Questions as text
Q1 · NOT TO SCALE l 0.7 cm h 16.5 cm 1.5 cm The diagram shows a pencil of length 18 cm 0580/41 May/June 2005
6 NOT TO SCALE l 0.7 cm h 16.5 cm 1.5 cm The diagram shows a pencil of length 18 cm. It is made from a cylinder and a cone. The cylinder has diameter 0.7 cm and length 16.5 cm. The cone has diameter 0.7 cm and length 1.5 cm. (a) Calculate the volume of the pencil. [The volume, V, of a cone of radius r and height h is given by V = 1 πr 2h.] [3] 3 (b) NOT TO SCALE 18 cm x cm w cm Twelve of these pencils just fit into a rectangular box of length 18 cm, width w cm and height x cm. The pencils are in 2 rows of 6 as shown in the diagram. (i) Write down the values of w and x. [2] (ii) Calculate the volume of the box. [2] (iii) Calculate the percentage of the volume of the box occupied by the pencils. [2] (c) Showing all your working, calculate (i) the slant height, l, of the cone, [2] (ii) the total surface area of one pencil, giving your answer correct to 3 significant figures. [The curved surface area, A, of a cone of radius r and slant height l is given by A = πrl .] [6]
17 marks
Mark scheme: 6 (a) Vol of cyl.= π × 0.352 × 16.5 (6.3…) M1 Use of radius = 0.7 loses all marks in (a) 0 . 35 2 M1 After that they can revert to 0.35 without Vol of cone = π x x 1.5 (0.19...) penalty 3 a.r.t. 6.54 (cm3) A1 Any later use of 0.7 after 0.35 penalty 2 from the marks gained using 0.7 (b)(i) 4.2 B1 8.4 B1 1.4 B1 2.8 B1 (ii) 18 × their 4.2 × their 1.4 M1 18 × their 8.4 × their 2.8 M1 106 (cm3) (105.84) A1 423 (cm3) (423.36) A1 (iii) 12 × their (a) ×100 M1 12 × their (a) ×100 M1 their (b)(ii) their (b)(ii) 74.(0) to 74.2 (%) c.a.o. A1 74.1 to 74.3 (%) A1 (c)(i) (l =) √( 1.52 + 0.352) M1 (l =) √( 1.52 + 0.72) M1 1.54 (cm) A1 1.66 (cm) A1 (ii) Circle = π × 0.352 M1 Circle = π × 0.72 M1 Cylinder = 2 × π × 0.35 × 16.5 M1 Cylinder = 2 × π × 0.7 × 16.5 M1 Cone = π × 0.35 × their (c)(i) M1 Cone = π × 0.7 × their (c)(i) M1 Any 2 correct areas B2 Any 2 correct areas B2 ( a.r.t. 0.385 a.r.t. 36.3 a.r.t. 1.69 ) (a.r.t. 1.54 72.5 to 72.6 a.r.t. 3.65) 0.1225π 11.55π 0.539π 0.49π 23.1π 1.162π 38.3 to 38.4 (cm2) c.a.o. A1 77.7 to 77.8 (cm2) A1 17 IGCSE – JUNE 2005 0580/0581 4
This question in 0580/41 May/June 2005
Q2 · P D C 3 cm NOT TO 5 cm SCALE M F A 6 cm B The diagram shows a pyramid on a rectangular… 0580/41 Oct/Nov 2005
6 P D C 3 cm NOT TO 5 cm SCALE M F A 6 cm B The diagram shows a pyramid on a rectangular base ABCD, with AB = 6 cm and AD = 5 cm. The diagonals AC and BD intersect at F. The vertical height FP = 3 cm. (a) How many planes of symmetry does the pyramid have? [1] (b) Calculate the volume of the pyramid. 1 [The volume of a pyramid is × area of base × height.] [2] 3 (c) The mid-point of BC is M. Calculate the angle between PM and the base. [2] (d) Calculate the angle between PB and the base. [4] (e) Calculate the length of PB. [2]
11 marks
Mark scheme: 6 (a) 2 B1 (b) 1 × 6 × 5 × 3 o.e. M1 3 30 A1 (c) 3 M1 Isos. triangle or invtan ( ) o.e. 3 45 A1 www2 (d) 2 2 M1 (BD ) = 6 + 5 o.e. M1 Dep. (BF = 3.905….) BF = 1 BD 2 3 M1 Dep on previous method angle = invtan their BF 37.5 to 37.54 A1 www4 (e) (l2) = 32 + (their FB)2 o.e. M1 Not for FB = 3 4.92 to 4.93 A1 ww2 [11] IGCSE – NOVEMBER 2005 0580/0581 4
This question in 0580/41 Oct/Nov 2005
Q3 · NOT TO B C SCALE B C 12 cm 18 cm A O 12 cm A 40 cm E 22 cm D E D Diagram 1 Diagram 2… 0580/41 May/June 2006
2 NOT TO B C SCALE B C 12 cm 18 cm A O 12 cm A 40 cm E 22 cm D E D Diagram 1 Diagram 2 Diagram 1 shows a closed box. The box is a prism of length 40 cm. The cross-section of the box is shown in Diagram 2, with all the right-angles marked. AB is an arc of a circle, centre O, radius 12 cm. ED = 22 cm and DC = 18 cm. Calculate (a) the perimeter of the cross-section, [3] (b) the area of the cross-section, [3] (c) the volume of the box, [1] (d) the total surface area of the box. [4]
11 marks
Mark scheme: 2 (a) π × 24 M1 Arc length = (18.8…) 4 Perimeter = 6 + 22 + 18 + 10 + their arc M1 74.8 to 74.9 (cm) A1 (b) π × 122 M1 Sector area = (113. …) 4 Area = (6 x 22) + (12 x 10) + their sector o.e. M1 365 to 365.2 (cm2) A1 (c) 14600 to 14605 (cm3) B1 (d) their (b) x 2 M1 indep. their (a) x 40 M1 indep. Addition M1 dep. 3720 to 3730 (cm2) A1 11 IGCSE – May/June 2006 0580 and 0581 04
This question in 0580/41 May/June 2006
Q4 · NOT TO SCALE 0.8 m 0.3 m 1.2 m The diagram shows water in a channel 0580/41 May/June 2007
7 NOT TO SCALE 0.8 m 0.3 m 1.2 m The diagram shows water in a channel. This channel has a rectangular cross-section, 1.2 metres by 0.8 metres. (a) When the depth of water is 0.3 metres, the water flows along the channel at 3 metres/minute. Calculate the number of cubic metres which flow along the channel in one hour. [3] (b) When the depth of water in the channel increases to 0.8 metres, the water flows at 15 metres/minute. Calculate the percentage increase in the number of cubic metres which flow along the channel in one hour. [4] (c) The water comes from a cylindrical tank. When 2 cubic metres of water leave the tank, the level of water in the tank goes down by 1.3 millimetres. Calculate the radius of the tank, in metres, correct to one decimal place. [4] (d) When the channel is empty, its interior surface is repaired. This costs $0.12 per square metre. The total cost is $50.40. Calculate the length, in metres, of the channel. [4]
15 marks
Mark scheme: 7 (a) 1.2 × 0.3 × 3 oe M1 (1.08) or 3 × 60 (180) × 60 oe M1dep × 1.2 × 0.3 (0.36) 64.8 cao A1 www 3 (b) 1.2 × 0.8 × 15 × 60 oe (= 864 seen) M1 Their (a) 8 3 × 5 oe seen Their 864 – their (a) M1ind or their 864 ÷ their (a) × 100 (1333.3..) ÷ their (a) × 100 M1dep subtract 100 (Dep on second M1) 1230 (%) or better (1233.3…) cao A1 www 4 (1330 or 1333.3…www M1M1M0) (c) πr2× figs13 = figs 2 oe M1 2 ÷ 0.0013 M1ind (implied by 1538.46…) 2 2 ( r ) = oe M1dep Dep on M2 (489.7..) π × .00013 22.1 or 22.12 – 22.14 cao A1 www 4 figs 221… imply first M1 (d) 0.8 + 1.2 + 0.8 = (2.8) M1 Accept 2.8 seen 50.40 = area × 0.12 oe M1ind Accept 420 seen Length × their perimeter = their area oe M1 150 cao A1 www 4 [15]
This question in 0580/41 May/June 2007
Q5 · [The surface area of a sphere of radius r is 4πr2 and the volume is 4 πr 3 .] 3 (a) A… 0580/41 Oct/Nov 2007
4 [The surface area of a sphere of radius r is 4πr2 and the volume is 4 πr 3 .] 3 (a) A solid metal sphere has a radius of 3.5 cm. One cubic centimetre of the metal has a mass of 5.6 grams. Calculate (i) the surface area of the sphere, [2] (ii) the volume of the sphere, [2] (iii) the mass of the sphere. [2] (b) NOT TO SCALE h 8 cm 16 cm 16 cm Diagram 1 Diagram 2 Diagram 1 shows a cylinder with a diameter of 16 cm. It contains water to a depth of 8 cm. Two spheres identical to the sphere in part (a) are placed in the water. This is shown in Diagram 2. Calculate h, the new depth of water in the cylinder. [4] (c) A different metal sphere has a mass of 1 kilogram. One cubic centimetre of this metal has a mass of 4.8 grams. Calculate the radius of this sphere. [3]
13 marks
Mark scheme: 4 (a) (i) 4 π 5.3 2 = 153.86 to 153.96 or 154 M1A1 www2 (ii) 4 3 π 5.3 3 = 179.5 to 179. 62 or 180 M1A1 www2 (iii) their (ii)× 5.6 M1 1005 to 1006 or 1008or 1010 (g) A1ft their (ii)× 5.6 correct to 3sf or better (allow in kg) (b) π 82 × 8 (1608-1609) M1 Alt π 8 2 d = 2 × their (ii) M1 2 M1dep 2 π 8 h = 2×their (ii) + π 82 × 8 (2×their (a)(ii)) ÷ (π 8 ) M1dep (2×their (ii) + π 82 × 8 ) ÷( π 8 2 ) M1dep add 8 M1dep 9.78 to 9.79 (cm) A1 www4 (c) 1000 (or 1) ÷4.8 ÷ 43 π M1 49.7….. (or 0.0497) 3 3 M1dep Dep on previous M1 ans (or 10 × ans ) 3.67 to 3.68 (cm) A1 www3 figs 368 or ans 3.7 gets M2 [13] IGCSE – October/November 2007 0580/0581 04
This question in 0580/41 Oct/Nov 2007
Q6 · For Examiner's NOT TO Use SCALE x cm x cm 250 cm A solid metal bar is in the shape of a… 0580/41 May/June 2009
7 For Examiner's NOT TO Use SCALE x cm x cm 250 cm A solid metal bar is in the shape of a cuboid of length of 250 cm. The cross-section is a square of side x cm. The volume of the cuboid is 4840 cm3. (a) Show that x = 4.4. Answer (a) [2] (b) The mass of 1 cm3 of the metal is 8.8 grams. Calculate the mass of the whole metal bar in kilograms. Answer(b) kg [2] (c) A box, in the shape of a cuboid measures 250 cm by 88 cm by h cm. 120 of the metal bars fit exactly in the box. Calculate the value of h. Answer(c) h = [2] (d) One metal bar, of volume 4840 cm3, is melted down to make 4200 identical small spheres. For Examiner's All the metal is used. Use (i) Calculate the radius of each sphere. Show that your answer rounds to 0.65 cm, correct to 2 decimal places. 4 3 [The volume, V, of a sphere, radius r, is given by V = πr .] 3 Answer(d)(i) [4] (ii) Calculate the surface area of each sphere, using 0.65 cm for the radius. [The surface area, A, of a sphere, radius r, is given by A = 4 πr 2 .] Answer(d)(ii) cm2 [1] (iii) Calculate the total surface area of all 4200 spheres as a percentage of the surface area of the metal bar. Answer(d)(iii) % [4]
15 marks
Mark scheme: 7 (a) 250x2 = 4840 o.e. M1 Allow M1 for 250 × 4.42 = 4840 x² = 19.36 or (x =) 4840 ÷ 250 (= 4.4) E1 Then E1 for 250 × 19.36 = 4840 (b) 42.6 (kg) cao (42.592 or 42.59) B2 SC1 for figures 426 or 4259… (c) 26.4 (cm) c.a.o. B2 If B0, M1 for any of following 88 ÷ 4.4 = 20 and 120 ÷ 20 = 6 (accept 6 bars high o.e.) or 88h = 4.42 × 120 or 250 × 88 × h = 120 × 4840 (d) (i) 4840 ÷ 4200 (implied by 1.15(2)) M1 4200 × 4 3 π r3 = 4840 ÷ 4 3 π (implied by 0.274 to 0.276) M1 (r3 =) 4840 ÷ (4200 × 4 3 π ) 3 (seen or implied by correct answer to M1 3 Third M dependent on M1M1 dep more than 2 dp) 0.649 – 0.651 A1 Must be 3dp or better (ii) 5.31 (5.306 – 5.31) (cm2) B1 4200 × their (ii) (iii) × 100 M3 If M0, M1 for 4200 × their (ii) (22299) 2 × 4.4 2 + 4 × 4.4 × 250 and M1 (independent) for correct method for surface area of solid cuboid (4438.72) 501.9 – 503 (%) c.a.o. www4 A1 [15]
This question in 0580/41 May/June 2009
Q7 · For A Examiner's NOT TO Use SCALE 60° O 24 cm B (a) The sector of a circle, centre O… 0580/41 Oct/Nov 2009
7 For A Examiner's NOT TO Use SCALE 60° O 24 cm B (a) The sector of a circle, centre O, radius 24 cm, has angle AOB = 60°. Calculate (i) the length of the arc AB, Answer(a)(i) cm [2] (ii) the area of the sector OAB. Answer(a)(ii) cm2 [2] (b) The points A and B of the sector are joined together to make a hollow cone as shown in the diagram. The arc AB of the sector becomes the circumference of the base of the cone. O NOT TO SCALE 24 cm A B Calculate For Examiner's (i) the radius of the base of the cone, Use Answer(b)(i) cm [2] (ii) the height of the cone, Answer(b)(ii) cm [2] (iii) the volume of the cone. 1 [The volume, V, of a cone of radius r and height h is V = πr2h.] 3 Answer(b)(iii) cm3 [2] (c) A different cone, with radius x and height y, has a volume W. Find, in terms of W, the volume of (i) a similar cone, with both radius and height 3 times larger, Answer(c)(i) [1] (ii) a cone of radius 2x and height y. Answer(c)(ii) [1]
12 marks
Mark scheme: 7 (a) (i) 60 × π × 2 × 24 oe M1 360 A1 Accept 8 π 25.1 (25.12 to 25.14) www2 (ii) 60 × π × 242 oe M1 360 A1 Accept 96 π 301 or 302 or 301.4 to 301.7 www2 (b) (i) πd = their (a) (i) oe M1 4 (3.99 – 4.01) cao www2 A1 (ii) 242 – (their radius)2 M1 Alt trig method for h explicit 23.7 (23.66 to 23.67) cao www2 A1 Accept 560 2, 140 4, 35 (iii) 1 × π × (their r)2 × (their h) M1 Not for h = 24 3 394 – 398 cao www2 A1 (c) (i) 27W B1 (ii) 4W B1 If B0, B0 in (c), SC1 for 27 and 4 alone [12]
This question in 0580/41 Oct/Nov 2009
Q8 · Calculate the volume of a cylinder of radius 31 centimetres and length 15 metres 0580/41 May/June 2010
7 (a) Calculate the volume of a cylinder of radius 31 centimetres and length 15 metres. Examiner's Give your answer in cubic metres. Use Answer(a) m3 [3] (b) A tree trunk has a circular cross-section of radius 31 cm and length 15 m. One cubic metre of the wood has a mass of 800 kg. Calculate the mass of the tree trunk, giving your answer in tonnes. Answer(b) tonnes [2] (c) NOT TO plastic SCALE sheet D C E The diagram shows a pile of 10 tree trunks. Each tree trunk has a circular cross-section of radius 31 cm and length 15 m. A plastic sheet is wrapped around the pile. C is the centre of one of the circles. CE and CD are perpendicular to the straight edges, as shown. For (i) Show that angle ECD = 120°. Examiner's Use Answer(c)(i) [2] (ii) Calculate the length of the arc DE, giving your answer in metres. Answer(c)(ii) m [2] (iii) The edge of the plastic sheet forms the perimeter of the cross-section of the pile. The perimeter consists of three straight lines and three arcs. Calculate this perimeter, giving your answer in metres. Answer(c)(iii) m [3] (iv) The plastic sheet does not cover the two ends of the pile. Calculate the area of the plastic sheet. Answer(c)(iv) m2 [1]
13 marks
Mark scheme: 7 (a) 4.53 or 4.526 – 4.530…. 3 SC2 for figs 453 or 4526 – 4530 If SC0, M1 for π × (figs 31)2 × 15 (b) 3.62 to 3.624 ft 2ft M1 for their (a) × figs 8 oe (c) (i) 360 – 2 × 90 – 60 oe 2 E2 The 90’s and the 60 must be clearly justified. Accept in diagram. SC1 for 60 or two 90’s soi in correct positions oe e.g 360 ÷ 3 scores 0 (ii) 0.649 (0.6492 to 0.6493) 2 M1 for π × figs 62 ÷ 3 (iii) 7.53 (7.527 or 7.528….) 3 M1 for their (ii) × 3 M1 (indep) for 18 × figs 31 This M is spoiled by extra lengths. (iv) 112.9 to 113 ft 1ft ft their (iii) × 15
This question in 0580/41 May/June 2010
Q9 · A spherical ball has a radius of 2.4 cm 0580/42 May/June 2010
6 A spherical ball has a radius of 2.4 cm. Examiner's Use (a) Show that the volume of the ball is 57.9 cm3, correct to 3 significant figures. 4 3 [The volume V of a sphere of radius r is V = πr . ] 3 Answer(a) [2] (b) NOT TO SCALE Six spherical balls of radius 2.4 cm fit exactly into a closed box. The box is a cuboid. Find (i) the length, width and height of the box, Answer(b)(i) cm, cm, cm [3] (ii) the volume of the box, Answer(b)(ii) cm3 [1] (iii) the volume of the box not occupied by the balls, Answer(b)(iii) cm3 [1] (iv) the surface area of the box. Answer(b)(iv) cm2 [2] For (c) Examiner's Use NOT TO SCALE The six balls can also fit exactly into a closed cylindrical container, as shown in the diagram. Find (i) the volume of the cylindrical container, Answer(c)(i) cm3 [3] (ii) the volume of the cylindrical container not occupied by the balls, Answer(c)(ii) cm3 [1] (iii) the surface area of the cylindrical container. Answer(c)(iii) cm2 [3]
16 marks
Mark scheme: 4 3 6 (a) π × 4.2 M1 Must see method 3 57.87 – 57.92 to at least 4 figures A1 (b) (i) 14.4, 9.6, 4.8 1, 1, 1 Any order (ii) 664 (663.5 – 663.6) ft 1ft (iii) 315 or 316 or 317 (315.2 – 316.8) ft 1ft ft their (b)(ii) – 6 × ‘57.9’ (only if positive) (iv) 507 (506.8 – 506.9) ft 2ft M1 for (14.4 × 9.6 + 14.4 × 4.8 + 9.6 × 4.8) × 2 or their 3 lengths. IGCSE – May/June 2010 0580 42 (c) (i) Height seen or implied as 6 × 4.8 M1 or better π × 2.42 × their height M1 Indep 521 (520.8 – 521.3) www 3 A1 (ii) 174 or 173 (173.2 – 174.1) ft 1ft ft their (c)(i) – 6 × ‘57.9’ only if positive (iii) 470 – 471 cao www 3 3 M1 for 2 × π × 2.42 (36.17 to 36.2), and M1 indep for π × 4.8 × their height from (c)(i)
This question in 0580/42 May/June 2010
Q10 · Examiner's Use NOT TO SCALE 3 cm 6 cm 10 cm A solid metal cuboid measures 10 cm by 6 cm… 0580/43 May/June 2010
8 Examiner's Use NOT TO SCALE 3 cm 6 cm 10 cm A solid metal cuboid measures 10 cm by 6 cm by 3 cm. (a) Show that 16 of these solid metal cuboids will fit exactly into a box which has internal measurements 40 cm by 12 cm by 6 cm. Answer(a) [2] (b) Calculate the volume of one metal cuboid. Answer(b) cm3 [1] (c) One cubic centimetre of the metal has a mass of 8 grams. The box has a mass of 600 grams. Calculate the total mass of the 16 cuboids and the box in (i) grams, Answer(c)(i) g [2] (ii) kilograms. Answer(c)(ii) kg [1] For (d) (i) Calculate the surface area of one of the solid metal cuboids. Examiner's Use Answer(d)(i) cm2 [2] (ii) The surface of each cuboid is painted. The cost of the paint is $25 per square metre. Calculate the cost of painting all 16 cuboids. Answer(d)(ii) $ [3] (e) One of the solid metal cuboids is melted down. Some of the metal is used to make 200 identical solid spheres of radius 0.5 cm. Calculate the volume of metal from this cuboid which is not used. 4 [The volume, V, of a sphere of radius r is V = π r 3.] 3 Answer(e) cm3 [3] (f) 50 cm3 of metal is used to make 20 identical solid spheres of radius r. Calculate the radius r. Answer(f) r = cm [3]
17 marks
Mark scheme: (iv) 8 ft 1ft ft a positive root –4 if positive answer − ( −)1 ± ( −)1 2 − 41()( −4) 2 (c) 2 B1 for ( −)1 − 41()( −4) or better 2 )1( p + q p − q If in form or r r then B1 for –(–1) and 2(1) or better Brackets and full line may be implied later –1.56, 2.56 2 B1 B1 If B0, SC1 for –1.6 or –1.562 to –1.561 and 2.6 or 2.561 to 2.562 10 (a) Dots all correctly placed in Diagram 4 1 (b) Column 4 16, 25, 16, 41 7 B2 or B1 for three correct Column 5 25, 41, 20, 61 B2 or B1 for three correct Column n: n2, 4n, n2 + (n + 1)2 oe B1 B1 B1 oe likely to be (n –1)2 + n2 + 4n or 2n2 + 2n + 1 After any correct answer for column n, apply isw (c)(i) 79 601 cao 1 (ii) 800 ft 1ft ft their 4n linear expression only
This question in 0580/43 May/June 2010
Q11 · For Examiner's 4 cm Use NOT TO SCALE 13 cm The diagram shows a cone of radius 4 cm and… 0580/42 Oct/Nov 2010
4 (a) For Examiner's 4 cm Use NOT TO SCALE 13 cm The diagram shows a cone of radius 4 cm and height 13 cm. It is filled with soil to grow small plants. Each cubic centimetre of soil has a mass of 2.3g. (i) Calculate the volume of the soil inside the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = π r h .] 3 Answer(a)(i) cm3 [2] (ii) Calculate the mass of the soil. Answer(a)(ii) g [1] (iii) Calculate the greatest number of these cones which can be filled completely using 50 kg of soil. Answer(a)(iii) [2] (b) A similar cone of height 32.5 cm is used for growing larger plants. Calculate the volume of soil used to fill this cone. Answer(b) cm3 [3] (c) For Examiner's Use NOT TO SCALE 12 cm Some plants are put into a cylindrical container with height 12 cm and volume 550 cm3. Calculate the radius of the cylinder. Answer(c) cm [3]
11 marks
Mark scheme: 4 (a) (i) 218 (217.7 to 218) 2 M1 for 1/3π × 42 × 13 (ii) 501 (500.7 to 501.4) 1ft ft their (a) × 2.3 (iii) 99 2ft ft 50 000 ÷ their (a)(ii) and truncated to whole number M1 for 50 000 ÷ their (a)(ii) oe or answers 99.8 or 100 3 325. (b) their (a)(i) × oe M2 or 1/3π × 102 × 32.5 13 or M1 for (32.5 ÷ 13)³ (=15.625) seen or (13 ÷ 32.5)³ (= 0.064) seen 3400 or 3410 (3401 to 3407) A1 www3 (c) (r² =) 550 ÷ 12π M2 (14.58 to 14.6) or M1 for 12π r² = 550 or better 3.82 (3.818 to 3.821) A1 www3 IGCSE – October/November 2010 0580 42
This question in 0580/42 Oct/Nov 2010
Q12 · For Examiner's NOT TO Use SCALE 3 cm 12 cm The diagram shows a solid made up of a… 0580/43 Oct/Nov 2010
8 For Examiner's NOT TO Use SCALE 3 cm 12 cm The diagram shows a solid made up of a hemisphere and a cylinder. The radius of both the cylinder and the hemisphere is 3 cm. The length of the cylinder is 12 cm. (a) (i) Calculate the volume of the solid. 4 3 [ The volume, V, of a sphere with radius r is V = πr .] 3 Answer(a)(i) cm3 [4] (ii) The solid is made of steel and 1 cm3 of steel has a mass of 7.9 g. Calculate the mass of the solid. Give your answer in kilograms. Answer(a)(ii) kg [2] (iii) The solid fits into a box in the shape of a cuboid, 15 cm by 6 cm by 6 cm. For Calculate the volume of the box not occupied by the solid. Examiner's Use Answer(a)(iii) cm3 [2] (b) (i) Calculate the total surface area of the solid. You must show your working. [ The surface area, A, of a sphere with radius r is A = 4πr 2 .] Answer(b)(i) cm2 [5] (ii) The surface of the solid is painted. The cost of the paint is $0.09 per millilitre. One millilitre of paint covers an area of 8 cm2. Calculate the cost of painting the solid. Answer(b)(ii) $ [2]
15 marks
Mark scheme: 2 8 (a) (i) 396 (395.6 – 396) 4 M1 for × π × 33 and M1 (independent) for 3 π × 32 × 12, M1 (dependent on M2) for adding 126 π implies M3 (ii) 3.13 (3.125 – 3.128….) ft 2ft ft their (i) × 7.9 ÷ 1000 . M1 for × 7.9 soi by figs 313 or 3125 – 3128… (iii) 144 (144 – 144.4) ft 2ft ft 15 × 6 × 6 – their (a)(i) M1 for 6 × 6 × 15 oe (b) (i) 311 (310.8 – 311.1) 5 M1 for 2 × π × 32 and M1 (independent) for π × 6 × 12 and M1 for π × 32, M1 (dependent on M3) for adding. (99π implies M4) (ii) 3.50 (3.496 to 3.50) ft 2ft ft their (b)(i) × 0.01125 M1 for their (b)(i) ÷ 8 and × figs 9 implied by figs 3496 to 350 9
This question in 0580/43 Oct/Nov 2010
Q13 · For F Examiner's Use NOT TO SCALE C D E 14 cm 36 cm A 19 cm B In the diagram, ABCDEF is a… 0580/41 May/June 2011
6 For F Examiner's Use NOT TO SCALE C D E 14 cm 36 cm A 19 cm B In the diagram, ABCDEF is a prism of length 36 cm. The cross-section ABC is a right-angled triangle. AB = 19 cm and AC = 14 cm. Calculate (a) the length BC, Answer(a) BC = cm [2] (b) the total surface area of the prism, Answer(b) cm2 [4] (c) the volume of the prism, Answer(c) cm3 [2] (d) the length CE, Answer(d) CE = cm [2] (e) the angle between the line CE and the base ABED. Answer(e) [3]
13 marks
Mark scheme: 6 (a) 23.6 (23.60…) 2 M1 for 142 + 192 (b) 2300 or 2303 to 2304 cao 4 M3 for 2 × ½ × 14 × 19 + 14 × 36 + 19 × 36 + their BC × 36 M2 for 4 of these added M1 for ½ × 14 × 19 (c) 4788 or 4790 cao 2 M1 their triangle area × 36 (d) 43(.0) or 43.04 to 43.05 cao 2 M1 for (their (a))2 + 362 or 362 + 192 + 142 14 (e) 18.9° to 19.02° cao 3 M2 for inv sin or their CE 14 inv tan 2 2 or 19 + 36 2 2 19 + 36 inv cos or complete longer their CE methods (M1 for clearly identifying angle CEA) IGCSE – May/June 2011 0580 41
This question in 0580/41 May/June 2011
Q14 · For V Examiner's V Use B C NOT TO SCALE B C A D 9.5 cm A D 2.5 cm 2.5 cm F 2.5 cm E F E F… 0580/42 May/June 2011
7 (a) For V Examiner's V Use B C NOT TO SCALE B C A D 9.5 cm A D 2.5 cm 2.5 cm F 2.5 cm E F E F E A solid pyramid has a regular hexagon of side 2.5 cm as its base. Each sloping face is an isosceles triangle with base 2.5 cm and height 9.5 cm. Calculate the total surface area of the pyramid. Answer(a) cm2 [4] (b) O 55° NOT TO 15 cm SCALE A B A sector OAB has an angle of 55° and a radius of 15 cm. Calculate the area of the sector and show that it rounds to 108 cm2, correct to 3 significant figures. Answer (b) [3] (c) For Examiner's Use 15 cm NOT TO SCALE The sector radii OA and OB in part (b) are joined to form a cone. (i) Calculate the base radius of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(c)(i) cm [2] (ii) Calculate the perpendicular height of the cone. Answer(c)(ii) cm [3] (d) 7.5 cm NOT TO SCALE A solid cone has the same dimensions as the cone in part (c). A small cone with slant height 7.5 cm is removed by cutting parallel to the base. Calculate the volume of the remaining solid. 1 [The volume, V, of a cone with radius r and height h is V = πr2h.] 3 Answer(d) cm3 [3]
15 marks
Mark scheme: 7 (a) 87.5 (87.45 to 87.52) www 4 4 M1 for ½ × 2.5 × 9.5 soi by 11.875 or 71.25 and M2 for ½ × 2.52 × sin60 × 6 oe (16.23 to 16.24) or M1 for ½ × 2.52 × sin60 (2.706..) or 1 trapezium (8.1189..) (b) 107.9 ….. to 108.0…..www3 3 Must see at least 4 figures 55 55 M2 for × π × 152 or M1 for seen 360 360 (c) (i) 2.29 (2.291 to 2.293) www 2 2 M1 for 108 = 15πr oe allow 107.9 to 108.0… for their 108 (ii) 14.8 (14.82 to 14.83) cao www 3 3 M2 for 15 2 − their 2.29 2 (M1 for h 2 + their 2.29 2 = 15 2 ) π (d) 70.9 to 71.5 cao www 3 3 M2 for (their 2.292 × their 14.8 – their 1.1452 3 × their 7.4) (not 15 or 7.5) 7 π or × × their 2.292 × their 14.8 8 3 5.7 3 or M1 for 1/8 oe e.g. 3 or 7/8 or (½ their R 15 and ½ their h) seen
This question in 0580/42 May/June 2011
Q15 · For r Examiner's Use NOT TO 8 cm s SCALE 2.7 cm 20 cm The diagram shows a plastic cup in… 0580/41 Oct/Nov 2011
4 For r Examiner's Use NOT TO 8 cm s SCALE 2.7 cm 20 cm The diagram shows a plastic cup in the shape of a cone with the end removed. The vertical height of the cone in the diagram is 20 cm. The height of the cup is 8 cm. The base of the cup has radius 2.7 cm. (a) (i) Show that the radius, r, of the circular top of the cup is 4.5 cm. Answer(a)(i) [2] (ii) Calculate the volume of water in the cup when it is full. 1 [The volume, V, of a cone with radius r and height h is V = πr2h.] 3 Answer(a)(ii) cm3 [4] (b) (i) Show that the slant height, s, of the cup is 8.2 cm. For Examiner's Answer(b)(i) Use [3] (ii) Calculate the curved surface area of the outside of the cup. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(b)(ii) cm2 [5]
14 marks
Mark scheme: 20 4 (a) (i) 2.7 × oe = 4.5 E2 M1 for (SF =) 20/12 or 12/20 (but not from 12 2.7/4.5 or 4.5/2.7) (ii) 1/3π × 4.52 × 20 – 1/3π × 2.72 × 12 M3 M1 for 1/3π × 4.52 × 20 (424 ... or 135π) or and M1 for 1/3π × 2.72 × 12 (91.6..or 29.16π) (1 – (3/5)3) × 1/3π × 4.52 × 20 oe 332.3 to 332.6 or 332 or 333 A1 (b) (i) 8² + (4.5 – 2.7)² oe M1 e.g. Alt: 20² + 4.5² and 12² + 2.7² sq root M1 Dep on 1st M1 Alt: 20.5 – 12.3 Other complete correct methods are M2 8.2 E1 No errors seen (ii) 185 or 186 or 185.5 or 185.45 5 M4 for π × 4.5 × 20.5 – π × 2.7 × 12.3 to 185.51 or other complete correct method or M3 for π × 4.5 × 20.5 or π × 2.7 × 12.3 (290 or 92.25π) (104.3...or 33.21π) or B2 for (slant height of large cone =) 20.5 or (slant height of removed cone =) 12.3 or M1 for 5.4 2 + 20 2 or 2.7 2 + 12 2 or 12/8 × 8.2 oe or 20/8 × 8.2 oe IGCSE – October/November 2011 0580 41
This question in 0580/41 Oct/Nov 2011
Q16 · For Examiner's Use NOT TO 24 cm SCALE 9 cm A solid metal cone has base radius 9 cm and… 0580/41 May/June 2012
10 For Examiner's Use NOT TO 24 cm SCALE 9 cm A solid metal cone has base radius 9 cm and vertical height 24 cm. (a) Calculate the volume of the cone. 1 [The volume, V, of a cone with radius r and height h is V = πr2h.] 3 Answer(a) cm3 [2] (b) NOT TO 16 cm SCALE 9 cm A cone of height 8 cm is removed by cutting parallel to the base, leaving the solid shown above. Show that the volume of this solid rounds to 1960 cm3, correct to 3 significant figures. Answer (b) [4] (c) The 1960 cm3 of metal in the solid in part (b) is melted and made into 5 identical cylinders, each of length 15 cm. Show that the radius of each cylinder rounds to 2.9 cm, correct to 1 decimal place. Answer (c) [4]
10 marks
Mark scheme: 10 (a) 2030 or 2040 or 2034 to 2036. (…) 2 (V = ) 13 × π × 9 2 × 24 Accept 648π for 2 marks if final answer 8 (b) (upper radius =) 3 B1 accept 9× oe 24 (vol cut off =) 13 × π × their 32 × 8 M1 (= 75.36 to 75.41) their r must be less than 9 their (a) − their 75.39 M1 [ alternate method M1 for ratio sides 1:3 dep M1 ratio vols 1 : 27 M1 their (a) × 26 ÷ 27 ] 624π implies B1 M2 or M3 1958 to 1964.(...) E1 must see a figure after decimal point if 1960 (c) 1960 = 5 × π × r 2 × 15 soi M1 r 2 = 1960 ÷ π ÷ 15 ÷ 5 M1 implied by 8.318... √ their 8.318 M1 dep on M1 M1 SC2 for 5 × π × 9.2 2 × 15 = 1980 to 1982 2.88 to 2.89 E1
This question in 0580/41 May/June 2012
Q17 · For NOT TO Examiner's Use SCALE 20 cm 24 cm 46 cm Jose has a fish tank in the shape of a… 0580/41 Oct/Nov 2012
5 (a) For NOT TO Examiner's Use SCALE 20 cm 24 cm 46 cm Jose has a fish tank in the shape of a cuboid measuring 46 cm by 24 cm by 20 cm. Calculate the length of the diagonal shown in the diagram. Answer(a) cm [3] (b) Maria has a fish tank with a volume of 20 000 cm3. Write the volume of Maria’s fish tank as a percentage of the volume of Jose’s fish tank. Answer(b) % [3] (c) Lorenzo’s fish tank is mathematically similar to Jose’s and double the volume. Calculate the dimensions of Lorenzo’s fish tank. Answer(c) cm by cm by cm [3] (d) A sphere has a volume of 20 000 cm3. Calculate its radius. 4 [The volume, V, of a sphere with radius r is V = πr3.] 3 Answer(d) cm [3]
12 marks
Mark scheme: 5 (a) 55.6 to 55.61 www 3 M2 for 46 2 + 24 2 + 20 2 oe [ 3092 ] or M1 for 46² + 24² oe [soi by 2692 or art 51.9] or 46² + 20² oe [soi by 2516 or art 50.2] or 24² + 20² oe [soi by 976 or art 31.2] (b) 90.6 or 90.57 to 90.58 3 20000 M2 for × 100 oe (20 × 24 × 46 ) or M1 for 20 × 24 × 46 [22080] (c) 25.19 to 25.21, 30.23 to 30.246 or 3 M2 for 20 × 3 2 or 24 × 3 2 or 46 × 3 2 30.2, 57.95 to 57.97 or 58[.0] 3 2 oe seen [1.259 to 1.261] M1 for 3 20000 (d) 16.8 to 16.842 3 M2 for oe or answer figs 168 to 4 3π 16842 3 20000 or M1 for [4770 – 4780] seen 4 3π IGCSE – October/November 2012 0580 41
This question in 0580/41 Oct/Nov 2012
Q18 · A rectangular piece of card has a square of side 2 cm removed from each corner 0580/41 Oct/Nov 2012
8 A rectangular piece of card has a square of side 2 cm removed from each corner. For Examiner's Use 2 cm 2 cm NOT TO SCALE (2x + 3) cm (x + 5) cm (a) Write expressions, in terms of x, for the dimensions of the rectangular card before the squares are removed from the corners. Answer(a) cm by cm [2] (b) The diagram shows a net for an open box. Show that the volume, V cm3, of the open box is given by the formula V = 4x2 + 26x + 30 . Answer(b) [3] (c) (i) Calculate the values of x when V = 75. For Show all your working and give your answers correct to two decimal places. Examiner's Use Answer(c)(i) x = or x = [5] (ii) Write down the length of the longest edge of the box. Answer(c)(ii) cm [1] Question 9 is printed on the next page.
11 marks
Mark scheme: 8 (a) 2x + 7 final answer 2 B1 for each, accept in either order x + 9 final answer After 0 scored allow SC1 mark for both correct but unsimplified (b) 2(2x + 3)(x + 5) at any stage M1 The × 2 could be embedded within one of the brackets e.g. (4x + 6)(x + 5) 2x2 + 3x + 10x + 15 or better B1 Expands brackets correctly 4x2 + 26x + 30 E1 No errors seen and two previous stages shown (c) (i) 4x² + 26x – 45 [= 0] soi B1 − 26 ± ( 26 ) 2 − 4( 4 )( −45) B1 ft ft their 4x² + 26x ± k [k ≠ 0] oe 2 ( 4 ) B1 ft In square root B1 ft for ( 26) 2 − 4( 4)( −45) or better (1396) p+ q p− q If in form or ; r r B1 ft for –26 and 2(4) or better –7.92, 1.42 final answers B1 B1 If B0, SC1 for –7.9 and 1.4 or both answers – 7.920…., 1.420….. or for–7.92 , 1.42 seen (ii) 6.42 [0…] 1 ft ft their greatest positive root If their x ≤ 2 then ft x + 5 If their x > 2 then ft 2x + 3 IGCSE – October/November 2012 0580 41 7
This question in 0580/41 Oct/Nov 2012
Q19 · A metal cuboid has a volume of 1080 cm3 and a mass of 8 kg 0580/43 Oct/Nov 2012
3 A metal cuboid has a volume of 1080 cm3 and a mass of 8 kg. For Examiner's (a) Calculate the mass of one cubic centimetre of the metal. Use Give your answer in grams. Answer(a) g [1] (b) The base of the cuboid measures 12 cm by 10 cm. Calculate the height of the cuboid. Answer(b) cm [2] (c) The cuboid is melted down and made into a sphere with radius r cm. (i) Calculate the value of r. 4 [The volume, V, of a sphere with radius r is V = πr 3.] 3 Answer(c)(i) r = [3] (ii) Calculate the surface area of the sphere. For Examiner's [The surface area, A, of a sphere with radius r is A = 4πr 2.] Use Answer(c)(ii) cm2 [2] (d) A larger sphere has a radius R cm. The surface area of this sphere is double the surface area of the sphere with radius r cm in part (c). R Find the value of . r Answer(d) [2]
10 marks
Mark scheme: 3 (a) 7.407….. or 7.41 1 (b) 9 2 M1 for 1080 ÷ (12 × 10) oe 1080 (c) (i) 6.36 to 6.37 www 3 M2 for 3 4 oe π 3 1080 or M1 for 4 oe [ 257.7 to 258.7] π 3 Accept 4.18 to 4.19 for 4/3 π (ii) 508 to 510 2 M1 for 4 × π × (their (c)(i))2 (d) 2 or 1.41 [1.414…] www 2 Allow over 1 or 2 : 1 etc M1 for (R / r)2 = 2 oe or [R2 =] (2 × their (c)(ii))/4π or [R2 =] 2 × (their (c)(i))²
This question in 0580/43 Oct/Nov 2012
Q20 · For Examiner′s I Use NOT TO SCALE H J F 7 cm 40 cm E 22 cm G EFGHIJ is a solid metal… 0580/43 May/June 2013
4 For Examiner′s I Use NOT TO SCALE H J F 7 cm 40 cm E 22 cm G EFGHIJ is a solid metal prism of length 40 cm. The cross section EFG is a right-angled triangle. EF = 7 cm and EG = 22 cm. (a) Calculate the volume of the prism. Answer(a) … cm3 [2] (b) Calculate the length FJ. Answer(b) FJ = … cm [4] (c) Calculate the angle between FJ and the base EGJH of the prism. For Examiner′s Use Answer(c) … [3] (d) The prism is melted and made into spheres. Each sphere has a radius 1.5 cm. Work out the greatest number of spheres that can be made. 4 [The volume, V, of a sphere with radius r is V = πr3.] 3 Answer(d) … [3] (e) (i) A right-angled triangle is the cross section of another prism. This triangle has height 4.5 cm and base 11.0 cm. Both measurements are correct to 1 decimal place. Calculate the upper bound for the area of this triangle. Answer(e)(i) … cm2 [2] (ii) Write your answer to part (e)(i) correct to 4 signifi cant fi gures. Answer(e)(ii) … cm2 [1] _____________________________________________________________________________________
15 marks
Mark scheme: 4 (a) 3080 2 M1 for ½ × 7 × 22 × 40 (b) 46.2 or 46.18 to 46.2 www 4 M3 for 7 2 + 22 2 + 40 2 or M2 for 72 + 222 + 402 soi by 2133 or M1 for correct Pythagoras on one face 7 (c) 8.7 or 8.7 to 8.72 www 3 M2 for sin– 1 their(b) oe 7 or M1 for sin = oe their(b) 4 (d) 217 3 M1 for ×π×1.53 soi by 14.1 to 14.14 3 and M1 dep for their (a) ÷ their 14.14 soi by 218. Dependent on M1 earned (e) (i) 25.13875 final answer 2 B1 for 4.55 and 11.05 seen or 25.13875 seen and then spoiled (ii) 25.14 1FT Strict FT their (e)(i) correct to 4s.f. if rounding is possible
This question in 0580/43 May/June 2013
Q21 · For Examiner′s O 8 cm A Use 42° NOT TO 8 cm SCALE B h cm A wedge of cheese in the shape… 0580/42 Oct/Nov 2013
4 For Examiner′s O 8 cm A Use 42° NOT TO 8 cm SCALE B h cm A wedge of cheese in the shape of a prism is cut from a cylinder of cheese of height h cm. The radius of the cylinder, OA, is 8 cm and the angle AOB = 42°. (a) (i) The volume of the wedge of cheese is 90 cm3. Show that the value of h is 3.84 cm correct to 2 decimal places. Answer(a)(i) [4] (ii) Calculate the total surface area of the wedge of cheese. Answer(a)(ii) … cm2 [5] (b) A mathematically similar wedge of cheese has a volume of 22.5 cm3. Calculate the height of this wedge. Answer(b) … cm [3] _____________________________________________________________________________________
12 marks
Mark scheme: 4 (a) (i) 90 ÷ (42/360 × π × 82) o.e. M3 M2 for 42/360 × π × 82 × h = 90 or M1 for 42/360 × π × 82 3.836 to 3.837 A1 (ii) 131 or 130.75 to 130.9 nfww 5 M2 for 42/360 × π × 2 × 8 × 3.84 oe [22.48 to 22.53] or M1 for 42/360 × π × 2 × 8 oe soi [5.86 to 5.87] and M1 for 2 × (8 × 3.84) [61.37 to 61.44] and M1 for 2 × (42/360 × π × 82) [46.88 to 47] 22 5. .3 84 3 × 22 5. (b) 2.42 or 2.416 to 2.419 3 M2 for 3.84 × 3 oe or h = 3 90 90 22 5. 90 or M1 for 3 oe or 3 oe seen 90 22 5. .3 84 3 90 or 3 = oe h 22 5. IGCSE – October/November 2013 0580 42
This question in 0580/42 Oct/Nov 2013
Q22 · A rectangular metal sheet measures 9 cm by 7 cm 0580/43 Oct/Nov 2013
3 A rectangular metal sheet measures 9 cm by 7 cm. For Examiner′s A square, of side x cm, is cut from each corner. Use The metal is then folded to make an open box of height x cm. 9 cm NOT TO x cm SCALE 7 cm x cm x cm (a) Write down, in terms of x, the length and width of the box. Answer(a) Length = … Width = … [2] (b) Show that the volume, V, of the box is 4x3 – 32x2 + 63x. Answer(b) [2] (c) Complete this table of values for V = 4x3 – 32x2 + 63x. x 0 0.5 1 1.5 2 2.5 3 3.5 V 0 35 36 30 9 0 [2] (d) On the grid opposite, draw the graph of V = 4x3 – 32x2 + 63x for 0 Y x Y 3.5 . Three of the points have been plotted for you. For V Examiner′s Use 40 35 30 25 20 15 10 5 x 0 0.5 1 1.5 2 2.5 3 3.5 [3] (e) The volume of the box is at least 30 cm3. Write down, as an inequality, the possible values of x. Answer(e) … [2] (f) (i) Write down the maximum volume of the box. Answer(f)(i) … cm3 [1] (ii) Write down the value of x which gives the maximum volume. Answer(f)(ii) … [1] _____________________________________________________________________________________
13 marks
Mark scheme: 3 (a) 9 – 2x, 7 – 2x oe 2 B1 for each, accept in any order (b) x(9 – 2x)(7 – 2x) M1FT 4x3 – 32x2 + 63x A1 Correct expansion and simplification with no errors (c) 24 20 2 B1 for each correct value (d) Correct curve 3 B2FT for 5 correct plots or B1FT for 3 or 4 correct plots (e) 0.65 to 0.75 ≤ x ≤ 2 oe 2 B1 for 0.65 to 0.75 seen (f) (i) 36 to 37 1 (ii) 1.2 to 1.4 1
This question in 0580/43 Oct/Nov 2013
Q23 · Sandra has designed this open container 0580/43 Oct/Nov 2013
6 Sandra has designed this open container. For Examiner′s The height of the container is 35 cm. Use NOT TO SCALE 35 cm The cross section of the container is designed from three semi-circles with diameters 17.5 cm, 6.5 cm and 24 cm. 17.5 cm 6.5 cm NOT TO SCALE (a) Calculate the area of the cross section of the container. Answer(a) … cm2 [3] (b) Calculate the external surface area of the container, including the base. Answer(b) … cm2 [4] (c) The container has a height of 35 cm. For Examiner′s Use Calculate the capacity of the container. Give your answer in litres. Answer(c) … litres [3] (d) Sandra’s container is completely fi lled with water. All the water is then poured into another container in the shape of a cone. The cone has radius 20 cm and height 40 cm. 20 cm NOT TO r SCALE 40 cm h (i) The diagram shows the water in the cone. h Show that r = . 2 Answer(d)(i) [1] (ii) Find the height, h, of the water in the cone. 1 [The volume, V, of a cone with radius r and height h is V = 3 πr 2h.] Answer(d)(ii) h = … cm [3] _____________________________________________________________________________________
14 marks
Mark scheme: 6 (a) 329.7 to 330 3 M2 for ½π(122 + 8.752 – 3.252) oe or M1 for ½π122 or ½π8.752 or ½π3.252 SC2 for answer 1318 to 1320 (b) 2970 or 2967 to 2969.[…] 4 M3 for ½π(24 + 17.5 + 6.5) × 35 + their (a) or M2 for ½π(24 + 17.5 + 6.5) × 35 or M1 for ½π × 24 or ½π × 17.5 or ½π × 6.5 SC3 for 3955 to 3960 dep on SC2 in (a) (c) 11.5 or 11.6 or 11.53 to 11.55 3FT M1 for their (a) × 35 A1 for 11500 or 11530 to 11550 IGCSE – October/November 2013 0580 43 r 20 r h Accept 20 : 40 = r : h leading to 40r = 20h [r = h/2] = = or (d) (i) 1 20 1 r 1 h 40 20 40 = and = 40 2 h 2 their 11545 × 12 (ii) 35.3 or 35.31 to 35.34 3 M2 for 3 oe or 2 × their r π or 1 h 2 M1 for their 11545 = × π × × h oe 3 2 1 2 or their 11545 = × π × r × 2 r oe 3 3 14 − ( − 4 ) 1 5
This question in 0580/43 Oct/Nov 2013
Q24 · The running costs for a papermill are $75 246 0580/41 May/June 2014
3 (a) The running costs for a papermill are $75 246. This amount is divided in the ratio labour costs : materials = 5 : 1. Calculate the labour costs. Answer(a) $ … [2] (b) In 2012 the company made a profi t of $135 890. In 2013 the profi t was $150 675. Calculate the percentage increase in the profi t from 2012 to 2013. Answer(b) … % [3] (c) The profi t of $135 890 in 2012 was an increase of 7% on the profi t in 2011. Calculate the profi t in 2011. Answer(c) $ … [3] (d) 2 cm NOT TO SCALE 21 cm 30 cm Paper is sold in cylindrical rolls. There is a wooden cylinder of radius 2 cm and height 21 cm in the centre of each roll. The outer radius of a roll of paper is 30 cm. (i) Calculate the volume of paper in a roll. Answer(d)(i) … cm3 [3] (ii) The paper is cut into sheets which measure 21 cm by 29.7 cm. The thickness of each sheet is 0.125 mm. (a) Change 0.125 millimetres into centimetres. Answer(d)(ii)(a) … cm [1] (b) Work out how many whole sheets of paper can be cut from a roll. Answer(d)(ii)(b) … [4] __________________________________________________________________________________________
16 marks
Mark scheme: 3 (a) 62 705 2 M1 for 75 246 ÷ 6 soi by 12 541 or 75 246 × 5 (b) 10.9 or 10.88… 3 (150 675 − 135 890) M2 for × 100 oe 135 890 or M1 for correct fraction soi by 0.1088… 150 675 or × 100 soi by 110.88… 135 890 IGCSE – May/June 2014 0580 41 Qu Answers Mark Part Marks (c) 127 000 3 M2 for 135 890 ÷ 1.07 oe or M1 for 135 890 associated with 107% (d) (i) 59 112 to 59 113 or 59 100 or 59 110 3 M2 for π × 21 × (302 – 22) oe or 59 119 to 59 120 or 59 100 Or nfww M1 for π × 21 × 302 or π × 21 × 22 (ii) (a) 0.0125 1 (b) 7580 or 7582 or 7581 or 7583 nfww 4 M1 for 21 × 29.7 × their 0.0125 [=7.796 or 7.8[0]] and M1 for their (d)(i) ÷ (21 × 29.7 × their 0.0125) A1 for 7580 to 7583.2 (non integer) If 0 then SC1 for their (d)(i) ÷ (21 × 29.7 × 0.125)
This question in 0580/41 May/June 2014
Q25 · 8 cm NOT TO SCALE r cm The three sides of an equilateral triangle are tangents to a… 0580/43 May/June 2014
10 (a) 8 cm NOT TO SCALE r cm The three sides of an equilateral triangle are tangents to a circle of radius r cm. The sides of the triangle are 8 cm long. Calculate the value of r. Show that it rounds to 2.3, correct to 1 decimal place. Answer(a) [3] (b) 8 cm NOT TO SCALE 12 cm The diagram shows a box in the shape of a triangular prism of height 12 cm. The cross section is an equilateral triangle of side 8 cm. Calculate the volume of the box. Answer(b) … cm3 [4] (c) The box contains biscuits. Each biscuit is a cylinder of radius 2.3 centimetres and height 4 millimetres. Calculate (i) the largest number of biscuits that can be placed in the box, Answer(c)(i) … [3] (ii) the volume of one biscuit in cubic centimetres, Answer(c)(ii) … cm3 [2] (iii) the percentage of the volume of the box not fi lled with biscuits. Answer(c)(iii) … % [3] __________________________________________________________________________________________ Question 11 is printed on the next page.
15 marks
Mark scheme: 10 (a) [r =] 2.30[9...] 3 B2 for [r =] 2.31 or M2 for 4 tan 30 r or M1 for = tan 30 4 (b) 333 or 332.5 to 332.6 4 M3 for 0.5 × 8 × 8 × sin 60 × 12 oe or M2 for 0.5 × 8 × 8 × sin 60 oe or M1 for their triangle area × 12 shown 1 dep on ‘ ’used within their area of triangle 2 method (c) (i) 30 3 M2 for 12 ÷ 0.4 or 120 ÷ 4 or SC1 for figs 3 (ii) 6.65 or 6.647 to 6.648[...] 2 M1 for π × 3.2 2 × 4.0 or SC1 for π × 3.2 2 × 4 soi by 66.5 or 66.47 to 66.48[…] their ( c )(i ) × their ( c )(ii ) (iii) 40[.0] or 40.1 or 40.0 to 40.2 nfww 3 M2 for 100 − × 100 their (b ) their (b ) − their ( c )(i ) × their ( c )(ii ) or × 100 their (b ) their ( c )(i ) × their ( c )(ii ) or M1 for × 100 their (b ) their ( b ) − their ( c )(i ) × their ( c )(ii ) or their ( b ) 1 1 1
This question in 0580/43 May/June 2014
Q26 · NOT TO SCALE 75 cm 55 cm 120 cm The diagram shows a water tank in the shape of a cuboid… 0580/42 Oct/Nov 2014
7 NOT TO SCALE 75 cm 55 cm 120 cm The diagram shows a water tank in the shape of a cuboid measuring 120 cm by 55 cm by 75 cm. The tank is fi lled completely with water. (a) Show that the capacity of the water tank is 495 litres. Answer(a) [2] (b) (i) The water from the tank fl ows into an empty cylinder at a uniform rate of 750 millilitres per second. Calculate the length of time, in minutes, for the water to be completely emptied from the tank. Answer(b)(i) … min [2] (ii) When the tank is completely empty, the height of the water in the cylinder is 112 cm. NOT TO SCALE 112 cm Calculate the radius of the cylinder. Answer(b)(ii) … cm [3] (c) x cm NOT TO SCALE 75 cm cm 145 55 cm 120 cm A rod of length 145 cm is placed inside the water tank. One end of the rod is in the bottom corner of the tank as shown. The other end of the rod is x cm below the top corner of the tank as shown. Calculate the value of x. Answer(c) x = … [4] (d) Calculate the angle that the rod makes with the base of the tank. Answer(d) … [3] __________________________________________________________________________________________
14 marks
Mark scheme: 7 (a) (i) 120 × 55 × 75 [= 495000] M1 M1 ÷ 1000 [= 495] or 495[l] × 1000 = 495000[ml] (b) (i) 11 2 M1 for 495000 ÷ 750 [÷ 60] oe [660] After 0 scored, SC1 for answer figs 11 figs 495 (ii) 37.5 or 37.50 to 37.51 3 M2 for oe 112π 2 figs 495 or M1 for [112r = ] or π 2 figs 495 [ πr = ] or better 112 (c) 15 4 B3 for answer 60 or M3 for 75 – 145 2 − (55 2 + 120 2 ) oe M2 for 145 2 − ( 55 2 + 120 2 ) oe or M1 for 55 2 + 120 2 2 + 120 2 /145) oe, e.g. (d) 24.4[4..] to 24.45 3 M2 for cos–1 ( 55 or sin −(751 – their (c))/145 or tan −((751 – their (c))/ 55 2 + 120 2 ) or M1 for cos = 55 2 + 120 2 /145 oe or sin = (75 – their (c))/145 or tan = (75 – their (c))/ 55 2 + 120 2
This question in 0580/42 Oct/Nov 2014
Q27 · The diagram shows a sector of a circle A with centre O and radius 24 cm 0580/42 Feb/March 2015
8 (a) The diagram shows a sector of a circle A with centre O and radius 24 cm. NOT TO SCALE x° O 24 cm (i) The total perimeter of the sector is 68 cm. B Calculate the value of x. Answer(a)(i) x = … [3] (ii) The points A and B of the sector are joined together to O make a hollow cone. The arc AB becomes the circumference of the base of the cone. NOT TO SCALE AB Calculate the volume of the cone. 1 [The volume, V, of a cone with radius r and height h is V = πr2h.] 3 Answer(a)(ii) … cm3 [6] (b) Q NOT TO SCALE M P O X Y 8 cm The diagram shows a shape made from a square, a quarter circle and a semi-circle. OPXY is a square of side 8 cm. OPQ is a quarter circle, centre O. The line OMQ is the diameter of the semi-circle. Calculate the area of the shape. Answer(b) … cm2 [5] __________________________________________________________________________________________
14 marks
Mark scheme: 8 (a) (i) 47.7 or 47.74 to 47.75 3 M1 for [arc =] 68 − 2 × 24 x or 24 + 24 + × 2 π × 24 = 68 360 M1 for [x =] their arc × 360 ÷ (2 × π × 24) 20 (ii) 252 or 252.3 to 252.4 … 6 M1 for r = or 2 π their 47 7. × 2 × π × 24 ÷ (2 π ) 360 10 A1 for r = 3.18 or 3.182 to 3.183... or π M1 for h 2 = 24 2 − their r 2 A1 for h = 23.8 or 23.78... to 23.79 M1dep on M1 earned for 1 2 V = π × their h × their r 3 2 2 1 2 1 8 (b) 139 or 139.3 to 139.4... nfww 5 M4 for 8 + π × 8 + π × 4 2 2 1 2 or M1 for π × 8 4 2 1 8 and M1 for π × 2 2 and M1 for 82 added to at least one term with π
This question in 0580/42 Feb/March 2015
Q28 · Luc is painting the doors in his house 0580/41 Oct/Nov 2015
1 (a) Luc is painting the doors in his house. He uses 34 of a tin of paint for each door. Work out the least number of tins of paint Luc needs to paint 7 doors. Answer(a) … [3] (b) Jan buys tins of paint for $17.16 each. He sells the paint at a profit of 25%. For how much does Jan sell each tin of paint? Answer(b) $ … [2] (c) The cost of $17.16 for each tin of paint is 4% more than the cost in the previous year. Work out the cost of each tin of paint in the previous year. Answer(c) $ … [3] (d) In America a tin of paint costs $17.16 . In Italy the same tin of paint costs €13.32 . The exchange rate is $1 = €0.72 . Calculate, in dollars, the difference in the cost of the tin of paint. Answer(d) $ … [2] (e) Paint is sold in cylindrical tins of height 11 cm. Each tin holds 750 ml of paint. (i) Write 750 ml in cm3. Answer(e)(i) … cm3 [1] (ii) Calculate the radius of the tin. Give your answer correct to 1 decimal place. Answer(e)(ii) … cm [3] (iii) A mathematically similar tin has a height of 22 cm. How many litres of paint does this tin hold? Answer(e)(iii) … litres [2] (f) The mass of a tin of paint is 890 grams, correct to the nearest 10 grams. Work out the upper bound of the total mass of 10 tins of paint. Answer(f) … g [1] (g) The probability that a tin of paint is dented is 0.07 . Out of 3000 tins of paint, how many would you expect to be dented? Answer(g) … [2] (h) Tins of paint are filled at the rate of 2 m3 per minute. How many 750 ml tins of paint can be filled in 1 hour? Answer(h) … [3] __________________________________________________________________________________________
22 marks
Mark scheme: Question Answer Mark Part marks 1 (a) 6 3 1 B2 for 5 or 5.25 shown in working isw 4 3 or M1 for × 7 soi by answer 5 4 (b) 21.45 cao final answer 2 M1 for 17.16 × 0.25 or 17.16 × 1.25 (c) 16.5[0] nfww 3 M2 for 17.16 ÷ 1.04 oe or M1 for 17.16 associated with 104[%] oe isw (d) 1.34 cao final answer 2 M1 for 13.32 ÷ 0.72 soi by 18.5[0] or for any correct complete longer method If zero scored, SC1 for 0.96 [euros] seen (e) (i) 750 1 (ii) 4.7 cao 3 B2 for 4.658 to 4.66 or M2 for their (e)(i) ÷ 11π or M1 for 11πr2 = their (e)(i) 1 (iii) 6 2 M1 for 23 or 3 oe seen 2 or for π × (2 × their (e)(ii))2 × 22 If zero scored, SC1 for answer 6 000 (f) 8 950 1 (g) 210 2 M1 for 0.07 × 3 000 (h) 160 000 3 M2 for 2 × 60 × 1003 ÷ 750 oe or M1 for figs 16 as answer or 1003 seen
This question in 0580/41 Oct/Nov 2015
Q29 · The diagram shows a horizontal water trough in the shape of a prism 0580/43 Oct/Nov 2015
3 The diagram shows a horizontal water trough in the shape of a prism. NOT TO 35 cm SCALE 12 cm 6 cm 120 cm 25 cm The cross section of this prism is a trapezium. The trapezium has parallel sides of lengths 35 cm and 25 cm and a perpendicular height of 12 cm. The length of the prism is 120 cm. (a) Calculate the volume of the trough. Answer(a) … cm3 [3] (b) The trough contains water to a depth of 6 cm. (i) Show that the volume of water is 19 800 cm3. Answer (b)(i) [2] (ii) Calculate the percentage of the trough that contains water. Answer(b)(ii) … % [1] (c) The water is drained from the trough at a rate of 12 litres per hour. Calculate the time it takes to empty the trough. Give your answer in hours and minutes. Answer(c) … h … min [4] (d) The water from the trough just fills a cylinder of radius r cm and height 3r cm. Calculate the value of r. Answer(d) r = … [3] (e) The cylinder has a mass of 1.2 kg. 1 cm3 of water has a mass of 1 g. Calculate the total mass of the cylinder and the water. Give your answer in kilograms. Answer(e) … kg [2] __________________________________________________________________________________________
15 marks
Mark scheme: 3 (a) 43 200 3 M2 for 0.5 × (35 + 25) × 12 × 120 oe or M1 for 0.5 × (35 + 25) × 12 oe (b) (i) 0.5 × (25 + 30) × 6 ×120 [= 19 800] M2 Dep on a valid method for obtaining the width of 30 cm B1 for 0.5 × (25 + 35) oe 19 800 (ii) 45.8 or 45.83… 1FT FT for × 100 their (a) (c) 1 hr 39 min 4 33 B3 for 1.65 [h] or 99 mins or 20 19 800 or M2 for oe 12 × 1000 19 800 19 800 or M1 for or or 12 × 1000 12 1000 If zero scored then SC1 for figs 165 and B1 for converting their time (in hours) into hours and minutes 19 800 (d) 12.8 or 12.80 to 12.81 3 M2 for 3 3 π or M1 for π r 2 3r = 19 800 19 800 (e) 21[.0] 2 M1 for + 2.1 1000
This question in 0580/43 Oct/Nov 2015
Q30 · Calculate the volume of a metal sphere of radius 15 cm and show that it rounds to 14 140… 0580/41 May/June 2016
4 (a) Calculate the volume of a metal sphere of radius 15 cm and show that it rounds to 14 140 cm3, correct to 4 significant figures. [The volume, V, of a sphere with radius r is V = 43 r r 3 .] [2] (b) (i) The sphere is placed inside an empty cylindrical tank of radius 25 cm and height 60 cm. The tank is filled with water. 25 cm NOT TO SCALE 60 cm Calculate the volume of water required to fill the tank. … cm3 [3] (ii) The sphere is removed from the tank. NOT TO SCALE d Calculate the depth, d, of water in the tank. d = … cm [2] (c) The sphere is melted down and the metal is made into a solid cone of height 54 cm. (i) Calculate the radius of the cone. [The volume, V, of a cone with radius r and height h is V = 13 r r 2 h .] … cm [3] (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 .] … cm2 [4]
14 marks
Mark scheme: 4 (a) 4 3 14 137 to 14 137.2 or 14 139 2 M1 for × π × 15 3 (b) (i) 104 000 or 103 600 to 103 700 3 M2 for π × 252 × 60 – 14140 or M1 for π × 252 × 60
This question in 0580/41 May/June 2016
Q31 · The diagram shows a cuboid 0580/42 May/June 2016
6 The diagram shows a cuboid. F G E H 30 cm NOT TO SCALE B C 35 cm A 60 cm D AD = 60 cm, CD = 35 cm and CG = 30 cm. (a) Write down the number of planes of symmetry of this cuboid. … [1] (b) (i) Work out the surface area of the cuboid. … cm2 [3] (ii) Write your answer to part (b)(i) in square metres. … m2 [1] (c) Calculate (i) the length AG, AG = … cm [4] (ii) the angle between AG and the base ABCD. … [3] (d) (i) Show that the volume of the cuboid is 63 000 cm3. [1] (ii) A cylinder of height 40 cm has the same volume as the cuboid. Calculate the radius of the cylinder. … cm [3]
16 marks
Mark scheme: 6 (a) 3 1 (b) (i) 9900 3 M2 for 2(60 × 35) + 2(60 × 30) + 2(30 × 35) oe or M1 for one correct rectangle (ii) 0.99 oe 1FT FT their(b)(i) ÷ 10 000
This question in 0580/42 May/June 2016
Q32 · A NOT TO SCALE 12 cm O 145° B The diagram shows a sector, centre O, and radius 12 cm 0580/43 May/June 2016
9 A NOT TO SCALE 12 cm O 145° B The diagram shows a sector, centre O, and radius 12 cm. (a) Calculate the area of the sector. … cm2 [3] (b) The sector is made into a cone by joining OA to OB. Calculate the volume of the cone. 1 2 r r h .] [The volume, V, of a cone with base radius r and height h is V = 3 … cm3 [6]
9 marks
Mark scheme: 360 145 2 9 (a) 270 or 270.17 to 270.22 3 M2 for × π12 oe 360 or B1 for 215 seen θ 2 or M1 for × π12 used 360 (b) 518 or 517.6 to 517.8 nfww 6 B4 for vertical height = 9.62 to 9.63 or B3 for radius = 7.166 to 7.17 or B2 for length of sector = 45.[0] or 45.02 to 45.04 360 − 145 or M1 for × 2 × π × 12 oe 360 or for 12 2 − their radius 2 and M1 indep for 1 2 π × their radius × their h 3 (h ≠ 12 or r ≠12)
This question in 0580/43 May/June 2016
Q33 · NOT TO SCALE 13 cm 25 cm The diagram shows a solid made up of a cylinder and two… 0580/41 Oct/Nov 2016
3 (a) NOT TO SCALE 13 cm 25 cm The diagram shows a solid made up of a cylinder and two hemispheres. The radius of the cylinder and the hemispheres is 13 cm. The length of the cylinder is 25 cm. (i) One cubic centimetre of the solid has a mass of 2.3 g. Calculate the mass of the solid. Give your answer in kilograms. 4 [The volume, V, of a sphere with radius r is V = r r3 .] 3 … kg [4] (ii) The surface of the solid is painted at a cost of $4.70 per square metre. Calculate the cost of painting the solid. [The surface area, A, of a sphere with radius r is A = 4 rr 2 .] $ … [4] (b) NOT TO 2x cm SCALE x cm The cone in the diagram has radius x cm and height 2x cm. The volume of the cone is 500 cm3. Find the value of x. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 x = … [3] (c) Two mathematically similar solids have volumes of 180 cm3 and 360 cm3. The surface area of the smaller solid is 180 cm2. Calculate the surface area of the larger solid. … cm2 [3]
14 marks
Mark scheme: 3 (a) (i) 51.7 or 51.69 to 51.70… 4 M3 for 2 3 2 (2 × × π × 13 + π × 13 × 25) × 2.3 [ ÷ 1000] oe 3 or SC3 for figs 517 or figs 5169 to 5170… 2 3 2 or M2 for (2 × × π × 13 + π × 13 × 25) oe 3 OR 2 3 M1 for 2 × × π × 13 seen 3 or π × 132 × 25 seen M1indep for their volume × 2.3 ÷ 1000 (ii) 1.96 or 1.957 to 1.958 … 4 M3 for (2 × 2 × π × 132 + π × 2 × 13 × 25)[ ÷ 100 2 ] × 4.7 oe or SC3 for figs 196 or figs 1957 to 1958… M2 for (2 × 2 × π × 132 + π × 2 × 13 × 25) oe OR M1 for 2 × 2 × π × 132 seen or π × 2 × 13 × 25 seen M1indep for their area divided by 100² soi
This question in 0580/41 Oct/Nov 2016
Q34 · NOT TO 0.8 cm SCALE 0.8 cm 1.1 cm 1.5 cm The diagram shows two sweets 0580/42 Oct/Nov 2016
6 (a) NOT TO 0.8 cm SCALE 0.8 cm 1.1 cm 1.5 cm The diagram shows two sweets. The cuboid has length 1.5 cm, width 1.1 cm and height 0.8 cm. The cylinder has height 0.8 cm and the same volume as the cuboid. (i) Calculate the volume of the cuboid. … cm3 [2] (ii) Calculate the radius of the cylinder. … cm [2] (iii) Calculate the difference between the surface areas of the two sweets. … cm2 [5] (b) A bag of sweets contains x orange sweets and y lemon sweets. Each orange sweet costs 2 cents and each lemon sweet costs 3 cents. The cost of a bag of sweets is less than 24 cents. There are at least 9 sweets in each bag. There are at least 2 lemon sweets in each bag. (i) One of the inequalities that shows this information is 2x + 3y 1 24 . Write down the other two inequalities. … … [2] (ii) On the grid, by shading the unwanted regions, show the region which satisfies the three inequalities. y 12 11 10 9 8 7 6 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 [4] (iii) Find the lowest cost of a bag of sweets. Write down the value of x and the value of y that give this cost. Lowest cost = … cents x = … y = … [3]
18 marks
Mark scheme: 6 (a) (i) 1.32 2 M1 for 0.8 × 1.5 × 1.1 (ii) 0.725 or 0.7246 to 0.7247… 2 M1 for πr 2 × 0.8 = their(a)(i) or πr 2 = 1.5 × 1.1 oe (iii) 0.513 to 0.518 nfww 5 M1 for 2(1.5 × 1.1 + 1.5 × 0.8 + 1.1 × 0.8) M1 for [2 ×] π × (their (a)(ii)) 2 M2 for π × 2 × ( their (a)(ii) ) × 0.8 or M1 for π × 2 × ( their (a)(ii) ) (b) (i) x + y . 9 oe 1 y . 2 oe 1 If zero scored, SC1 for x + y > 9 and y > 2 (ii) Fully correct diagram with unwanted region shaded 4 B1 for 2x + 3y = 24 ruled B1 for x + y = 9 ruled B1 for y = 2 ruled (iii) 20 1 [x = ] 7 1 [y =] 2 1 If zero scored, SC1 for 2x + 3y evaluated from integers
This question in 0580/42 Oct/Nov 2016
Q35 · The diagram shows a cylindrical container used to serve coffee in a hotel 0580/41 May/June 2017
5 (a) The diagram shows a cylindrical container used to serve coffee in a hotel. NOT TO 18 cm SCALE 50 cm The container has a height of 50 cm and a radius of 18 cm. (i) Calculate the volume of the cylinder and show that it rounds to 50 900 cm3, correct to 3 significant figures. [2] (ii) 30 litres of coffee are poured into the container. Work out the height, h, of the empty space in the container. NOT TO SCALE h h = … cm [3] (iii) Cups in the shape of a hemisphere are filled with coffee from the container. The radius of a cup is 3.5 cm. NOT TO SCALE 3.5 cm Work out the maximum number of these cups that can be completely filled from the 30 litres of coffee in the container. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 … [4] (b) The hotel also uses glasses in the shape of a cone. r NOT TO 8.4 cm SCALE The capacity of each glass is 95 cm3. (i) Calculate the radius, r, and show that it rounds to 3.3 cm, correct to 1 decimal place. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 [3] (ii) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] … cm2 [4]
16 marks
Mark scheme: 5(a)(i) 50890 or 50893 to 50900.4 2 M1 for π × 182 × 50 5(a)(ii) 20.5 or 20.52 to 20.534 3 B2 for answer 29.5 or 29.46 to 29.48 OR oe π × 18 2 M2 for ( 50900 − 30000 ) ÷ ( ) or M1 for (figs 50.9 –figs 30) ÷ ( π× figs182) h π × 18 2 or M1 for ( 50900 − 30000 ) = ( ) oe OR alternative method 30000 M2 for 50 − oe π × 18 2 M1 for figs 30 = π × figs 182 × ( 50 −h ) oe figs 30 or for oe π × figs18 2 OR alternative method (50.9 − 30) M2 for × 50 oe 50.9 (50.9 − 30) 30 or M1 for or × 50 oe 50.9 50.9 or M1 for (figs 50.9 − figs 30) × 50 oe figs 50.9 5(a)(iii) 334 nfww 4 2 3 M2 for figs 30 ÷ π × 3.5 oe 3 1 4 3 or M1 for × π × 3.5 oe 2 3 and B1 for 30 000 5(b)(i) 3.28[6..] or 3.29 3 95 × 3 M2 for [ r2 = ] oe 8.4π 1 2 or M1 for π × r × 8.4 [ = 95] 3 5(b)(ii) 93.1 to 93.6 4 2 2 M3 for π × 3.3 × 3.3 + 8.4 or M2 for 3.32 + 8.4 2 or M1 for 3.32 + 8.42
This question in 0580/41 May/June 2017
Q36 · NOT TO SCALE 10 cm 3 cm The diagram shows a hollow cone with radius 3 cm and slant height… 0580/42 May/June 2017
5 NOT TO SCALE 10 cm 3 cm The diagram shows a hollow cone with radius 3 cm and slant height 10 cm. (a) (i) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [2] (ii) Calculate the perpendicular height of the cone. … cm [3] (iii) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 … cm3 [2] (b) O O NOT TO SCALE x 10 cm 3 cm P The cone is cut along the line OP and is opened out into a sector as shown in the diagram. Calculate the sector angle x. x = … [4] (c) O NOT TO SCALE The diagram shows the same sector as in part (b). Calculate the area of the shaded segment. … cm2 [4]
15 marks
Mark scheme: 5(a)(i) 94.2 or 94.3 or 94.24 to 94.26 2 M1 for π × 3 × 10 5(a)(ii) 9.54 or 9.539… 3 2 2 M2 for 10 − 3 or M1 for h 2 + 32 = 10 2 oe 5(a)(iii) 89.9 or 89.90 to 89.92… 2 1 2 M1 for × π × 3 × their (a)(ii) 3 5(b) 108 or 107.9 to 108.1 nfww 4 π × 3 × 10 their (a)(i) M3 for × 360 oe or × 360 oe or π × 10 2 π × 10 2 2 × π × 3 × 360 oe 2 × π × 10 x 2 or M2 for × π × 10 = their (a)(i) oe 360 x or × 2 × π × 10 = 2 × 3 × π oe 360 x 2 or M1 for × π × 10 seen 360 x or × 2 × π × 10 seen 360 5(c) 46.6 to 46.8 4 their (b) 2 1 M3 for × π × 10 − × 10 × 10 × sin(their (b)) oe 360 2 their (b) 2 or M1 for × π × 10 or their (a)(i) soi 360 1 and M1 for × 10 × 10 × sin(their (b)) soi 2
This question in 0580/42 May/June 2017
Q37 · The diagram shows a solid metal prism with cross section ABCDE 0580/43 May/June 2017
4 (a) The diagram shows a solid metal prism with cross section ABCDE. G 2 cm F B A NOT TO SCALE K 7 cm 4 cm E H J 8 cm C D 4 cm (i) Calculate the area of the cross section ABCDE. … cm2 [6] (ii) The prism is of length 8 cm. Calculate the volume of the prism. … cm3 [1] (b) A cylinder of length 13 cm has volume 280 cm3. (i) Calculate the radius of the cylinder. … cm [3] (ii) The cylinder is placed in a box that is a cube of side 14 cm. Calculate the percentage of the volume of the box that is occupied by the cylinder. … % [3]
13 marks
Mark scheme: 4(a)(i) 17.5 or 17.46 … nfww 6 B3 for triangle height 3.46[4...] or 12 oe or M2 for 4 2 − 2 2 or M1 for h 2 + 2 2 = 4 2 and M2 for 2 × 7 + 12 × 2 × their h oe or M1 for 12 × 2 × their h 4(a)(ii) 140 or 139.6 to 139.7... 1FT FT their (a) × 8 4(b)(i) 2.62 or 2.618… 3 280 M2 for [r2 = ] oe 13π or M1 for 280 = π × r2 × 13 4(b)(ii) 10 3 280 10.2 or 10.20… or 10 M2 for 3 [×100] oe 49 14 or B1 for 2 744 or 143 seen
This question in 0580/43 May/June 2017
Q38 · L h NOT TO SCALE 5 mm The diagram shows a solid made from a hemisphere and a cone 0580/41 Oct/Nov 2017
8 l h NOT TO SCALE 5 mm The diagram shows a solid made from a hemisphere and a cone. The base diameter of the cone and the diameter of the hemisphere are each 5 mm. 115r (a) The total surface area of the solid is mm2. 4 Show that the slant height, l, is 6.5 mm. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl.] [The surface area, A, of a sphere with radius r is A = 4rr2.] [4] (b) Calculate the height, h, of the cone. h = … mm [3] (c) Calculate the volume of the solid. 1 [The volume, V, of a cone with radius r and height h is V = rr2h.] 3 4 [The volume, V, of a sphere with radius r is V = rr3.] 3 … mm3 [4] (d) The solid is made from gold. 1 cubic centimetre of gold has a mass of 19.3 grams. The value of 1 gram of gold is $38.62 . Calculate the value of the gold used to make the solid. $ … [3]
14 marks
Mark scheme: 8(a) 2 M2 2 5 4 5 115π 5 4 5 π × × l + × π × = oe M1 for π × × l or × π × 2 2 2 4 2 2 2 115π 4 5 2 5 or – × π × = π × × l oe 4 2 2 2 5πl 65π B1 nfww = oe oe both terms must be written in terms of π 2 4 115π 2 or [ l = ] − 2 × π × 2.5 ÷ 2.5π oe nfww 4 or correct complete method for l with decimals 65π × 2 65π A1 [l =] or oe = 6.5 Correct calculation with no errors and B1 earned 4 × 5π 10π 8(b) 6 3 2 2 M2 for 6.5 − 2.5 or M1 for h2 + 2.52 = 6.52 If zero scored, SC2dep for answer 4.15[3]… 8(c) 72[.0…] or 71.99… nfww 4 2 3 π 5 1 4π 5 M3 for × × their 6 + × × 3 2 2 3 2 oe π 5 2 or M1 for × × their 6 oe 3 2 1 4π 5 3 and M1 for × × oe 2 3 2 If zero scored, SC3dep for π 2 1 4π 3 × ( 5 ) × their 4.15 + × × ( 5 ) oe 3 2 3 or π 2 SC1dep for × ( 5 ) × their 4.15 oe 3 1 4π 3 SC1dep for × × ( 5 ) oe 2 3 8(d) 53.7 or 53.65 to 53.67 3 M1 for figs (their (c)) × 19.3 × 38.62 or better M1 for ÷ 1000 soi
This question in 0580/41 Oct/Nov 2017
Q39 · NOT TO r 2r SCALE A sphere of radius r is inside a closed cylinder of radius r and height… 0580/42 Oct/Nov 2017
2 (a) NOT TO r 2r SCALE A sphere of radius r is inside a closed cylinder of radius r and height 2r. r r .] [The volume, V, of a sphere with radius r is V = 43 3 (i) When r = 8 cm, calculate the volume inside the cylinder which is not occupied by the sphere. … cm3 [3] (ii) Find r when the volume inside the cylinder not occupied by the sphere is 36 cm3. r = … cm [3] (b) 12 cm NOT TO SCALE 5 cm The diagram shows a solid cone with radius 5 cm and perpendicular height 12 cm. (i) The total surface area is painted at a cost of $0.015 per cm2. Calculate the cost of painting the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] $ … [4] (ii) The cone is made of metal and is melted down and made into smaller solid cones with radius 1.25 cm and perpendicular height 3 cm. Calculate the number of smaller cones that can be made. … [3]
13 marks
Mark scheme: 2(a)(i) 1070 or 1072. .. 3 M1 for π × 82 × 2 × 8 4 3 M1 for × π × 8 3 or M2 for 23 πr 3 or M1 for π r 2 2 r − 43 π r 3 2(a)(ii) 2.58 or 2.580 to 2.581 3 3 36 × 3 B2 for r = or better 2π 2 4 3 or M1 for π × r × 2 × r – × π × r = 36 oe 3 2(b)(i) 4.24 or 4.241 to 4.242 4 2 2 2 M3 for (π × 5 + π × 5 × 5 + 12 ) or M2 for π × 5 × 5 2 + 12 2 or M1 for 5 2 + 12 2 or π × 52 2(b)(ii) 64 cao final answer 3 [ kπ ] × 5 2 × 12 M2 for [ kπ ] × 1.25 2 × 3 or M1 for 13 × π × 5 2 × 12 or 13 × π × 1.25 2 × 3 OR 3 1 3 M2 for 4 or seen 4 1 or M1 for factor 4 or soi 4
This question in 0580/42 Oct/Nov 2017
Q40 · R h NOT TO SCALE 10 cm The diagrams show a cube, a cylinder and a hemisphere 0580/43 Oct/Nov 2017
6 (a) r h NOT TO SCALE 10 cm The diagrams show a cube, a cylinder and a hemisphere. The volume of each of these solids is 2000 cm3. (i) Work out the height, h, of the cylinder. h = … cm [2] (ii) Work out the radius, r, of the hemisphere. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 r = … cm [3] (iii) Work out the surface area of the cube. … cm2 [3] (b) NOT TO 7 cm SCALE 40º 10 cm (i) Calculate the area of the triangle. … cm2 [2] (ii) Calculate the perimeter of the triangle and show that it is 23.5 cm, correct to 1 decimal place. Show all your working. [5] (c) NOT TO SCALE cº 9 cm The perimeter of this sector of a circle is 28.2 cm. Calculate the value of c. c = … [3]
18 marks
Mark scheme: 6(a)(i) 25.5 or 25.46… 2 M1 for π × 52 × h = 2000 oe 6(a)(ii) 9.85 or 9.847… 3 2 M2 for [r3=] 2000 ÷ π oe 3 2 or M1 for πr3 = 2000 oe 3 6(a)(iii) 952 or 952.4…. 3 3 2 M2 for [6 ×] 2000 or M1 for 3 2000 or 6 times their area of one face 6(b)(i) 22.5 or 22.49… 2 1 M1 for × 7 × 10 × sin40 2 6(b)(ii) √(102 + 72 – 2 × 10 × 7 cos40) + 7 M3 M2 for 102 + 72 – 2 × 10 × 7 cos40 + 10 or M1 for correct implicit cosine rule 23.46… A2 A1 for 6.46… or 41.7 to 41.8 6(c) 64.9 or 64.92 to 64.94 3 c M2 for 28.2 – 2 × 9 = × 2 × π × 9 oe 360 c or M1 for × 2 × π × 9 soi 360
This question in 0580/43 Oct/Nov 2017
Q41 · A solid hemisphere has volume 230 cm3 0580/41 May/June 2018
6 A solid hemisphere has volume 230 cm3. (a) Calculate the radius of the hemisphere. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 … cm [3] (b) A solid cylinder with radius 1.6 cm is attached to the hemisphere to make a toy. NOT TO SCALE The total volume of the toy is 300 cm3. (i) Calculate the height of the cylinder. … cm [3] (ii) A mathematically similar toy has volume 19 200 cm3. Calculate the radius of the cylinder for this toy. … cm [3]
9 marks
Mark scheme: 6(a) 4.79 or 4.788 to 4.789 3 230 × 3 M2 for 3 oe 2 × π 2 3 or M1 for 230 = × π ×r oe 3 If 0 scored SC1 for answer 3.8[0…] 6(b)(i) 8.7[0] or 8.702 to 8.704 3 M2 for (300 − 230) ÷ (1.6 2 π) or M1 for π × 1.6 2 × h 6(b)(ii) 6.4 3 19200 M2 for 1.6 × 3 oe 300 19200 300 or M1 for sf 3 or 3 oe 300 19200 1.6 3 300 or for = r 19200
This question in 0580/41 May/June 2018
Q42 · NOT TO SCALE 1.5 cm Water flows through a cylindrical pipe at a speed of 8 cm/s 0580/43 May/June 2018
7 (a) NOT TO SCALE 1.5 cm Water flows through a cylindrical pipe at a speed of 8 cm/s. The radius of the circular cross-section is 1.5 cm and the pipe is always completely full of water. Calculate the amount of water that flows through the pipe in 1 hour. Give your answer in litres. … litres [4] (b) NOT TO SCALE 12 cm y cm x cm 6 cm The diagram shows three solids. The base radius of the cone is 6 cm and the slant height is 12 cm. The radius of the sphere is x cm and the radius of the hemisphere is y cm. The total surface area of each solid is the same. (i) Show that the total surface area of the cone is 108r cm2. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] [2] (ii) Find the value of x and the value of y. [The surface area, A, of a sphere with radius r is A = 4 r r 2 .] x = … y = … [4]
10 marks
Mark scheme: 7(a) 204 or 203.5 to 203.6… nfww 4 M2 for π × 1.5 2 × 8 × 60 × 60 or M1 for π × 1.52 M1 for dividing their volume by 1000 If 0 scored SC1 for an answer figs 204 or figs 2035 to 2036 without working 7(b)(i) π × 6 × 12 + π × 62 = 108π M2 M1 for π × 6 × 12 7(b)(ii) [x = ] 5.2[0] or 5.196… 4 B2 or M1 for 4πx 2 = 108π seen [y = ] 6 B2 or M1 for ½(4πy2) + πy2 or better seen
This question in 0580/43 May/June 2018
Q43 · NOT TO SCALE 18 cm h cm x° 6 cm The diagram shows a prism with length 18 cm and volume… 0580/41 Oct/Nov 2018
5 NOT TO SCALE 18 cm h cm x° 6 cm The diagram shows a prism with length 18 cm and volume 253.8 cm3. The cross-section of the prism is a right-angled triangle with base 6 cm and height h cm. (a) (i) Show that the value of h is 4.7 . [3] (ii) Calculate the value of x. x = … [2] (b) Calculate the total surface area of the prism. … cm2 [6]
11 marks
Mark scheme: 5(a)(i) 6 3 For M3 no errors at any stage [h =] 253 8. ÷ 18 ÷ or 1 2 M2 for 253.8 = × 6 × h × 18 oe (no 2 253.8 × 2 [h =] or previous errors) 6 × 18 1 253.8 or M1 for triangle area = ×6 × h soi [h =] 2 6 18 × 2 5(a)(ii) 38.1 or 38.06 to 38.08 2 7.4 M1 for tan = oe 6 5(b) 358 or 357.9 to 358 6 M1 for 6 2 + 7.4 2 M1 for 6 2 + 4.7 2 × 18 [× 2] M1 for 6 × 18 [× 2] M1 for 4.7 × 18 1 M1 for 2 × × 6 × 4.7 oe 2
This question in 0580/41 Oct/Nov 2018
Q44 · NOT TO 17 cm SCALE 8 cm The diagram shows a solid cone 0580/43 Oct/Nov 2018
3 (a) NOT TO 17 cm SCALE 8 cm The diagram shows a solid cone. The radius is 8 cm and the slant height is 17 cm. (i) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [2] (ii) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 … cm3 [4] (iii) The cone is made of wood and 1 cm3 of the wood has a mass of 0.8 g. Calculate the mass of the cone. … g [1] (iv) The cone is placed in a box. The total mass of the cone and the box is 1.2 kg. Calculate the mass of the box. Give your answer in grams. … g [1] (b) NOT TO 8r r SCALE 3r The diagram shows a solid cylinder and a solid sphere. The cylinder has radius 3r and height 8r. The sphere has radius r. (i) Find the volume of the sphere as a fraction of the volume of the cylinder. Give your answer in its lowest terms. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 … [4] (ii) The surface area of the sphere is 81r cm2. Find the curved surface area of the cylinder. Give your answer in terms of r. [The surface area, A, of a sphere with radius r is A = 4 r r 2 .] … cm2 [4]
16 marks
Mark scheme: 3(a)(i) 427 or 427.2 to 427.3… 2 M1 for π × 8 × 17 3(a)(ii) 1010 or 1005…. 4 2 2 M2 for 17 − 8 oe or M1 for h 2 + 82 = 17 2 oe 1 2 M1 for × π × 8 × their h oe 3 3(a)(iii) 804 or 804.2 to 804.4 or 808 1 FT their (ii) × 0.8 3(a)(iv) 396 or 395.6 to 395.8 or 392 1 FT 1200 – their (iii) 3(b)(i) 1 4 4 3 πr 54 3 B3 for or better 72πr 3 4 3 × π × r 3 or M2 for or 72 × π × r3 π × (3r ) 2 × 8 r or M1 for π × (3r ) 2 × 8r 1 If 0 scored, SC2 for answer of 18 3(b)(ii) 972π final answer 4 9 B2 for r = oe 2 or M1 for 4πr 2 = 81π or better M1 for 2 × π × (3 × their r) × (8 × their r) isw
This question in 0580/43 Oct/Nov 2018
Q45 · NOT TO 1.2 m SCALE 3 m The diagram shows the surface of a garden pond, made from a… 0580/41 May/June 2019
5 NOT TO 1.2 m SCALE 3 m The diagram shows the surface of a garden pond, made from a rectangle and two semicircles. The rectangle measures 3 m by 1.2 m. (a) Calculate the area of this surface. … m2 [3] (b) The pond is a prism and the water in the pond has a depth of 20 cm. Calculate the number of litres of water in the pond. … litres [3] (c) After a rainfall, the number of litres of water in the pond is 1007. Calculate the increase in the depth of water in the pond. Give your answer in centimetres. … cm [3]
9 marks
Mark scheme: 5(a) 4.73 or 4.730 to 4.731... 3 M2 for 3 × 1.2 + π × 0.62 oe 2 1 2 or M1 for π × 0.6 or × π × 0.6 or 2 3 × 1.2 5(b) 946 or 946.0 to 946.2... 3 M2 for their (a) × 0.2 × 1000 oe or M1 for their (a) × 0.2 or 20 implied by figs 946[0] to 9462 5(c) 1.28 or 1.29 or 1.284 to 1.290 3 (1007 − their (b)) ÷ 1000 M2 for × 100 oe their (a) 1007 − their ( b ) or for × 20 oe their ( b ) 1007 − their ( b ) or M1 for figs or their ( a ) 1007 figs their ( a ) 1007 − their ( b ) or for or their ( b ) 1007 × 20 oe their ( b )
This question in 0580/41 May/June 2019
Q46 · The volume of each of the following solids is 1000 cm3 0580/41 May/June 2019
10 The volume of each of the following solids is 1000 cm3. Calculate the value of x for each solid. (a) A cube with side length x cm. x = … [1] (b) A sphere with radius x cm. 4 3 [The volume, V, of a sphere with radius r is V = r r . ] 3 x = … [3] (c) NOT TO x 5cm SCALE x cm A cone with radius x cm and slant height x 5cm. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h. ] 3 x = … [4] (d) x NOT TO cm 2 SCALE x cm 27x cm 2 A prism with a right-angled triangle as its cross-section. x = … [4] Question 11 is printed on the next page.
12 marks
Mark scheme: 10(a) 10 1 10(b) 6.2[0] or 6.203 to 6.204 3 4 M2 for [x3 = ] 1000 ÷ π oe or better 3 4 3 or M1 for πx = 1000 3 10(c) 7.82 or 7.815 to 7.816 4 3 1 B3 for [ x = ]1000 ÷ π ÷ 2 oe or better 3 2 or M1 for x 5 − x 2 soi by 4x2 or 2x ( ) 1 2 M1dep for π × x × theirh[ = 1000] 3 10(d) 2 4 3 27 3 x 6 or 6.67 or 6.666 to 6.667 B3 for [ x = ]1000 ÷ oe or = 10 or 3 8 2 better 1 x 27 x or M2 for × x × × = 1000 oe 2 2 2 1 x or M1 for × x × 2 2 If 0 scored, SC2 for answer 5.29 or 5.291..
This question in 0580/41 May/June 2019
Q47 · The volume of a solid metal sphere is 24 430 cm3 0580/42 May/June 2019
10 (a) The volume of a solid metal sphere is 24 430 cm3. (i) Calculate the radius of the sphere. 4 3 [The volume, V, of a sphere with radius r is V = r r . ] 3 … cm [3] (ii) The metal sphere is placed in an empty tank. The tank is a cylinder with radius 50 cm, standing on its circular base. Water is poured into the tank to a depth of 60 cm. Calculate the number of litres of water needed. … litres [3] (b) A different tank is a cuboid measuring 1.8 m by 1.5 m by 1.2 m. Water flows from a pipe into this empty tank at a rate of 200 cm3 per second. Find the time it takes to fill the tank. Give your answer in hours and minutes. … hours … minutes [4] (c) NOT TO SCALE Area = 159.5 cm2 Area = 295 cm2 17 cm The diagram shows two mathematically similar shapes with areas 295 cm2 and 159.5 cm2. The width of the larger shape is 17 cm. Calculate the width of the smaller shape. … cm [3]
13 marks
Mark scheme: 10(a)(i) 18[.0] or 17.99 to 18.00… 3 24430 × 3 M2 for 3 oe 4 π 4 3 or M1 for πr = 24430 3 10(a)(ii) 447 or 446.8 to 446.9... 3 M2 for π × 50 2 × 60 − 24430 oe or M1 for π × 502 × 60 oe 10(b) 4 [hours] 30 [ mins] nfww 4 B3 for 16200 or 4.5 or 270 figs 18 × figs 15 × figs 12 or M2 for oe figs 2 or M1 for figs 18 × figs 15 × figs 12 oe 10(c) 12.5 or 12.50… 3 159 5. M2 for 17 × oe 295 159 5. 295 or M1 for or seen 295 159 5. 159.5 x 2 or for = oe 295 17 2
This question in 0580/42 May/June 2019
Q48 · 5.6 cm NOT TO 10 cm SCALE The diagram shows a hemispherical bowl of radius 5.6 cm and a… 0580/43 May/June 2019
4 (a) 5.6 cm NOT TO 10 cm SCALE The diagram shows a hemispherical bowl of radius 5.6 cm and a cylindrical tin of height 10 cm. (i) Show that the volume of the bowl is 368 cm3, correct to the nearest cm3. 4 3 [The volume, V, of a sphere with radius r is V = r r . ] 3 [2] (ii) The tin is completely full of soup. When all the soup is poured into the empty bowl, 80% of the volume of the bowl is filled. Calculate the radius of the tin. … cm [4] (b) NOT TO SCALE 6 cm 1.75 cm The diagram shows a cone with radius 1.75 cm and height 6 cm. (i) 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 = r rl . ] … cm2 [5] (ii) NOT TO SCALE 4.5 cm 1.75 cm The cone contains salt to a depth of 4.5 cm. The top layer of the salt forms a circle that is parallel to the base of the cone. (a) Show that the volume of the salt inside the cone is 18.9 cm3, correct to 1 decimal place. 1 2 [The volume, V, of a cone with radius r and height h is V = r r h. ] 3 [4] (b) The salt is removed from the cone at a constant rate of 200 mm3 per second. Calculate the time taken for the cone to be completely emptied. Give your answer in seconds, correct to the nearest second. … s [3]
18 marks
Mark scheme: 4(a)(i) 1 4 3 M1 × × π × 5.6 2 3 367.8... to 367.9 A1 4(a)(ii) 3.06 or 3.060 to 3.061... 4 M1 for 0.8 × 368 [= 294.4] their 294.4 M2 for [r2 =] oe 10π or M1 for πr2 × 10 = their 294.4 oe 4(b)(i) 44[.0] or 43.98 to 43.99 nfww 5 25 B2 for [slant height = ] oe 4 or M1 for [l2 = ] 62 + 1.752 oe M2 for π × 1.75 × theirl + π × 1.75 2 or M1 for π × 1.75 ×theirl or π × 1.75 2 4(b)(ii)(a) 1 B1 SF = oe soi 4 1 2 1 2 M2 M1 for π × 1.75 × 6 − π × their 0.4375 × 1.5 3 3 1 2 1 2 π × 1.75 × 6 or π × their 0.4375 × 1.5 OR 3 3 OR 3 1 1 3 π × 1.75 2 × 6 × 1 − oe 1 3 4 oe M1 for 1– 4 18.94 or 18.939 to18.944… A1 4(b)(ii)(b) 95 final answer 3 B2 for 94.5 or 94.69 to 94.722 OR M2 for 18.9 ×103 ÷ 200 oe or M1 for 18.9 × 103 or 200 ÷ 103 or figs 189..÷ 200 or 18.9.. ÷ figs 2
This question in 0580/43 May/June 2019
Q49 · Calculate the external curved surface area of a cylinder with radius 8 m and height 19 m 0580/41 Oct/Nov 2019
4 (a) (i) Calculate the external curved surface area of a cylinder with radius 8 m and height 19 m. … m2 [2] (ii) This surface is painted at a cost of $0.85 per square metre. Calculate the cost of painting this surface. $ … [2] (b) A solid metal sphere with radius 6 cm is melted down and all of the metal is used to make a solid cone with radius 8 cm and height h cm. (i) Show that h = 13.5 . 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 [2] (ii) Calculate the slant height of the cone. … cm [2] (iii) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [1] (c) Two cones are mathematically similar. The total surface area of the smaller cone is 80 cm2. The total surface area of the larger cone is 180 cm2. The volume of the smaller cone is 168 cm3. Calculate the volume of the larger cone. … cm3 [3] (d) The diagram shows a pyramid with a P square base ABCD. DB = 8 cm. NOT TO P is vertically above the centre, X, of SCALE the base and PX = 5 cm. D C X A B Calculate the angle between PB and the base ABCD. … [3]
15 marks
Mark scheme: 4(a)(i) 955 or 955.0 to 955.2 2 M1 for 2 × π × 8 × 19 oe 4(a)(ii) 812 or 811.7 to 811.9... 2 FT their (i) × 0.85 M1 for their (i) × 0.85 or their (i) × 85 4(b)(i) 4 3 M2 4 3 1 2 × π × 6 M1 for × π × 6 = × π × 8 × h 3 3 3 or cancelling clearly 1 2 × π × 8 3 seen to reach 13.5 4(b)(ii) 15.7 or 15.69... 2 M1 for 82 + 13.52 or better 4(b)(iii) 394 or 395 or 394.3 to 394.6... 1 FT π × 8 × their (b)(ii) 4(c) 567 3 3 168 80 2 M2 for = oe or better V 180 1 1 180 2 80 2 or M1 for or oe seen or 80 180 better 4(d) 51.3 or 51.34... 3 5 M2 for tan = oe 4 or M1 for recognition of angle PBX
This question in 0580/41 Oct/Nov 2019
Q50 · NOT TO SCALE 6 cm The diagram shows a hemisphere with radius 6 cm 0580/42 Oct/Nov 2019
4 (a) NOT TO SCALE 6 cm The diagram shows a hemisphere with radius 6 cm. Calculate the volume. Give the units of your answer. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 … … [3] (b) A B NOT TO 5.2 cm SCALE 10 cm F C E D 18 cm The diagram shows a prism ABCDEF. The cross-section is a right-angled triangle BCD. BD = 10 cm, BC = 5.2 cm and ED = 18 cm. (i) (a) Work out the volume of the prism. … cm3 [6] (b) Calculate angle BEC. Angle BEC = … [4] (ii) The point G lies on the line ED and GD = 7 cm. Work out angle BGE. Angle BGE = … [3]
16 marks
Mark scheme: 4(a) 452 or 452.2 to 452.4… 2 1 4 3 M1 for × × π× 6 2 3 cm3 1 4(b)(i)(a) 400 or 399.6 to 399.9 6 B3 for [CD =] 72.96 or [angle CBD =] 58.7 or 58.66 to 58.67 or M2 for 10 2 − 5.2 2 oe or 2.5 [CBD = ] cos-1 oe 10 or M1 for (CD)2 + 5.22 = 102 oe or 5.2 cos [CBD] = oe 10 5.2 or sin [CDB] = oe 10 5.2 × their CD M1dep for oe 2 or 12 × 5.2 ×10 × sin(their CBD) oe M1 for their area × 18 oe 4(b)(i)(b) 14.6 or 14.62 to 14.63… 4 2.5 M3 for sin BEC = oe 10 2 + 18 2 or M2 for [BE=] 10 2 + 18 2 oe seen or [EC = ] 18 2 + 10 2 − 5.2 2 oe seen or M1 for [BE2 =] 102 + 182 oe seen or [EC2=] 182 + 102 – 5.22 seen 4(b)(ii) 125 or 124.9 to 125.0… 3 B2 for 55[.0…] seen 10 or M2 for 180 – tan-1 oe 7 112 + (10 2 + 7 2 ) − (10 2 + 18) 2 or cos EGB = oe 2 × 11 × 10 2 + 7 2 10 or M1 for tan[ ] = oe 7 or for (102 + 182) = 112 + (102 + 72) – 2×11× 10 2 + 7 2 cos EGB oe
This question in 0580/42 Oct/Nov 2019
Q51 · Manjeet uses 220 litres of water each day 0580/42 Feb/March 2020
3 (a) Manjeet uses 220 litres of water each day. She reduces the amount of water she uses by 15%. Calculate the number of litres of water she now uses each day. … litres [2] (b) Manjeet has two mathematically similar bottles in her bathroom. The large bottle holds 1.35 litres and is 29.7 cm high. The small bottle holds 0.4 litres. Calculate the height of the small bottle. … cm [3] (c) Water from Manjeet’s shower flows at a rate of 12 litres per minute. The water from the shower flows into a tank that is a cuboid of length 90 cm and width 75 cm. Calculate the increase in the level of water in the tank when the shower is used for 7 minutes. … cm [3]
8 marks
Mark scheme: 3(a) 187 2 15 M1 for 220 × 1 − oe 100 or B1 for 33 seen 3(b) 19.8 3 0.4 M2 for 29.7 × 3 oe 1.35 0.4 1.35 or M1 for 3 or 3 oe seen 1.35 0.4 29.7 3 1.35 or for = oe x 3 0.4 3(c) 12.4 or 12.44… 3 M1 for 90 × 75 × h = 7 × figs 12 B1 for 1000 cm3 = 1 litre soi
This question in 0580/42 Feb/March 2020
Q52 · A solid metal cone has radius 1.65 cm and slant height 4.70 cm 0580/42 Feb/March 2020
4 A solid metal cone has radius 1.65 cm and slant height 4.70 cm. (a) 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 .] … cm2 [2] (b) Find the angle the slant height makes with the base of the cone. … [2] (c) (i) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [4] (ii) A metal sphere with radius 5 cm is melted down to make cones identical to this one. Calculate the number of complete identical cones that are made. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3
12 marks
Mark scheme: 4(a) 32.9 or 32.91 to 32.92… 2 M1 for π × 1.65 × 4.7 + π × 1.652 4(b) 69.4 or 69.44 to 69.45 2 M1 for cos = 1.65 ÷ 4.7 oe 4(c)(i) 12.5 or 12.54 to 12.55 4 1 2 2 2 M3 for × π × 1.65 × 4.7 − 1.65 oe 3 or M2 for 4.7 2 − 1.65 2 oe or for 4.7 × sin(their (b)) oe or M1 for 1.65 2 + h 2 = 4.7 2 oe h or for = sin(their (b)) oe 4.7 4(c)(ii) 41 nfww 4 B3 for 41.7… to 41.9 4 3 or M2 for × π × 5 ÷their 12.5 3 4 3 or M1 for × π × 5 3 After M2 scored, M1 for truncating their decimal number of cones seen to an integer answer
This question in 0580/42 Feb/March 2020
Q53 · C R NOT TO SCALE A B 8 cm P Q 12 cm Triangle ABC is mathematically similar to triangle PQR 0580/42 May/June 2020
8 (a) C R NOT TO SCALE A B 8 cm P Q 12 cm Triangle ABC is mathematically similar to triangle PQR. The area of triangle ABC is 16 cm2. (i) Calculate the area of triangle PQR. … cm2 [2] (ii) The triangles are the cross-sections of prisms which are also mathematically similar. The volume of the smaller prism is 320 cm3. Calculate the length of the larger prism. … cm [3] (b) A cylinder with radius 6 cm and height h cm has the same volume as a sphere with radius 4.5 cm. Find the value of h. 4 3 [The volume, V, of a sphere with radius r is V = rr . ] 3 h = … [3] (c) A solid metal cube of side 20 cm is melted down and made into 40 solid spheres, each of radius r cm. Find the value of r. 4 3 [The volume, V, of a sphere with radius r is V = rr . ] 3 r = … [3] 7x(d) A solid cylinder has radius x cm and height cm. 2 The surface area of a sphere with radius R cm is equal to the total surface area of the cylinder. Find an expression for R in terms of x. [The surface area, A, of a sphere with radius r is A = 4rr 2 . ] R = … [3]
14 marks
Mark scheme: 8(a)(i) 36 2 2 2 8 12 M1 for or oe 12 8 8(a)(ii) 30 3 12 M2 for 320 ÷ 16 × oe 8 or M1 for 320 ÷ 16 8(b) 3.375 cao 3 4 3 π × 4.5 3 M2 for or better π × 6 2 2 4 3 or M1 for π × 6 × h = × π × 4.5 3 8(c) 3.63 or 3.627 to 3.628 3 20 3 M2 for 4 40 × π 3 4 3 3 or M1 for 40 × × π × r = 20 3 8(d) 3x 1 3 B2 for 4 R 2 = 9 x 2 oe or better or 1.5x or x 2 12 2 2 7 x or M1 for 4πR = 2πx + π × 2 x × 2
This question in 0580/42 May/June 2020
Q54 · H 5 cm G NOT TO D SCALE C E F 70° 12 cm A B 8 cm The diagram shows a prism with a… 0580/42 Oct/Nov 2020
9 H 5 cm G NOT TO D SCALE C E F 70° 12 cm A B 8 cm The diagram shows a prism with a rectangular base, ABFE. The cross-section, ABCD, is a trapezium with AD = BC. AB = 8 cm, GH = 5 cm, BF = 12 cm and angle ABC = 70°. (a) Calculate the total surface area of the prism. … cm2 [6] (b) The perpendicular from G onto EF meets EF at X. (i) Show that EX = 6.5 cm. [1] (ii) Calculate AX. AX = … cm [2] (iii) Calculate the angle between the diagonal AG and the base ABFE. … [2]
11 marks
Mark scheme: 9(a) 315 or 314.5 to 315.0 6 height M1 for tan70 = oe or better seen 1 ( 8 - 5 ) 2 1 M1dep for ( 8 + 5 ) ×their height or better 2 seen dep on trig attempt for height 1 ( 8 − 5 ) 2 M2 for 12 × oe or better seen cos70 1 ( 8 − 5 ) 2 or M1 for oe or better seen cos70 M1 for 8 × 12 oe isw and 5 × 12 oe isw 9(b)(i) 8 – ½ (8 – 5) or 5 + ½ (8 – 5) M1 9(b)(ii) 13.6 or 13.64 to 13.65 2 M1 for 122 + (6.5)2 oe 9(b)(iii) 16.8 or 16.9 or 16.79 to 16.91… 2 M1 for identifying angle GAX from a nfww diagram or from working or better
This question in 0580/42 Oct/Nov 2020
Q55 · H G F E 11 cm NOT TO SCALE D C 5 cm A 8 cm B ABCDEFGH is a cuboid 0580/43 Oct/Nov 2020
6 H G F E 11 cm NOT TO SCALE D C 5 cm A 8 cm B ABCDEFGH is a cuboid. AB = 8 cm, BC = 5 cm and CG = 11 cm. (a) Work out the volume of the cuboid. … cm3 [2] (b) Ivana has a pencil of length 13 cm. Does this pencil fit completely inside the cuboid? Show how you decide. [4] (c) (i) Calculate angle CAB. Angle CAB = … [2] (ii) Calculate angle GAC. Angle GAC = … [2]
10 marks
Mark scheme: 6(a) 440 2 M1 for 8 × 5 × 11 6(b) 2 2 2 M3 M2 for 82 + 52 + 112 or 82 + 112 oe 8 + 5 + 11 oe or M1 for 82 + 52 or 52 + 112 oe or 82 + 52 + 112 and 132 ALTERNATIVE 8 2 + 112 or 82 + 112 and 132 Yes and 14.5 or 14.4 or 14.49… A1 Accept equivalent conclusion or Yes and 13.6[0…] 6(c)(i) 32.0[…] 2 5 M1 for tan[..] = oe 8 6(c)(ii) 49.4 or 49.38 to 49.39 2 11 M1 for sin[..] = oe their AG
This question in 0580/43 Oct/Nov 2020
Q56 · A box is a cuboid with length 45 cm, width 30 cm and height 42 cm 0580/42 Feb/March 2021
10 (a) A box is a cuboid with length 45 cm, width 30 cm and height 42 cm. The box is completely filled with 90.72 kg of sand. Calculate the density of this sand in kg/m3. [Density = mass ÷ volume] … kg/m3 [3] (b) A bag contains 15000 cm 3 of sand. Some of this sand is used to completely fill a hole in the shape of a cylinder. The hole is 30 cm deep and has radius 10 cm. Calculate the percentage of the sand from the bag that is used. … % [3] (c) Sand costs $98.90 per tonne. This cost includes a tax of 15%. Calculate the amount of tax paid per tonne of sand. $ … [3] (d) Raj buys some sand for 3540 rupees. Calculate the cost in dollars when the exchange rate is $1 = 70.8 rupees. $ … [2]
11 marks
Mark scheme: 10(a) 1600 3 B2 for answer figs 16 or M2 for 90.72 ÷ (figs45 × figs3 × figs42) or M1 for volume = figs 45 × figs 3 × figs 42 isw 10(b) 62.8 or 62.83 to 62.84 3 π× 10 2 × 30 M2 for × 100 15000 or M1 for π × 10 2 × 30 10(c) 12.9[0] 3 B2 for 86 OR 98.9 98.9 M2 for × 0.15 oe or 98.9 − oe 15 15 1 + 1 + 100 100 15 or M1 for 1 + a = 98.9 oe isw 100 10(d) 50 2 M1 for 3540 ÷ 70.8
This question in 0580/42 Feb/March 2021
Q57 · NOT TO SCALE 6.3 cm R cm 2.4 cm The diagram shows a solid cone and a solid hemisphere 0580/41 May/June 2021
3 (a) NOT TO SCALE 6.3 cm R cm 2.4 cm The diagram shows a solid cone and a solid hemisphere. The cone has radius 2.4 cm and slant height 6.3 cm. The hemisphere has radius R cm. The total surface area of the cone is equal to the total surface area of the hemisphere. Calculate the value of R. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] [The curved surface area, A, of a sphere with radius r is A = 4rr 2 .] R = … [4] (b) NOT TO SCALE 16 cm 12 cm 7.6 cm 7.6 cm The diagram shows a solid cone with radius 7.6 cm and height 16 cm. A cut is made parallel to the base of the cone and the top section is removed. The remaining solid has height 12 cm, as shown in the diagram. Calculate the volume of the remaining solid. r r h .] [The volume, V, of a cone with radius r and height h is V = 13 2 … cm3 [4]
8 marks
Mark scheme: 3(a) 2.64 or 2.638… 4 2 π × 2.4 2 + π × 2.4 × 6.3 M3 for [ R = ] oe π + 2π or M2 for 2 2 1 2 π× 2.4 + π× 2.4 × 6.3 = πR + × 4πR 2 or M1 for [ π × 2.4 2 ] + π× 2.4 × 6.3 oe 2 1 2 or [ πR ] + × 4πR oe 2 3(b) 953 or 952.6 to 952.8 4 16 − 12 3 1 2 M3 for × π× 7.6 × 16 × 1 − 3 16 1 2 1 2 or × π× 7.6 × 16 − × π× 1.9 × (16 − 12 ) 3 3 OR 16 − 12 3 B1 for top radius = 1.9 or oe 16 M2 for 1 2 1 2 × π× 7.6 × 16 − × π× (their 1.9) × (16 − 12 ) 3 3 16 − 12 3 1 2 or × π× 7.6 × 16 × 1 − their 3 16 1 2 or M1 for × π× 7.6 × 16 3 1 2 or for × π× (their 1.9) × (16 − 12 ) 3
This question in 0580/41 May/June 2021
Q58 · D 9 NOT TO A SCALE 13 cm 20 cm E F 24 cm B C The diagram shows a prism, ABCDEF 0580/41 May/June 2021
D 9 NOT TO A SCALE 13 cm 20 cm E F 24 cm B C The diagram shows a prism, ABCDEF. AB = 13 cm, AC = 20 cm, CF = 24 cm and angle ABC = 90°. (a) Calculate the total surface area of the prism. … cm2 [6] (b) Calculate the volume of the prism. … cm3 [1] (c) Calculate the angle that AF makes with the base BCFE. … [4]
11 marks
Mark scheme: 9(a) 1350 or 1354…. 6 M2 for 20 2 − 132 or M1 for BC2 + 132 = 202 A1 for 231 or 15.2 or 15.19 to 15.20 M1 for 20 × 24 and 13 × 24 and their 15.2 × 24 M1 for [½ ×] their 15.2 × 13 9(b) 2370 or 2369 to 2371… cao 1 9(c) 24.6 or 24.58 to 24.59 4 13 M3 for sin [...] = oe 20 2 + 24 2 or M2 for 20 2 + 24 2 or 24 2 + 20 2 − 132 or M1 for AF2 = 202 + 242 or 242 + 202 - 132 or M1 for correct angle identified
This question in 0580/41 May/June 2021
Q59 · A solid cuboid measures 20 cm by 12 cm by 5 cm 0580/43 May/June 2021
8 (a) A solid cuboid measures 20 cm by 12 cm by 5 cm. (i) Calculate the volume of the cuboid. … cm3 [1] (ii) (a) Calculate the total surface area of the cuboid. … cm2 [3] (b) The surface of the cuboid is painted. The cost of the paint used is $1.52 . Find the cost to paint 1 cm 2 of the cuboid. Give your answer in cents. … cents [1] 9x (b) A solid metal cylinder with radius x and height is melted. 2 All the metal is used to make a sphere with radius r. Find r in terms of x. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 r = … [3] (c) NOT TO SCALE 20 cm 5 cm 150 cm The diagram shows a cylinder of length 150 cm on horizontal ground. The cylinder has radius 20 cm. The cylinder contains water to a depth of 5 cm, as shown in the diagram. Calculate the volume of water in the cylinder. Give your answer in litres. … litres [7]
15 marks
Mark scheme: 8(a)(i) 1200 1 8(a)(ii)(a) 800 3 M2 for [2 ×] (20 × 12 + 20 × 5 + 12 × 5) or M1 for 20 × 12 or 20 × 5 or 12 × 5 8(a)(ii)(b) 0.19 1 FT 152 ÷ their 800 8(b) 3 x 3 3 27 x 3 [π] or 1.5x B2 for r = or better 2 8[π] 4 3 2 9 x or M1 for πr = πx × 3 2 8(c) 13.6 or 13.59 to 13.61 7 If chord is AB and O is centre of the cross section −1 20 − 5 M2 for 2 × cos oe 20 20 − 5 or M1 for cos = oe 20 theirAOB 2 M1 for × π × 20 360 1 82.8π or (20)2 2 180 1 M1 for × 202 × sin(their AOB) oe 2 M1 for their area × 150 M1 for their volume ÷ 1000
This question in 0580/43 May/June 2021
Q60 · The diagram shows a container for storing grain 0580/42 Oct/Nov 2021
7 (a) The diagram shows a container for storing grain. The container is made from a hemisphere, a cylinder and a cone, each with radius 2 m. 2 m The height of the cylinder is 5.2 m and the height of the cone is h m. 5.2 m NOT TO SCALE 2 m h m (i) Calculate the volume of the hemisphere. Give your answer as a multiple of r. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … m3 [2] 88 r (ii) The total volume of the container is m3 . 3 Calculate the value of h. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 h = … [4] (iii) The container is full of grain. Grain is removed from the container at a rate of 35 000 kg per hour. 1m3 of grain has a mass of 620 kg. Calculate the time taken to empty the container. Give your answer in hours and minutes. … h … min [3] (b) O NOT TO r cm 140° B SCALE A A and B are points on a circle, centre O, radius r cm. The area of the shaded segment is 65cm2. Calculate the value of r. r = … [4]
13 marks
Mark scheme: 7(a)(i) 16π 2 1 4 3 or 51 π final answer M1 for × π × 2 oe 3 3 2 3 7(a)(ii) 2.4[0] 4 B3 for answer in range 2.396… to 2.40… OR 16π M3 for their + π × 22 × 5.2 + 3 1 2 88π π× 2 × h = oe 3 3 88π 16π or M2 for – their – π × 22 × 5.2 3 3 oe or M1 for π × 22 × 5.2 oe 1 2 or π× 2 × h oe soi 3 7(a)(iii) 1 hour 38 min or 1 hour 37.8 min to 1 3 B2 for 1.63[2…] or 98 [mins] or 97.8 to hour 37.9… min 97.9… ] 88π × 620 3 or M1 for [× 60] oe 35000 7(b) 8.5[0] or 8.496 to 8.497 4 65 M3 for [r= ] oe 140 1 π − sin140 360 2 140 1 or M2 for π × r2 – r2 × sin140 [=65] 360 2 oe or M1 for either area expression seen
This question in 0580/42 Oct/Nov 2021
Q61 · 5 cm NOT TO SCALE 4 cm C D A B 10 cm The diagram shows a prism 0580/43 Oct/Nov 2021
3 (a) 5 cm NOT TO SCALE 4 cm C D A B 10 cm The diagram shows a prism. The cross-section of the prism is a trapezium with CD parallel to AB and AC = BD. AB = 10 cm, CD = 4 cm and the height of the trapezium is 5 cm. The volume of the prism is 525 cm3. (i) The prism is made of iron. 1 cm3 of iron has a mass of 7.8 g. Calculate the mass of the prism. Give your answer in kilograms. … kg [2] (ii) Calculate the length of the prism. … cm [3] (iii) Calculate the total surface area of the prism. … cm2 [6] (iv) In a mathematically similar prism, the height of the trapezium is 10 cm. Calculate the volume of this prism. … cm3 [3] (b) A cuboid measures 10 cm by 4 cm by 6 cm. Each side is measured correct to the nearest centimetre. Complete the inequality for the volume, V, of this cuboid. … cm 3 G V 1 … cm3 [3]
17 marks
Mark scheme: 3(a)(i) 4.095 2 B1 for figs 4095 525 × 7.8 or M1 for 1000 3(a)(ii) 15 3 B2 for 35 OR 1 M2 for (10 + 4) × 5 × L = 525 oe 2 1 M1 for (10 + 4) × 5 oe 2 3(a)(iii) 455 or 454.9... 6 2 2 M3 for their [BD =] 3 + 5 × (their 15) [× 2] or B2 for 34 or 5.83 or 5.830 to 5.831 2 1 2 or M1 for 5 + (10 − 4 ) 2 and M1 for their 35 × 2 M1 for (their 15) × 10 and (their 15) × 4 3(a)(iv) 4200 3 3 10 M2 for 525 × oe 5 10 3 5 3 or M1 for or oe 5 10 3(b) 182.875 ... 307.125 final answer 3 B2 for either seen or M1 for 10 ± 0.5 or 6 ± 0.5 or 4 ± 0.5 oe
This question in 0580/43 Oct/Nov 2021
Q62 · ABCDEFGH is a regular octagon with sides of length 6 cm 0580/41 May/June 2022
5 (a) ABCDEFGH is a regular octagon with sides of length 6 cm. The diagram shows part of the octagon. O is the centre of the octagon and M is the midpoint of AB. A M B NOT TO SCALE O (i) (a) Show that angle OAM is 67.5°. [2] (b) Calculate the area of the octagon. … cm2 [4] (ii) Find the area of the circle that passes through the vertices of the octagon. … cm2 [3] (b) NOT TO SCALE 4 m 0.45 m The diagram shows a horizontal container for water with a uniform cross-section. The cross-section is a semicircle. The radius of the semicircle is 0.45 m and the length of the container is 4 m. (i) Calculate the volume of the container. … m3 [2] (ii) NOT TO SCALE 0.3 m The greatest depth of the water in the container is 0.3 m. The diagram shows the cross-section. Calculate the number of litres of water in the container. Give your answer correct to the nearest integer.
17 marks
Mark scheme: 5(a)(i)(a) 8 2 180 8 2 oe M2 8 2 180 8 or 360 2 8 4 90 8 5(a)(i)(b) 174 or 173.8 … 4 M3 for 1 6 2 OM oe or 2 1 2 OA sin45 oe or 1 6 67.5 2 OA sin oe where OA and OM are as in the M2 or M2 for 3 tan67.5 OM oe or for 3 67.5 OA cos or 6 67.5 45 sin sin oe or M1 for tan67.5 3 OM oe or for 3 67.5 cos OA oe or for 45 67.5 6 sin sin OA oe 5(a)(ii) 193 or 193.0 to 193.1 3 M2 for 2 3 67.5 cos oe or M1 for 3 67.5 cos r or 45 67.5 6 sin sin r Question Answer Marks Partial Marks 5(b)(i) 1.27 or 1.272 to 1.273 2 M1 for 2 1 0.45 4 2 or 2 1 0.45 4 2 5(b)(ii) 742 or 743 6 M5 for a method leading to the volume of water e.g. 2 0.15 cos 0.45 4 {2 0.45 360 inv 2 1 0.15 0.45 sin 2 cos 2 0.45 inv } oe OR M2 2 0.15 cos 0.45 2 0.45 360 inv oe or 2 0.15 90 cos 0.45 2 0.45 360 inv oe or M1 for use of 2 0.45 360 oe M2 for 2 1 0.15 0.45 sin 2 cos 2 0.45 inv oe or 1 0.15 0.15 0.45 sin cos 2 2 0.45 inv oe Question Answer Marks Partial Marks 5(b)(ii) or M1 for use of 2 1 0.45 2 × sinθ oe or 1 2 0.15 0.45 sinβ 2 oe If 0 scored, SC1 for invcos 0.15 0.45 or invsin 0.15 0.45 or 2 2 0.45 0.15 soi
This question in 0580/41 May/June 2022
Q63 · 28 cm A D AD NOT TO SCALE 20 cm N BC B C A rectangular sheet of paper ABCD is made into… 0580/42 May/June 2022
11 (a) 28 cm A D AD NOT TO SCALE 20 cm N BC B C A rectangular sheet of paper ABCD is made into an open cylinder with the edge AB meeting the edge DC. AD = 28 cm and AB = 20 cm. (i) Show that the radius of the cylinder is 4.46 cm, correct to 3 significant figures. [2] (ii) Calculate the volume of the cylinder. … cm3 [2] (iii) N is a point on the base of the cylinder, such that BN is a diameter. Calculate the angle between AN and the base of the cylinder. … [3] (b) The volume of a solid cone is 310 cm 3. The height of the cone is twice the radius of its base. Calculate the slant height of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm [5]
12 marks
Mark scheme: 11(a)(i) 4.455 to 4.456… [= 4.46] 2 28 M1 for [r =] oe 2π 11(a)(ii) 1250 or 1247 to 1249.9… 2 M1 for 20 4.46 2 oe 11(a)(iii) 66[.0] or 65.95 to 66.02 3 20 M2 for [tan] = oe 2 4.46 or B1 for identifying angle ANB on cylinder not on rectangle 11(b) 11.8 or 11.82 to 11.83 5 310 3 M2 for [ r ] 3 oe 2π 310 3 4 or [ h ] 3 oe π or M1 for 310 13 r 2 2 r 1 h 2 or 310 π h 3 2 M2 for (their r ) 2 2 their r 2 oe or M1 for [l 2 ] their r 2 2 their r 2 oe
This question in 0580/42 May/June 2022
Q64 · Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2]… 0580/41 Oct/Nov 2022
1 (a) Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2] (ii) a solid hemisphere with radius 6 cm. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm3 [2] (b) NOT TO SCALE 14 cm 6 cm The cylinder and hemisphere in part (a) are joined to form the solid in the diagram. The solid is made of steel and 1 cm 3 of steel has a mass of 7.85 g. (i) Show that 1 cm 3 of steel has a mass of 0.007 85 kg. [1] (ii) Calculate the total mass of the solid. … kg [2] (c) 2000 cm 3of iron is melted down and some of it is used to make 50 spheres with radius 2 cm. (i) Calculate the percentage of iron that is left over. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3] (ii) The iron left over is then made into a cube. Calculate the length of an edge of the cube. … cm [1] (d) A solid cone has radius 3R cm and slant height 9R cm. A solid cylinder has radius x cm and height 7x cm. The total surface area of the cone is equal to the total surface area of the cylinder. Given that R = kx , find the value of k. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] k = … [4]
15 marks
Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1580 or 1583 to 1584 2 M1 for π 6 2 14 1(a)(ii) 452 or 452.3 to 452.4... 2 3 M1 for 1 4 π 6 2 3 1(b)(i) 7.85 ÷ 1000 [= 0.00785] M1 1(b)(ii) 16[.0] or 15.95 to 15.99 2 FT {their (a)(i) + their (a)(ii)} 0.00785 evaluated to 3 sig fig or better M1 for (their (a)(i) + their (a)(ii)) × 0.00785 1(c)(i) 16.2 or 16.21 to 16.23 3 4 3 2000 − 50 π 2 3 M2 for 100 2000 4 3 50 π 2 3 or for 100 2000 4 3 50 π 2 3 or M1 for 2000 1(c)(ii) 6.87 or 6.870 to 6.872 1 4 3 FT 3 2000 − their 50 π 2 3 evaluated to 3sf or better 1(d) 2 4 M1 for [π](3 R ) 2 + [π]3 R 9 R oe oe 3 M1 for 2[π]x 2 + 2[π]x 7 x oe M1 for their area of cone = their area of cylinder seen
This question in 0580/41 Oct/Nov 2022
Q65 · NOT TO 50 cm SCALE 40 cm 1.2 m 36 cm The diagram shows a water trough in the shape of a… 0580/43 Oct/Nov 2022
5 NOT TO 50 cm SCALE 40 cm 1.2 m 36 cm The diagram shows a water trough in the shape of a prism. The prism has a cross-section in the shape of an isosceles trapezium. The trough is completely filled with water. (a) Show that the volume of water in the trough is 206.4 litres. [3] (b) The water from the trough is emptied at a rate of 600 ml per second. Calculate the time taken, in minutes and seconds, for the trough to be emptied. … minutes … seconds [3] (c) All the water from the trough is emptied into a vertical cylindrical tank. The depth of the water in the tank is 84 cm. (i) Calculate the radius of the tank. … cm [3] (ii) The tank is 60% full. Calculate the height of the tank. … cm [2] (d) M NOT TO 50 cm SCALE 40 cm 1.2 m A 36 cm A steel rod AM is placed inside the empty water trough as shown in the diagram. A is a vertex at the base of the isosceles trapezium and M is the midpoint of the top edge on the opposite face. Calculate the length of the steel rod, AM. AM = … cm [4]
15 marks
Mark scheme: 5(a) (36 + 50) 40 120 oe M2 2 (36 + 50) 40 (0.36 + 0.5) 0.4 or M1 for oe or oe 2 2 (0.36 + 0.5) 0.4 1.2 oe 2 206400 ÷ 1000 = 206.4 A1 Must see an explicit conversion or 0.2064 × 1000 = 206.4 nfww 5(b) 5 [minutes] 44 seconds 3 B2 for 344 [seconds] oe 5.73…[mins] or M1 for figs206.4 ÷ figs 6 oe 5(c)(i) 28[.0] or 27.96 to 27.97 3 figs 2064 M2 for [r2=] ( figs84) or M1 for r 2 figs 84 = figs 2064 5(c)(ii) 140 cao 2 M1 for 0.6h = 84 oe ALT method 2 M1 for ( their (c)(i) ) h = figs 206400 0.6 oe 5(d) 128 or 127.7 to 127.8 4 B3 for 40 2 + 120 2 + 18 2 oe OR B1 for horizontal length 18 soi M1 for any correct attempt at 2-dimensional Pythagoras’ 182 + 1202, 1202 + 402, 182 + 402
This question in 0580/43 Oct/Nov 2022
Q66 · 12 cm NOT TO SCALE 3 cm The diagram shows a cylinder containing water 0580/42 Feb/March 2023
3 12 cm NOT TO SCALE 3 cm The diagram shows a cylinder containing water. There is a solid metal sphere touching the base of the cylinder. Half of the sphere is in the water. The radius of the cylinder is 12 cm and the radius of the sphere is 3 cm. (a) The sphere is removed from the cylinder and the level of the water decreases by h cm. Show that h = 0.125 . 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 [3] (b) The water in the cylinder is poured into another cylinder of radius R cm. The depth of the water in this cylinder is 18 cm. Calculate the value of R. R = … [3] (c) The sphere is melted down and some of the metal is used to make 30 cubes with edge length 1.5 cm. Calculate the percentage of metal not used. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3]
9 marks
Mark scheme: 3(a) 1 4 3 M3 π 3 2 3 [ h = ] oe π 12 2 leading to 0.125 or 2 1 4 3 12 −3 π 3 2 3 3 – oe π 12 2 2 1 4 3 M2 for π 12 h = π 3 oe 2 3 leading to 0.125 or for π ×122 × 3 = π×122 × x + ⅔×π×33 oe 1 4 3 π 3 2 3 h or for = oe π 12 2 3 3 2 1 4 3 or M1 for π 12 h or π 3 oe 2 3 or 12 2 3 3(b) 4.8[0] or 4.795 to 4.796 3 M2 for π 12 2 (3 − 0.125) = π R 2 18 oe or π ×122 × 3 – ⅔×π×33 = π R 2 18 or B1 for 3 – 0.125 or for 414 oe 3(c) 10.5 or 10.47 to 10.49 3 4 3 3 π 3 − 30 1.5 3 3 30 1.5 M2 for or 100 oe 4 3 4 3 π 3 π 3 3 3 4 3 3 30 1.53 or M1 for π 3 − 30 1.5 or oe 3 4 3 π 3 3
This question in 0580/42 Feb/March 2023
Q67 · NOT TO 15 cm SCALE 8 cm A cone has base diameter 8 cm and perpendicular height 15 cm 0580/42 May/June 2023
5 (a) NOT TO 15 cm SCALE 8 cm A cone has base diameter 8 cm and perpendicular height 15 cm. (i) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [2] (ii) A label completely covers the curved surface area of the cone. Calculate the area of the label as a percentage of 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 .] … % [5] (b) NOT TO SCALE 0.45 m An empty cylindrical container has radius 0.45 m. 300 litres of water is poured into the container at a rate of 375 ml per second. (i) Find the time taken, in minutes and seconds, for all the water to be poured into the container. … min … s [3] (ii) Calculate the height of the water in the container. … m [3]
13 marks
Mark scheme: 5(a)(i) 251 or 251.3 to 251.4 2 1 2 M1 for π 4 15 oe 3 5(a)(ii) 79.5 or 79.51… 5 2 2 M3 for π 4 4 15 oe or M2 for 15 2 4 2 oe or M1 for [l2 = ] 42 + 152 oe or π×4× theirl M1 for their curved surfacearea [ 100] their curved surfacearea π 4 2 oe 5(b)(i) 13 min 20 sec 3 40 B2 for 800 or oe seen 3 or M1 for figs 3 ÷ figs 375 or figs 3 ÷ 22 500 5(b)(ii) 0.472 or 0.4715 to 0.4716… 3 M2 for π 0.452 h 0.3 or π 45 2 h 300000 oe or M1 for π figs45 2 h figs3 oe
This question in 0580/42 May/June 2023
Q68 · A shop sells shirts for $x and jackets for $(x + 27) 0580/43 May/June 2023
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]
12 marks
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) 2x 6x + 2x2 or 2x(6x + x) M2 or M1 for 2x 6x or 2x2 Their (2x 6x + 2x2) = 4r2 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
This question in 0580/43 May/June 2023
Q69 · 3.63.6 cmcm 6.5 cm NOT TO SCALE 5.4 cm The diagram shows a solid formed by joining two… 0580/41 Oct/Nov 2023
8 (a) 3.63.6 cmcm 6.5 cm NOT TO SCALE 5.4 cm The diagram shows a solid formed by joining two hemispheres and a cylinder. The radius of the large hemisphere is 5.4 cm. The radius of the small hemisphere and the radius of the cylinder are both 3.6 cm. The height of the cylinder is 6.5 cm. (i) Show that the volume of the solid is 692 cm 3, correct to the nearest cubic centimetre. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 [4] (ii) A mathematically similar solid is made of silver. In this solid, the cylinder has radius 0.6 cm. 1 cm 3 of silver has a mass of 10.49 grams. Calculate the total mass of this silver solid. … g [4] (b) A 10 cm NOT TO O 216° SCALE B AOB is a sector of a circle, centre O. AO = 10 cm and the sector angle is 216°. (i) Calculate the length of the arc of this sector. Give your answer as a multiple of r. … cm [2] (ii) A cone is made from this sector by joining OA to OB. Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [4]
14 marks
Mark scheme: 8(a)(i) 2 3 2 3 M3 2 3 2 3 3(3.6) + 3(5.4) + M1 for either 3(3.6) or 3(5.4) (3.6) 2 6.5 M1 for (3.6) 2 6.5 692.1 to 692.2… A1 8(a)(ii) 33.6 or 33.60 to 33.62 4 3 0.6 M3 for 692 10.49 oe 3.6 0.6 3 or M2 for 692 oe 3.6 0.6 3 3.6 3 or M1 for or oe 3.6 0.6 If 0 scored, SC1 for their volume 10.49 8(b)(i) 12π final answer 2 216 M1 for 2 10 oe 360 After 0 scored SC1 for final answer 8π or 12π + 20 8(b)(ii) 302 or 301.5 to 301.6… 4 M1 for 2πr = their (b)(i) oe or for 216 2 π 10 = π r 10 oe 360 and M1 for [h =] 10 2 − their 6 2 oe and 1 2 M1 for [V =] (their 6) (their 8) 3
This question in 0580/41 Oct/Nov 2023
Q70 · O O NOT TO SCALE x° 7.5 cm 7.5 cm A B B 1.5 cm A The diagram shows a sector of a circle… 0580/42 Oct/Nov 2023
4 (a) O O NOT TO SCALE x° 7.5 cm 7.5 cm A B B 1.5 cm A The diagram shows a sector of a circle that is made into a cone by joining OA to OB. The sector angle is x° and the radius of the sector is 7.5 cm. The base radius of the cone is 1.5 cm. Calculate the value of x. x = … [3] (b) NOT TO SCALE The diagram shows a cylinder with radius 8 cm inside a sphere with radius 17 cm. Both ends of the cylinder touch the curved surface of the sphere. (i) Show that the height of the cylinder is 30 cm. [2] (ii) Calculate the volume of the cylinder as a percentage of the volume of the sphere. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [4] (c) 15 cm NOT TO SCALE The diagram shows a solid sphere with radius 6 cm inside a cube with side length 20 cm. The cube contains water to a depth of 15 cm. The sphere is removed. Calculate the new depth of water in the cube. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm [3]
12 marks
Mark scheme: 4(a) 72 or 72.0 cao nfww 3 x M2 for 2 π 7.5=2 π 1.5 oe 360 x or M1 for 2 π 7.5 or for 360 2 π 1.5 oe OR x 2 M2 for π 7.5 =π 1.5 7.5 oe 360 x 2 or M1 for π 7.5 or for 360 π 1.5 7.5 oe 4(b)(i) M2 M1 for 17 2 = 82 + d 2 or 342 = 162 + k2 2 172 − 82 or 342 − 162 oe 4(b)(ii) 29.3 or 29.30 to 29.31 4 2 4 3 M3 for ( [π] 8 30 ) ÷ [π] 17 [× 3 100] oe OR M1 for π 82 30 oe 4 3 M1 for π 17 oe 3 4(c) 12.7 or 12.73 to 12.74 3 B2 for 2.26 or 2.261 to 2.262…. soi 2 4 3 2 or M2 for 20 15 − π 6 20 oe 3 4 3 2 or for 15 – π 6 20 oe 3 2 4 3 or M1 for 20 15 − π 6 oe 3 2 4 3 or 20 D = π 6 oe 3 If 0 scored, SC1 for answer 11[.0] or 10.97 to 10.98
This question in 0580/42 Oct/Nov 2023
Q71 · F 9 cm NOT TO SCALE D C 12 cm M E B 12 cm The diagram shows a pyramid with a square base… 0580/42 Feb/March 2024
4 (a) F 9 cm NOT TO SCALE D C 12 cm M E B 12 cm 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. (i) Calculate the volume of the pyramid. 1 [The volume, V, of a pyramid with base area A and height h is V = Ah .] 3 … cm3 [2] (ii) Calculate the total surface area of the pyramid. … cm2 [5] (b) NOT TO SCALE 3r r 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 cm 2. Calculate the value of r. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] [The curved surface area, A, of a sphere with radius r is A = 4rr 2 .] r = … [4]
11 marks
Mark scheme: 4(a)(i) 432 2 M1 for 12 × 12 × 9 ÷ 3 oe 4(a)(ii) 404 or 403.5 to 403.7 5 2 1 2 2 M4 for 12 + 4 12 6 + 9 oe 2 1 2 2 or M3 for 12 6 + 9 oe 2 or M2 for explicit method to find height of triangular face e.g. 62 + 92 oe or M1 for implicit method to find height of triangular face or for 6 2 + 9 2 oe seen or B1 for slant height of triangle FC 153 or 3 17 or 12.4 or 12.36 to 12.37 soi 4(b) 4.4[0] or 4.398 to 4.399... nfww 4 304 M3 for oe ( 2 + 3 ) π 4πr 2 or M2 for + πr 3r = 304 oe 2 4π r 2 or M1 for oe seen or πr 3r oe seen 2
This question in 0580/42 Feb/March 2024
Q72 · NOT TO SCALE 50° 12 cm The diagram shows a circle of radius 12 cm, with a sector removed 0580/42 Feb/March 2024
12 (a) NOT TO SCALE 50° 12 cm The diagram shows a circle of radius 12 cm, with a sector removed. Calculate the perimeter of the remaining shaded shape. … cm [4] (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 cm 3. Calculate the height of the cake. … cm [3]
7 marks
Mark scheme: 12(a) 88.9 or 88.92 to 88.93... 4 360 − 50 M3 for 2 12 + 2 π 12 oe 360 ( 360 − 50 ) or M2 for 2 π 12 oe isw 360 50 or M1 for 2 π 12 oe isw 360 12(b) 9.01 or 9.009 to 9.010… 3 ( 360 − 50 ) 2 M2 for π 12 h = 3510 360 k 2 or M1 for π 12 h oe seen 360 with k = 50 or 360 – 50
This question in 0580/42 Feb/March 2024
Q73 · NOT TO SCALE 12 cm 1 m The diagram shows a tank in the shape of a half-cylinder of radius… 0580/42 May/June 2024
4 (a) NOT TO SCALE 12 cm 1 m The diagram shows a tank in the shape of a half-cylinder of radius 12 cm and length 1 metre. The tank is fixed horizontally and is completely filled with water. (i) Calculate the volume of water in the tank. Give your answer correct to the nearest 10 cm3. … cm3 [3] (ii) NOT TO 6 cm SCALE Water is removed from the tank until the level of water is 6 cm below the top of the tank. The diagram shows the cross-section of the tank. Calculate the volume of water that is now in the tank. … cm3 [5] (b) A rectangular fish tank with length 42 cm and width 35 cm is full of water. A stone lies at the bottom of the tank. When the stone is removed from the tank, the depth of the water decreases by 0.2 cm. The density of the stone is 2.2 g/cm3. Calculate the mass of the stone in grams. [ Density = mass ' volume] … g [3] (c) H G E F 15 cm NOT TO SCALE D C 12 cm A 8 cm B The diagram shows a cuboid, ABCDEFGH. Calculate the angle that AG makes with the base of the cuboid. … [4]
15 marks
Mark scheme: 4(a)(i) 22 620 cao 3 B2 for 7200 or 22 608 to 22 629 1 2 or M1 for 12 [ figs 1] oe 2 4(a)(ii) 8840 or 8850 or 8836 to 8850. 5 6 M1 for cos COM = oe 12 6 or sin AOC = oe 12 theirCOD 2 M1 for 12 oe M 360 1 2 oe M1 for 12 sin theirCOD 2 M1dep for (their area of sector COD– their area of triangle COD) 100 dep on at least M1M1 oe 4(b) 647 or 646.8 3 m M2 for 2.2 oe 42 35 0.2 or M1 for [vol of stone =] 42×35×0.2 oe If 0 scored SC1 for answer figs 647 or figs 6468 4(c) 46.1 or 46.12 to 46.14 4 15 M3 for tan oe 8 2 12 2 or M2 for 82 + 122 oe or 82 + 122 + 152 oe or M1 for identifying the angle GAC
This question in 0580/42 May/June 2024
Q74 · 5 (a) Simplify 25x 6 2 0580/42 May/June 2024
3 5 (a) Simplify 25x 6 2 . ` j … [2] (b) These are the first five terms of a sequence. 1 1 6 36 216 6 Find the nth term of the sequence. … [2] (c) Expand and simplify. ( x + 4)( x - 3)( 3x - 1) … [3] 7 2(d) (i) Show that ( 3x + 5) + = x simplifies to 2x + x - 3 = 0 . x - 2 [4] (ii) Solve by factorisation 2x 2 + x - 3 = 0 . x = … or x = … [3] (e) A solid cylinder has base radius x and height 3x. The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y. 2 75y 2 Show that x = . 8 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]
18 marks
Mark scheme: 5(a) 125x 9 final answer 2 B1 for answer 125 kx or m x 9 or for correct answer seen then spoilt 5(b) 6 n 2 oe final answer 2 B1 for answer of form 6k oe k 1 or answer of the form oe 6 or for correct answer seen 5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three brackets 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 5(d)(i) eliminates the fraction correctly M1 eg (3x + 5) (x – 2) + 7 = x (x – 2) 3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2 B1 for 3x2 + 5x –6x –10 [ + 7] oe seen with at least 3 terms correct leading to 2x2 + x – 3 = 0 A1 dep on M1 B2 with no errors or omissions 5(d)(ii) 2 x 3 x 1 M2 or M1 for (2x + a)(x + b) where ab = −3 or 2b + a = [+]1 or for partial factors 2x(x – 1) + 3(x – 1) or x(2x + 3) –[1](2x + 3) −1.5 oe and +1 B1 5(e) 2 M1 [TSA cylinder =] 2x 2x 3 x 2 4(5 y ) 2 M1 [TSA hemisphere=] (5 y ) 2 Leading to M1 dep M1M1 2x 2 6x 2 50y 2 25y 2 oe 2 75 y 2 A1 dep on M1M1M1 x 8
This question in 0580/42 May/June 2024
Q75 · O NOT TO 60° 10 cm SCALE 17 cm D C A B OAB is a sector of a circle, centre O, radius 17 cm 0580/42 May/June 2024
9 (a) O NOT TO 60° 10 cm SCALE 17 cm D C A B OAB is a sector of a circle, centre O, radius 17 cm. OCD is a sector of a circle, centre O, radius 10 cm. OCA and ODB are straight lines and angle AOB = 60° . The perimeter of the shaded shape ABDC can be written in the form ( a r+ b ) cm. Find the value of a and the value of b. a = … b = … [3] (b) NOT TO SCALE The diagram shows a regular hexagon. The area of the hexagon is 127.3 cm2. (i) Show that the length of one side of the hexagon is 7.0 cm , correct to 1 decimal place. [4] (ii) The hexagon is the cross-section of a prism of length 10 cm. 127.3 cm2 NOT TO SCALE 10 cm 7.0 cm (a) Find the volume of the prism. … cm3 [1] (b) Calculate the surface area of the prism. … cm2 [2]
10 marks
Mark scheme: 9(a) [a =] 9 3 B2 for a =9 [b =] 14 OR M2 for 60 60 2 17 2 10 7 7 360 360 oe or M1 for 60 60 2 17 oe or 2 10 oe 360 360 If 0 scored SC1 for b =14 9(b)(i) 60° at centre B1 or interior angle = 120° 1 2 M1 [6] d sin60 oe 2 2 127.3 M1 [ d ] 1 6 sin60 2 6.99[9…] to 7.00[…] A1 Dep on M1M1 9(b)(ii)(a) 1273 1 9(b)(ii)(b) 675 or 674.5 to 674.6 2 M1 for 2 ×127.3 oe or 6 × 7 × 10 oe
This question in 0580/42 May/June 2024
Q76 · NOT TO SCALE 16 cm 1.5 cm The diagram shows a solid made from a cylinder and a cone 0580/43 May/June 2024
8 (a) NOT TO SCALE 16 cm 1.5 cm The diagram shows a solid made from a cylinder and a cone. The height of the cylinder is 16 cm and the height of the cone is 1.5 cm. The radius of the cylinder and the base radius of the cone are each 0.35 cm. (i) Calculate the total surface area of the solid. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] … cm2 [5] (ii) Calculate the volume of the solid. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [3] (iii) NOT TO 1.4 cm SCALE 3.5 cm 10 of the solids are placed in a box in the shape of a cuboid of length 17.5 cm. The diagram shows one end of the box. Calculate the volume of the empty space in the box. … cm3 [3] (b) NOT TO SCALE The diagram shows two mathematically similar solids. The surface area of the larger solid is 200 cm 2 and the surface area of the smaller solid is 98 cm 2. The volume of the larger solid is 450 cm 3. Calculate the volume of the smaller solid. … cm3 [3]
14 marks
Mark scheme: 8(a)(i) 37.3 or 37.26 to 37.27 5 M2 for 0.35 0.35 2 1.5 2 oe or M1 for 0.35 2 1.5 2 or better M1 for 0.352 M1 for 2 0.35 16 8(a)(ii) 6.35 or 6.349 to 6.351 3 M1 for π 0.352 16 1 M1 for π 0.352 1.5 3 8(a)(iii) 22.2 or 22.3 or 22.24 to 22.26 3 M2 for 17.5 × 3.5 × 1.4 – 10 × their(a)(ii) or M1 for 17.5 × 3.5 × 1.4 8(b) 154 or 154.3 to 154.4 3 3 98 M2 for 450 oe 200 98 3 200 3 or M1 for or oe 200 98 450 2 200 3 or for oe V 98
This question in 0580/43 May/June 2024
Q77 · H G F E NOT TO 17 cm SCALE D C 8 cm A 10 cm B ABCDEFGH is a solid cuboid 0580/41 Oct/Nov 2024
9 H G F E NOT TO 17 cm SCALE D C 8 cm A 10 cm B ABCDEFGH is a solid cuboid. AB = 10 cm, BC = 8 cm and CG = 17 cm. (a) Work out the volume of the cuboid. … cm3 [1] (b) Work out the total surface area of the cuboid. … cm2 [3] (c) Calculate the angle between GA and the base ABCD. … [4] (d) A straight rod PQ is placed inside the cuboid. One end of the rod, P, is placed at the midpoint of AB. The other end of the rod, Q, rests on GH. HQ : QG = 4 : 1 . Q H G F E NOT TO 17 cm SCALE D C 8 cm A P B 10 cm Calculate the length of the rod PQ. … cm [4]
12 marks
Mark scheme: 9(a) 1360 1 9(b) 772 3 M2 for [2 ×] (10 × 8 + 10 × 17 + 8 × 17) oe or M1 for 10 × 8 oe or 10 × 17 oe or 8 × 17 oe 9(c) 53 or 53.0 to 53.01 4 17 M3 for tan [GAC] = oe 10 2 + 8 2 or M2 for 102 + 82 oe or for 102 + 82 + 172 oe or M1 for recognising angle GAC is required 9(d) 19[.0] or 19.02 to 19.03 4 M3 for 32 + 82 + 172 oe OR B1 for QG = 2 soi or HQ = 8 M1 for (5 – 2) 2 + 82 or (5 – 2) 2 + 172
This question in 0580/41 Oct/Nov 2024
Q78 · A box contains 50 cuboids 0580/43 Oct/Nov 2024
4 (a) A box contains 50 cuboids. Each cuboid has a mass of 135 g. The total mass of the cuboids and the box is 7 kg. Calculate the mass of the box. Give your answer in grams. … g [2] (b) A solid cube of side 4 cm is fixed to the base inside an empty cube of side 6 cm. Water is poured into the larger cube until it reaches the top of the smaller cube. Calculate the amount of water poured into the larger cube. … cm3 [2] (c) 4 cm NOT TO SCALE 20 cm The diagram shows a solid triangular prism of length 20 cm. The cross-section is an equilateral triangle with side length 4 cm. The prism is made of wood with a density of 0.85 g/cm3. Calculate the mass of the prism. [Density = mass ÷ volume] … g [4] (d) NOT TO SCALE 24 cm 10 cm The diagram shows a solid cone with base radius 10 cm and height 24 cm. (i) Show that the total surface area of the cone is 1131 cm2, correct to the nearest cm2. [The curved surface area of a cone with base radius r and slant height l is A = r rl .] [4] (ii) The total surface area of the cone is painted. (a) The cost to paint the cone is $1.71 . Calculate the cost to paint 1 cm2 of the cone. Give your answer in cents. … cents [1] (b) One tin of paint has enough paint to cover 2.5 m2. Calculate the number of these cones that can be painted completely using one tin of paint. … [2]
15 marks
Mark scheme: 4(a) 250 2 B1 for 6750 or M1 for 7000 – 50 × 135 or for 7 – 50 × 0.135 4(b) 80 2 M1 for 6 × 6 × 4 or for 43 oe OR M1 for (6 6) – (4 4) oe 4(c) 118 or 117.7 to 117.8 4 1 M3 for 4 4 sin60 × 20 × 0.85 oe 2 OR 1 M1 for 4 4 sin60 or 2 1 2 2 4 4 − 2 oe 2 M1 for 20 × their area of triangle M1 dep for 0.85 × their volume, dependent on previous M1 If 0 scored SC1 for height = 3.46... 4(d)(i) 2 2 2 M3 π 24 + 10 10 + π 10 or 2 2 2 2 2 2 π 10 2 M2 for π 24 + 10 10 or π 24 + 10 + π 10 2 ( ) 2 π 26 2 2 2 π 10 π 24 + 10 ( ) 2 π 26 or M1 for 24 2 + 10 2 or 102 1130.9 to 1131.1... A1 Must see at least 5 sf 4(d)(ii)(a) 0.151 or 0.1511 to 0.1512... 1 4(d)(ii)(b) 22 2 B1 for figs 22[1…] 2.5 100 2 or M1 for 1131
This question in 0580/43 Oct/Nov 2024
Q79 · NOT TO 3 m SCALE O 2.5 m The diagram shows the major segment of a circle, centre O… 0580/42 Feb/March 2025
23 NOT TO 3 m SCALE O 2.5 m The diagram shows the major segment of a circle, centre O, radius 2.5 m. The segment is the cross section of a tunnel with height 3 m. The length of the tunnel is 800 m and it has the same cross section throughout its length. Calculate the volume of the tunnel. … m3 [7]
7 marks
Mark scheme: 23 9835 to 9844 nfww 7 B5 for [area of segment =] 12.29 to 12.31 nfww M1 for 12.29 to 12.31 × 800 OR B2 for angle at centre = 157 or 156.9 or 156.92 to 156.93 or for [reflex angle =] 203 to 203.1 3 − 2.5 or M2 for 2 × cos–1 oe 2.5 3 − 2.5 or M1 for cos x = oe 2.5 360 − their 2 M2dep for π 2.5 360 [360 −]their 2 or M1dep for π 2.5 seen 360 1 M1dep for 2.5 2.5 sin(their) oe 2 or 2 × 0.5 × 2.52 − 0.52 × 0.5 oe M1 for their area × 800 leading to answer
This question in 0580/42 Feb/March 2025
Q80 · O NOT TO 24 cm SCALE D C M 10.5 cm A 10.5 cm B The diagram shows a pyramid OABCD 0580/41 May/June 2025
20 O NOT TO 24 cm SCALE D C M 10.5 cm A 10.5 cm B The diagram shows a pyramid OABCD. The pyramid has a square base, ABCD, with sides 10.5 cm. The vertex O is vertically above the centre of the base, M. The height of the pyramid is 24 cm. (a) Calculate the angle that OA makes with the base. … [4] (b) NOT TO SCALE 16 cm D C M 10.5 cm A 10.5 cm B The diagram shows a frustum of the pyramid OABCD. The height of the frustum is 16 cm. Calculate the volume of the frustum. … cm3 [5] Question 21 is on page 16.
9 marks
Mark scheme: 20(a) 72.8 or 72.81… 4 24 M3 for oe or better 1 2 2 10.5 + 10.5 2 1 2 2 or M2 for AM = 10.5 + 10.5 oe 2 or AM = 10.5 cos45 oe or M1 for AC2 = 10.52 + 10.52 or AM 2 = 5.252 +5.252 AM or = cos45 oe 10.5 If 0 scored, SC1 for identifying OAM 20(b) 1 5 Method 1 849 or 849 or 849.3… 3 B2 for side of small square = 3.5 10.5 24 or M1 for = or better x 24 − 16 1 1 M2 for 10.52 24 – (their 3.5)2 3 3 (24 – 16) 1 or M1 for 10.52 24 3 1 or for (their 3.5)2 (24 – 16) 3 Method 2 1 26 B2 for or 27 27 24 − 16 3 or M1 for volume scale factor = 24 M2 for 1 1 2 1 − their 10.5 24 oe 27 3 1 2 or M1 for 3 10.5 24 1 1 2 or their 10.5 24 oe 27 3
This question in 0580/41 May/June 2025
Q81 · A solid wooden cone has base radius 4 cm and height 12 cm 0580/42 May/June 2025
6 A solid wooden cone has base radius 4 cm and height 12 cm. The density of the wood is 0.74 g/cm 3. Calculate the mass of the cone. [ Density = Mass ' Volume] … g [3]
3 marks
Mark scheme: 6 149 or 148.7 to 148.8… 3 1 2 M1 for 3π 4 12 oe M1 for 0.74 × their volume
This question in 0580/42 May/June 2025
Q82 · 8 cm NOT TO SCALE 4.5 cm 13.2 cm The diagram shows a solid cuboid with sides of length… 0580/43 May/June 2025
18 8 cm NOT TO SCALE 4.5 cm 13.2 cm The diagram shows a solid cuboid with sides of length 4.5 cm, 8 cm and 13.2 cm. (a) Calculate the volume of the cuboid. … cm3 [1] (b) Calculate the total surface area of the cuboid. … cm2 [3] (c) B 8 cm NOT TO SCALE 4.5 cm A 13.2 cm Calculate the angle between AB and the horizontal base of the cuboid. … [4]
8 marks
Mark scheme: 18(a) 475.2 1 18(b) 402 3 M2 for [2 ×] (13.2 × 4.5 + 13.2 × 8 + 4.5 × 8) or M1 for [2 ×] 13.2 × 4.5 or [2 ×] 13.2 × 8 or [2 ×] 4.5 × 8 18(c) 29.8 or 29.83 to 29.84… 4 8 M3 for tan = oe 13.2 2 + 4.5 2 or M2 for 13.22 + 4.52 or for 13.22 + 4.52 + 82 or M1 for identifying correct angle
This question in 0580/43 May/June 2025
Q83 · 4.2 cm NOT TO 4.5 cm SCALE 7.5 cm The diagram shows a frustum made by removing a small… 0580/43 May/June 2025
23 4.2 cm NOT TO 4.5 cm SCALE 7.5 cm The diagram shows a frustum made by removing a small cone from a large cone. The height of the small cone is 7.5 cm. The height of the frustum is 4.5 cm. The radius of the large cone is 4.2 cm. Work out the volume of the frustum. … cm3 [4]
4 marks
Mark scheme: 23 168 or 167.5 to 167.6 4 M3 for π 2 7.5 2 × 4.2 × ( 4.5 + 7.5 ) − 4.2 × × 7.5 3 4.5 + 7.5 oe or M2 for 3 7.5 1 2 × × π × 4.2 × ( 4.5 + 7.5 ) 7.5 + 4.5 3 1 7.5 2 or for × π × 4.2 × × 7.5 3 4.5 + 7.5 1 2 or M1 for × π × 4.2 × ( 4.5 + 7.5 ) 3 OR 1 2 M1 for × π × 4.2 × ( 4.5 + 7.5 ) 3 4.2 radius M1 for = oe or better 4.5 + 7.5 7.5 1 2 M1 for × π × ( their 2.625 ) × 7.5 3
This question in 0580/43 May/June 2025
Q84 · A cube contains a solid metal sphere 0580/41 Oct/Nov 2025
14 A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm. 256 (a) Show that the volume of the sphere is rcm 3. 3 [1] (b) Calculate the percentage of the cube that is not occupied by the sphere. … % [3] (c) The density of the metal of the sphere is 7.86 g/cm3. Calculate the mass of the sphere. Give your answer in kilograms. [Density = mass ' volume] … kg [2] (d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder. … cm2 [4]
10 marks
Mark scheme: 14(a) 4 3 256 1 π 4 [= π ] 3 3 14(b) 47.6 nfww or 3 50 B2 for 52.4 or 52.35 to 52.37 or π nfww 47.63 to 47.64… nfww 3 OR 3 256 8 − π 3 M2 for 3 100 oe 8 3 256 256 8 − π π 3 3 or M1 for 3 [ 100] oe or 3 100 8 8 oe 14(c) 2.11 or 2.107… 2 256 M1 for π 7.86 3 14(d) 233 or 233.3 to 233.4 4 2 256 M1 for π 3.1 h = π 3 M2dep for 2 π 3.12 + 2 π 3.1 their h or M1dep for 2 π 3.1 theirh or M1 for 2π 3.12
This question in 0580/41 Oct/Nov 2025
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