Skill Align TASC Mathematics Methods - Foundation Level 3 Free Online Pack 0 Section B - 2026 Edition
Current for 2026 | Free-online Pack 0 v2.0 booklet 2 of 2 | 100 marks | approximately 100 minutes | calculators permitted.
- Paper
- Section B - Calculator Permitted
- Reading
- Included in the shared 15-minute preparation time
- Writing
- Approximately 100 minutes
- Assessment
- 100 marks
Section B conditions: TASC-approved calculators are permitted only when the supervisor allows calculator use. A candidate who begins this booklet before the end of the 80-minute Section A period must not use a calculator until then. Supply the current TASC Mathematics Methods - Foundation Level 3 information sheet with this booklet. Internet access and external communication are not permitted.
Part 1 - Algebra - Criterion 4 (20 marks)
Answer all four questions in this part. The question marks vary and total 20 marks. One-mark items do not require working. For items worth 2 or more marks, show relevant working. Extended items require the governing relation, relevant intermediate results and a check against the stated conditions.
Question 1
2 marksQuestion 2
6 marksQuestion 3
3 marksQuestion 4
9 marksPart 2 - Linear, Quadratic and Cubic Functions - Criterion 5 (20 marks)
Answer all four questions in this part. The question marks vary and total 20 marks. One-mark items do not require working. For items worth 2 or more marks, show relevant working. Extended items require the governing relation, relevant intermediate results and a check against the stated conditions.
Question 5
3 marksQuestion 6
6 marksQuestion 7
4 marksQuestion 8
7 marksPart 3 - Logarithmic, Exponential and Trigonometric Functions - Criterion 6 (20 marks)
Answer all four questions in this part. The question marks vary and total 20 marks. One-mark items do not require working. For items worth 2 or more marks, show relevant working. Extended items require the governing relation, relevant intermediate results and a check against the stated conditions.
Question 9
7 marksQuestion 10
3 marksQuestion 11
4 marksQuestion 12
6 marksPart 4 - Differential Calculus - Criterion 7 (20 marks)
Answer all four questions in this part. The question marks vary and total 20 marks. One-mark items do not require working. For items worth 2 or more marks, show relevant working. Extended items require the governing relation, relevant intermediate results and a check against the stated conditions.
Question 13
3 marksQuestion 14
4 marksQuestion 15
5 marksQuestion 16
8 marksPart 5 - Statistics and Probability - Criterion 8 (20 marks)
Answer all four questions in this part. The question marks vary and total 20 marks. One-mark items do not require working. For items worth 2 or more marks, show relevant working. Extended items require the governing relation, relevant intermediate results and a check against the stated conditions.
Question 17
6 marksQuestion 18
3 marksQuestion 19
7 marksQuestion 20
4 marksWorked Solutions And Marking Guide
Question 1
Indicative answer: x
(27x^6)^(1/3) = 3x^2, so (3x^2)/(3x) = x.
Mark allocation
- TASC-style partial-mark policy: one-mark items may receive 0.5 for valid working; items worth two or more marks receive consequential partial credit for correct later work after an earlier arithmetic slip.
- A correct answer without required working is capped at 1.5 / 2 for a two-mark item and at half marks for an item worth three or more marks.
- Accept algebraically equivalent methods and exact forms. Unless a question states otherwise, accept a correctly rounded decimal within + / -0.01 of the listed value.
Detailed marking criteria
Integrated response (2 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 2
Indicative answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81
Using coefficients 1, 4, 6, 4, 1 gives 16x^4 - 96x^3 + 216x^2 - 216x + 81.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 3
Indicative answer: x = 4 and y = 3
Adding the equations gives 3x = 12, so x = 4. Substitution gives y = 3.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 4
(a) 2a+b=2 and 3a-b=13.
From P(2)=0, 8+4a+2b-12=0, so 2a+b=2. From P(-3)=0, -27+9a-3b-12=0, so 3a-b=13.
(b) a=3 and b=-4.
Adding the equations gives 5a=15, so a=3. Substitution into 2a+b=2 gives b=-4.
(c) P(x)=x³+3x²-4x-12=(x-2)(x+3)(x+2).
Substitute a=3 and b=-4. The two supplied factors have product x²+x-6; the remaining monic factor is x+2.
(d) The zeros are 2, -3 and -2; for example, P(-2)=0.
Set each linear factor equal to zero. Direct substitution gives -8+12+8-12=0 for x=-2.
Detailed marking criteria
Part 4(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 4(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 4(c) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 4(d) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 5
Indicative answer: y = -(2/3)x + 19/3
The given line has gradient 3/2, so the perpendicular gradient is -2/3. Substituting (2,5) gives c=19/3.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 6
Indicative answer: Zeros 1 and 7; axis x=4; vertex (4,-9); range f(x) >= -9
The zeros are 1 and 7, so the axis is their midpoint x=4. Then f(4)=3(-3)=-9. The parabola opens upward.
Detailed marking criteria
Integrated response (6 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 7
Indicative answer: Zero -2 (double), zero 3 (single); y-intercept -12; negative for x< -2 and -2<x<3, positive for x>3
The factors give a double zero at -2 and a single zero at 3. f(0)=4(-3)=-12. The double root does not change sign; the single root does.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 8
Indicative answer: Initial 0 m; maximum 18 m at 3 s; lands at 6 s; 0 <= h <= 18
h(0)=0. Vertex form gives maximum 18 at t=3. Solving h=0 gives (t-3)^2=9, so t=0 or 6; the later time is landing.
Detailed marking criteria
Integrated response (7 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 9
(a) V(5)=$1605.87.
1200(1.06)⁵=1605.8707ldots, which rounds to $1605.87.
(b) The doubling time is 11.9 years.
Solve 2400=1200(1.06)^t, or 2=(1.06)^t. Thus t=log2 / log1.06=11.8956ldots, which is 11.9 years to one decimal place.
(c) The first whole year is year 9.
V(8)=$1912.62 and V(9)=$2027.37, so year 9 is the first whole year above $2000.
Detailed marking criteria
Part 9(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 9(b) (3 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 9(c) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 10
Indicative answer: x=17, with x>1
The logarithm requires x-1>0. Converting to exponential form gives x-1=2^4=16, so x=17.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 11
(a) The y-intercept is (0,3) and the horizontal asymptote is y=0.
Read the intercept and asymptote directly from the graph.
(b) a=3, b=2, so y=3(2^x).
At x=0, a=3. The output doubles from 3 to 6 when x increases from 0 to 1, so b=2.
Detailed marking criteria
Part 11(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 11(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 12
(a) Amplitude 2, period 2π / 3, midline y=-1, and range -3leq yleq1.
The amplitude is |2|=2, the period is 2π / 3, and the vertical translation gives midline y=-1 and range [-3,1].
(b) (0,-1), (π / 6,1), (π / 3,-1), (π / 2,-3) and (2π / 3,-1).
The quarter-period is π / 6. Starting at the midline, the sine cycle moves through maximum, midline, minimum and back to the midline.
(c) A smooth sine cycle through the five listed points on the stated interval.
Plot the five quarter-period points and join them with the standard smooth sine shape.
Detailed marking criteria
Part 12(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 12(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 12(c) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 13
Indicative answer: 2
f(1)=4 and f(5)=12. The average rate is (12-4)/(5-1)=8/4=2.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 14
Indicative answer: f'(x)=12x^2-10x+7; f'(2)=35
Apply the power rule: f'(x)=12x^2-10x+7. Then f'(2)=48-20+7=35.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 15
Indicative answer: y-4=10(x-2), or y=10x-16
At x=2, y=8-4=4. The derivative is 3x^2-2, giving gradient 10. Use point-gradient form.
Detailed marking criteria
Integrated response (5 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 16
Indicative answer: The stationary points are ((-5,104)), a local maximum, and ((1,-4)), a local minimum. The function increases on (-6 <= x < -5) and (1 < x <= 2), decreases on (-5 < x < 1), and has range ([-4,104]) on the restricted domain.
The derivative is (f'(x)=3x^2+12x-15=3(x+5)(x-1)). Its signs are positive, negative and positive across the zeros (-5) and (1), so the first point is a local maximum and the second a local minimum. Evaluating gives (f(-5)=104) and (f(1)=-4). The endpoint values are (f(-6)=94) and (f(2)=6), so the restricted range is ([-4,104]).
Detailed marking criteria
Integrated response (8 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 17
(a) P(E)=66 / 120=0.55 and P(Bcap E)=42 / 120=0.35.
Use the table totals and the bus-and-early cell over the 120 observed students.
(b) P(E|B)=42 / 60=0.70. About 70% of similar bus travellers are estimated to arrive early.
Condition on the 60 observed bus travellers, then describe the result as an estimate based on the survey.
(c) They do not appear independent because P(E|B)=0.70ne0.55=P(E).
For independent events, conditioning on bus travel would not change the early-arrival probability.
Detailed marking criteria
Part 17(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 17(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 17(c) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 18
Indicative answer: 5/14
P(RR)=5/8 times 4/7=20/56=5/14.
Detailed marking criteria
Integrated response (3 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 19
(a) First branches: A 0.60 and B 0.40. From A: good 0.80 and not good 0.20. From B: good 0.65 and not good 0.35.
Use complementary branch probabilities at each split so that each pair sums to one.
(b) P(G)=0.60(0.80)+0.40(0.65)=0.74.
Add the two mutually exclusive good-component paths in the tree.
(c) P(A|G)=0.48 / 0.74=24 / 37approx0.649.
The A-and-good path has probability 0.60(0.80)=0.48; divide it by the total good probability 0.74.
(d) About 1110 good components.
1500(0.74)=1110.
Detailed marking criteria
Part 19(a) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 19(b) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 19(c) (2 marks)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Part 19(d) (1 mark)
Award one mark for each independently observable checkpoint. Apply consequential marking when a correct later method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically or trigonometrically equivalent exact form where applicable.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Question 20
Indicative answer: 120 total; 36 include the captain
Total committees C(10,3)=120. If the captain is included, choose the other two from nine: C(9,2)=36.
Detailed marking criteria
Integrated response (4 marks)
Award one mark for each independently observable checkpoint or grouped outcome. Apply consequential marking when a correct method uses an earlier incorrect value.
Acceptable alternatives: Accept any mathematically equivalent method with sufficient working. Accept an algebraically equivalent exact form or a correct calculator-supported method when sufficient working is shown. Unless the question states another tolerance, accept a correctly rounded decimal within + / -0.01 of the listed answer.
Do not credit by itself: An unsupported answer or a correct value obtained from inconsistent working is not full-credit evidence.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Section B | Part 1 | Criterion 4 | Algebra | Skill: Apply index laws to a fractional power. | Expected evidence: x | Q1 | 2 | ___ | Rework Question 1: Rewrite the cube root as a power before cancelling common factors. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Skill: Use binomial coefficients with signed terms. | Expected evidence: 16x^4 - 96x^3 + 216x^2 - 216x + 81 | Q2 | 6 | ___ | Rework Question 2: Write the five binomial terms before simplifying them. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Skill: Solve a pair of linear equations exactly. | Expected evidence: x = 4 and y = 3 | Q3 | 3 | ___ | Rework Question 3: Eliminate one variable, then verify both values in the original equations. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Question 4(a) | Skill: Form simultaneous equations with the factor theorem | Expected result: \(2a+b=2\) and \(3a-b=13\). | Mark-by-mark evidence: Forms the equation 2a+b=2; Forms the equation 3a-b=13 | Q4(a) | 2 | ___ | Rework Question 4(a): Use the factor theorem to form two simultaneous equations in (a) and (b). Then compare each checkpoint with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Question 4(b) | Skill: Solve a two-variable linear system | Expected result: \(a=3\) and \(b=-4\). | Mark-by-mark evidence: Finds a=3; Finds b=-4 | Q4(b) | 2 | ___ | Rework Question 4(b): Solve the simultaneous equations to find (a) and (b). Then compare each checkpoint with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Question 4(c) | Skill: Complete and factorise a cubic polynomial | Expected result: \(P(x)=x^3+3x^2-4x-12=(x-2)(x+3)(x+2)\). | Mark-by-mark evidence: Writes the completed polynomial; Uses both supplied factors; Obtains the complete factorisation \(x-2\)\(x+3\)\(x+2\) | Q4(c) | 3 | ___ | Rework Question 4(c): Write the completed polynomial and factorise it completely. Then compare each checkpoint with the worked solution. |
| Section B | Part 1 | Criterion 4 | Algebra | Question 4(d) | Skill: State and verify the zeros of a cubic | Expected result: The zeros are \(2\), \(-3\) and \(-2\); for example, \(P(-2)=0\). | Mark-by-mark evidence: States all three zeros; Correctly verifies one zero in the completed polynomial | Q4(d) | 2 | ___ | Rework Question 4(d): State every zero and verify one of them by substitution in the completed polynomial. Then compare each checkpoint with the worked solution. |
| Section B | Part 2 | Criterion 5 | Linear, Quadratic and Cubic Functions | Skill: Use perpendicular gradients and a point. | Expected evidence: y = -(2/3)x + 19/3 | Q5 | 3 | ___ | Rework Question 5: Rearrange the given line first, then use the negative reciprocal gradient. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 2 | Criterion 5 | Linear, Quadratic and Cubic Functions | Skill: Analyse a factored quadratic. | Expected evidence: Zeros 1 and 7; axis x=4; vertex (4,-9); range f(x) >= -9 | Q6 | 6 | ___ | Rework Question 6: Use the midpoint of the roots, then evaluate the function and inspect the leading coefficient. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 2 | Criterion 5 | Linear, Quadratic and Cubic Functions | Skill: Interpret a factored cubic including multiplicity. | Expected evidence: Zero -2 (double), zero 3 (single); y-intercept -12; negative for x< -2 and -2<x<3, positive for x>3 | Q7 | 4 | ___ | Rework Question 7: Use factor signs on one test value in each interval. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 2 | Criterion 5 | Linear, Quadratic and Cubic Functions | Skill: Interpret a quadratic model in vertex form. | Expected evidence: Initial 0 m; maximum 18 m at 3 s; lands at 6 s; 0 <= h <= 18 | Q8 | 7 | ___ | Rework Question 8: Read the vertex, then solve the domain-valid intercept equation. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 9(a) | Skill: Evaluate an exponential financial model | Expected result: \(V(5)=$1605.87\). | Mark-by-mark evidence: Substitutes t=5 in the model; Rounds the value to $1605.87 | Q9(a) | 2 | ___ | Rework Question 9(a): Calculate (V(5)), giving the value to the nearest cent. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 9(b) | Skill: Solve an exponential equation with logarithms | Expected result: The doubling time is 11.9 years. | Mark-by-mark evidence: Forms the doubling equation; Uses logarithms to isolate t; Rounds the time to 11.9 years | Q9(b) | 3 | ___ | Rework Question 9(b): Find the doubling time, giving the answer to one decimal place. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 9(c) | Skill: Apply an exponential model to a threshold | Expected result: The first whole year is year 9. | Mark-by-mark evidence: Calculates or identifies the neighbouring year values; Concludes that year 9 is the first whole year above $2000 | Q9(c) | 2 | ___ | Rework Question 9(c): Find the first whole year for which the investment value exceeds ($2000). Support the answer with neighbouring values. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Skill: Convert between logarithmic and exponential form. | Expected evidence: x=17, with x>1 | Q10 | 3 | ___ | Rework Question 10: State the log argument restriction before solving. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 11(a) | Skill: Interpret key features of an exponential graph | Expected result: The y-intercept is \((0,3)\) and the horizontal asymptote is \(y=0\). | Mark-by-mark evidence: States the y-intercept \(0,3\); States the horizontal asymptote y=0 | Q11(a) | 2 | ___ | Rework Question 11(a): State the y-intercept and horizontal asymptote. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 11(b) | Skill: Determine an exponential rule from a graph | Expected result: \(a=3\), \(b=2\), so \(y=3(2^x)\). | Mark-by-mark evidence: Finds a=3 and b=2; Writes the rule y=3\(2^x\) | Q11(b) | 2 | ___ | Rework Question 11(b): Use the labelled points to determine (a) and (b), and hence write the function rule. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 12(a) | Skill: Identify transformed sine-graph features | Expected result: Amplitude \(2\), period \(2\pi/3\), midline \(y=-1\), and range \(-3\leq y\leq1\). | Mark-by-mark evidence: States the amplitude and period; States the midline and range | Q12(a) | 2 | ___ | Rework Question 12(a): State the amplitude, period, midline and range. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 12(b) | Skill: Calculate exact key points for a sine graph | Expected result: \((0,-1)\), \((\pi/6,1)\), \((\pi/3,-1)\), \((\pi/2,-3)\) and \((2\pi/3,-1)\). | Mark-by-mark evidence: Uses quarter-period spacing of pi/6; States all five correct key points | Q12(b) | 2 | ___ | Rework Question 12(b): Give the five quarter-period key points for one complete cycle on the interval. Then compare each checkpoint with the worked solution. |
| Section B | Part 3 | Criterion 6 | Logarithmic, Exponential and Trigonometric Functions | Question 12(c) | Skill: Sketch a transformed sine graph | Expected result: A smooth sine cycle through the five listed points on the stated interval. | Mark-by-mark evidence: Plots and labels the five key points; Draws a smooth sine cycle on the stated domain | Q12(c) | 2 | ___ | Rework Question 12(c): Sketch one complete cycle on labelled axes, showing the five key points. Then compare each checkpoint with the worked solution. |
| Section B | Part 4 | Criterion 7 | Differential Calculus | Skill: Calculate average rate of change. | Expected evidence: 2 | Q13 | 3 | ___ | Rework Question 13: Evaluate both endpoints before forming the difference quotient. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 4 | Criterion 7 | Differential Calculus | Skill: Differentiate a cubic polynomial. | Expected evidence: f'(x)=12x^2-10x+7; f'(2)=35 | Q14 | 4 | ___ | Rework Question 14: Differentiate term by term, then substitute only after simplifying. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 4 | Criterion 7 | Differential Calculus | Skill: Find a tangent to a cubic. | Expected evidence: y-4=10(x-2), or y=10x-16 | Q15 | 5 | ___ | Rework Question 15: Calculate both the point and derivative gradient at the stated x-value. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 4 | Criterion 7 | Differential Calculus | Skill: analysing a cubic over a restricted domain | Expected evidence: The stationary points are ((-5,104)), a local maximum, and ((1,-4)), a local minimum. The function increases on (-6 <= x < -5) and (1 < x <= 2), decreases on (-5 < x < 1), and has range ([-4,104]) on the restricted domain. | Q16 | 8 | ___ | differentiate and factor the cubic, use derivative signs for intervals and classifications, then compare stationary and endpoint values for the range. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 17(a) | Skill: Calculate probabilities from a two-way frequency table | Expected result: \(P(E)=66/120=0.55\) and \(P(B\cap E)=42/120=0.35\). | Mark-by-mark evidence: Calculates P\(E\)=0.55; Calculates P(B intersection E)=0.35 | Q17(a) | 2 | ___ | Rework Question 17(a): Find (P(E)) and (P(Bcap E)). Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 17(b) | Skill: Calculate and interpret a conditional relative frequency | Expected result: \(P(E|B)=42/60=0.70\). About \(70%\) of similar bus travellers are estimated to arrive early. | Mark-by-mark evidence: Calculates P(E given B)=0.70; Interprets 0.70 as an estimated early-arrival rate for similar bus travellers | Q17(b) | 2 | ___ | Rework Question 17(b): Find (P(E|B)) and interpret this experimental probability in context. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 17(c) | Skill: Assess independence from observed frequencies | Expected result: They do not appear independent because \(P(E|B)=0.70\ne0.55=P(E)\). | Mark-by-mark evidence: Compares the relevant conditional and overall probabilities; Concludes with justification that the events are not independent | Q17(c) | 2 | ___ | Rework Question 17(c): Use the table to decide whether (B) and (E) appear independent. Justify the decision numerically. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Skill: Calculate sequential probability without replacement. | Expected evidence: 5/14 | Q18 | 3 | ___ | Rework Question 18: Update both numerator and denominator after the first draw. Then compare the setup, intermediate result and final presentation with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 19(a) | Skill: Construct a two-stage probability tree | Expected result: First branches: A 0.60 and B 0.40. From A: good 0.80 and not good 0.20. From B: good 0.65 and not good 0.35. | Mark-by-mark evidence: Draws and labels the supplier branches; Draws and labels both good/not-good branch pairs | Q19(a) | 2 | ___ | Rework Question 19(a): Draw a labelled probability tree showing every branch probability. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 19(b) | Skill: Calculate a total probability from a tree | Expected result: \(P(G)=0.60(0.80)+0.40(0.65)=0.74\). | Mark-by-mark evidence: Calculates both good-path probabilities; Adds them to obtain 0.74 | Q19(b) | 2 | ___ | Rework Question 19(b): Find the probability that a randomly selected component is good. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 19(c) | Skill: Calculate a reverse conditional probability | Expected result: \(P(A|G)=0.48/0.74=24/37\approx0.649\). | Mark-by-mark evidence: Forms the conditional probability 0.48/0.74; Rounds the result to 0.649 | Q19(c) | 2 | ___ | Rework Question 19(c): Given that a component is good, find the probability that it came from supplier A. Give the answer to three decimal places. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Question 19(d) | Skill: Use a probability to estimate a count | Expected result: About 1110 good components. | Mark-by-mark evidence: Calculates 1110 good components | Q19(d) | 1 | ___ | Rework Question 19(d): Estimate the number of good components in a batch of 1500 made under the same conditions. Then compare each checkpoint with the worked solution. |
| Section B | Part 5 | Criterion 8 | Statistics and Probability | Skill: Count unordered selections with a required member. | Expected evidence: 120 total; 36 include the captain | Q20 | 4 | ___ | Rework Question 20: Separate the fixed captain from the remaining selections. Then compare the setup, intermediate result and final presentation with the worked solution. |