IGCSE Mathematics 0580 Paper 2 Practice Paper (Extended, Non-calculator) — 19 questions, 100 marks, worked solutions

Original IGCSE Mathematics 0580 Extended Paper 2 practice paper: 2 hours, 100 marks, non-calculator, 19 questions with worked solutions and marking notes.

Section A — Short questions (1–10)

Section A holds the shorter questions on core techniques, 4–5 marks each; aim to finish them in 50 minutes without losing accuracy.

1.　Answer all three parts without a calculator.

(a)　Write 0.000 046 in standard form.　[1]

(b)　Work out 3/4 + 5/6, giving your answer as a fraction in its simplest form.　[2]

(c)　Work out (2.4 × 10⁶) × (5 × 10⁻³), giving your answer in standard form.　[1]

2.　A laptop has a normal price of $640. In a sale the price is reduced by 15%.

(a)　Work out the reduction, in dollars.　[1]

(b)　Work out the sale price of the laptop.　[1]

(c)　After the sale the shop raises the sale price by 10%. Work out the new price and the percentage change from the normal price.　[2]

3.　A map is drawn to a scale of 1 : 25 000.

(a)　On the map a road is 6.4 cm long. Work out the real length, in km.　[2]

(b)　A field is 1.8 km long. Work out its length on the map, in cm.　[2]

(c)　Work out the real area, in m², represented by 1 cm² on the map.　[1]

4.　Do not use a calculator.

(a)　Work out 27^(2/3).　[2]

(b)　Simplify √50 + √18, giving your answer in the form k√2.　[2]

5.　The first four terms of a sequence are 5, 11, 17, 23, …

(a)　Find the nth term of the sequence.　[2]

(b)　Find the 40th term.　[1]

(c)　Explain whether 200 is a term of this sequence.　[1]

6.　Show your working for each part.

(a)　Expand and simplify (2x − 3)(x + 5).　[2]

(b)　Factorise completely 6x² − 15x.　[1]

(c)　Factorise x² − 9.　[1]

7.　Show all your working.

(a)　Solve the simultaneous equations 3x + 2y = 16 and x − y = 2.　[3]

(b)　Make x the subject of 5y − 3x = 10.　[2]

8.

(a)　Work out the interior angle of a regular pentagon.　[2]

(b)　Write down the exterior angle of a regular pentagon.　[1]

(c)　The interior angle of a regular polygon is 156°. Work out the number of sides.　[2]

9.　All lengths are in cm. Give exact answers where possible.

(a)　A right-angled triangle has hypotenuse 13 cm and one side 5 cm. Work out the third side.　[2]

(b)　In a right-angled triangle the hypotenuse is 14 cm and one angle is 30°. Work out the side opposite the 30° angle.　[1]

(c)　Work out the side adjacent to the 30° angle in part (b), in the form k√3.　[2]

10.　A bag contains 5 red, 3 blue and 2 green counters. All the counters are different. Two counters are taken at random without replacement.

(a)　Write down the probability that the first counter is red.　[1]

(b)　Work out the probability that both counters are red.　[2]

(c)　Work out the probability that at least one counter is green.　[2]

Section B — Extended questions (11–19)

Section B holds the extended questions and needs full method: quadratics, mensuration, a frequency table, triangle area, functions, vectors and proof.

11.　Show your method for both parts.

(a)　Solve 2x² − 7x + 3 = 0, showing your factorisation.　[3]

(b)　Write x² − 6x + 5 in the form (x − a)² + b, and state the coordinates of the minimum point of y = x² − 6x + 5.　[3]

12.　The line L passes through A(2, 5) and B(6, 13).

(a)　Work out the gradient of L.　[2]

(b)　Find the equation of L in the form y = mx + c.　[2]

(c)　Find the x-intercept of L.　[1]

(d)　Explain whether the point (10, 21) lies on L.　[1]

13.　A closed cylinder has radius 7 cm and height 10 cm. Use π = 22/7.

(a)　Work out the volume of the cylinder, in cm³.　[3]

(b)　Work out the curved surface area, in cm².　[2]

(c)　Work out the total surface area of the closed cylinder, in cm².　[2]

14.　In a sale a bicycle is sold for $272 after a 20% discount.

(a)　Work out the normal price of the bicycle.　[3]

(b)　A shop assistant adds 20% to $272 and says this is the normal price. Explain, with figures, why this is wrong.　[3]

15.　The table shows the number of goals scored by a football team in each of its matches this season.

(a)　Work out the mean number of goals per match.　[3]

(b)　Work out the median number of goals.　[2]

(c)　Write down the mode.　[1]

(d)　How many matches had 3 or more goals?　[1]

Goals per match (x) and number of matches (f)

16.　Triangle ABC has AB = AC = 8 cm and angle BAC = 60°.

(a)　Show that BC = 8 cm.　[2]

(b)　Work out the area of triangle ABC, in the form k√3 cm².　[2]

(c)　The triangle is the cross-section of a prism of length 12 cm. Work out the volume of the prism, correct to 3 significant figures.　[3]

17.　f(x) = 3x − 2 and g(x) = x² + 1.

(a)　Find f⁻¹(x).　[2]

(b)　Work out gf(2).　[2]

(c)　Explain whether f(x) = g(x) has any real solutions.　[2]

18.　In triangle OAB, OA = a and OB = b. P is the midpoint of OA. Q lies on AB with AQ : QB = 1 : 3.

(a)　Find PQ in terms of a and b, in its simplest form.　[3]

(b)　M is the midpoint of AB. Show that PQ is parallel to OM and state the ratio PQ : OM.　[3]

19.　Answer both parts.

(a)　Solve the inequality 4 − 3x < 13.　[2]

(b)　n is an integer. Prove that n² + n is always an even number.　[2]

Marking summary (out of 100)

Answers, worked solutions and marking notes

Q1 Answer　(a) 4.6 × 10⁻⁵　(b) 19/12 (= 1 7/12)　(c) 1.2 × 10⁴

·　(a) Move the point five places: 0.000 046 = 4.6 × 10⁻⁵.

·　(b) Common denominator 12: 9/12 + 10/12 = 19/12 = 1 7/12.

·　(c) 2.4 × 5 = 12 and 10⁶ × 10⁻³ = 10³, so 12 × 10³ = 1.2 × 10⁴.

Marking note　(b) 2 marks: 1 for the common denominator 12, 1 for 19/12; do not accept 1.58…

Q2 Answer　(a) $96　(b) $544　(c) $598.40; 6.5% below the normal price (−6.5%)

·　(a) 10% of 640 = 64 and 5% of 640 = 32, so 15% = 96.

·　(b) 640 − 96 = 544.

·　(c) 10% of 544 = 54.4, so the new price is 598.40.

·　Change = 598.40 − 640 = −41.60; −41.60 ÷ 640 × 100% = −6.5%.

Marking note　(c) 1 mark for $598.40, 1 for −6.5% (accept “6.5% decrease”); −10% or −15% scores 0.

Q3 Answer　(a) 1.6 km　(b) 7.2 cm　(c) 62 500 m² (6.25 hectares)

·　(a) 6.4 × 25 000 = 160 000 cm = 1 600 m = 1.6 km.

·　(b) 1.8 km = 180 000 cm; 180 000 ÷ 25 000 = 7.2 cm.

·　(c) Area scale factor = 25 000² = 625 000 000, so 1 cm² = 625 000 000 cm² = 62 500 m² = 6.25 hectares.

Marking note　(c) The common error is using 25 000 instead of 25 000²; the answer must be in m².

Q4 Answer　(a) 9　(b) 8√2

·　(a) 27^(1/3) = 3, then 3² = 9.

·　(b) √50 = 5√2 and √18 = 3√2, so the sum is 8√2.

Marking note　(a) 1 mark for 27^(1/3) = 3, 1 for 9; (b) 1 for either surd simplified, 1 for 8√2.

Q5 Answer　(a) 6n − 1　(b) 239　(c) no: 6n − 1 = 200 gives n = 33.5, not a whole number

·　(a) The terms go up by 6, so the nth term is 6n + c with 6(1) + c = 5, giving c = −1: 6n − 1.

·　(b) 6(40) − 1 = 239.

·　(c) 6n − 1 = 200 gives n = 33.5, which is not a whole number, so 200 is not a term.

Marking note　(c) The reason must say n is not an integer; “no” alone scores 0.

Q6 Answer　(a) 2x² + 7x − 15　(b) 3x(2x − 5)　(c) (x − 3)(x + 3)

·　(a) 2x(x + 5) − 3(x + 5) = 2x² + 10x − 3x − 15 = 2x² + 7x − 15.

·　(b) Common factor 3x: 3x(2x − 5).

·　(c) Difference of two squares: (x − 3)(x + 3).

Marking note　(a) 1 mark for three correct terms before simplifying, 1 for 2x² + 7x − 15; (b) must take out 3x, not just x.

Q7 Answer　(a) x = 4, y = 2　(b) x = (5y − 10)/3

·　(a) From x − y = 2, x = y + 2. Substituting: 3(y + 2) + 2y = 16, so 5y = 10, y = 2 and x = 4. Check: 3(4) + 2(2) = 16 ✓.

·　(b) 5y − 10 = 3x, so x = (5y − 10)/3 (or 5y/3 − 10/3).

Marking note　(a) 1 mark for a correct substitution/elimination step, 1 for y = 2, 1 for x = 4; (b) 1 for collecting the x term, 1 for dividing the whole expression by 3.

Q8 Answer　(a) 108°　(b) 72°　(c) 15 sides

·　(a) Sum of interior angles = (5 − 2) × 180° = 540°, so each angle is 540° ÷ 5 = 108°.

·　(b) 180° − 108° = 72° (or 360° ÷ 5).

·　(c) Exterior angle = 180° − 156° = 24°, and 360° ÷ 24° = 15 sides.

Marking note　(c) Working with the exterior angle is the reliable method; using 156° ÷ 360° scores 0.

Q9 Answer　(a) 12 cm　(b) 7 cm　(c) 7√3 cm (≈ 12.1 cm)

·　(a) √(13² − 5²) = √(169 − 25) = √144 = 12.

·　(b) sin 30° = 1/2, so the opposite side is 14 × 1/2 = 7.

·　(c) cos 30° = √3/2, so the adjacent side is 14 × √3/2 = 7√3 ≈ 12.1.

Marking note　(c) Exact surd form is required because the paper is non-calculator; accept 12.1 only if the exact form is also shown.

Q10 Answer　(a) 1/2　(b) 2/9　(c) 17/45 (≈ 0.378)

·　(a) 5 out of 10, so 1/2.

·　(b) 5/10 × 4/9 = 20/90 = 2/9.

·　(c) P(no green) = 8/10 × 7/9 = 56/90 = 28/45, so P(at least one green) = 1 − 28/45 = 17/45 ≈ 0.378.

Marking note　(b) The second draw must use 4/9; using 5/10 twice scores 1 mark only.

Q11 Answer　(a) x = 3 or x = 1/2　(b) (x − 3)² − 4; minimum point (3, −4)

·　(a) 2x² − 7x + 3 = (2x − 1)(x − 3) = 0, so x = 1/2 or x = 3.

·　(b) x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4, so a = 3 and b = −4.

·　The squared term is never negative, so the minimum is −4 when x = 3: (3, −4).

Marking note　(a) 1 mark for the correct pair of factors, 1 for each root; (b) 2 marks for (x − 3)² − 4, 1 for the coordinates.

Q12 Answer　(a) 2　(b) y = 2x + 1　(c) x = −1/2　(d) yes: 2(10) + 1 = 21

·　(a) (13 − 5) ÷ (6 − 2) = 8 ÷ 4 = 2.

·　(b) 5 = 2(2) + c gives c = 1, so y = 2x + 1.

·　(c) 0 = 2x + 1 gives x = −1/2.

·　(d) 2(10) + 1 = 21, so (10, 21) does lie on L.

Marking note　(d) The substitution must be shown; a bare “yes” scores 0.

Q13 Answer　(a) 1 540 cm³　(b) 440 cm²　(c) 748 cm²

·　(a) πr²h = 22/7 × 49 × 10 = 22 × 7 × 10 = 1 540 cm³.

·　(b) 2πrh = 2 × 22/7 × 7 × 10 = 440 cm².

·　(c) Two circular ends: 2 × 154 = 308, so total = 440 + 308 = 748 cm².

Marking note　(a) 1 mark for πr²h with the numbers substituted, 1 for 22/7 × 49 = 154, 1 for 1 540; (c) the ends must be counted twice.

Q14 Answer　(a) $340　(b) 272 × 1.2 = $326.40, not $340: the two 20% amounts use different bases (68 and 54.40)

·　(a) 272 is 80% of the normal price, so 1% is 3.40 and 100% is $340. Check: 340 × 0.8 = 272 ✓.

·　(b) 20% of 272 is 54.40, giving 326.40, but 20% of 340 is 68. The two percentages are taken from different amounts, so adding 20% does not undo a 20% discount.

Marking note　(a) Dividing by 0.8 (or by 80 then × 100) is the expected method; 272 × 1.2 scores 0. (b) 2 marks for both figures, 1 for “different base amount”.

Q15 Answer　(a) 5/3 ≈ 1.67　(b) 2　(c) 2　(d) 5 matches

·　(a) Σf = 30 and Σfx = 4×0 + 9×1 + 12×2 + 3×3 + 2×4 = 50, so mean = 50 ÷ 30 = 5/3 ≈ 1.67.

·　(b) The 15th and 16th values both fall in the “2 goals” row (cumulative 4, 13, 25, 28, 30), so the median is 2.

·　(c) The highest frequency is 12, for 2 goals, so the mode is 2.

·　(d) 3 + 2 = 5 matches.

Marking note　(a) 2 marks for Σfx = 50 and Σf = 30, 1 for 5/3 (accept 1.67); dividing by 5 (the number of rows) instead of 30 scores 0.

Q16 Answer　(a) BC² = 64, so BC = 8 cm　(b) 16√3 cm² (≈ 27.7)　(c) 192√3 ≈ 333 cm³

·　(a) BC² = 8² + 8² − 2(8)(8)cos 60° = 64 + 64 − 64 = 64, so BC = 8 cm (the triangle is equilateral).

·　(b) Area = ½ × 8 × 8 × sin 60° = 32 × √3/2 = 16√3 ≈ 27.7 cm².

·　(c) Volume = 16√3 × 12 = 192√3 ≈ 332.55 ≈ 333 cm³.

Marking note　(a) The cosine rule with cos 60° = 1/2 is the expected route, and “show that” requires the working. (c) 2 marks for 192√3, 1 for rounding to 3 s.f.

Q17 Answer　(a) f⁻¹(x) = (x + 2)/3　(b) 17　(c) no real solutions (discriminant −3 < 0)

·　(a) y = 3x − 2, so x = (y + 2)/3 and f⁻¹(x) = (x + 2)/3.

·　(b) f(2) = 4, then g(4) = 17.

·　(c) 3x − 2 = x² + 1 gives x² − 3x + 3 = 0. The discriminant is 9 − 12 = −3 < 0, so there are no real solutions.

Marking note　(b) gf means g(f(x)), so f is applied first; giving 37 scores 0. (c) The discriminant must be calculated and interpreted.

Q18 Answer　(a) PQ = ¼a + ¼b　(b) OM = ½(a + b) and PQ = ½OM, so PQ is parallel to OM; PQ : OM = 1 : 2

·　(a) OP = ½a and OQ = a + ¼(b − a) = ¾a + ¼b, so PQ = OQ − OP = ¼a + ¼b.

·　(b) OM = ½(a + b) = ½a + ½b. Since PQ = ½OM, the two vectors are parallel, and PQ : OM = 1 : 2.

Marking note　(a) 1 mark for OQ, 1 for the subtraction, 1 for ¼a + ¼b; (b) the scalar multiple ½ is the proof of parallelism.

Q19 Answer　(a) x > −3　(b) n² + n = n(n + 1); one of two consecutive integers is even, so the product is even

·　(a) 4 − 3x < 13 gives −3x < 9; dividing by −3 reverses the sign, so x > −3.

·　(b) n² + n = n(n + 1), the product of two consecutive integers. One of them is even, so the product is even.

Marking note　(a) 1 mark for −3x < 9, 1 for x > −3 (x < −3 scores 0). (b) The factorised form and the consecutive-integer argument are both needed.

Grade descriptors (out of 100)

Suggested timing (120 minutes)

Notices

•　This paper is 2 hours, 100 marks, non-calculator, matching Cambridge IGCSE Mathematics 0580 Extended Paper 2. All questions are original, not Cambridge past-paper questions.

•　There are 19 questions worth 100 marks in total. The number in brackets is the marks for that part, and parts worth more marks usually need more steps.

•　Because no calculator is allowed, leave answers as integers, fractions or surds; when a question asks for a given accuracy, state it (for example 333 cm³).

•　For Questions 11–19 you must show your method; a correct answer alone does not earn the method marks.

•　Use π = 22/7 for the cylinder question, and learn the exact trigonometric values (sin 30° = 1/2, cos 30° = √3/2).

•　Do Section A first and keep the harder Section B questions for when you have enough time left.

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