IGCSE Maths Past Paper Solutions: 0580/02 Specimen Paper 2025 (Paper 2 Extended Non-calculator)

0.01 × 0.01: 1 × 1 = 1, and there are 2 + 2 = 4 decimal places, so the answer is 0.0001

The first 4 significant figures are 5, 7, 3, 9. The next digit is 9, so round the 4th figure up. 57.39 + 0.01 = 57.40, so the answer is 57.40

250 × 1.10 = 250 + 25 = $275

AB = AC, so triangle ABC is isosceles and its base angles are equal: angle ACD = angle ABD = 61°.

Angles in triangle ADC add to 180°: angle DAC = 180 − 81 − 61 = 38°

1 m = 100 cm, so 1 m² = 100 × 100 = 10 000 cm².

0.17 × 10 000 = 1700 cm²

Convert the mass to grams first: 4 kg = 4000 g.

Density = 4000 ÷ 600 = 40 ÷ 6 = 6⅔ g/cm³ (or 20/3)

1. (a) 2v = −2410, so u − 2v = 3 − (−24)−2 − 10 = (27, −12) as a column vector
2. (b) |v| = √((−12)² + 5²) = √(144 + 25) = √169 = 13

The perimeter is the curved arc (half a circumference) plus the straight diameter.

Half circumference = ½ × π × 9 = 4.5π cm

Perimeter = 4.5π + 9 = (4.5π + 9) cm

1. (a) Each term is 5 less than the one before: T₅ = 2 − 5 = −3
2. (b) The difference is −5, so T_n starts −5n. When n = 1, −5n = −5, but T₁ = 17, so add 22.
   T_n = 22 − 5n
   Check: n = 4 gives 22 − 20 = 2 ✓

Method 1 (improper fractions): 2⅔ = ⁸⁄₃ and 3½ = ⁷⁄₂. With denominator 6: ¹⁶⁄₆ + ²¹⁄₆ = ³⁷⁄₆.

37 ÷ 6 = 6 remainder 1, so 6⅙

Method 2: 2 + 3 = 5, and ⁴⁄₆ + ³⁄₆ = ⁷⁄₆ = 1⅙, so 5 + 1⅙ = 6⅙.

The denominator 3 means cube root; the numerator 2 means square. 64^⅔ = (∛64)² = 4² = 16

1. (a) Multiply the numbers and add the powers: 7.1 × 2 = 14.2 and 10⁻¹⁵ × 10³ = 10⁻¹².
   14.2 × 10⁻¹² is not standard form (14.2 ≥ 10), so write 14.2 = 1.42 × 10¹.
   1.42 × 10⁻¹¹
2. (b) Make the powers match: 5.2 × 10⁷ = 52 × 10⁶.
   52 × 10⁶ + 5.2 × 10⁶ = 57.2 × 10⁶ = 5.72 × 10⁷
   Check: 52 000 000 + 5 200 000 = 57 200 000 ✓

Interior and exterior angles add to 180°: exterior angle = 180 − 162 = 18°.

The exterior angles of any polygon add to 360°: number of sides = 360 ÷ 18 = 20

Build the set step by step. Median 10 means the middle value is 10; mode 10 suggests repeating it: _, 10, 10, 10, _. Mean 10 means the total is 50, so the two outer numbers add to 20. Range 10 means they differ by 10: 5 and 15.

• Range = 15 − 5 = 10 ✓
• Mode = 10 ✓
• Median = 10 ✓
• Mean = 50 ÷ 5 = 10 ✓

5, 10, 10, 10, 15 (other sets are possible, e.g. 6, 8, 10, 10, 16)

T and A are the same size, but the long side has turned from horizontal to vertical, so this is a 90° rotation. Tracing paper (allowed in the exam) confirms the direction: clockwise.

To find the centre, follow one vertex: (4, 1) maps to (2, 3). The centre must be the same distance from both points with a right angle between them — that point is (4, 3).

Rotation, 90° clockwise, centre (4, 3)

Check all three vertices: (4, 1) → (2, 3), (1, 1) → (2, 6), (4, 2) → (3, 3) ✓

1. (a)(i) Median: half of 400 is 200. Read across from 200 to the curve and down: 30 cm
2. (a)(ii) Lower quartile at ¼ × 400 = 100 → h ≈ 17 cm. Upper quartile at ¾ × 400 = 300 → h ≈ 37 cm.
   IQR = 37 − 17 = 20 cm
3. (a)(iii) 80th percentile at 0.8 × 400 = 320 → 39 cm (approximately)
4. (a)(iv) At h = 60 the cumulative frequency is about 363, so 363 plants are 60 cm or shorter.
   Taller than 60 cm: 400 − 363 = 37 plants
5. (b) Frequency density = frequency ÷ class width.

   Class	Width	Frequency	Frequency density
   0 < h ≤ 20	20	120	6 (given)
   20 < h ≤ 30	10	80	8
   30 < h ≤ 40	10	124	12.4
   40 < h ≤ 80	40	76	1.9

1. (a)(i) Angles BDA and BCA both stand on chord AB, so they are equal (angles in the same segment): angle BDA = 34°
2. (a)(ii) Opposite angles of a cyclic quadrilateral add to 180°: angle BCD = 180 − 74 = 106°
3. (a)(iii) Angles in triangle ABD add to 180°: angle ABD = 180 − 74 − 34 = 72°
4. (b) The order of letters gives the matching sides: A↔B, D↔C, X↔X. So AD matches BC and DX matches CX.
   Scale factor = AD ÷ BC = 9 ÷ 4.5 = 2.
   XD = 2 × 3.3 = 6.6 cm

1. (a)(i) f(−3) = 3 − 2(−3) = 3 + 6 = 9
2. (a)(ii) gf(−3) means apply f first, then g: g(9) = 2(9) + 3 = 21
3. (b) Write y = 3 − 2x and rearrange for x: 2x = 3 − y, so x = ^{3 − y}⁄₂.
   f⁻¹(x) = (3 − x) / 2
4. (c) gg(x) = g(2x + 3) = 2(2x + 3) + 3 = 4x + 9.
   4x + 9 = 7 → 4x = −2 → x = −½
5. (d) h⁻¹(x) = 5 means x = h(5) = 2⁵ = 32

1. (a) Find square factors: √32 = √(16 × 2) = 4√2 and √98 = √(49 × 2) = 7√2.
   4√2 + 7√2 = 11√2
2. (b) Multiply top and bottom by √2 − 1:
   ^{√2 − 1}⁄_{(√2 + 1)(√2 − 1)} = ^{√2 − 1}⁄_{2 − 1} = √2 − 1

Write y = ^k⁄_√x. Substitute y = 8, x = 4: 8 = ^k⁄₂, so k = 16.

When x = 49: y = ¹⁶⁄_√49 = ¹⁶⁄₇ = 16/7 = 2²⁄₇

1. (a) Rectangle area = length × height, so 39 = (x + 1) × height.
   Height of rectangle = 39 / (x + 1)
2. (b) From the triangle: ½xh = 11, so h = ²²⁄_x. The rectangle height is also h + 2, so:
   ³⁹⁄_{x + 1} = ²²⁄_x + 2
   Multiply every term by x(x + 1): 39x = 22(x + 1) + 2x(x + 1)
   39x = 22x + 22 + 2x² + 2x
   0 = 2x² + 24x − 39x + 22
   2x² − 15x + 22 = 0 ✓
3. (c) Find two numbers that multiply to 2 × 22 = 44 and add to −15: −4 and −11.
   2x² − 4x − 11x + 22 = 0
   2x(x − 2) − 11(x − 2) = 0
   (2x − 11)(x − 2) = 0 → x = 5.5 or x = 2
   Then h = ²²⁄_x: x = 5.5 gives h = 4; x = 2 gives h = 11.
   h = 4 or h = 11

The 8 cm side is adjacent to the 30° angle and x is the hypotenuse, so use cosine: cos 30° = ⁸⁄_x.

cos 30° = ^√3⁄₂, so ^√3⁄₂ = ⁸⁄_x → x = ¹⁶⁄_√3

Rationalise: x = ^16√3⁄₃ → x = 16√3 / 3

Use the common denominator (x − 4)(x + 3):

^{3(x + 3) − 4(x − 4)}⁄_{(x − 4)(x + 3)} = ^{3x + 9 − 4x + 16}⁄_{(x − 4)(x + 3)}

(25 − x) / ((x − 4)(x + 3))

1. (a) Halve the coefficient of x: (x − 2)² = x² − 4x + 4. We need +7, so add 3: (x − 2)² + 3
2. (b) (x − 2)² is smallest (zero) when x = 2, giving y = 3. Turning point: (2, 3)

For similar shapes, area scale factor = (length scale factor)².

Area ratio small : large = 25 : 36, so length ratio = √25 : √36 = 5 : 6.

x = 9 × ⁵⁄₆ = ⁴⁵⁄₆ = 7.5

1. (a) f(x) = 0 when x = 0, x = −2 or x = 3, and the curve passes through the origin, so the y-intercept is 0. The x³ term is positive, so the curve rises from bottom-left to top-right, with a maximum between −2 and 0 and a minimum between 0 and 3. The ends of the axes match the curve: f(−3) = −18 and f(4) = 24.
   Crosses x-axis at −2, 0, 3; y-axis at 0
2. (b) (x + 2)(x − 3) = x² − 3x + 2x − 6 = x² − x − 6. Multiply by x: x³ − x² − 6x
3. (c) Differentiate: ^dy⁄_dx = 3x² − 2x − 6. At x = 1: gradient = 3 − 2 − 6 = −5.
   Tangent through (1, −6) with gradient −5: y + 6 = −5(x − 1) → y = −5x − 1.
   It meets the y-axis where x = 0: y = −1.
   B = (0, −1)