Cambridge IGCSE Mathematics 0580 — 2024 May/June Paper 4 · Variant 1

1 (a) The table shows the areas, in km2, of the four largest rainforests in the world. Rainforest Area (km2) Amazon 5 500 000 Congo 2 000 000 Atlantic 1 315 000 Valdivian 250 000 (i) Find the area of the Valdivian rainforest as a percentage of the area of the Amazon rainforest. ..............................................% [1] (ii) Write, in its simplest form, the ratio of the areas of the rainforests Valdivian : Atlantic : Congo. ....................... : ...................... : ...................... [2] (iii) The Amazon rainforest has 60% of its area in Brazil and 10% of its area in Colombia. 43 1% of the remaining area of the rainforest is in Peru. 3 Find the percentage of the Amazon rainforest that is in Brazil, Colombia and Peru. ..............................................% [3] 27 (iv) The area of the Amazon rainforest represents of the total area of rainforest in the world. 50 Calculate the total area of rainforest in the world. Give your answer correct to the nearest 100 000 km2. ......................................... km 2 [3] (v) In the world, 60.7 hectares of rainforest are lost every minute. Calculate the total area, in hectares, of rainforest that is lost in 365 days. Give your answer in standard form. ................................... hectares [3] (b) The Amazon river has a length of 6440 km, correct to the nearest 10 km. The Congo river has a length of 4400 km, correct to the nearest 100 km. Calculate the upper bound of the difference between the lengths of the Amazon river and the Congo river. ........................................... km [3]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 4.55 or 4.545… 1 1(a)(ii) 50 : 263 : 400 cao 2 M1for a correct simplification from 250 000 : 1 315 000 : 2 000 000 1(a)(iii) 83 cao 3 1 433 M2 for  100  60  10  oe 100 or M1 for 100 – 60 – 10 seen 1(a)(iv) 10 200 000 cao 3 B2 for 10 185 185 to 10 185 200 or M1 for 5 500 000 ÷ 27 [× 50] 1(a)(v) 3.19 × 107 or 3.190… × 107 3 B2 for 31903920 or M1 for 60.7 × 60 × 24 × 365 If B0 scored SC1 for correctly converting their number seen to standard form to 3sf or better 1(b) 2095 nfww 3 M2 for 6445 – C where 4300 ⩽ C < 4400 oe or A – 4350 where 6440 < A ⩽ 6450 oe or M1 for 6440 +5 or 6440 –5 or 4400 + 50 or 4400 – 50 seen oe

More questions on Percentages

2 (a) y 6 4 2 x – 6 – 4 – 2 0 2 4 6 – 2 T – 4 – 6 On the grid, draw the image of (i) triangle T after a reflection in the x-axis [1] - 5 (ii) triangle T after a translation by the vector [2] e- 2o 1 (iii) triangle T after an enlargement by scale factor - with centre - ,1 1 [2] 2 ` j. (b) A shape P is enlarged by scale factor 3 to give shape Q. 2 Shape Q is then enlarged by scale factor to give shape R. 5 The area of shape P is 10 cm2. Calculate the area of shape R. ......................................... cm 2 [3]

Mark scheme: 2(a)(i) Triangle at (2, 1) (1, 3) (5, 3) 1 2(a)(ii) Triangle at (–4, –5) (–3, –3) 2   5   k  B1 for translation by   or   (0, –5)  k    2  2(a)(iii) Triangle at (–2.5, 2) (–4, 3) 2 (–2, 3) 1 B1 for enlargement by sf  with any 2 centre 2(b) 14.4 3 2  2  M2 for [10 ×] 32 ×   oe  5   2  2 or M1 for 32 or   soi  5 

More questions on Transformations

1 23 (a) C = xy 4 (i) Find C when x = 5 and y = 8 . C = ................................................ [2] (ii) Find the positive value of y when C = 15 and x = 2.4 . y = ................................................ [2] (b) Write as a single fraction in its simplest form. 4 3 - x - 1 2x + 5 ................................................. [3] (c) Expand and simplify. 2 2x + 3 4 - x ` `j j ........................................................................................ [3] (d) Simplify. 8 - 43 y 16 f 16x p ................................................. [3]

Mark scheme: 3(a)(i) 80 2 1 2 M1 for 5 8 4 3(a)(ii) 5 2 2 15  4 M1 for [ y  ] oe 2.4 3(b) 5 x  23 5 x  23 3 or final ( x  1)(2 x  5) 2 x 2  3 x  5 B1 for 4(2x + 5) –3(x – 1) oe isw answer B1 for common denominator = (x – 1) (2x + 5) oe isw 3(c) 2x3 –13x2 + 8x + 48 final answer 3 B2 for correct expansion of 3 brackets but unsimplified or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 3(d) 8x12 12 6 3 B2 for two elements correct in final or 8 x y final answer 6 answer y or for correct answer seen then spoiled or for correct expression where all parts of the power have been dealt with 3 1  2 x 4  or for  or  2   y  or B1 for 8 or y6 or y 6 or x12 correct in final answer 3 3 4  y 2   16 x16  or for  8  or  4   y   2 x 

More questions on Introduction to algebra

4 (a) Jianyu records the time, in seconds, that some cars take to travel 195 m. The box and whisker plot shows this information. 5 6 7 8 9 10 11 12 13 14 Time (s) (i) Find the median time. ............................................... s [1] (ii) Find the interquartile range. ............................................... s [1] (iii) Find the difference between the average speed of the fastest car and the average speed of the slowest car. Give your answer in kilometres per hour. ........................................ km/h [5] (b) Matilda records the distances that 80 different cars can travel with a full tank of fuel. The table shows this information. Distance (d km) 250 1 d G 300 300 1 d G 400 400 1 d G 420 420 1 d G 450 450 1 d G 500 Frequency 7 13 19 21 20 (i) Write down the class interval that contains the median. ................... 1 d G ................... [1] (ii) Calculate an estimate of the mean. ........................................... km [4] (iii) A histogram is drawn to show the information in the table. The height of the bar for the interval 250 1 d G 300 is 2.8 cm. Calculate the height of the bar for each of the following intervals. 300 1 d G 400 ........................................... cm 400 1 d G 420 ........................................... cm 420 1 d G 450 ............................................ cm [3] (iv) Two of the 80 cars are chosen at random. Find the probability that, with a full tank of fuel, one of the cars can travel more than 450 km and the other car can travel not more than 300 km. ................................................. [3]

Mark scheme: 4(a)(i) 9.3 1 4(a)(ii) 3.4 1 4(a)(iii) 63 5 195 3600 195 3600 M4 for    oe 6 1000 13 1000 195 3600 195 3600 or M3 for  oe or  6 1000 13 1000 oe 195 or for (  195) k  oe 6 13 OR 195 195 M1 for or or their speed 6 13 3600  seen 1000 M1 for selecting 6 and 13 4(b)(i) 420 < d ⩽ 450 1 4(b)(ii) 411.25 4 M1 for 275, 350, 410, 435, 475 soi M1 for fx M1 dep for their fx ÷ 80 4(b)(iii) 2.6 3 B1 for each 19 If 0 scored, SC1 for 3 of 0.14, 0.13, 0.95 14 or 0.7 oe 4(b)(iv) 7 3 oe 158 7 M2 for  2  20  oe 80 79 20 7 7 20 or M1 for or or or oe 80 79 80 79 seen 7 After 0 scored, SC1 for oe 160

More questions on Averages and measures of spread

5 (a) P is the point (1, 7). Q is the point (5, –5). y P NOT TO SCALE O x Q (i) Find PQ . PQ = [2] f p (ii) Show that OP = OQ . [3] (iii) PQ is a chord of a circle with centre O. Calculate the circumference of this circle. ................................................. [2] (iv) PQ is the diameter of a different circle with centre R. Find the coordinates of R. ( ...................... , ...................... ) [2] (v) Find the equation of the perpendicular bisector of PQ. Give your answer in the form y = mx + c . y = ................................................ [4] (b) The position vector of A is a. The position vector of B is b. M is a point on AB such that AM : MB = 2 : 3. Find, in terms of a and b, the position vector of M. Give your answer in its simplest form. ................................................. [4]

Mark scheme: 5(a)(i)  4  2 B1 for each    12  5(a)(ii) 12 + 72 M1 52 + ([–]5)2 M1 Both 50 oe A1 With no errors seen If M0M0A0 scored SC1 for 50 oe for each 5(a)(iii) 44.4 or 44.42[8…] to 44.435 2 FT their (a)(ii) correct to 3sf or better M1 for 2 ×  × their 50 oe 5(a)(iv) (3, 1) 2 B1 for each 5(a)(v) 1 4 [y =] x B3 for a correct equation in the wrong 3 form as final answer Or B2 for 1/3 stated or used as perpendicular gradient OR 7 5 M1 for [grad PQ] = oe 1  5 1 M1 for their grad PQ M1dep for substituting their(a)(iv) or (0,0) into y = their mx + c oe dep on the 2nd M1 or B2 5(b) 3 2 4 a + b final answer 5 5 B3 for an unsimplified correct answer 2 or B2 for AM   b  a  soi 5 3 or B M   a  b  soi 5 or B1 for AB = b – a or BA = a – b or for a correct route for OM or for correct diagram

More questions on Coordinates

6 North B C 11 km 25° NOT TO A SCALE 32 km 14 km 55° H The diagram shows the positions of two lighthouses A and B, a boat C and a harbour H . C is due east of B. (a) Find the bearing of the harbour from boat C. ................................................. [1] (b) (i) Show that angle CBH = 100c . [1] (ii) Show that BH = 13.7 km, correct to 1 decimal place. [3] (c) Calculate the bearing of A from B. ................................................. [5] (d) At 1 pm boat C sails 32 km directly to the harbour at a speed of 10 knots. (i) Calculate the time when boat C arrives at the harbour. Give this time correct to the nearest minute. [1 knot = 1.852 km/h] ................................................. [4] (ii) Calculate the distance of boat C to the harbour when boat C is at the shortest distance from lighthouse B. ........................................... km [3]

Mark scheme: 6(a) 245 1 6(b)(i) 180 – (55 + 25) [=100] M1 6(b)(ii) 32  sin 25 M2 sin 25 sin100 oe M1 for  oe sin100 BH 32 13.73… A1 6(c) 258 or 257.9 to 258.0… 5 B4 for 67.9 to 68.0… OR  112  13.7 2  14 2  M2 for [cos =]   2  13.7  11   A1 for 0.3738 to 0.376 or M1 for 142 = 112 + 13.72 – 2× 11 × 13.7 × cos B M1dep on at least M1 for 190 + their angle B 6(d)(i) 2 44 pm or 14 44 cao 4 B3 for 1 hour 44 or 1 hour 43.6 to 1 hour 43.8 or 104 or 103.6 to 103.8 or B2 for 1.727 to 1.73 32 or M2 for  60 10  1.852 or M1 for 32 ÷ (10 × 1.852) 6(d)(ii) 7.857 to 7.88 3 x M2 for  cos55 oe 13.7 or M1 for dist to H occurs when perpendicular from B meets CH soi

More questions on Right-angled triangles

7 (a) NOT TO SCALE 40 cm 30 cm 70 cm The diagram shows a box in the shape of a cuboid. The box is open at the top. (i) Work out the surface area of the inside of the open box. ......................................... cm 2 [3] (ii) Cylinders with height 20 cm and diameter 15 cm are placed in the box. Work out the maximum number of these cylinders that can completely fit inside the box. ................................................. [3] (b) A solid bronze cone has a mass 750 g. The density of the bronze is 8.9 g/cm3. The ratio radius of cone : height of cone = 1 : 3. (i) Show that the radius of the cone is 2.99 cm, correct to 3 significant figures. [Density = mass ÷ volume] 1 2 [The volume, V , of a cone with radius r and height h is V = rr h .] 3 [4] (ii) Calculate the total surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] ......................................... cm 2 [5]

Mark scheme: 7(a)(i) 10 100 3 M2 for 30  70  2  30  40  2  40  70 or M1 for 30 × 40 or 30 × 70 or 40 × 70 7(a)(ii) 16 3 M2 for 2 fit width, 2 fit height and 4 fit length soi or M1 for 70, 30 or 40 ÷ 15 or 20 7(b)(i) 1 2 M2 M1 for using 750 and 8.9 correctly in πr  3r = their (750 ÷ 8.9 ) oe v  m / d oe or 750 ÷ 8.9 3 their  750  8.9  M1dep r3 = oe  r = 2.993… A1 7(b)(ii) 117 or 116.9 to 117.2 5 M4 for π  2.99 2  π  2.99  2.99 2  (3  2.99) 2 oe or M3 for π  2.99  2.99 2  (3  2.99) 2 or M2 for 2.99 2  (3  2.99) 2 or M1 for 2.992 + (3 × 2.99)2 or for π  2.99 2

More questions on Compound shapes and parts of shapes

8 (a) On the axes, sketch the graph of y = x 2 + 7x - 18 . On your sketch, write the values where the graph meets the x-axis and the y-axis. y O x [4] (b) (i) Find the derivative of y = x 2 - 3x - 28 . ................................................. [2] (ii) Find the coordinates of the turning point of y = x 2 - 3x - 28 . ( ...................... , ...................... ) [3] (c) The line y = 5 - 2x intersects the graph of y = x 2 - 3x - 28 at point P and point Q. Find the coordinates of P and Q. You must show all your working and give your answers correct to 2 decimal places. ( ...................... , ...................... ) ( ...................... , ...................... ) [6]

Mark scheme: 8(a) Correct sketch with roots indicated at 4 B1 for U shaped parabola x = –9 and x = 2 and y intercept = –18 Minimum should be in 3rd quadrant B2 for roots at –9 and 2 on diagram or M1 for (x + 9) (x – 2) [= 0] B1 for y – intercept at –18 on diagram Maximum 3 marks if sketch not fully correct 8(b)(i) 2x – 3 2 B1 for 2x + k or kx[p] – 3 8(b)(ii) (1.5, –30.25) oe 3 B2 for x = 1.5 or M1 for their (b)(i) = 0 or for (x – 1.5)2 8(c) x2 – x – 33 [ = 0] seen B1 2 B2FT FT their quadratic dep on no factors [  ]1   [  ]1  4 1 33  oe 2  1 2 B1 for  [  ]1  4 1  33  or better [  ]1  q [  ]1  q or B1 for oe or 2(1) 2(1) oe –5.27 or –5.267 to –5.266 and B2 B1 for each 6.27 or 6.266 to 6.267 If 0 scored, SC1 for –6.27 and 5.27 (–5.27, 15.53 or 15.54) and B1 (6.27, –7.53 or 7.54)

More questions on Equations

9 f ( )x = 4 x + 1 g ( )x = 6 - 2 x h ( )x = 3 x - 2 (a) Find (i) f ( 3) ................................................. [1] (ii) gf ( 3) . ................................................. [1] (b) Find g -1 ( )x . g -1 ( )x = ................................................. [2] (c) Find x when f ( x) = g ( 2x - 7) . x = ................................................. [4] (d) Find the value of hh(2). ................................................. [2] (e) Find x when h -1 ( )x = 10 . x = ................................................. [2]

Mark scheme: 9(a)(i) 13 1 9(a)(ii) –20 1 FT 6 – 2(their (a)(i)) 9(b) 6 x 2 M1 for correct first step oe final answer 2 y x = 6 – 2y, y – 6 = – 2x,  3  x 2 9(c) 2.375 oe 4 B1 for 6 – 2(2x – 7) oe B1 for 4x + 1 = 6 – 4x + 14 M1 for 8x = 19 FT their linear equation rearranged correctly from ax  b  cx  d to form ex = f 9(d) 1 2 M1 for h(1) or 3^(3x-2 – 2) or 3^(32-2 – 2) or 0.333… or better 3 9(e) 6561 2 M1 for 310–2 or x = h(10)

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