2026 released paper · teacher edition · 23 October 2026
Paper 2: worked solutions & mark scheme
Checked solutions for all nine questions, with a proposed M/B/A allocation for every subpart.
Rechecked 24 October 2026: all 43 numbered subparts, covering 45 answer checks. Each computed result was compared with the answer key. Q7(d) is corrected to 103.1°; the diagram and acute inverse-sine value are corrected too. No other answer discrepancy was found. The M/B/A allocations are unchanged.
90marks fully allocated
31M: method marks
31B: independent marks
28A: accuracy marks
Independent teacher scheme, not the official Cambridge/SEAB mark scheme. Printed subpart totals are taken from the paper. The distribution of M, B and A marks, follow-through guidance and drawing tolerances are proposed for teaching and review.
Code
How to use it
M1
One method mark for a valid, relevant method. A later arithmetic slip does not automatically remove an earned method mark.
B1
One independent mark for a correct fact, result, statement, plot or justified reason; no preceding method mark is required.
A1
One accuracy mark for the indicated correct result or completion of a valid method. Dependencies are within that subpart.
FT
Follow through where noted, if a prior error leaves a meaningful problem of comparable difficulty. Do not award impossible lengths/probabilities or contradictory claims.
Each displayed mark is worth one. Award equivalent valid methods unless the question specifies a method (for example, factorisation in Q7(b)). Exact answers are acceptable where appropriate; otherwise use 3 significant figures and 1 decimal place for angles unless specified. Do not penalise the same numerical slip repeatedly without considering follow-through. Mark explanatory or “show that” parts from the reasoning, not the final line alone.
Quick answer key
Part
Answer
Marks
1(a)(i)
2.4
1
1(a)(ii)
Two branches; no value at x = 0
3
1(a)(iii)
(1, 6)
1
1(b)
(−4, 0), (2, 0), (0, 8); opens downwards
2
2(a)
a = 3, b = 196
1
2(b)
92
1
2(c)(i)
148
1
2(c)(ii)
Overestimate
1
2(d)(i)
6n − 8
2
2(d)(ii)
6n − 8 = 3(2n − 3) + 1
1
2(d)(iii)(a)
17 408
2
2(d)(iii)(b)
17 410
2
3(a)
x < 2.5
2
3(b)
4 + 48x⁻²
2
3(c)
x = −0.19 or 1.79
5
3(d)(i)
t = 4y + 3
1
3(d)(ii)
y = (7 − 4x²)/(4x² − 8)
4
4(a)(i)
105 m³
3
4(a)(ii)
31.5 minutes
3
4(b)(i)
d = 22.56
3
4(b)(ii)(a)
0.7744 : 1
1
4(b)(ii)(b)
0.88 : 1
1
Part
Answer
Marks
5(a)
Accurate PQRS construction
2
5(b)(i)
△ABC ≅ △AED (AAS)
3
5(b)(ii)
∠ACD = (x + y)/2
2
6(a)(i)
276 children
3
6(a)(ii)
295/23 = 12.8 years (3 s.f.)
3
6(b)(i)
T = (368.40, 693.55)ᵀ
2
6(b)(ii)
Total profit from cake sales over both days
1
6(c)(i)
1/1024
1
6(c)(ii)
195/256
1
7(a)
16x² − 10x − 21 = 0
2
7(b)
x = 1.5 or −0.875
3
7(c)
Reject x = −7/8
1
7(d)
∠ACB = 103.1°
5
8(a)(i)
PQ = (3, 5)ᵀ
1
8(a)(ii)
3√34/7 = 2.50
2
8(b)(i)
AB = −2a + b
1
8(b)(ii)
MN = (5/9)MC; M, N, C are collinear
3
8(b)(iii)
MN : NC = 5 : 4
1
9(a)
$32 000
1
9(b)
Example estimate: 10 900 people
2
9(c)
Claim is correct for 10 900 people; interest ≈ $162 000 (3 s.f.)
7
No matching question. Clear the filters to see all nine.
Q1 · Two function graphs
7 marks
Source: printed pages 4–5. M0, B7, A0
1(a)(i) · Substitute into a rational function 1 marks
2.4
Worked solution and mark allocation
Mark
Working / award criterion
B1
At x = −3, y = 4/(−3) − 2/(−3)² + 4 = 22/9 = 2.444…; to 1 decimal place, 2.4.
1(a)(ii) · Plot a rational function with two disconnected branches 3 marks
Two branches; no value at x = 0
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
B1
Accurately plot the four negative-x points: (−3, 2.4), (−2, 1.5), (−1, −2), (−0.5, −12).
B1
Accurately plot the five positive-x points: (0.25, −12), (0.5, 4), (1, 6), (2, 5.5), (3, 5.1).
B1
Join each branch smoothly; the right branch rises to a maximum near (1, 6) and then falls. Both branches descend out of the grid as x approaches 0. Do not join across x = 0.
Marking note: Proposed plot credit: judge placement against the printed grid (about half a small square). The formula is undefined at x = 0. The illustration uses the exact function and exact points; students plot the table values rounded as requested.
1(a)(iii) · Read a turning point 1 marks
(1, 6)
Worked solution and mark allocation
Mark
Working / award criterion
B1
The maximum point on the positive branch is (1, 6).
1(b) · Sketch a factorised quadratic and label intercepts 2 marks
(−4, 0), (2, 0), (0, 8); opens downwards
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
B1
From y = −(x − 2)(x + 4), the x-intercepts are (−4, 0) and (2, 0), correctly labelled.
B1
Draw a downward-opening parabola through the labelled y-intercept (0, 8). Its maximum is at (−1, 9), useful for checking the shape but not separately required.
Q2 · Numbers, estimation and sequences
11 marks
Source: printed pages 6–7. M3, B5, A3
2(a) · Integer powers and square numbers 1 marks
a = 3, b = 196
Worked solution and mark allocation
Mark
Working / award criterion
B1
3³ = 27 and 223 − 27 = 196 = 14², so a = 3, b = 196. Both values are required.
2(b) · Rounding bounds with an integer constraint 1 marks
92
Worked solution and mark allocation
Mark
Working / award criterion
B1
4.5 ≤ ∛n < 5.5, hence n ≥ 4.5³ = 91.125. Since n is an integer, its smallest possible value is 92.
2(c)(i) · Estimate using 1 significant figure 1 marks
148
Worked solution and mark allocation
Mark
Working / award criterion
B1
Round each number first: (4³ + √100)/0.5 = (64 + 10)/0.5 = 148. Show this working.
2(c)(ii) · Explain the direction of an estimate 1 marks
Overestimate
Worked solution and mark allocation
Mark
Working / award criterion
B1
The positive numerator is increased (3.99→4 and 95.2→100), while the positive denominator is decreased (0.517→0.5). Both changes make the quotient larger, so this is an overestimate.
2(d)(i) · Find the nth term of an arithmetic sequence 2 marks
6n − 8
Worked solution and mark allocation
Mark
Working / award criterion
M1
The common difference is 28 − 22 = 6. Use Sₙ = 22 + (n − 5) × 6.
A1
Simplify to Sₙ = 6n − 8.
2(d)(ii) · Prove a divisibility property 1 marks
6n − 8 = 3(2n − 3) + 1
Worked solution and mark allocation
Mark
Working / award criterion
B1
Sₙ = 3(2n − 3) + 1. Since 2n − 3 is an integer, each term is one greater than a multiple of 3.
2(d)(iii)(a) · Use a positive common ratio 2 marks
17 408
Worked solution and mark allocation
Mark
Working / award criterion
M1
17r² = 272 gives r² = 16. The ratio is positive, so r = 4.
A1
T₆ = 17 × 4⁵ = 17 408.
2(d)(iii)(b) · Find the first sequence term above a threshold 2 marks
17 410
Worked solution and mark allocation
Mark
Working / award criterion
M1
6n − 8 > 17 408 gives n > 2902⅔, so the first possible integer position is n = 2903.
A1
S₂₉₀₃ = 6 × 2903 − 8 = 17 410. The previous term is 17 404.
Marking note: The question asks for the term, not its position. Follow through a consistent positive sixth-term value from the preceding part where appropriate.
Q3 · Algebra, indices and equations
14 marks
Source: printed pages 8–9. M6, B1, A7
3(a) · Solve a linear inequality 2 marks
x < 2.5
Worked solution and mark allocation
Mark
Working / award criterion
M1
Multiply by 2: 8x + 2x − 15 < 10, so 10x < 25.
A1
x < 2.5.
3(b) · Expand and use index laws 2 marks
4 + 48x⁻²
Worked solution and mark allocation
Mark
Working / award criterion
M1
Distribute 12x⁻³: 12x⁻³ × ⅓x³ + 12x⁻³ × 4x.
A1
This is 4 + 48x⁻² (x ≠ 0), in the required a + bxⁿ form.
3(c) · Clear algebraic fractions and solve a genuine quadratic 5 marks
Use the quadratic formula: x = [32 ± √(32² + 4 × 20 × 7)]/40 = (8 ± √99)/10.
A1
The negative root is −0.194987… = −0.19 to 2 decimal places.
A1
The positive root is 1.794987… = 1.79 to 2 decimal places.
Marking note: Both roots are allowed: neither is 2 nor −¼. Keep full precision until final rounding. Substituting the rounded values back will not give exactly 5.
3(d)(i) · Rearrange a linear formula 1 marks
t = 4y + 3
Worked solution and mark allocation
Mark
Working / award criterion
B1
4y = t − 3, hence t = 4y + 3.
3(d)(ii) · Eliminate a parameter from a square-root formula 4 marks
y = (7 − 4x²)/(4x² − 8)
Worked solution and mark allocation
Mark
Working / award criterion
M1
Square the formula for x and substitute t = 4y + 3: x² = [2(4y + 3) + 1]/[(4y + 3) + 1].
Cross-multiply and collect y terms: 4x²y + 4x² = 8y + 7, so (4x² − 8)y = 7 − 4x². Divide by the coefficient of y.
A1
Give the single fraction y = (7 − 4x²)/(4x² − 8), or (4x² − 7)/(8 − 4x²).
Marking note: The accuracy marks reward two distinct results: the correctly simplified substitution and the final single fraction. Do not award two accuracy marks just for rewriting the same final fraction. An equivalent elimination method is valid. The original relation requires x ≥ 0 and x ≠ √2; these restrictions are explanatory, not additional marks.
Q4 · A prism, filling rate and circular scaling
11 marks
Source: printed pages 10–11. M5, B3, A3
4(a)(i) · Composite cross-section and prism volume 3 marks
105 m³
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
M1
The lower trapezium has height 4 − 1.5 = 2.5 m. Its parallel sides are 5 m and 3 m; area = ½(5 + 3) × 2.5 = 10 m².
M1
Add the rectangle: cross-sectional area = 5 × 1.5 + 10 = 17.5 m². Multiply by the prism length 6 m.
A1
Volume = 17.5 × 6 = 105 m³.
4(a)(ii) · Convert a filling rate into total time 3 marks
31.5 minutes
Worked solution and mark allocation
Mark
Working / award criterion
B1
1 minute 30 seconds = 1.5 minutes.
M1
The tank requires 105/5 = 21 lots of 5 m³. Time = 21 × 1.5 minutes.
A1
Time = 31.5 minutes.
Marking note: Follow through the volume from 4(a)(i) for a correct rate method; 31 minutes 30 seconds is equivalent but the requested unit is minutes.
4(b)(i) · Successive percentage changes and area scale factor 3 marks
d = 22.56
Worked solution and mark allocation
Mark
Working / award criterion
M1
Circumference P / circumference R = 1.10 × 0.80 = 0.88. Radii have the same ratio.
M1
Areas scale as radius squared: area P / area R = 0.88² = 0.7744.
A1
d = (1 − 0.7744) × 100 = 22.56.
Marking note: 22.56 is exact, so it need not be rounded to 3 significant figures. The circles are similar; do not add or subtract the two percentage changes.
4(b)(ii)(a) · Compare cylinder volumes with equal heights 1 marks
0.7744 : 1
Worked solution and mark allocation
Mark
Working / award criterion
B1
V = πr²h. The common height cancels, so Vₚ : Vᵣ = 0.7744 : 1.
4(b)(ii)(b) · Compare curved areas with equal heights 1 marks
0.88 : 1
Worked solution and mark allocation
Mark
Working / award criterion
B1
Curved area = 2πrh. The common height cancels, so the ratio is 0.88 : 1.
Marking note: These two cylinders are not similar: their heights are equal while their radii differ. Do not cube 0.88 for volume or square it for curved surface area.
Q5 · Construction and geometric proof
7 marks
Source: printed pages 12–13. M1, B5, A1
5(a) · Construct a quadrilateral from two sides and two angles 2 marks
Accurate PQRS construction
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
B1
At P, draw a ray making 62° above PQ. Mark S on it with PS = 6.4 cm.
B1
At Q, draw a ray making 80° above QP (100° anticlockwise from the positive horizontal). Mark R with QR = 5.2 cm, then join R to S.
Marking note: Use a protractor and ruler. Suggested teacher tolerance: about ±1° and ±1 mm, judged on the original paper; this is not an official tolerance. The on-screen diagram is proportional, not a centimetre measurement template.
5(b)(i) · Prove congruence using AAS 3 marks
△ABC ≅ △AED (AAS)
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
B1
∠ABC = ∠AED (given), and AB = AE (given).
B1
∠ACB = ∠CAD because AD ∥ BC (alternate angles), and ∠CAD = ∠ADE because AC ∥ ED (alternate angles). Thus ∠ACB = ∠ADE.
B1
Two corresponding angles and a corresponding non-included side are equal, so △ABC ≅ △AED by AAS.
Marking note: Correspondence is A↔A, B↔E, C↔D. An ASA proof using the third angles is also valid. Parallel sides alone do not establish equal lengths.
5(b)(ii) · Use congruence and an isosceles triangle to prove an angle 2 marks
∠ACD = (x + y)/2
Worked solution and mark allocation
Mark
Working / award criterion
M1
In △AED, ∠ADE = 180° − x − y (angle sum). Since AC ∥ ED, ∠CAD = ∠ADE = 180° − x − y (alternate angles).
A1
AC = AD (corresponding sides of the congruent triangles), so △ACD is isosceles. Its equal base angles are [180° − (180° − x − y)]/2 = (x + y)/2.
Marking note: Give the parallel-line, triangle-angle-sum and equal-base-angle reasons. Do not assume AC = AD from the sketch.
Q6 · Festival statistics, matrices and probability
11 marks
Source: printed pages 14–15. M5, B3, A3
6(a)(i) · Interpret a pie chart using a difference 3 marks
276 children
Worked solution and mark allocation
Mark
Working / award criterion
M1
The men–women angle difference is 135° − 110° = 25°, representing 60 people, so 1° represents 60/25 = 2.4 people.
M1
The children’s angle is 360° − 135° − 110° = 115°. Multiply by 2.4.
A1
Number of children = 115 × 2.4 = 276.
6(a)(ii) · Recover a subgroup mean from a weighted mean 3 marks
295/23 = 12.8 years (3 s.f.)
Worked solution and mark allocation
Mark
Working / award criterion
M1
Women = 110 × 2.4 = 264; men = 135 × 2.4 = 324; total = 864. Total age of everyone = 27 × 864 = 23 328.
M1
Children’s total age = 23 328 − 32 × 264 − 35 × 324 = 3540. Divide by 276.
A1
Mean age = 3540/276 = 295/23 = 12.826086… = 12.8 years.
Marking note: Equivalent angle-weighted method: [27 × 360 − 32 × 110 − 35 × 135]/115. Do not take an unweighted average of the three group means.
6(b)(i) · Subtract a cost matrix and multiply by quantities 2 marks
T = (368.40, 693.55)ᵀ
Worked solution and mark allocation
Mark
Working / award criterion
M1
D − C is the column matrix (3.95, 6.10, 4.75)ᵀ. Multiply each row of P by this column.
A1
T = (368.40, 693.55)ᵀ: 21(3.95)+32(6.10)+19(4.75)=368.40; 18(3.95)+67(6.10)+45(4.75)=693.55.
Marking note: Write a 2 × 1 column matrix, not a row matrix. Equivalent PD − PC working earns the method mark.
6(b)(ii) · Interpret a matrix result 1 marks
Total profit from cake sales over both days
Worked solution and mark allocation
Mark
Working / award criterion
B1
The sum is Robert’s total profit, in dollars, from selling all three cake types over the two days, using the costs given. It is $1061.95; the calculation is not required.
6(c)(i) · Repeated independent events 1 marks
1/1024
Worked solution and mark allocation
Mark
Working / award criterion
B1
P(all five blue) = (¼)⁵ = 1/1024.
6(c)(ii) · Use the complement of two extreme outcomes 1 marks
195/256
Worked solution and mark allocation
Mark
Working / award criterion
B1
P(at least one of each colour) = 1 − (¼)⁵ − (¾)⁵ = 1 − 1/1024 − 243/1024 = 195/256.
Marking note: “At least one of each” excludes both all-blue and all-red; subtracting just one of them is insufficient.
Q7 · A quadratic inside a trigonometry problem
11 marks
Source: printed pages 16–17. M4, B2, A5
7(a) · Form an equation using ½ab sin C 2 marks
16x² − 10x − 21 = 0
Worked solution and mark allocation
Mark
Working / award criterion
M1
Use 3 = ½(2x + 1)(8x − 9)sin 30°, so (2x + 1)(8x − 9) = 12.
Split the middle term: 16x² + 14x − 24x − 21 = 0, or obtain valid factors directly.
A1
Factorise as (2x − 3)(8x + 7) = 0.
A1
Thus x = 3/2 or −7/8 (1.5 or −0.875).
Marking note: Factorisation is explicitly requested. In this proposed allocation, both accuracy marks depend on a valid factorisation method. Formula-only working is not credited under this proposed factorisation allocation. The official treatment of partial credit is not available.
7(c) · Reject a root in a length context 1 marks
Reject x = −7/8
Worked solution and mark allocation
Mark
Working / award criterion
B1
Reject x = −7/8: it makes AB = −¾ cm and AC = −16 cm, which cannot be side lengths. x = 3/2 gives positive lengths.
7(d) · Use cosine and sine rules, choosing the obtuse angle 5 marks
∠ACB = 103.1°
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
B1
With x = 3/2, AB = 4 cm and AC = 3 cm.
M1
BC² = 4² + 3² − 2(4)(3)cos 30° = 25 − 12√3.
A1
BC = √(25 − 12√3) = 2.05314157… cm.
M1
Sine rule: sin C / 4 = sin 30° / BC, hence sin C = 2/BC = 0.9741169477…. Since C is obtuse, C = 180° − 76.9356865705…°.
A1
∠ACB = 103.064313…° = 103.1°.
Marking note: Alternative: cosine rule for C gives cos C = [3² + BC² − 4²]/(6BC), directly returning the obtuse angle. Use unrounded BC. The acute inverse-sine result 76.9° is not the requested angle.
8(b)(i) · Express a displacement in a and b 1 marks
AB = −2a + b
Worked solution and mark allocation
Mark
Working / award criterion
B1
AB = OB − OA = (a + 2b) − (3a + b) = −2a + b.
8(b)(ii) · Prove collinearity using a common point 3 marks
MN = (5/9)MC; M, N, C are collinear
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
M1
OC = OA + ⁷⁄₅AB = ⅕a + ¹²⁄₅b. OM = ³⁄₂a + ½b and ON = ⁷⁄₉a + ¹⁴⁄₉b.
M1
MN = ON − OM = −¹³⁄₁₈a + ¹⁹⁄₁₈b. MC = OC − OM = −¹³⁄₁₀a + ¹⁹⁄₁₀b.
A1
Therefore MN = ⁵⁄₉MC. The vectors are parallel and share M, so M, N and C lie on one straight line.
Marking note: An equivalent proof with NC = ⅘MN is valid. State the shared-point argument; parallel vectors on unrelated lines are not enough.
8(b)(iii) · Read a section ratio from scalar multiples 1 marks
MN : NC = 5 : 4
Worked solution and mark allocation
Mark
Working / award criterion
B1
MN is 5/9 of MC, leaving NC as 4/9 of MC. Thus MN : NC = 5 : 4.
Q9 · Population planning and compound interest
10 marks
Source: printed pages 20–21. M4, B2, A4
9(a) · Calculate the council’s share of a unit cost 1 marks
$32 000
Worked solution and mark allocation
Mark
Working / award criterion
B1
⅕ × $160 000 = $32 000 for each new housing unit.
9(b) · Extrapolate a population graph and interpret thousands 2 marks
Example estimate: 10 900 people
Worked solution and mark allocation
Geometry-defined illustration; calculations use the stated data. Swipe the figure sideways on a small screen if needed.
Mark
Working / award criterion
M1
Use a reasonable continuation of the plotted trend. One defensible model continues the recent 2021–2026 rise from 10.75 to 10.80 thousand: +0.05 thousand every five years.
A1
Two more five-year periods give 10.90 thousand = 10 900 people in 2036.
Marking note: This is an estimate, not a unique exact answer. Credit a sensible, explained graphical extrapolation and consistent subsequent working. The suggested value is a worked example, not an official required reading.
9(c) · Build a cost model, account for existing capacity and test a loan claim 7 marks
Claim is correct for 10 900 people; interest ≈ $162 000 (3 s.f.)
Worked solution and mark allocation
Mark
Working / award criterion
M1
Using 10 900 from (b), extra people = 10 900 − 10 800 = 100. Housing units = 100/2 = 50 (round up to whole units if necessary).
A1
Housing cost paid by the planning office = 50 × $32 000 = $1 600 000.
Water capacity = 2 × 7500 = 15 000, so no extra plant. Electricity capacity = 4 × 2700 = 10 800, so one extra substation costs $300 000.
M1
Total loan P = 1 600 000 + 740 740.741… + 300 000 = $2 640 740.741…. Interest only = P[(1.015)⁴ − 1].
A1
Interest = $162 045.228… = $162 000 (3 significant figures), which is below $1 000 000. The claim is correct, under this population estimate.
Marking note: Follow through a reasonable estimate from (b), recalculating housing, transport and required services. Charge only additional plants/substations, not existing ones. Subtract the principal to obtain interest; do not compare the final loan balance with $1 million. Retain full precision during the calculation. The paper requests 3 significant figures for a non-exact answer unless otherwise stated; the interactive HTML calculator also shows $162 045.23 to the nearest cent for checking.
Q9: check a different population estimate
A sensible estimate must be justified from the graph. This calculator checks the cost model; it does not declare every entered estimate acceptable.
Assumptions: new housing covers the increase from 10 800; two people per unit; whole housing units/plants/substations are rounded up; existing service capacity is retained. Interest uses 1.5% compounded annually for four years.