IGCSE Maths Past Paper Solutions: 0580/04 Specimen Paper 2025 (Paper 4 Extended Calculator)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 4 Extended Specimen Paper for examination from 2025 (0580/04/SP/25) — 24 questions, 100 marks, calculator allowed. Attempt each question yourself first, then click “Show solution” to check your working step by step.
Write down the integer values of x that satisfy the inequality −2 ≤ x < 2.
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The integers from −2 up to (but not including) 2 are:
−2, −1, 0, 1
In triangle PQR, QR = 10 cm and PR = 11 cm. Using a ruler and compasses only, construct triangle PQR. The line PQ has been drawn for you.
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Method: Open the compasses to 10 cm and draw an arc centred at Q (since QR=10 cm, R must lie on this arc). Open the compasses to 11 cm and draw an arc centred at P (since PR=11 cm, R must also lie on this arc). Where the two arcs cross is point R. Join QR and PR with straight lines. Leave all construction arcs visible.
Simplify (x⁸y⁷) ÷ (x−1y³).
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Subtract the powers of matching bases: x-power = 8−(−1)=9; y-power = 7−3=4.
x⁹y⁴
f(x) = 3x − 5. The domain of f(x) is {−3, 0, 2}. Find the range of f(x).
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f(−3) = 3(−3)−5 = −9−5 = −14
f(0) = 3(0)−5 = −5
f(2) = 3(2)−5 = 6−5 = 1
{−14, −5, 1}
Two towns, A and B, are shown on a map. The scale of the map is 1 cm to 3 km.
- (a) Find the actual distance between A and B.
- (b) Measure the bearing of B from A.
- (c) Calculate the bearing of A from B. You must show all your working.
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- (a) Measure the length of line AB on the printed map with a ruler, then multiply by 3 (since 1 cm represents 3 km) to get the actual distance in km.
- (b) Place a protractor at A, with 0° aligned along the North line, and measure clockwise round to the line AB to get the bearing of B from A.
- (c) Since the North lines at A and B are parallel, the bearing of A from B is exactly 180° different from the bearing of B from A (using co-interior/alternate angle facts with the parallel North lines): bearing of A from B = (bearing of B from A) ± 180°, adjusted to stay within 0°–360°.
A solid metal cuboid has a volume of 600 cm³.
- (a) The base of the cuboid is 10 cm by 12 cm. Calculate the height of the cuboid.
- (b) The solid metal cuboid is melted and made into 1120 spheres, each with radius 0.45 cm. Find the volume of metal not used in making these spheres.
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- (a) Volume = base area × height, so height = 600 ÷ (10×12) = 600÷120 = 5 cm
- (b) Volume of one sphere = &frac43;πr³ = &frac43; × π × (0.45)³ = &frac43; × π × 0.091125 = 0.3817 cm³.
Volume used for 1120 spheres = 1120 × 0.3817 = 427.5 cm³.
Volume not used = 600 − 427.5 = 172 cm³ (3 s.f.)
On any day the probability that it rains is ⅓. When it rains the probability that Amira goes fishing is ⅗. When it does not rain the probability that Amira goes fishing is ¾.
- (a) In a period of 60 days, on how many days is it expected to rain?
- (b) Complete the tree diagram.
- (c) Find the probability that on any day Amira goes fishing.
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- (a) 60 × ⅓ = 20 days
- (b) Each pair of branches must add to 1:
- P(No rain) = 1 − ⅓ = ⅔
- P(No fishing | rain) = 1 − ⅗ = ⅖
- P(Fishing | no rain) = ¾
- P(No fishing | no rain) = 1 − ¾ = ¼
- (c) P(fishing) = P(rain)×P(fish|rain) + P(no rain)×P(fish|no rain)
= (⅓)(⅗) + (⅔)(¾) = ⅕ + ½ = &frac2{10} + &frac5{10} = &frac7{10}
0.7 (7/10)
The grid shows axes from 0 to 8 for both x and y.
- (a) On the grid, draw the lines y = x and x + y = 7.
- (b) Region R satisfies the three inequalities y ≥ 0, y ≤ x and x + y ≥ 7. On the grid, label the region R.
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- (a) y = x is the diagonal line through (0,0) and (8,8). x + y = 7 passes through (0,7) and (7,0).
- (b) The three lines y=0, y=x and x+y=7 create a boundary. Testing which side of each line satisfies the inequality:
• y ≥ 0: above the x-axis
• y ≤ x: below/right of the line y=x
• x+y ≥ 7: on the upper-right side of the line x+y=7
Within the grid (0 to 8), the region satisfying all three is the quadrilateral with corners (3.5, 3.5), (8, 8), (8, 0), (7, 0) — label this area R.
The diagram shows the speed–time graph of part of a car journey: speed rises to 8 m/s and stays constant until 10 seconds, then decelerates to 0 by 13 seconds.
- (a) Find the deceleration of the car between 10 and 13 seconds.
- (b) Calculate the total distance travelled during the 13 seconds.
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- (a) Deceleration = change in speed ÷ time = (8−0) ÷ (13−10) = 8÷3 = 2.67 m/s² (3 s.f.)
- (b) Distance = area under the graph.
Rectangle (0 to 10s at 8 m/s): 8 × 10 = 80 m
Triangle (10 to 13s, decelerating): ½ × 3 × 8 = 12 m
Total = 80 + 12 = 92 m
Factorise: 2x + 6 − 3xy − 9y
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Group the terms: (2x + 6) − (3xy + 9y) = 2(x+3) − 3y(x+3)
Factor out the common bracket: (x + 3)(2 − 3y)
The Venn diagram shows sets A and B within 𝒰. n(𝒰) = 20, n(A∪B)′ = 3, n(A) = 10 and n(B) = 13.
- (a) Find n(A ∩ B).
- (b) Find n(A′ ∩ B).
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- (a) n(A∪B) = n(𝒰) − n(A∪B)′ = 20 − 3 = 17.
n(A∩B) = n(A) + n(B) − n(A∪B) = 10 + 13 − 17 = 6 - (b) A′∩B is the part of B that is outside A: n(B) − n(A∩B) = 13 − 6 = 7
The height, h cm, of each of 100 students is measured. The table shows the results.
| Height (h cm) | 100 < h ≤ 150 | 150 < h ≤ 160 | 160 < h ≤ 165 | 165 < h ≤ 185 |
|---|---|---|---|---|
| Frequency | 7 | 30 | 41 | 22 |
Calculate an estimate of the mean.
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Using the midpoint of each class:
| Midpoint | 125 | 155 | 162.5 | 175 |
|---|---|---|---|---|
| Frequency | 7 | 30 | 41 | 22 |
| Midpoint × Frequency | 875 | 4650 | 6662.5 | 3850 |
Total = 875 + 4650 + 6662.5 + 3850 = 16 037.5, over 100 students.
Mean = 16 037.5 ÷ 100 = 160 cm (3 s.f., exact value 160.375 cm)
The diagram shows a quadrilateral, ABCD, formed from two triangles, ABC and ACD. ABC is a right-angled triangle (right angle at B), with AB=12 cm, BC=14 cm. In triangle ACD, AD=15 cm and angle DAC=62°.
- (a) Calculate angle BAC.
- (b) Calculate BD.
- (c) Calculate the shortest distance from D to AC.
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- (a) In right triangle ABC (right angle at B), the side opposite angle BAC is BC=14, and the side adjacent is AB=12.
tan(BAC) = 14÷12 = 1.1667, so angle BAC = 49.4° (1 d.p.) - (b) First find AC using Pythagoras in triangle ABC: AC² = 12²+14² = 144+196 = 340, so AC = √340 = 18.439 cm.
Angle BAD = angle BAC + angle DAC = 49.399° + 62° = 111.399°.
Using the cosine rule in triangle ABD: BD² = AB²+AD²−2(AB)(AD)cos(BAD)
= 144+225−2(12)(15)cos(111.399°) = 369−360×(−0.3648) = 369+131.3 = 500.3
BD = √500.3 = 22.4 cm (3 s.f.) - (c) The shortest distance from D to AC is the perpendicular height of triangle ACD from D.
Area of triangle ACD = ½ × AD × AC × sin(DAC) = ½ × 15 × 18.439 × sin(62°) = 122.1 cm²
Height = (2 × Area) ÷ AC = (2 × 122.1) ÷ 18.439 = 13.2 cm (3 s.f.)
(a) Hong has $4000 to invest. She invests $2000 at a rate of 2.5% per year simple interest. She also invests $2000 at a rate of 2% per year compound interest.
- (i) Find the value of each investment at the end of 8 years.
- (ii) Find the overall percentage increase in the $4000 investment at the end of 8 years.
- (iii) Find the number of complete years it takes for the compound interest investment of $2000 to become greater than $2500.
(b) Alain invests $5000 at a rate of r% per year compound interest. At the end of 15 years, the value of the investment is $7566. Find the value of r.
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- (a)(i) Simple interest: interest = 2000 × 0.025 × 8 = $400. Value = 2000+400 = $2400
Compound interest: 2000 × (1.02)⁸ = 2000 × 1.171659 = $2343.32 - (a)(ii) Total final value = 2400 + 2343.32 = $4743.32. Increase = 4743.32−4000 = $743.32.
Percentage increase = (743.32÷4000) × 100 = 18.6% (3 s.f.) - (a)(iii) Need 2000(1.02)ⁿ > 2500, i.e. (1.02)ⁿ > 1.25.
Testing: (1.02)¹&sup9; = 1.2434 (year 11, value $2486.75 — not yet over $2500); (1.02)¹² = 1.2682 (year 12, value $2536.48 — now over $2500).
12 complete years - (b) 5000(1+r/100)¹⁵ = 7566
(1+r/100)¹⁵ = 1.5132
1+r/100 = 1.51321/15 = 1.028
r = 2.8
y = √(u²x)
- (a) Find the value of y when u = 7 and x = 25.
- (b) Rearrange the formula to write x in terms of u and y.
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- (a) y = √(7² × 25) = √(49 × 25) = √1225 = 35
- (b) Square both sides: y² = u²x. Divide by u²: x = y² ÷ u²
A is the point (7, 2) and B is the point (−5, 8).
- (a) Calculate the length of AB.
- (b) Find the equation of the line that is perpendicular to AB and that passes through the point (−1, 3). Give your answer in the form y = mx + c.
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- (a) AB = √[(7−(−5))² + (2−8)²] = √(12²+(−6)²) = √(144+36) = √180 = 6√5 ≈ 13.4 cm (3 s.f.)
- (b) Gradient of AB = (8−2)÷(−5−7) = 6÷(−12) = −0.5.
Perpendicular gradient = negative reciprocal = 2.
Using y−3 = 2(x−(−1)): y = 2x+2+3
y = 2x + 5
A triangle has two sides 18 cm and 13 cm, with included angle x°. The area of the triangle is 50 cm².
Calculate the value of sin x.
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Area = ½ × 18 × 13 × sin(x) = 50
117 × sin(x) = 50
sin(x) = 50÷117 = 0.427 (3 s.f.)
Solve: 3y⁄2y−1 = ¾
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Cross-multiply: 4(3y) = 3(2y−1)
12y = 6y − 3
6y = −3
y = −0.5
The cross-section of a prism is an equilateral triangle of side 6 cm. The length of the prism is 20 cm. Calculate the total surface area of the prism.
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Area of one equilateral triangular face = √3⁄4 × 6² = √3⁄4 × 36 = 9√3 = 15.588 cm². There are 2 triangular faces: 2 × 15.588 = 31.18 cm².
Lateral (side) surface area = perimeter of triangle × length of prism = (6×3) × 20 = 18 × 20 = 360 cm²
Total surface area = 31.18 + 360 = 391 cm² (3 s.f.)
y = 2xk + ux⁷ and dy⁄dx = 18xk−1 + 21x⁶. Find the value of k and the value of u.
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Differentiating y=2xk+ux⁷ gives dy⁄dx = 2kxk−1 + 7ux⁶.
Comparing coefficients: 2k = 18, so k = 9
7u = 21, so u = 3
Simplify: 5p² − 20p⁄2p² − 32
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Numerator: 5p²−20p = 5p(p−4)
Denominator: 2p²−32 = 2(p²−16) = 2(p−4)(p+4)
Cancel the common factor (p−4): 5p / [2(p + 4)]
The diagram shows triangle OPT. OT = t and OP = p. OK:KT = 2:1 and TL:LP = 2:1.
- (a) Find, in terms of t and p, in its simplest form:
- PL
- KL
- (b) KL is extended to the point M. KM = −⅔t + &frac43;p. Show that M lies on OP extended.
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- (a)(i) Since TL:LP=2:1, L divides TP so that L = t + ⅔(p−t) = ⅓t + ⅔p.
PL = L − p = ⅓t + ⅔p − p = ⅓t − ⅓p
PL = ⅓(t − p) - (a)(ii) Since OK:KT=2:1, K = ⅔t.
KL = L − K = (⅓t+⅔p) − ⅔t = −⅓t + ⅔p
KL = ⅓(2p − t) - (b) Notice KM = −⅔t+&frac43;p = 2 × (−⅓t+⅔p) = 2KL, so M lies on the line through K and L extended (as expected, since KL was extended to M).
Position of M from O: OM = OK + KM = ⅔t + (−⅔t+&frac43;p) = &frac43;p
Since OM = &frac43;p is simply a scalar multiple of OP = p, point M lies on the straight line through O and P — and since the scalar (&frac43;) is greater than 1, M lies beyond P, i.e. M lies on OP extended.
Serge walks 7.9 km, correct to the nearest 100 metres. The walk takes 133 minutes, correct to the nearest minute. Calculate the maximum possible average speed of Serge’s walk. Give your answer in kilometres/hour.
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For the maximum possible speed, use the largest possible distance and the smallest possible time.
Upper bound of distance = 7.9 + 0.05 = 7.95 km
Lower bound of time = 133 − 0.5 = 132.5 minutes = 132.5÷60 hours
Speed = 7.95 ÷ (132.5÷60) = 7.95 × 60 ÷ 132.5 = 477÷132.5 = 3.6 km/h
The straight line y = 2x + 1 intersects the curve y = x² + 3x − 4 at the points A and B. Find the coordinates of A and B. Give your answers correct to 2 decimal places.
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Set the two expressions for y equal: 2x+1 = x²+3x−4
Rearranging: 0 = x²+x−5, i.e. x²+x−5 = 0
Using the quadratic formula: x = [−1 ± √(1+20)] ÷ 2 = [−1 ± √21] ÷ 2
√21 = 4.5826, so x = 1.7913 or x = −2.7913
Substituting into y=2x+1:
y(1.7913) = 2(1.7913)+1 = 4.5826
y(−2.7913) = 2(−2.7913)+1 = −4.5826
A (1.79, 4.58) B (−2.79, −4.58)