IGCSE Maths (0580) · Extended · Paper 4

IGCSE Maths Past Paper Solutions: 0580/42 February/March 2024 (Paper 4 Extended)

Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 4 Extended exam paper from February/March 2024 (0580/42/F/M/24) — 12 questions, 130 marks, 2 hours 30 minutes. Attempt each question yourself first, then click “Show solution” to check your working step by step.

112 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)

A grocer sells potatoes, mushrooms and carrots.

  1. (a) A customer buys 3 kg of mushrooms at $1.04 per kg and 4 kg of carrots at $1.28 per kg. Calculate the total cost. [2]
  2. (b) In one week, the ratio of the masses of vegetables sold by the grocer is potatoes : mushrooms : carrots = 11 : 8 : 6.
    1. Work out the mass of mushrooms sold as a percentage of the total mass. [2]
    2. The total mass of potatoes, mushrooms and carrots sold is 1500 kg. Find the mass of carrots the grocer sells this week. [2]
    3. The profit the grocer makes selling 1 kg of carrots is $0.75. Find the total profit the grocer makes selling carrots this week. [1]
    4. On the last day of the week, the grocer reduces the price of 1 kg of potatoes by 8% to $1.15. Calculate the original price of 1 kg of potatoes. [2]
  3. (c) The grocer buys 620 kg of onions, correct to the nearest 20 kg. He packs them into bags each containing 5 kg of onions, correct to the nearest 1 kg. Calculate the upper bound for the number of bags of onions that he packs. [3]
Show solution
  1. (a) 3 × 1.04 = 3.12. 4 × 1.28 = 5.12. Total = 3.12 + 5.12 = $8.24
  2. (b)(i) Total parts = 11+8+6 = 25. Mushrooms fraction = 8/25 = 32%
  3. (b)(ii) Carrots fraction = 6/25. Mass = (6/25) × 1500 = 360 kg
  4. (b)(iii) Profit = 360 × 0.75 = $270
  5. (b)(iv) The reduced price is 92% of the original: original × 0.92 = 1.15 → original = 1.15 ÷ 0.92 = $1.25
  6. (c) Upper bound of total mass (nearest 20 kg) = 620+10 = 630 kg. Lower bound of bag size (nearest 1 kg) = 5−0.5 = 4.5 kg. The upper bound for the number of bags is the largest total divided by the smallest bag size: 630 ÷ 4.5 = 140
28 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)

A, B, C and D are points on a circle. ADX and BCX are straight lines. Angle BAD = x° and angle DCX = y°.

Question 2 — circle with points A, B, C, D, and external point X on lines ADX and BCX
  1. (a) Explain why x = y. Give a geometrical reason for each statement you make. [2]
  2. (b) Show that triangle ABX is similar to triangle CDX. [2]
  3. (c) AD = 15 cm, DX = 9 cm and CX = 12 cm.
    1. Find BC. [3]
    2. Complete the statement: the ratio area of triangle ABX : area of triangle CDX = …… : 1. [1]
Show solution
  1. (a) ABCD is a cyclic quadrilateral, so opposite angles sum to 180°: angle BAD + angle BCD = 180° (opposite angles of a cyclic quadrilateral). Also angle BCD + angle DCX = 180° (angles on a straight line BCX). Since both sums equal 180° and share angle BCD, it follows that angle BAD = angle DCX, i.e. x = y
  2. (b) Angle AXB = angle CXD (it is the same angle at X, common to both triangles since A,D,X and B,C,X are straight lines). Angle XAB = angle BAD = x (same angle, as D lies on AX), and from part (a), x = y = angle DCX = angle XCD. So triangle ABX and triangle CDX have two pairs of equal angles, and are therefore similar (AA).
  3. (c)(i) AX = AD+DX = 15+9 = 24 cm. Since the triangles are similar with correspondence A↔C, B↔D, X↔X, the scale factor is AX/CX = 24/12 = 2. So BX = 2 × DX = 2 × 9 = 18 cm, and BC = BX − CX = 18−12 = 6 cm
  4. (c)(ii) The ratio of areas of similar triangles equals the square of the scale factor: 2² = 4 : 1
Please double-check against the original diagram: the exact positions of A, B, C, D and X around the circle should be confirmed against the printed figure, since the labelling determines which angles correspond in part (a).
313 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) The table shows information about the marks gained by each of 10 students in a test.
Mark151617181920
Frequency412102
    1. Calculate the range. [1]
    2. Calculate the mean. [3]
    3. Find the median. [1]
    4. Write down the mode. [1]
  2. (b) Paulo’s mean mark for 7 homework tasks is 17. After completing the 8th task, his mean mark is 17.5. Calculate Paulo’s mark for the 8th task. [3]
  3. (c) The table shows the percentage scored by each of 100 students in their final exam.
Percentage (p)0 < p ≤ 3030 < p ≤ 5050 < p ≤ 6060 < p ≤ 7070 < p ≤ 100
Frequency1218352015

On the grid, draw a histogram to show this information. [4]

Question 3(c) — blank histogram grid, frequency density 0 to 4, percentage 0 to 100
Show solution
  1. (a)(i) Range = highest − lowest = 20−15 = 5
  2. (a)(ii) Mean = (15×4 + 16×1 + 17×2 + 18×1 + 19×0 + 20×2) ÷ 10 = (60+16+34+18+0+40)÷10 = 168÷10 = 16.8
  3. (a)(iii) Ordering the 10 marks: 15,15,15,15,16,17,17,18,20,20. The median is the mean of the 5th and 6th values: (16+17)÷2 = 16.5
  4. (a)(iv) The most frequent mark is 15, occurring 4 times: mode = 15
  5. (b) Total after 7 tasks = 17×7 = 119. Total after 8 tasks = 17.5×8 = 140. 8th mark = 140−119 = 21
  6. (c) Frequency density = frequency ÷ class width for each bar:
    1. 0<p≤30 (width 30): fd = 12÷30 = 0.4
    2. 30<p≤50 (width 20): fd = 18÷20 = 0.9
    3. 50<p≤60 (width 10): fd = 35÷10 = 3.5
    4. 60<p≤70 (width 10): fd = 20÷10 = 2.0
    5. 70<p≤100 (width 30): fd = 15÷30 = 0.5
    Draw bars at these heights across each stated interval on the grid.
411 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) The diagram shows a pyramid with a square base BCDE. The diagonals CE and BD intersect at M, and the vertex F is directly above M. BE = 12 cm and FM = 9 cm.
Question 4(a) — square-based pyramid BCDE with apex F above centre M, BE=12cm, FM=9cm
    1. Calculate the volume of the pyramid. [The volume, V, of a pyramid with base area A and height h is V = ⅓Ah.] [2]
    2. Calculate the total surface area of the pyramid. [5]
  2. (b) The diagram shows a toy made from a cone and a hemisphere. The base radius of the cone and the radius of the hemisphere are both r cm. The slant height of the cone is 3r cm. The total surface area of the toy is 304 cm².
Question 4(b) — toy made of a cone (slant height 3r) on top of a hemisphere, both radius r

Calculate the value of r. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] [The curved surface area, A, of a sphere with radius r is A = 4πr².] [4]

Show solution
  1. (a)(i) Base area = 12×12 = 144 cm². Volume = ⅓ × 144 × 9 = 432 cm³
  2. (a)(ii) M is the centre of the square base, so the distance from M to the midpoint of any base edge is half of 12, i.e. 6 cm. The slant height of each triangular face (from F to the midpoint of a base edge) is found using Pythagoras: l = √(9²+6²) = √117 = 10.817 cm. Area of one triangular face = ½ × 12 × 10.817 = 64.90 cm². Four faces = 4 × 64.90 = 259.6 cm². Total surface area = base + 4 faces = 144 + 259.6 = 404 cm² (3 s.f.)
  3. (b) Curved surface of cone = πr(3r) = 3πr². Curved surface of hemisphere = half of a sphere’s curved surface = 2πr² (the flat circular face is where the two solids join, so it isn’t part of the outer surface). Total surface = 3πr² + 2πr² = 5πr² = 304. So r² = 304÷(5π) = 19.353, and r = √19.353 = r = 4.40 cm (3 s.f.)
Please double-check against the original diagram: confirm from the printed figure that M is indeed the centre of the square base (intersection of the diagonals) as stated, and that the cone in part (b) sits directly on top of the hemisphere along matching radii, before sharing these solutions with students.
520 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a)
    1. Factorise. x² − x − 12 [2]
    2. Simplify. (x²−16) / (x²−x−12) [2]
  2. (b) Simplify. (2x−3)² − (x+1)² [3]
  3. (c) Write as a single fraction in its simplest form. (2x+4)/(x+1) − x/(x−3) [4]
  4. (d) Expand and simplify. (x−3)(x−5)(2x+1) [3]
  5. (e) Solve the simultaneous equations. You must show all your working.
    x − 3y = 13
    2x² − 9y = 116 [6]
Show solution
  1. (a)(i) Need two numbers multiplying to −12 and adding to −1: these are −4 and 3. (x−4)(x+3)
  2. (a)(ii) x²−16 = (x−4)(x+4) (difference of two squares). Dividing by (x−4)(x+3) and cancelling the common factor (x−4): (x+4)/(x+3)
  3. (b) (2x−3)² = 4x²−12x+9. (x+1)² = x²+2x+1. Subtracting: 4x²−12x+9−x²−2x−1 = 3x² − 14x + 8
  4. (c) Common denominator (x+1)(x−3): numerator = (2x+4)(x−3) − x(x+1) = (2x²−2x−12) − (x²+x) = x²−3x−12. (x² − 3x − 12) / [(x+1)(x−3)]
  5. (d) (x−3)(x−5) = x²−8x+15. Multiplying by (2x+1): x²(2x+1) −8x(2x+1)+15(2x+1) = 2x³+x²−16x²−8x+30x+15 = 2x³ − 15x² + 22x + 15
  6. (e) From the first equation, x = 13+3y. Substitute into the second: 2(13+3y)²−9y = 116. Expanding (13+3y)² = 169+78y+9y²: 2(169+78y+9y²)−9y = 116 → 18y²+147y+338 = 116 → 18y²+147y+222 = 0. Dividing by 3: 6y²+49y+74 = 0. Using the quadratic formula: discriminant = 49²−4(6)(74) = 2401−1776 = 625, √625 = 25. y = (−49±25)/12, giving y = −2 or y = −37/6. For y=−2: x = 13+3(−2) = 7. For y=−37/6: x = 13−37/2 = −5.5.
    x = 7, y = −2   x = −5.5, y = −37/6 (−6.17)
69 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)

The diagram shows triangle ABC with AB = 17.2 cm. Angle ABC = 54° and angle ACB = 68°. M lies on BC.

Question 6 — triangle ABC, AB=17.2cm, angle B=54 degrees, angle C=68 degrees, M on BC with MC=12.8cm
  1. (a) Calculate AC. [3]
  2. (b) M lies on BC and MC = 12.8 cm. Calculate AM. [3]
  3. (c) Calculate the shortest distance from A to BC. [3]
Show solution
  1. (a) Angle BAC = 180−54−68 = 58°. By the sine rule: AC/sin(ABC) = AB/sin(ACB) → AC = 17.2 × sin54° ÷ sin68° = 17.2 × 0.80902 ÷ 0.92718 = 15.0 cm (3 s.f.)
  2. (b) In triangle AMC, angle ACM = angle ACB = 68° (same angle, as M lies on BC), with AC ≈ 15.008 cm and MC = 12.8 cm. Using the cosine rule: AM² = AC²+MC²−2(AC)(MC)cos68° = 225.24+163.84−2(15.008)(12.8)(0.37461) = 225.24+163.84−143.94 = 245.14. AM = √245.14 = 15.7 cm (3 s.f.)
  3. (c) The shortest distance from A to line BC is the perpendicular height, which can be found using the right-angled triangle formed with angle B: height = AB × sin(ABC) = 17.2 × sin54° = 17.2 × 0.80902 = 13.9 cm (3 s.f.)
77 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) p = (8, −5)   q = (−4, 5)
    1. Find 3q. [1]
      1. Find p − q. [1]
      2. Find |p − q|. [2]
  2. (b) In triangle OMN, O is the origin, OM→ = a and ON→ = b. S is a point on MN such that MS:SN = 5:3. Find, in terms of a and/or b, the position vector of S. Give your answer in its simplest form. [3]
Question 7(b) — triangle OMN with O the origin, OM=a, ON=b, and S on MN with MS:SN=5:3
Show solution
  1. (a)(i) 3q = 3×(−4, 5) = (−12, 15)
  2. (a)(ii)(a) p−q = (8−(−4), −5−5) = (12, −10)
  3. (a)(ii)(b) |p−q| = √(12²+(−10)²) = √(144+100) = √244 = 15.6 (3 s.f.)
  4. (b) Since MS:SN = 5:3, S divides MN so that MS→ = &frac58;MN→. Position vector of S = OM→ + &frac58;(ON→−OM→) = a + &frac58;(b−a) = &frac38;a + &frac58;b
812 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) On the axes, sketch the graph of y = 4−3x. [2]
Question 8(a) — blank axes for sketching y = 4 - 3x
  1. (b) On the axes, sketch the graph of y = −x². [2]
Question 8(b) — blank axes for sketching y = -x squared
  1. (c)
    1. Find the coordinates of the turning points of the graph of y = 10+9x²−2x³. You must show all your working. [5]
    2. Determine whether each turning point is a maximum or a minimum. Show how you decide. [3]
Show solution
  1. (a) y=4−3x is a straight line with gradient −3 and y-intercept 4 (so it passes through (0,4) and crosses the x-axis at x=&frac43;), sloping downward from left to right.
  2. (b) y=−x² is a downward-opening parabola with vertex at the origin (0,0), symmetric about the y-axis, and lying entirely on or below the x-axis.
  3. (c)(i) dy/dx = 18x−6x² = 6x(3−x). Setting this to zero: x=0 or x=3. At x=0: y=10. At x=3: y=10+9(9)−2(27) = 10+81−54 = 37.
    (0, 10) and (3, 37)
  4. (c)(ii) d²y/dx² = 18−12x. At x=0: 18−0=18>0, so this is a minimum. At x=3: 18−36=−18<0, so this is a maximum.
99 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) Janna and Kamal each invest $8000. At the end of 12 years, they each have $12 800.
    1. Janna invests in an account that pays simple interest at a rate of r% per year. Calculate the value of r. [3]
    2. Kamal invests in an account that pays compound interest at a rate of R% per year. Calculate the value of R. [3]
  2. (b) The population of a city is growing exponentially at a rate of 1.8% per year. The population now is 260 000. Find the number of complete years from now when the population will first be more than 300 000. [3]
Show solution
  1. (a)(i) Total simple interest earned = 12800−8000 = 4800. Using I = Prt/100: 4800 = 8000 × r × 12 ÷ 100 = 960r. r = 5
  2. (a)(ii) Using 8000(1+R/100)¹² = 12800: (1+R/100)¹² = 1.6, so 1+R/100 = 1.61/12 = 1.03996. R = 4.00 (3 s.f.)
  3. (b) Solve 260000(1.018)n > 300000, i.e. (1.018)n > 1.15385. Testing values: at n=8, 1.018⁸ ≈ 1.1534 (still below); at n=9, 1.018⁹ ≈ 1.1742 (above 1.15385). So the population first exceeds 300 000 after 9 years
1012 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)

The table shows some values for y = 2x³+6x²−2.5.

x−3−2.5−2−1.5−1−0.500.51
y?3.755.54.251.5?−2.5−0.75?
  1. (a) Complete the table. [3]
  2. (b) On the grid, draw the graph of y = 2x³+6x²−2.5 for −3 ≤ x ≤ 1. [4]
Question 10(b) — blank grid, x axis -3 to 1, y axis -4 to 6, for plotting y = 2x^3+6x^2-2.5
  1. (c) By drawing a suitable line on the graph, solve the equation 2x³+6x² = 4.5. [3]
  2. (d) The equation 2x³+6x²−2.5 = k has exactly two solutions. Write down the two possible values of k. [2]
Show solution
  1. (a) At x=−3: y = 2(−27)+6(9)−2.5 = −54+54−2.5 = −2.5. At x=−0.5: y = 2(−0.125)+6(0.25)−2.5 = −0.25+1.5−2.5 = −1.25. At x=1: y = 2(1)+6(1)−2.5 = 2+6−2.5 = 5.5.
    x=−3: y=−2.5   x=−0.5: y=−1.25   x=1: y=5.5
  2. (b) Plot the nine points from the table and join them with a smooth curve. The curve has a local maximum near (−2, 5.5) and a local minimum near (0, −2.5).
  3. (c) Rearranging, 2x³+6x² = 4.5 is the same as 2x³+6x²−2.5 = 2, so draw the horizontal line y=2 and read off where it crosses the curve. Solving numerically confirms three crossings, at approximately
    x = −2.7   x = −1.1   x = 0.8 (readings from the graph; small variation is acceptable)
  4. (d) The curve has exactly two solutions for a given k only when the horizontal line y=k is tangent to the curve at a turning point — i.e. when k equals the local maximum or local minimum y-value: k = 5.5 or k = −2.5
Please double-check against the original diagram: part (c)’s answers are graph readings, so please verify by drawing the curve accurately on the printed grid and confirming the exact crossing points with the line y=2 — the values above (−2.7, −1.1, 0.8) are the calculated roots of the underlying cubic and should match closely, but allow for normal graph-reading tolerance (typically ±0.1 to ±0.2).
1110 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)

f(x) = 1/x, x≠0   g(x) = 3x−5   h(x) = 2x

  1. (a) Find.
    1. gf(2) [2]
    2. g−1(x) [2]
  2. (b) Find in its simplest form g(x−2). [2]
  3. (c) Find the value of x when
    1. fg(x) = 0.1 [2]
    2. h(x)−g(7) = 0 [2]
Show solution
  1. (a)(i) f(2) = 1/2. g(1/2) = 3(1/2)−5 = 1.5−5 = −3.5
  2. (a)(ii) Let y=3x−5, so x=(y+5)/3. g−1(x) = (x+5)/3
  3. (b) g(x−2) = 3(x−2)−5 = 3x−6−5 = 3x − 11
  4. (c)(i) fg(x) = 1/(3x−5) = 0.1 → 3x−5 = 10 → 3x = 15 → x = 5
  5. (c)(ii) g(7) = 3(7)−5 = 16. So 2x = 16 = 2⁴ → x = 4
127 marksExtendedFeb/Mar 2024 · Paper 4 (0580/42)
  1. (a) The diagram shows a circle of radius 12 cm, with a sector of angle 50° removed. Calculate the perimeter of the remaining shaded shape. [4]
Question 12(a) — circle radius 12cm with a 50 degree sector removed, remaining shape shaded
  1. (b) The diagram in part (a) shows the top of a cylindrical cake with a slice removed. The volume of cake that remains is 3510 cm³. Calculate the height of the cake. [3]
Show solution
  1. (a) The remaining shape is the major sector, with angle 360−50 = 310°. Its perimeter consists of two straight radii (12 cm each) plus the major arc: arc length = (310/360) × 2π(12) = 0.86111 × 75.398 = 64.93 cm. Perimeter = 2(12) + 64.93 = 24 + 64.93 = 88.9 cm (3 s.f.)
  2. (b) The remaining cake is (310/360) of a full cylinder of radius 12 cm and height h: Volume = (310/360) × π(12)² × h = 0.86111 × 452.39 × h = 389.6h. Setting this equal to 3510: h = 3510 ÷ 389.6 = 9.01 cm (3 s.f.)

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