5 marks)
Part a.1 (1 mark)
Uses A = π r² with r = 0.9 m and obtains 2.5434 m².
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Rounds the base area to 2.54 m² as requested.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.3 (1 mark)
Multiplies by the height and converts 6.10416 m³ to 6104.16 L.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.4 (1 mark)
Applies 70% to the unrounded capacity and obtains 4272.912 L.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.5 (1 mark)
Rounds to the nearest 10 litres and states about 4270 L.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 9
(a) 1 : 12.
Divide 0.6 : 7.2 by 0.6.
(b) Ramp length 7.22 m; edging cost $404.60.
The length is √(7.2² + 0.6²) = √52.2 = 7.2249... m. The edging cost is 2 x √52.2 x $28 = $404.596..., or $404.60.
Detailed marking criteria
Part (a) (1 mark)
Part a.1 (1 mark)
Divides both terms by 0.6 and states the ratio 1 : 12.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Uses Pythagoras' theorem with 7.2 m and 0.6 m as perpendicular sides.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Calculates the ramp length and rounds it to 7.22 m.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Uses the unrounded length for two sides and states the edging cost as $404.60.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 10
(a) Subtotal $154.00; GST $15.40.
Four tyres cost 4 x $38.50 = $154.00. GST is 10% of this amount, or $15.40.
(b) First invoice $169.40; second supplier is cheaper by $4.40.
The first invoice is $154.00 + $15.40 = $169.40. Comparing this with $165.00 shows that the second supplier is cheaper by $4.40.
Detailed marking criteria
Part (a) (2 marks)
Part a.1 (1 mark)
Calculates the four-tyre subtotal as 4 x $38.50 = $154.00.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Calculates 10% GST on the subtotal as $15.40.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Adds subtotal and GST to obtain the first supplier's invoice total of $169.40.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Compares both GST-inclusive totals and identifies the second supplier as cheaper.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Subtracts the totals and states a saving of $4.40.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 11
(a) 215 minutes; 12:55 pm.
The stage times total 45 + 110 + 35 + 25 = 215 minutes, or 3 hours 35 minutes. Adding this to 9:20 am gives 12:55 pm.
(b) 1:15 pm; the deadline is missed by 15 minutes.
The break changes the finishing time from 12:55 pm to 1:15 pm. This is 15 minutes after the deadline.
Detailed marking criteria
Part (a) (2 marks)
Part a.1 (1 mark)
Converts the assembly time and totals all four stages as 215 minutes.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part a.2 (1 mark)
Adds 3 hours 35 minutes to the start time and states 12:55 pm.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Adds the 20-minute break to obtain a 1:15 pm finishing time.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Compares 1:15 pm with the 1:00 pm deadline and concludes that it is missed.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
States that the job finishes 15 minutes late.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Section B Question 12
(a) 12 parts.
Add 7 + 3 + 2 = 12.
(b) A: 280; B: 120; C: 80. Centre A has 56 boxes, or 20%, remaining.
Each ratio part is 480 / 12 = 40 boxes, giving allocations 280, 120 and 80. Centre A distributes 224 boxes, leaving 56. The remaining percentage is 56 / 280 x 100 = 20%.
Detailed marking criteria
Part (a) (1 mark)
Part a.1 (1 mark)
Adds the ratio terms and states 12 total parts.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part (b) (3 marks)
Part b.1 (1 mark)
Uses 40 boxes per ratio part and states allocations A = 280, B = 120 and C = 80.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.2 (1 mark)
Subtracts seven days of 32 boxes from Centre A's allocation and obtains 56 boxes remaining.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Part b.3 (1 mark)
Uses Centre A's allocation as the denominator and states 20% remaining.
Acceptable alternatives: Accept an equivalent mathematically correct method. Apply valid consequential marking when an earlier arithmetic error is carried consistently and the assessed method remains correct.
Do not credit by itself: Do not award this mark for an unsupported final answer when the criterion requires working, units, rounding, comparison or interpretation.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Discrete mathematics: Financial and consumer mathematics — fixed and usage charges | Section A Q1 | 1 | ___ | Multiply consumption by the unit rate, then add the fixed supply charge to obtain the total bill. |
| Discrete mathematics: Financial and consumer mathematics — percentage discount | Section A Q2 | 1 | ___ | Rework this exact question or part and verify each step: The discount is 0.15 x 220 = $33, so the price is 220 - 33 = $187. |
| Discrete mathematics: Financial and consumer mathematics — compound-interest growth | Section A Q3 | 1 | ___ | Rework this exact question or part and verify each step: The balance is 2500(1.03)² = $2,652.25, so the interest is $152.25. |
| Algebra, number and structure — formula substitution and interpretation | Section A Q4 | 1 | ___ | Rework this exact question or part and verify each step: Substitute C = 25: F = 1.8(25) + 32 = 45 + 32 = 77. |
| Algebra, number and structure — linear-equation solving | Section A Q5 | 1 | ___ | Rework this exact question or part and verify each step: Solve 48 + 6h = 72 + 3h. Then 3h = 24, so h = 8. |
| Data analysis, probability and statistics — summary statistics | Section A Q6 | 1 | ___ | Rework this exact question or part and verify each step: The total is 100 and 100 / 5 = 20. |
| Data analysis, probability and statistics — summary statistics | Section A Q7 | 1 | ___ | Rework this exact question or part and verify each step: The cumulative frequency is 8 by rating 2 and 17 by rating 3, so the 10th and 11th values are both 3. |
| Data analysis, probability and statistics — probability and relative frequency | Section A Q8 | 1 | ___ | Rework this exact question or part and verify each step: There are 15 counters and 10 are not black, so the probability is 10 / 15 = 2 / 3. |
| Data analysis, probability and statistics — largest decrease between consecutive graph values | Section A Q9 | 1 | ___ | Read each consecutive pair of graph values: 42 to 38 decreases by 4 kWh, 38 to 35 by 3 kWh, 35 to 29 by 6 kWh, and 29 to 31 increases, so the largest decrease is 6 kWh. |
| Data analysis, probability and statistics — sampling, association and evidence | Section A Q10 | 1 | ___ | Rework this exact question or part and verify each step: A convenience sample taken at one time may not represent members with different attendance patterns. |
| Space and measurement — L-shaped courtyard area | Section A Q11 | 1 | ___ | Calculate the complete rectangle area as 9 × 7 = 63 m², subtract the removed 3 × 2 = 6 m² corner, and state the courtyard area as 57 m². |
| Space and measurement — volume, capacity and unit conversion | Section A Q12 | 1 | ___ | Rework this exact question or part and verify each step: The volume is 2.4 x 1.5 x 0.8 = 2.88 cubic metres, which is 2880 litres. |
| Space and measurement — scale drawing conversion | Section A Q13 | 1 | ___ | Rework this exact question or part and verify each step: 7.2 x 50 000 = 360 000 cm = 3.6 km. |
| Space and measurement — similarity and scale factors | Section A Q14 | 1 | ___ | Rework this exact question or part and verify each step: The scale factor is 5 / 0.8 = 6.25, so the height is 1.2 x 6.25 = 7.5 m. |
| Space and measurement — circle circumference | Section A Q15 | 1 | ___ | Use the supplied value of π with circumference = π x diameter, retaining the unrounded value until the requested precision. |
| Discrete mathematics: Financial and consumer mathematics — GST and invoice totals | Section A Q16 | 1 | ___ | Rework this exact question or part and verify each step: GST is 0.10 x 360 = $36. The invoice total is $360 + $36 = $396. |
| Space and measurement — elapsed time from 1:25 pm to 3:10 pm | Section A Q17 | 1 | ___ | Count from 1:25 pm to 2:25 pm as 1 hour, then from 2:25 pm to 3:10 pm as 45 minutes, giving a total duration of 1 hour 45 minutes. |
| Algebra, number and structure — rates and unit rates | Section A Q18 | 1 | ___ | Rework this exact question or part and verify each step: There are 5 x 6 = 30 worker-hours, so the rate is 180 / 30 = 6 kits per worker-hour. |
| Algebra, number and structure — rates and unit rates | Section A Q19 | 1 | ___ | Rework this exact question or part and verify each step: Average speed = 135 / 2.25 = 60 km / h. |
| Algebra, number and structure — arithmetic sequences | Section A Q20 | 1 | ___ | Rework this exact question or part and verify each step: The common difference is 5, so the next term is 22 + 5 = 27. |
| Discrete mathematics: Financial and consumer mathematics — expense total and remaining income | Section B Q1(a) | 3 | ___ | Add all four listed expenses, then subtract their total from the $1,680 starting amount to find the amount of income remaining. |
| Discrete mathematics: Financial and consumer mathematics — remaining-income percentage and savings projection | Section B Q1(b) | 3 | ___ | Divide the $430 remainder by the original $1,680 income and convert to a percentage, then take 70% of the remainder and multiply that monthly saving by eight. |
| Discrete mathematics: Financial and consumer mathematics — simple interest and time conversion | Section B Q2(a) | 2 | ___ | Rework this exact question or part and verify each step: The time is 18 / 12 = 1.5 years. Then I = Prt = 9800 x 0.055 x 1.5 = $808.50. |
| Discrete mathematics: Financial and consumer mathematics — equal repayment total and monthly payment | Section B Q2(b) | 3 | ___ | Add the $9,800 principal and $808.50 interest to obtain a repayment total of $10,608.50, then divide by 30 and round each monthly payment to $353.62. |
| Algebra, number and structure — linear-model construction | Section B Q3(a) | 1 | ___ | Rework this exact question or part and verify each step: The fixed fee is 65 and each hour adds 18. |
| Algebra, number and structure — linear-model substitution and solving | Section B Q3(b) | 3 | ___ | Substitute the stated duration to find one cost, then form and solve the inverse cost equation for the second booking duration. |
| Data analysis, probability and statistics — median from an odd ordered list | Section B Q4(a) | 1 | ___ | Identify the fifth value in the ordered list of nine delivery times and state the median as 23 minutes. |
| Data analysis, probability and statistics — range and mean from delivery times | Section B Q4(b) | 2 | ___ | Subtract 16 from 35 to obtain the range of 19 minutes, then divide the total of 220 by 9 and state the mean as 24.4 minutes to one decimal place. |
| Data analysis, probability and statistics — updated mean and median after adding a value | Section B Q4(c) | 2 | ___ | Add the new 68-minute delivery to obtain a total of 288, divide by 10 to obtain a new mean of 28.8 minutes, and average the fifth and sixth ordered values to obtain a new median of 23.5 minutes. |
| Data analysis, probability and statistics — attendance graph values and increase | Section B Q5(a) | 2 | ___ | Read the Week 4 attendance as 108, then subtract the Week 1 attendance of 84 from the Week 6 attendance of 126 to obtain an increase of 42. |
| Data analysis, probability and statistics — graph percentage change and moving mean | Section B Q5(b) | 4 | ___ | Read the exact plotted values, use week 1 as the percentage denominator, average weeks 2 to 4, then compare that mean with week 3. |
| Data analysis, probability and statistics — probability and relative frequency | Section B Q6(a) | 1 | ___ | Rework this exact question or part and verify each step: There are 4 green tokens among 15 tokens. |
| Data analysis, probability and statistics — probability without replacement | Section B Q6(b) | 3 | ___ | Rework this exact question or part and verify each step: Both blue has probability 6 / 15 x 5 / 14 = 1 / 7. One yellow and one green has probability 5 / 15 x 4 / 14 + 4 / 15 x 5 / 14 = 4 / 21. |
| Space and measurement — planted area after subtracting a pond | Section B Q7(a) | 2 | ___ | Calculate 10 × 8 = 80 m², subtract the 3 × 4 = 12 m² pond, and state 68 m² to be planted. |
| Space and measurement — soil area with a percentage allowance | Section B Q7(b) | 2 | ___ | Multiply 68 m² by 1.05 to obtain 71.4 m² of soil. |
| Space and measurement — soil cost and budget remainder | Section B Q7(c) | 3 | ___ | Multiply 71.4 by $22 to obtain $1,570.80, compare it with the $1,600 budget and state that $29.20 remains. |
| Space and measurement — volume, capacity and unit conversion | Section B Q8(a) | 5 | ___ | Rework this exact question or part and verify each step: The base area is 3.14 x 0.9² = 2.5434 m². The volume is 2.5434 x 2.4 = 6.10416 m³ = 6104.16 L. Seventy per cent is 4272.912 L, which rounds to 4270 L. |
| Space and measurement — rise-to-run ratio | Section B Q9(a) | 1 | ___ | Form the ratio 0.6 : 7.2 and divide both terms by 0.6 to obtain 1 : 12. |
| Space and measurement — Pythagoras and distance | Section B Q9(b) | 3 | ___ | Rework this exact question or part and verify each step: The length is √(7.2² + 0.6²) = √52.2 = 7.2249... m. The edging cost is 2 x √52.2 x $28 = $404.596..., or $404.60. |
| Discrete mathematics: Financial and consumer mathematics — GST and invoice totals | Section B Q10(a) | 2 | ___ | Rework this exact question or part and verify each step: Four tyres cost 4 x $38.50 = $154.00. GST is 10% of this amount, or $15.40. |
| Discrete mathematics: Financial and consumer mathematics — supplier-price comparison | Section B Q10(b) | 3 | ___ | Bring both quotes to comparable GST-inclusive totals, identify the lower total, and subtract to calculate the saving. |
| Space and measurement — elapsed-time total and finishing time | Section B Q11(a) | 2 | ___ | Convert 1 hour 50 minutes to 110 minutes, add all four stage durations to obtain 215 minutes, and add 3 hours 35 minutes to 9:20 am to obtain a finishing time of 12:55 pm. |
| Space and measurement — elapsed time | Section B Q11(b) | 3 | ___ | Convert all intervals to one time unit, add the durations, then convert back to a clock time and compare explicitly with the deadline. |
| Algebra, number and structure — ratio allocation | Section B Q12(a) | 1 | ___ | Rework this exact question or part and verify each step: Add 7 + 3 + 2 = 12. |
| Algebra, number and structure — ratio allocation and percentage remaining | Section B Q12(b) | 3 | ___ | Allocate boxes from the ratio, subtract seven days of distribution from Centre A, then use Centre A's allocation as the percentage denominator. |