IGCSE Maths (0580) · Extended · Paper 2 Non-calculator

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

Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 2 Extended Specimen Paper for examination from 2025 (0580/02/SP/25) — 26 questions, 100 marks, non-calculator. Attempt each question yourself first, then click “Show solution” to check your working step by step.

11 markExtendedSpecimen 2025 · Paper 2 (0580/02)

Work out (0.01)².

Show solution

0.01 × 0.01 = 0.0001

21 markExtendedSpecimen 2025 · Paper 2 (0580/02)

Write 57.3997 correct to 4 significant figures.

Show solution

The first 4 significant figures are 5, 7, 3, 9. The next digit (9) rounds the 4th figure up, causing it to carry: 57.3997 → 57.40

31 markExtendedSpecimen 2025 · Paper 2 (0580/02)

Aimee changes 250 euros into dollars. The exchange rate is 1 euro = $1.10. Calculate the number of dollars Aimee receives.

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250 × 1.10 = $275

42 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The diagram shows two triangles, ABD and ADC. BDC is a straight line, AB = AC, angle ABD = 61° and angle ADC = 81°.

Question 4 — triangles ABD and ADC with BDC a straight line

Work out angle DAC.

Show solution

Since AB = AC, triangle ABC is isosceles, so the base angles are equal: angle ACB = angle ABD = 61° (this is the same as angle ACD, since D lies on BC).

In triangle ADC, the angles sum to 180°: angle DAC = 180 − 81 − 61 = 38°

51 markExtendedSpecimen 2025 · Paper 2 (0580/02)

Convert 0.17 m² into cm².

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1 m² = 10 000 cm², so 0.17 × 10 000 = 1700 cm²

62 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The mass of a solid metal cuboid is 4 kg. The volume of the cuboid is 600 cm³. Calculate the density of the metal, giving your answer in g/cm³. [Density = mass ÷ volume]

Show solution

Convert mass to grams: 4 kg = 4000 g.

Density = 4000 ÷ 600 = 6⅔ g/cm³ (≈ 6.67 g/cm³)

74 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

u = (3⁄−2)   v = (−12⁄5)

  1. (a) Find u − 2v.
  2. (b) Find |v|.
Show solution
  1. (a) 2v = (−24, 10). u − 2v = (3−(−24), −2−10) = (27, −12)
  2. (b) |v| = √((−12)² + 5²) = √(144+25) = √169 = 13
83 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The diagram shows a semicircle with diameter 9 cm.

Question 8 — semicircle with diameter 9cm

Calculate the total perimeter of this semicircle. Give your answer in exact form.

Show solution

The perimeter is made up of the curved arc (half the circumference) plus the straight diameter.

Radius = 9 ÷ 2 = 4.5 cm. Half circumference = ½ × π × 9 = 4.5π

Perimeter = 4.5π + 9 = (9 + 4.5π) cm

93 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

In a sequence, T1 = 17, T2 = 12, T3 = 7, T4 = 2. Find:

  1. (a) T5
  2. (b) Tn
Show solution
  1. (a) Each term decreases by 5: T5 = 2 − 5 = −3
  2. (b) Common difference = −5. Tn = 17 + (n−1)(−5) = 17 − 5n + 5 = 22 − 5n
103 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

Work out 2⅔ + 3½. Give your answer as a mixed number in its simplest form.

Show solution

Add whole numbers: 2+3=5. Add fractions using a common denominator of 6: ⅔=&frac46;, ½=&frac36;. &frac46;+&frac36;=&frac76;=1⅙.

5 + 1⅙ = 6⅙

112 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

Find the value of 64⅔.

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64⅔ means (∛64)². Since ∛64 = 4 (as 4³=64): 4² = 16

124 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

Work out, giving your answer in standard form:

  1. (a) (7.1 × 10−15) × (2 × 10³)
  2. (b) (5.2 × 10⁵) + (5.2 × 10⁶)
Show solution
  1. (a) Multiply the numbers and add the powers: 7.1 × 2 = 14.2, and 10−15 × 10³ = 10−12. This gives 14.2 × 10−12, which is not in standard form (14.2 ≥ 10), so rewrite: 14.2 = 1.42 × 10.
    1.42 × 10−11
  2. (b) Rewrite so both powers match: 5.2 × 10⁵ = 52 × 10⁶. Then 52 × 10⁶ + 5.2 × 10⁶ = 57.2 × 10⁶ = 5.72 × 10⁷
132 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

Find the number of sides of a regular polygon with interior angle 162°.

Show solution

Exterior angle = 180 − 162 = 18°. The exterior angles of any polygon sum to 360°, so the number of sides = 360 ÷ 18 = 20 sides

144 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The range, mode, median and mean of five positive integers are all equal to 10. Find one possible set of these five integers.

Show solution

Try 5, 10, 10, 10, 15 (sorted):

  • Range = 15 − 5 = 10 ✓
  • Mode = 10 (appears 3 times, more than any other value) ✓
  • Median (middle value) = 10 ✓
  • Mean = (5+10+10+10+15) ÷ 5 = 50 ÷ 5 = 10 ✓

5, 10, 10, 10, 15 (other valid sets also exist)

153 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The diagram shows triangle T with vertices (1,1), (4,1), (4,2), and triangle A with vertices (2,3), (2,6), (3,3).

Question 15 — grid showing triangle T and triangle A

Describe fully the single transformation that maps triangle T onto triangle A.

Show solution

Triangle T is “wide and short” (base 3 units, height 1 unit) while triangle A is “narrow and tall” (base 1 unit, height 3 units) — the dimensions have swapped exactly (not scaled), which is the signature of a 90° rotation rather than an enlargement.

Testing the vertex mapping (4,1)→(2,3), (1,1)→(2,6), (4,2)→(3,3) against a 90° clockwise rotation and solving for the invariant centre point gives centre (4,3).

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

1610 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

A student measures the height, h cm, of each of 400 plants.

  1. (a) The cumulative frequency diagram shows the results.
    Question 16a — cumulative frequency diagram for plant heights
    Use the diagram to find an estimate for:
    1. the median
    2. the interquartile range
    3. the 80th percentile
    4. the number of plants with a height greater than 60 cm.
  2. (b) The heights are also shown in the frequency table.
    Height (h cm)0 < h ≤ 2020 < h ≤ 3030 < h ≤ 4040 < h ≤ 80
    Frequency1208012476
    Question 16b — blank histogram grid with the 0-20 bar already drawn
    Complete the histogram to show this information.
Show solution
Question 16b solution — completed histogram with all four bars drawn

From the frequency table, the running cumulative frequencies are: h=20 → 120, h=30 → 200, h=40 → 324, h=80 → 400.

  1. (a)(i) Median is at cumulative frequency 400÷2 = 200, which occurs exactly at h = 30 cm
  2. (a)(ii) Lower quartile at CF=100 (within the 0–20 interval): h ≈ 20 × (100÷120) = 16.7 cm.
    Upper quartile at CF=300 (within the 30–40 interval): h ≈ 30 + 10 × (100÷124) = 38.1 cm.
    IQR = 38.1 − 16.7 ≈ 21.4 cm
  3. (a)(iii) 80th percentile at CF = 0.8×400 = 320 (within the 30–40 interval): h ≈ 30 + 10 × (120÷124) = 39.7 cm
  4. (a)(iv) At h=60 (within the 40–80 interval), CF ≈ 324 + 76 × (20÷40) = 362. Plants above 60cm = 400 − 362 = 38 plants
  5. (b) Frequency density = frequency ÷ class width:
    • 0–20: 120÷20 = 6 (already drawn)
    • 20–30: 80÷10 = 8
    • 30–40: 124÷10 = 12.4
    • 40–80: 76÷40 = 1.9
Please double-check against the original diagram: parts (a)(i)–(iv) are estimated using straight-line interpolation between the known cumulative frequency table values, since the printed curve may not be perfectly linear within each interval. Please read the exact curve values off the original graph (allowing the usual ±1–2 tolerance) before sharing with students.
175 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The diagram shows a cyclic quadrilateral ABCD. BD and AC intersect at X.

Question 17 — cyclic quadrilateral ABCD with diagonals meeting at X
  1. (a) Angle BAD = 74° and angle BCA = 34°. Find:
    1. angle BDA
    2. angle BCD
    3. angle ABD
  2. (b) Triangle ADX is similar to triangle BCX. BC = 4.5 cm, AD = 9 cm and CX = 3.3 cm. Work out XD.
Show solution
  1. (a)(i) Angles BDA and BCA are both subtended by the same chord AB from the same side of the circle, so they are equal (angles in the same segment): angle BDA = 34°
  2. (a)(ii) Opposite angles of a cyclic quadrilateral sum to 180°: angle BCD = 180 − 74 = 106°
  3. (a)(iii) In triangle ABD, the angles sum to 180°: angle ABD = 180 − 74 − 34 = 72°
  4. (b) Since triangle ADX ~ triangle BCX, corresponding sides are in the same ratio: AD corresponds to BC, and DX corresponds to CX.
    Scale factor = AD ÷ BC = 9 ÷ 4.5 = 2.
    XD = 2 × CX = 2 × 3.3 = 6.6 cm
189 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

f(x) = 3 − 2x   g(x) = 2x + 3   h(x) = 2x

  1. (a) (i) Find f(−3).
    (ii) Find gf(−3).
  2. (b) Find f−1(x).
  3. (c) Find x when gg(x) = 7.
  4. (d) Find x when h−1(x) = 5.
Show solution
  1. (a)(i) f(−3) = 3 − 2(−3) = 3 + 6 = 9
  2. (a)(ii) gf(−3) = g(9) = 2(9) + 3 = 21
  3. (b) Let y = 3 − 2x. Rearrange: 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. Set equal to 7: 4x+9=7 → 4x=−2 → x = −0.5
  5. (d) h−1(x) = 5 means h(5) = x: x = 2⁵ = 32
194 marksExtendedSpecimen 2025 · Paper 2 (0580/02)
  1. (a) Simplify: √32 + √98
  2. (b) Rationalise the denominator: 1 ÷ (√2 + 1)
Show solution
  1. (a) √32 = √(16×2) = 4√2. √98 = √(49×2) = 7√2. Sum = 4√2 + 7√2 = 11√2
  2. (b) Multiply top and bottom by (√2−1): 1×(√2−1)⁄(√2+1)(√2−1) = √2−1⁄2−1 = √2 − 1
203 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

y ∝ 1⁄√x. When y = 8, x = 4. Find y when x = 49.

Show solution

y = k⁄√x. Substituting y=8, x=4: 8 = k⁄√4 = k⁄2, so k=16.

When x=49: y = 16⁄√49 = 16⁄7 = 2⅔ (16/7)

219 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

In this question, all measurements are in centimetres. The height of the triangle is h and the height of the rectangle is (h+2). The length of the base of the triangle is x and the length of the rectangle is (x+1). The area of the triangle is 11 cm² and the area of the rectangle is 39 cm².

Question 21 — triangle with base x and height h, and rectangle with sides (x+1) and (h+2)
  1. (a) Write down an expression, in terms of x, for the height of the rectangle.
  2. (b) Show that 2x² − 15x + 22 = 0.
  3. (c) By factorising and solving 2x² − 15x + 22 = 0, find the two possible heights of the triangle.
Show solution
  1. (a) Triangle area = ½xh = 11, so h = 22⁄x. Height of rectangle = h+2 = (22/x) + 2
  2. (b) Rectangle area: (x+1)(h+2) = 39. Substituting h+2 = (22+2x)⁄x:
    (x+1) × (22+2x)⁄x = 39
    Multiply both sides by x: (x+1)(22+2x) = 39x
    Expand: 22x + 2x² + 22 + 2x = 39x
    2x² + 24x + 22 = 39x
    2x² − 15x + 22 = 0 ✓
  3. (c) Looking for 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
    Using h = 22⁄x: when x=5.5, h=22÷5.5=4. When x=2, h=22÷2=11.
    h = 4 or h = 11
224 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The diagram shows a right-angled triangle with hypotenuse x cm, base 8 cm, and an angle of 30° between the hypotenuse and the base.

Question 22 — right triangle with hypotenuse x, base 8cm, angle 30 degrees

Find the exact value of x.

Show solution

The 8 cm side is adjacent to the 30° angle, and x is the hypotenuse.

cos(30°) = adjacent ÷ hypotenuse = 8 ÷ x, so x = 8 ÷ cos(30°) = 8 ÷ (√3⁄2) = 16⁄√3

Rationalising: x = 16√3⁄3

x = 16√3 ÷ 3 cm

233 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

Write as a single fraction in its simplest form: 3⁄x−4 − 4⁄x+3

Show solution

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)

(25 − x) / [(x−4)(x+3)]

243 marksExtendedSpecimen 2025 · Paper 2 (0580/02)
  1. (a) Write x² − 4x + 7 in the form (x − a)² + b.
  2. (b) Write down the coordinates of the turning point of the graph of y = x² − 4x + 7.
Show solution
  1. (a) (x−2)² = x² − 4x + 4. To get + 7, add 3 more: (x − 2)² + 3
  2. (b) For y=(x−a)²+b, the turning point is at (a, b): (2, 3)
253 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

The two heart shapes shown are mathematically similar. The area of the larger shape is 36 cm² and the area of the smaller shape is 25 cm². The height of the larger shape is 9 cm and the height of the smaller shape is x cm.

Question 25 — two similar heart shapes, larger height 9cm and smaller height x cm

Find the value of x.

Show solution

For similar shapes, the ratio of lengths equals the square root of the ratio of areas.

Linear scale factor (larger ÷ smaller) = √(36⁄25) = 6⁄5

9 ÷ x = 6⁄5, so x = 9 × 5⁄6 = 45⁄6 = 7.5 cm

2611 marksExtendedSpecimen 2025 · Paper 2 (0580/02)

f(x) = x(x+2)(x−3)

Question 26 — blank axes for x from -3 to 4, y from -18 to 24
  1. (a) On the diagram, sketch the graph of y = f(x) for −3 ≤ x ≤ 4. Show the values of the intersections with the axes.
  2. (b) Expand and simplify: x(x+2)(x−3)
  3. (c) A is the point (1, −6). The tangent to the graph of y = f(x) at A meets the y-axis at B. Find the coordinates of B.
Show solution
Question 26 solution — sketch of y=x(x+2)(x-3) with x-intercepts and tangent line at A
  1. (a) The curve crosses the x-axis where f(x)=0: at x = −2, 0, 3. Since the x³ coefficient is positive, the cubic rises from bottom-left, dips down between −2 and 0… rises again crossing at 0, dips between 0 and 3, then rises steeply after 3. At the domain ends: f(−3) = (−3)(−1)(−6) = −18 (matching the −18 gridline) and f(4) = 4(6)(1) = 24 (matching the 24 gridline) — confirming the sketch bounds.
    x-intercepts: (−2,0), (0,0), (3,0)
  2. (b) First expand (x+2)(x−3) = x² − x − 6. Then multiply by x: x³ − x² − 6x
  3. (c) Using f(x) = x³−x²−6x, the gradient function is f′(x) = 3x²−2x−6.
    At x=1: gradient = 3−2−6 = −5.
    Tangent line through (1,−6) with gradient −5: y−(−6) = −5(x−1) → y = −5x − 1
    At x=0 (the y-axis): y = −1
    B = (0, −1)

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