IGCSE MATH PAST PAPER REVIEW
FEB/MARCH 2026 PAPERS
📐 IGCSE Mathematics 0580/42
February/March 2026
📌 Mark Scheme
Official mark scheme based on Cambridge IGCSE conventions (M/A/B marks).
B1 for correct figures.
✔ x = 106° , y = 164°
M1: ∠ACD = 180 − 90 − 74 = 16°
A1: y = 180 − 16 = 164
A1: x = 180 − 74 = 106
✔ 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
✔ 13/100 or 0.13 or 13%
M1: 1 − (27/100 + 60/100) A1: 13/100
✔ p = ±13
M1: p² = 225 − 56 = 169 A1: ±13
✔ t = (r² − p²)/8
M1: 8t = r² − p² A1 for final expression.
✔ Regular hexagon
B1: a regular hexagonal prism has 6 planes of symmetry.
✔ 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
✔ 16 cm
M1: cube root of volume ratio (1.35/0.4)
M1: k = 1.5 A1: 24 ÷ 1.5 = 16
✔ {4, 10, 14}
B1: two correct values B1: all three correct.
✔ f⁻¹(x) = (10 − x)/2
M1: swap x and y & solve A1: correct inverse.
✔ 8x² − 80x + 203
M1: gf(x) = 2(10−2x)² + 3 M1: expand (10−2x)² = 100−40x+4x²
A1: simplify → 8x² − 80x + 203
✔ a = 80 , b = (0.25)^(1/30)
B1: a = 80 M1: 20 = 80·b³⁰ M1: b³⁰ = ¼ A1: b = (¼)^(1/30)
✔ –0.75
M1: dy/dx = 3x² − 5x M1: substitute x = 1.5 A1: –0.75
M1: OE = OA+AB+BC+CD A1: p + 2q
M1: ½ × 150 × 600 × sin57° A1: 37700–37800
M1: sin∠ACD /150 = sin34°/290 M1: ∠ACD≈16.2° A1: ∠CAD = 180−34−16.2 = 85.8°
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.
📘 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.
Total marks: 100 | Prepared according to official assessment structure.