Cambridge IGCSE Mathematics 0580 — 2023 May/June Paper 4 · Variant 3

1 (a) Tomas sells a computer, a bike and a phone. The amounts he receives are in the ratio computer : bike : phone = 14 : 17 : 9. (i) Calculate the amount he receives for the phone as a percentage of the total. ............................................. % [2] (ii) The total amount he receives is $560. Calculate how much he receives for the bike. $ ................................................ [2] (iii) Tomas originally bought the bike for $195. He wanted to make a profit of at least 25% when he sold it. Does Tomas make a profit of at least 25%? You must show all your working to support your decision. [3] (b) Ulla invests $725 for 6 years in an account paying simple interest at a rate of 1.3% per year. Calculate the total interest earned at the end of 6 years. $ ................................................ [2] (c) In a sale, all prices are reduced by 24%. Victor pays $36.86 for a pair of shoes in the sale. Calculate the original price of the shoes. $ ................................................ [2]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 22.5 2 9 M1 for [ 100] 14  17  9 1(a)(ii) 238 2 FT their 14 + 17 + 9 = N seen in (a)(i) 560 M1 for  k, their 14  17  9  where k = 1, 9, 14 or 17 1(a)(iii) METHOD 1 M2 25 M1 for  195 100 1.25  195 oe 243[.75] and No oe A1 Strict FT yes if their (a)(ii) > 243.75 If M0 scored, then SC1 for 243.75 and a correct conclusion. METHOD 2 (M2) their 238 M1 for = 1.22... oe 195 their 238 – 1 = 0.22… oe 195 22[%] (or better) and No oe (A1) Strict FT yes if their (a)(ii) gives answer > 25 If M0 scored, then SC1 for 22.05 and a correct conclusion. METHOD 3 (M2) M1 for 0.25  195 195  0.25 = 48.75 oe and their 238 – 195 = 43 43 and 48.75 and NO (A1) Strict FT yes if their (a)(ii) gives profit > 48.75 If M0 scored, then SC1 for 43 and 48.75 and a correct conclusion. METHOD 4 (M2)  25  M1 for x   1   = their 238  100  their 238  100 125 190.4 and NO (A1) Strict FT yes if their (a)(ii) gives answer > 195 If M0 scored then SC1 for 190.4 and a correct conclusion. 1(b) 56.55 2 725  1.3[ 6] M1 for oe 100 1(c) 48.5[0] 2  24  M1 for x   1   = 36.86 oe  100 

More questions on Ratio and proportion

2 (a) Anna records the number of text messages she receives for 14 days. 17 15 31 38 31 22 13 18 21 27 28 21 31 29 (i) Complete the stem-and-leaf diagram. 1 2 3 Key: ........................................................... [3] (ii) Find the median. ................................................. [1] (iii) Find the mode. ................................................. [1] (iv) Find the range. ................................................. [1] (b) In a shop, there are 4 red and 8 grey phones. Anna and Pete each pick one of these phones at random. Work out the probability that they both pick a grey phone. ................................................. [2]

Mark scheme: 2(a)(i) 3 B2 for fully correct stem-and-leaf 1 3 5 7 8 diagram 2 1 1 2 7 8 9 OR 3 1 1 1 8 B1 for two rows correct or for fully correct unordered stem-and-leaf diagram 17 represents 17 [messages] or for a correct diagram with one error or omission B1 for correct key 2(a)(ii) 24.5 1 2(a)(iii) 31 1 2(a)(iv) 25 1 2(b) 14 2 8 7 oe M1 for  33 12 11

More questions on Statistical charts and diagrams

3 (a) The scale drawing shows two sides, AB and BC, of a field. The scale is 5 centimetres represents 200 metres. A B C Scale: 5 cm to 200 m (i) Measure angle ABC. Angle ABC = ................................................ [1] (ii) X is a point on BC. BX = 332 m. Mark the point X on the diagram. [2] (iii) Find the scale in the form 1 : n. 1: ................................................ [2] (b) A bronze statue is 4.5 m high and has a mass of 195 200 kg. The density of bronze is 8000 kg / m 3. The volume of a mathematically similar model of the statue is 0.385 m 3. Calculate the height of the model. [Density = Mass ÷ Volume] ............................................. m [5]

Mark scheme: 3(a)(i) 118 1 3(a)(ii) X is 8.3 cm from B 2 M1 for (332 ÷ 200)  5 oe 3(a)(iii) 1 : 4000 2 M1 for 200 ÷ 5 or 200  100, both soi 3(b) 1.13 or 1.128 to 1.129 5 0.385  8000 M4 for 4.5  3 oe 195200 4.53  0.385  8000 or 3 oe 195200 0.385 their 3080 or M3 for 3 or 3 their 24.4 195200 0.385 l 3 or  oe their 24.4 4.53 their 24.4 0.385 or M2 for or oe 0.385 their 24.4 or B2 for 24.4 or 3080 seen or M1 for 195 200 ÷ 8000 or for 0.385  8000

More questions on Scale drawings

4 (a) North 114° A NOT TO SCALE C B A, B and C are three towns and the bearing of C from A is 114°. B is due south of A and AC = BC. Calculate the bearing of B from C. ................................................. [3] (b) R S 74° NOT TO SCALE Q 58° 27° M P N P, Q, R and S lie on a circle. MPN is a tangent to the circle at P. Angle MPS = 58°, angle PSR = 74° and angle QPN = 27°. (i) Find angle PRS. Angle PRS = ................................................ [1] (ii) Find angle PQR. Angle PQR = ................................................ [1] (iii) Find angle RPQ. Angle RPQ = ................................................ [2] (c) N C B NOT TO SCALE O 34° M A T A, B and C lie on a circle, centre O, with diameter AC. TAM and TBN are tangents to the circle and angle ATO = 34°. Using values and geometrical reasons, complete these statements to show that CB is parallel to OT. In triangles AOT and BOT, OT is common. Angle OAT = angle OBT = 90° because .................................................................................... ..................................................................................................................................................... AT = BT because ........................................................................................................................ ..................................................................................................................................................... Triangle AOT is congruent to triangle BOT because of congruence criterion ........................... Angle AOT = angle BOT = 56° because angles in a triangle add up to 180°. Angle BOC = ...................... ° because ...................................................................................... Angle OBC = ...................... ° because ...................................................................................... ..................................................................................................................................................... CB is parallel to OT because ...................................................................................................... [6]

Mark scheme: 4(a) 246 3 B2 for BCS(outh) = 66 or BCA = 48 and ACN(orth) = 66 or BCW(est) = 24 or ACS(outh) = 114 or B1 for ABC = 66 or BAC = 66 or BCA = 48 or ACN(orth) = 66 4(b)(i) 58 1 4(b)(ii) 106 1 4(b)(iii) 47 2 B1 for PRQ = 27 or B1FT for SPR, either = 48 or = 106 – their (b)(i) or B1FT for RPQ = their (b)(i) – 11 4(c) Radius perpendicular to tangent 1 Tangents to circle from a/same point oe 1 RHS 1 68 angles on a [straight] line add up/sum to 1 180 oe 56 [base angles of] isosceles triangle 1 OBC = BOT Alternate angles 1 Angles and reason required and dependent on OBC and BOT correct

More questions on Geometrical terms

5 (a) X Y NOT TO SCALE 4.8 m A 20.4° C 5.6 m B ABC is a scalene triangle on horizontal ground. AYX is a straight vertical post, held in place by two straight wires XB and YC. AC = 4.8 m, BC = 5.6 m and angle ACB = 20.4°. (i) Calculate AB. AB = ............................................ m [3] (ii) Angle XBA = 64°. Calculate AX. AX = ............................................ m [2] (iii) AY = 2.9 m. Calculate the area of triangle YAC. ........................................... m2 [2] (b) R 30° NOT TO SCALE 8 cm 75° P M Q In triangle PQR, M is the midpoint of PQ. RM = 8 cm, angle PRM = 30° and angle RMQ = 75°. Calculate PQ. PQ = ........................................... cm [5]

Mark scheme: 5(a)(i) 2[.00] or 2.002 to 2.003 nfww 3 M2 for 4.8 2  5.6 2  2  4.8  5.6  cos20.4 OR M1 for 4.82 + 5.62 – 2  4.8  5.6  cos 20.4 A1 for 4.01[17…] or 4.012 5(a)(ii) 4.1[0] or 4.11 or 4.100 to 4.107 2 AX M1 for tan64  cao their (a)(i) AX their (a)(i) or for  oe sin64 sin(90  64) 5(a)(iii) 6.96 2 1 M1 for  4.8  2.9 oe 2 5(b) 11.3 or 11.31.. 5 8 M4 for 2×  sin30 sin(45) or B4 for PM = 5.65[685...] or 5.66 or better OR B1 for angle RPM = 45° 8 M2 for  sin30 sin(their 45) or M1 for implicit form

More questions on Non-right-angled triangles

6 (a) The cumulative frequency table shows information about the speed of each of 200 cars as they pass a speed camera. Speed v G 70 v G 80 v G 90 v G 95 v G 100 v G 120 (v km/h) Cumulative 12 46 115 155 177 200 frequency (i) On the grid, draw the cumulative frequency diagram. 200 180 160 140 120 Cumulative 100 frequency 80 60 40 20 0 v 60 70 80 90 100 110 120 Speed (km/h) [3] (ii) Use your cumulative frequency diagram to find an estimate of (a) the median ........................................ km/h [1] (b) the interquartile range ........................................ km/h [2] (c) the number of cars with a speed greater than 110 km/h. ................................................. [2] (b) The frequency table shows information about the mass of each of 50 trucks. Mass 2000 1 m G 2600 2600 1 m G 3500 3500 1 m G 5000 5000 1 m G 5700 (m kg) Frequency 12 15 16 7 (i) Calculate an estimate for the mean mass of the trucks. ............................................ kg [4] (ii) In a histogram showing this information, the height of the first block is 6 cm. Calculate the heights of the remaining three blocks. Height of block for 2600 1 m G 3500 .......................................... cm Height of block for 3500 1 m G 5000 .......................................... cm Height of block for 5000 1 m G 5700 ........................................... cm [3]

Mark scheme: 6(a)(i) Correct curve 3 B1 for correct horizontal placement for 6 plots B1 for correct vertical placement for 6 plots B1 dep on at least B1 for reasonable increasing curve through their 6 points If 0 scored, SC1 for 4 out of 6 points correctly plotted 6(a)(ii)(a) 87 to 89.5 1 6(a)(ii)(b) 12.5 to 14 2 B1 for [LQ =] 80.5 to 81.5 or [UQ =] 94 to 94.5 6(a)(ii)(c) Strict FT, 200 – their cumul freq reading 2 B1FT for correct cumul freq at 110 seen from their graph at 110 given to nearest or for non-integer answer integer 6(b)(i) 3576 4 M1 for midpoints soi M1 for use of fx where x is in the  correct interval including boundaries M1 (dep on 2nd M1) for fx ÷ 50  6(b)(ii) 5 3.2 3 3 B1 for each If 0 scored, SC1 for 3 frequency densities 12 15 16 7 , , , seen oe to 3sf 600 900 1500 700 or better or multiplier 3 or 300

More questions on Histograms

7 (a) The diagram shows the graph of a function. y x O Put a ring around the word which correctly identifies the type of function. reciprocal quadratic cubic exponential linear [1] (b) (i) y x O 1 On the diagram, sketch the graph of y = , x ! 0 . [2] 2 x 1 (ii) Solve the equation = 2x . 2 x x = ........................... and x = ........................... [2] (c) (i) y 1 0 180° 360° x – 1 On the diagram, sketch the graph of y = sin x for 0° G x G 360 ° . [2] (ii) Solve the equation 3 sinx + 1 = 0 for 0° G x G 360° . x = ........................... and x = ........................... [3]

Mark scheme: 7(a) Cubic 1 7(b)(i) Correct sketch 2 B1 for one branch correct or an attempt at the correct shape y Maximum 1 mark if sketch crosses x- axis or y-axis O x 7(b)(ii) 1 2 M1 for 4 x 2  1 oe  nfww 2 1 1 or B1 for or  nfww 2 2 7(c)(i) Correct sketch through (0, 0) (180, 0) and 2 B1 for correct sine curve shape, starting (360, 0) with max and min at 1 and –1 resp. at the origin, with minimum of 1 cycle. 180 360 7(c)(ii) 199.5 or 199.47... 3 B2 for one correct 1 and or M1 for sin x =  oe 3 340.5... If 0 scored, SC1 for two reflex angles with a sum of 540 or 2 non-reflex angles with a sum of 180

More questions on Trigonometric functions

8 (a) A shop sells shirts for $x and jackets for $(x + 27). The shop sells 4 shirts and 3 jackets for a total of $194.75 . Write down and solve an equation to find the cost of one shirt. $ ................................................ [3] (b) Solve the simultaneous equations. You must show all your working. x 2 + 4y = 37 5x + y =- 8 x = ..................... , y = ..................... x = ..................... , y = ..................... [5] (c) A solid cylinder has radius x and height 6x. A sphere of radius r has the same surface area as the total surface area of the cylinder. 2 7 2 Show that r = x . 2 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]

Mark scheme: 8(a) 4x + 3(x + 27) = 194.75 M1 or 4x + 3x + 81 = 194.75 16.25 cao B2 M1 for 7x = k where k < 194.75 or B1 for answer 16.3 8(b) x 2  20 x  69[  0] oe M2 M1 for x 2  4  8 5 x   37 oe or y 2  116 y  861[  0] oe  8 y  2 or for 37  4 y   oe  5  or for x2 + 4y = 37 and 20x + 4y = –32 subtracted with no more than one error (x + 3)(x – 23) [= 0] oe M1 correct method to solve their quadratic or 2 ( 20)  ( 20)  4 1 ( 69) (y – 7)(y + 123) [= 0] oe e.g. x = 2  1 or x – 10 =  13 or x – 10 = 169 x = − 3 y = 7 B2 B1 for one correct pair or two correct x = 23 y = −123 final answer x values or two correct y values 8(c) 2x  6x + 2x2 or 2x(6x + x) M2 or M1 for 2x  6x or 2x2 Their (2x  6x + 2x2) = 4r2 M1 Dep on at least on M1 earned Their LHS must be an area in terms of x only At least one further stage of working A1 with no error seen 2 7 2 leading to r  x 2

More questions on Equations

9 (a) M A B NOT TO SCALE D C N The diagram shows a shape made from a square ABCD and two equal sectors of a circle. The square has side 11 cm. MAB and DCN are straight lines. (i) Calculate the area of the shape. ......................................... cm2 [3] (ii) Calculate the perimeter of the shape. ............................................ cm [3] (b) H G E F NOT TO SCALE D C A B The diagram shows a cube ABCDEFGH of edge 7 cm. Calculate the angle between AG and the base of the cube. ................................................. [4]

Mark scheme: 9(a)(i) 311 or 311.0 to 311.1 3 1 M2 for 11  11 + 2     112 oe 4 1 or M1 for [2 ]    112 or 11  11 4 oe 9(a)(ii) 78.6 or 78.55 to 78.56... 3 1 M2 for 4  11 + 2   2    11 oe 4 1 or M1 for [2 ]  2    11 or 4  11 4 oe 9(b) 35.2 or 35.3 or 35.239… to 35.28 4 7 M3 for [tan =] 7 2  7 2 7 or [sin =] 7 2  7 2  7 2 7 2  7 2 or [cos =] 7 2  7 2  7 2 OR M2 for AG = 7 2  7 2  7 2 2  7  2 or for 7   oe  sin 45  2 2 7 or for AC = 7  7 or oe sin45 OR M1 for 72 + 72 or for implicit trigonometry or identifying correct angle

More questions on Pythagoras’ theorem and trigonometry

10 The table shows some values for y = 2 x - 3 . x -2 -1 0 0.5 1 1.5 2 2.5 y -2.75 -1.58 -0.17 1 2.66 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 2 x - 3 for - 2 G x G 2.5 . y 3 2 1 – 2 – 1 0 1 2 x – 1 – 2 – 3 [4] (c) Use your graph to solve the equation 2 x - 3 = 2 . x = ................................................ [1] (d) By drawing a suitable straight line, solve the equation 2 x - x - 1.5 = 0 . x = ......................... or x = ......................... [4]

Mark scheme: 10(a) 2.5 2 1 3 B1 for each 10(b) Correct curve 4 B3 FT for 8 or 7 correct plots B2 FT for 6 or 5 correct plots B1 FT for 4 or 3 correct plots y O x 10(c) 2.3 to 2.4 1 10(d) ruled line y  x  1.5 M2 M1 for y = x – 1.5 soi or for 2x – 3 = x – 1.5 seen. or y  x  k or y  kx  1.5 drawn Do not accept y = –1.5 1 and 1.55 to 1.7 A2 A1 for each

More questions on Graphs of functions

11 M has coordinates (4, 1) and N has coordinates ( -2, -7). (a) Find the length of MN. ................................................. [3] (b) Find the gradient of MN. ................................................. [2] (c) Find the equation of the perpendicular bisector of MN. ................................................. [4] Question 12 is printed on the next page.

Mark scheme: 11(a) 10 3 M2 for (1 – –7)2 + (4 – –2)2 oe or M1 for (1 – –7) or (4 – –2) oe 11(b) 4 8 2 1 7 or M1 for oe 3 6 4 2 11(c) 3 9 4 3 9 y  x  B3 for  x  4 4 4 4 or 4 y  3 x  9  0 oe OR final answers B1 for midpoint (1, − 3) 3 1 M1 for gradient  or  4 their (b) M1 for substituting their (1, −3) into y = (their m)x + c or for y 3 their m = oe x  1

More questions on Perpendicular lines

12 The equation of a curve is y = x 4 - 8x 2 + 5 . d y 4 2 (a) Find the derivative, of y = x - 8x + 5 . e d x o, ................................................. [2] (b) Find the coordinates of the three turning points. You must show all your working. (........... , ...........) and (........... , ...........) and (........... , ...........) [4] (c) Determine which one of these turning points is a maximum. Justify your answer. [2]

Mark scheme: 12(a) 4 x 3  16 x cao 2 M1 for 4x 3 +kx or kx 3  16 x or 4 x 3  16 x +k or 4 x 3  16 as final answers 12b dy dy B1 Their = 0 or stating = 0 dx dx Correct method to solve their 4x3 – 16x = 0 M1 e.g. 4x(x2 – 4) or 4x(x – 2)(x + 2) oe [x =] 0, –2, 2 A1 Or B1 for (–2, –11) and (2, –11) (0, 5) (–2, –11) (2, –11) A1 12(c) (0, 5) with correct reasoning 2 M1 for any of  correct use of 2nd derivative 12x2 −16  evaluates correctly both values of y on either side  evaluates correctly the gradient on either side  reasonable correct sketch

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