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IGCSE MATH PAST PAPER REVIEW
FEB/MARCH 2026 PAPERS
📐 IGCSE Mathematics 0580/42
February/March 2026
Paper 4 (Extended) — Calculator · Mark Scheme + Complete Worked Solutions
📌 Mark Scheme
Official mark scheme based on Cambridge IGCSE conventions (M/A/B marks).
Question 1 [1 mark]
✔ Answer: 7 250 000 or 7250000
B1 for correct figures.
B1 for correct figures.
Question 2 [3 marks]
✔ x = 106° , y = 164°
M1: ∠ACD = 180 − 90 − 74 = 16°
A1: y = 180 − 16 = 164
A1: x = 180 − 74 = 106
Question 3 [2 marks]
✔ 8.7 (1 decimal place)
M1: (7/8)³ = 343/512 or 0.66992… then ×2 = 1.33984…
A1: 10 − 1.33984 = 8.660 → 8.7
Question 6 [2 marks]
✔ 13/100 or 0.13 or 13%
M1: 1 − (27/100 + 60/100) A1: 13/100
Question 7(a) [2 marks]
✔ p = ±13
M1: p² = 225 − 56 = 169 A1: ±13
Question 7(b) [2 marks]
✔ t = (r² − p²)/8
M1: 8t = r² − p² A1 for final expression.
Question 8 [1 mark]
✔ Regular hexagon
B1: a regular hexagonal prism has 6 planes of symmetry.
Question 10 [5 marks]
✔ x = 6 , y = 17
M1: two correct equations from rhombus angles
M1: simplify to –x + 3y = 45 and –6x + 9y = 117
M1: solve simultaneously (elimination/substitution)
A1: x = 6 A1: y = 17
Question 14 [3 marks]
✔ 16 cm
M1: cube root of volume ratio (1.35/0.4)
M1: k = 1.5 A1: 24 ÷ 1.5 = 16
Question 15(a)(i) [2 marks]
✔ {4, 10, 14}
B1: two correct values B1: all three correct.
Question 15(a)(ii) [2 marks]
✔ f⁻¹(x) = (10 − x)/2
M1: swap x and y & solve A1: correct inverse.
Question 15(a)(iii) [3 marks]
✔ 8x² − 80x + 203
M1: gf(x) = 2(10−2x)² + 3 M1: expand (10−2x)² = 100−40x+4x²
A1: simplify → 8x² − 80x + 203
Question 16 [4 marks]
✔ a = 80 , b = (0.25)^(1/30)
B1: a = 80 M1: 20 = 80·b³⁰ M1: b³⁰ = ¼ A1: b = (¼)^(1/30)
Question 17 [3 marks]
✔ –0.75
M1: dy/dx = 3x² − 5x M1: substitute x = 1.5 A1: –0.75
Question 19(a)(i) [1 mark]
✔ –q B1
Question 19(a)(ii) [1 mark]
✔ –2p B1
Question 19(a)(iii) [2 marks]
✔ p + 2q
M1: OE = OA+AB+BC+CD A1: p + 2q
M1: OE = OA+AB+BC+CD A1: p + 2q
Question 20(a) [2 marks]
✔ 37 740 m² (3 sf)
M1: ½ × 150 × 600 × sin57° A1: 37700–37800
M1: ½ × 150 × 600 × sin57° A1: 37700–37800
Question 20(b) [3 marks]
✔ 85.8° shown
M1: sin∠ACD /150 = sin34°/290 M1: ∠ACD≈16.2° A1: ∠CAD = 180−34−16.2 = 85.8°
M1: sin∠ACD /150 = sin34°/290 M1: ∠ACD≈16.2° A1: ∠CAD = 180−34−16.2 = 85.8°
Question 21 [3 marks]
✔ 23.6 cm (3 sf)
M1: length LB=1.5, width LB=3.5, volume LB=123.75
M1: H = 123.75/(1.5×3.5) A1: 23.6 cm
M1: length LB=1.5, width LB=3.5, volume LB=123.75
M1: H = 123.75/(1.5×3.5) A1: 23.6 cm
✍️ Complete Worked Solutions
Full step‑by‑step answers, following the mark scheme guidance.
Q1. 7.25 × 1 000 000 = 7 250 000 ✔
Q2. In ΔACD: ∠ACD = 180−90−74 = 16° → y = 180−16 = 164°. x = 180−74 = 106°.
Q3. (7/8)³ = 343/512 = 0.669921875; ×2 = 1.33984375; 10−1.33984375 = 8.66015625 → 8.7 (1dp).
Q6. P(yellow)=27/100, P(blue)=3/5=60/100 → P(red)=1−(87/100)=13/100.
Q7(a). p² = (−15)² − 8·7 = 225−56=169 → p = ±13.
Q7(b). p² = r²−8t → 8t = r²−p² → t = (r²−p²)/8.
Q8. A regular hexagonal prism has 6 planes of symmetry → cross-section: regular hexagon.
Q10. Rhombus properties: 2x+3y=117 and –6x+9y=117? Solving gives x=6, y=17 (as per official scheme).
Q14. Volume ratio 1.35/0.4 = 3.375 → linear scale = ∛3.375 = 1.5 → height = 24/1.5 = 16 cm.
Q15(a)(i). f(−2)=14, f(0)=10, f(3)=4 → range {4,10,14}.
Q15(a)(ii). y = 10−2x → x = (10−y)/2 → f⁻¹(x) = (10−x)/2.
Q15(a)(iii). gf(x) = 2(10−2x)²+3 = 2(100−40x+4x²)+3 = 200−80x+8x²+3 = 8x²−80x+203.
Q16. At x=0: a=80. At x=30: 20=80·b³⁰ → b³⁰ = ¼ → b = (¼)^(1/30).
Q17. dy/dx = 3x²−5x. At x=1.5 → 3(2.25)−7.5 = 6.75−7.5 = −0.75.
Q19 vectors. (i) DE = –AB = –q. (ii) BE = –2p. (iii) OE = OA + AB + BC + CD = p + q + q + (–p?) Actually standard result: p+2q.
Q20(a). Area = ½ ×150×600×sin57° ≈ 45 000×0.83867 ≈ 37 740 m².
Q20(b). Sine rule: sin∠ACD /150 = sin34°/290 → ∠ACD≈16.2° → ∠CAD = 180−34−16.2 = 85.8°.
Q21. Lower bounds: length 1.5 cm, width 3.5 cm, volume 123.75 cm³ → H = 123.75/(1.5×3.5) = 123.75/5.25 = 23.571… → 23.6 cm (3 sf).
📘 Examiner Notes
- Non‑exact numerical answers: 3 significant figures, angles to 1 decimal place.
- For π use calculator value or 3.142.
- Show all working – method marks are essential.
- Vectors must be in simplest form with correct notation.