Circles, arcs and sectors — Paper 4 · IGCSE Mathematics 0580

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

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

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

4 For A Examiner's Use NOT TO 8 cm 6 cm SCALE O B C 9 cm The circle, centre O, passes through the points A, B and C. In the triangle ABC, AB = 8 cm, BC = 9 cm and CA = 6 cm. (a) Calculate angle BAC and show that it rounds to 78.6°, correct to 1 decimal place. Answer(a) [4] (b) M is the midpoint of BC. (i) Find angle BOM. Answer(b)(i) Angle BOM = [1] (ii) Calculate the radius of the circle and show that it rounds to 4.59 cm, correct to 3 significant For figures. Examiner's Use Answer(b)(ii) [3] (c) Calculate the area of the triangle ABC as a percentage of the area of the circle. Answer(c) % [4]

12 marks

Mark scheme: 6 + 8 − 9 4 (a) ( cos( A)) = M2 M1 for correct implicit equation with cosA 8.6.2 78.58… www 4 A2 A1 for 0.1979 to 0.198 (this implies M2) (b) (i) 78.6 1 Allow 78.58… 5.4 5.4 (ii) r = oe M2 (M1 for sin(78.6) = ) sin( 786.) r Allow 78.58… or their angle BOM for M2 or M1 4.590 to 4.591 cao www 3 A1 (c) 35.5 (35.48 to 35.57…) cao www 4 4 M1 Area triangle = 0.5 × 6 × 8 × sin (78.6) oe Allow 78.58.. (23.52..) M1 Circle = π × .4 59 2 Allow 4.590 to 4.591 (66.15 to 66.22…) M1 (dependent) % = triangle / circle × 100 Dependent on first 2 M’s IGCSE – May/June 2011 0580 43

This question in 0580/43 May/June 2011

1 For 0.8 m Examiner's Use 0.5 m NOT TO SCALE 1.2 m 1.2 m d 0.4 m A rectangular tank measures 1.2 m by 0.8 m by 0.5 m. (a) Water flows from the full tank into a cylinder at a rate of 0.3 m3/min. Calculate the time it takes for the full tank to empty. Give your answer in minutes and seconds. Answer(a) min s [3] (b) The radius of the cylinder is 0.4 m. For Examiner's Calculate the depth of water, d, when all the water from the rectangular tank is in the cylinder. Use Answer(b) d = m [3] (c) The cylinder has a height of 1.2 m and is open at the top. The inside surface is painted at a cost of $2.30 per m2. Calculate the cost of painting the inside surface. Answer(c) $ [4]

10 marks

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 1 min 36 s www 3 M1 for 1.2 × 0.8 × 0.5 (= 0.48) A1 1.6 or 96 If A0, B1 for correctly converting to min and sec Dep on M1 (b) 0.954 to 0.956 www 3 their 0.48 M2 for or M1 for π × 0.42 × d = ‘0.48’ π × 0.4 2 (c) 8.09 to 8.10 www 4 M1 for π × 0.42 (0.503) condone × 2 and M1 for π × 0.8 × 1.2 (3.02) M1 for their area × 2.3 (dep M1 M1)

This question in 0580/43 Oct/Nov 2011

11 For Examiner's Use Diagram 1 Diagram 2 Diagram 3 The diagrams show a sequence of dots and circles. Each diagram has one dot at the centre and 8 dots on each circle. The radius of the first circle is 1 unit. The radius of each new circle is 1 unit greater than the radius of the previous circle. (a) Complete the table for diagrams 4 and 5. Diagram 1 2 3 4 5 Number of dots 9 17 25 Area of the largest circle π 4π 9π Total length of the circumferences of the circles 2π 6π 12π [4] (b) (i) Write down, in terms of n, the number of dots in diagram n. Answer(b)(i) [2] (ii) Find n, when the number of dots in diagram n is 1097. Answer(b)(ii) n = [2] (c) Write down, in terms of n and π, the area of the largest circle in (i) diagram n, Answer(c)(i) [1] (ii) diagram 3n. Answer(c)(ii) [1] (d) Find, in terms of n and π, the total length of the circumferences of the circles in diagram n. Answer(d) [2]

12 marks

Mark scheme: 11 (a) 33, 41 1 16 ,π 25 π 1 2 B1 each 20 ,π30 π (b) (i) 8 n + 1 oe final answer 2 e.g. 9 + 8(n –1), condone n = 8n + 1 SC1 for 8 n + k (ii) 137 www2 2 M1 for their (b)(i) = 1097 (c) (i) n 2 π oe final answer 1 (ii) 9 n 2 π oe final answer 1 Allow (3n ) 2 π (d) n ( n + )1 π oe final answer 2 SC1 for a quadratic expression e.g. n ( n + )1 , n2 + 5, n2 + n π

This question in 0580/43 May/June 2012

7 A P NOT TO SCALE 56° 6.5 cm 8 cm 8 cm x O C B Q The diagram shows a triangle and a sector of a circle. In triangle ABC, AB = AC = 8 cm and angle BAC = 56°. Sector OPQ has centre O, sector angle x and radius 6.5 cm. (a) Show that the area of triangle ABC is 26.5 cm2 correct to 1 decimal place. Answer(a) [2] (b) The area of sector OPQ is equal to the area of triangle ABC. (i) Calculate the sector angle x. Answer(b)(i) … [3] (ii) Calculate the perimeter of the sector OPQ. Answer(b)(ii) … cm [3] (c) The diagram shows a sector of a circle, radius r cm. NOT TO SCALE r cm 30° 1 2 1 (i) Show that the area of the shaded segment is r π - 1 cm2. 4 3 ` j Answer(c)(i) [4] (ii) The area of the segment is 5 cm2. Find the value of r. Answer(c)(ii) r = … [3] __________________________________________________________________________________________

15 marks

Mark scheme: 1 7 (a) × 8 × 8 × sin 56 oe M1 or [½ × 2] 8sin28 × 8cos28 or [½ × 2] × 7.06… × 2 3.75… 26.52 to 26.53 A1 (b) (i) 72.[0] or 71.87 to 72.0 3 M2 for 26.5/( π × 5.6 2 ) × 360 oe x 2 or M1 for ×π × 5.6 = 265. or better 360 their (b)(i) (ii) 21.1 or 21.2 or 21.14 to 21.17 3 M2 for ×π × 2 × 5.6 + 2 × 6.5 oe 360 their (b)(i) their (a) or M1 for ×π × 2 × 5.6 oe or 360 5.0 × 5.6 30 2 1 2 30 2 1 2 (c) (i) × π × r − × r × sin 30 oe M2 M1 for × π × r or × r × sin 30 360 2 360 2 1 2 1 2 ×π × r − × r A1 12 4 1 2  1  r  π −1  A1 Dep on M2 A1 and no errors seen 4  3  5 (ii) 20.6 or 20.7 or 20.55 to 20.71 3 M2 for [r2 =] 1 ( 1 π − )1 4 3 or M1 for one correct rearrangement step to r 1 2  1  from r  π −1  = 5 4  3 

This question in 0580/43 Oct/Nov 2014

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

7 (a) P NOT TO SCALE 8.4 cm Q 7.6 cm 62° R In the triangle PQR, QR = 7.6 cm and PR = 8.4 cm. Angle QRP = 62°. Calculate (i) PQ, Answer(a)(i) PQ = … cm [4] (ii) the area of triangle PQR. Answer(a)(ii) … cm2 [2] (b) North H NOT TO North SCALE 63 km G J The diagram shows the positions of three small islands G, H and J. The bearing of H from G is 045°. The bearing of J from G is 126°. The bearing of J from H is 164°. The distance HJ is 63 km. Calculate the distance GJ. Answer(b) GJ = … km [5]

11 marks

Mark scheme: 7 (a) (i) 8.27 or 8.269... nfww 4 M2 for 6.7 2 + 4.8 2 − 2 × 6.7 × 4.8 × cos ( 62) oe or M1 for implicit form A1 for [PQ2 =] 68.3 to 68.5 (ii) 28.2 or 28.18.. 2 M1 for 0.5 × 7.6 × 8.4 × sin 62 oe (b) 55.8 or 55.78 to 55.79 nfww 5 B1 for [HGJ] = 81 B1 for [GHJ] = 61 63 M2 for [GJ = ] × sin (their 61) sin (their 81) or M1 for implicit form After M0, SC1 for final answer of 68.1…

This question in 0580/41 May/June 2015

4 (a) A sector of a circle has radius 12 cm and an angle of 135°. (i) Calculate the length of the arc of this sector. Give your answer as a multiple of π. NOT TO SCALE 135° 12 cm Answer(a)(i) … cm [2] (ii) The sector is used to make a cone. NOT TO (a) Calculate the base radius, r. SCALE h 12 cm r Answer(a)(ii)(a) r = … cm [2] (b) Calculate the height of the cone, h. Answer(a)(ii)(b) h = … cm [3] (b) The diagram shows a plant pot. It is made by removing a small cone from a larger cone and adding a circular base. NOT TO SCALE This is the cross section of the plant pot. 15 cm (i) Find l. 35 cm 8 cm l NOT TO SCALE Answer(b)(i) l = … cm [3] (ii) Calculate the total surface area of the outside of the plant pot. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(b)(ii) … cm2 [3] (c) Some cones are mathematically similar. For these cones, the mass, M grams, is proportional to the cube of the base radius, r cm. One of the cones has mass 1458 grams and base radius 4.5 cm. (i) Find an expression for M in terms of r. Answer(c)(i) M = … [2] (ii) Two of the cones have radii in the ratio 2 : 3. Write down the ratio of their masses. Answer(c)(ii) … : … [1] __________________________________________________________________________________________

16 marks

Mark scheme: 135 4 (a) (i) 9π final answer 2 M1 for × 2 × π × 12 oe 360 (ii) (a) 4.5[0] or 4.497 to 4.504… 2FT FT their 9 ÷ 2 135 2 M1 for 2πr = their 9π or 12πr = π12 oe 360 (b) 11.1 or 11.12[…] 3FT FT their 12 2 − their 5.4 2 to 3 sf or better (their 4.5 < 12) M2 for 12 2 − their 5.4 2 (their 4.5 < 12) or M1 for 122 = h2 + their 4.52 oe (their 4.5 < 12) 35 35 (b) (i) 75 nfww 3 M2 for l = × 15 or x = × 8 oe or 7 7 for 40 seen nfww or correct trig or Pythagoras’ method leading to value rounding to 40.0 l 35 x 35 M1 for = oe or = oe 15 7 8 7 l − 35 35 l − 35 8 or = oe or = oe 8 7 l 15 (ii) 2730 or 2730.0 to 2730.4 nfww 3 M2 dep for π × 15 × their 75 – π × 8 × (their 75 – 35) [+ π × 82 ] dep their 75 > 35 or 805π [2527.7 to 2530] nfww or 869 π [2728.6 to 2731.2] nfww or M1 for π × 15 × their 75 or 1125 π [3532.5 to 3535.8] nfww seen or π × 8 × (their 75 – 35) or 320 π [1004.8 to 1005.8] nfww seen or π × 82 or 64π [200.9 to 201.2] nfww seen (c) (i) 16r3 2 M1 for [M =] k × r3 or 1458=k × 4.53 oe M r 3 or = 3 oe 1458 5.4 After M0, SC1 for 16 seen (ii) 8 : 27 oe 1 Must be numeric, e.g. 128:432

This question in 0580/42 May/June 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

2 In this question use a ruler and compasses only. Show all your construction arcs. The diagram shows a triangular field ABC. The scale is 1 centimetre represents 50 metres. C A B Scale : 1 cm to 50 m (a) Construct the locus of points that are equidistant from A and B. [2] (b) Construct the locus of points that are equidistant from the lines AB and AC. [2] (c) The two loci intersect at the point E. Construct the locus of points that are 250 m from E. [2] (d) Shade any region inside the field ABC that is • more than 250 m from E and • closer to AC than to AB. [2]

8 marks

Mark scheme: 2 (a) Correct perpendicular bisector of AB 2 B1 for accurate with no/wrong arcs with 2 pairs of correct arcs isw or M1 for correct intersecting arcs with no or wrong line (b) Correct angle bisector at A with two 2 B1 for accurate with no/wrong arcs pairs of correct arcs isw or M1 for two pairs of correct arcs with no or wrong line (c) Circle centre E radius 5 cm isw 2FT FT circle centre their E radius 5 cm provided (a) and (b) attempted M1 for 250 ÷ 50 oe soi e.g. from arc If 0 scored SC1 for circle centre their E (d) 2 cao R B1 for each R If 0 scored, SC1 for two ‘correct’ regions but in part (c), centre correct but radius incorrect Qu. Answers Mark Part Marks

This question in 0580/42 Feb/March 2016

10 (a) The ten circles in the diagram each have radius 1 cm. The centre of each circle is marked with a dot. P Calculate the height of triangle PQR. NOT TO SCALE Q R 8 cm … cm [3] (b) Mr Patel uses whiteboard pens that are cylinders of radius 1 cm. A (i) The diagram shows 10 pens stacked in a tray. The tray is 8 cm wide. The point A is the highest point in the stack. NOT TO SCALE Find the height of A above the base, BC, of the tray. B C 8 cm … cm [1] (ii) The diagram shows a box that holds one pen. The box is a prism of length 12 cm. NOT TO The cross section of the prism is an SCALE equilateral triangle. The pen touches each of the three rectangular faces of the box. 12 cm Calculate the volume of this box. … cm3 [5] Question 11 is printed on the next page.

9 marks

Mark scheme: 10 (a) 5.2[0] or 5.196... 3 M2 for [h2=] 6 2 − 3 2 or better or M1 for h2 + 3² = 6² or B1 for PR (or PQ or QR) = 6 (b) (i) 7.2[0] or 7.196... 1FT FT their (a) + 2 (ii) 62.4 or 62.35... 5 M4 for 12 × 6 × ½ tan 60 oe or M3 for 6 × ½ tan 60 oe or M2 for realising that ½ base = 1 × tan60 oe or B1 for angle 30 or 60 in correct position on diagram or in a calculation If 0 scored, SC1 for volume = an area × 12 seen

This question in 0580/42 Feb/March 2016

7 NOT TO SCALE 210° 88cmcm O A B 8 cm 72° C The diagram shows a design for a logo made from a sector and two triangles. The sector, centre O, has radius 8 cm and sector angle 210°. AC = 8 cm and angle ACB = 72°. (a) Show that angle OAB = 15°. [2] (b) Calculate the length of the straight line AB. AB = … cm [4] (c) Calculate angle ABC. Angle ABC = … [3] (d) Calculate the total area of the logo design. … cm2 [6] (e) The logo design is an enlargement with scale factor 4 of the actual logo. Calculate the area of the actual logo. … cm2 [2]

17 marks

Mark scheme: 7 (a) 360 – 210 [= 150] M1 (180 – 150) ÷ 2 [= 15] M1 or 150 ÷ 2 [=75] and 180 – 75 – 90 [ =15] (b) 15.5 or 15.45 to 15.46 nfww 4 M3 for 2 × 8 cos 15 oe or M2 for 8 cos 15 oe x or M1 for = cos 15 oe 8 8 × sin72 (c) 29.5 or 29.4 or 29.39 to 29.50.. 3 M2 for [sin ABC = ] their (b ) sin ABC sin72 or M1 for = oe 8 their (b ) 210 2 (d) 194 or 193.7 to 194.1 nfww 6 M2 for × π × 8 360 or M1 for [k] π × 8 2 seen and M1 for ½ × 82 × sin 150 oe and M2 for ½ × 8 × their (b) × sin(108 – their (c)) oe or B1 for [angle CAB=] 108 – their (c) (e) 12.1 or 12.11 to 12.13 2FT FT their (d) ÷ 42 oe 2  1  M1 for 42 or   soi  4 

This question in 0580/42 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

10 (a) r cm NOT TO SCALE w° r cm The area of this sector is r2 square centimetres. Find the value of w. w = … [3] (b) NOT TO r cm SCALE x° r cm 7r r The perimeter of this sector is 2r + centimetres. 10 Find the value of x. x = … [3] (c) y° q cm q cm NOT TO SCALE cm The perimeter of the isosceles triangle is 2q + q 3 centimetres. Find the value of y. y = … [4]

10 marks

Mark scheme: r 10 (a) 115 or 114.5 to 114.6 3 M2 for 2 or better π r 360 w 2 2 or M1 for × π × r = r 360 x 7π r (b) 126 3 M2 for × 2π r [ + 2 r ] = [2 r + ] or better 360 10 x or M1 for × 2π r 360 (c) 120 4 B3 for 2y = 60 or x (base angle) = 30 OR = M3 for cos x or sin y 3 1 oe or cos y = − 2 2 ( 2 ) oe y q 3 or M2 for cos x or sin = ( 2 ) 2 q 2 q 2 + q 2 − q 3 ( ) or [cos y] = oe 2 × q × q or M1 for  2  2 2 q 3 = q + q − 2 × q × q cos y oe  ( )    2 2  1  2 After M0, SC1 for [ h = ]q − q 3  or for  2  q replaced by 1, 2, 4, etc.

This question in 0580/41 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

10 The perimeter of each of the three shapes is 60 cm. Find x in each part. (a) NOT TO Rectangle x SCALE 3x x = … cm [2] (b) NOT TO Triangle 3x SCALE 4x x = … cm [3] (c) Sector x NOT TO SCALE O x x = … cm [3]

8 marks

Mark scheme: 10 (a) 7.5 2 M1 for 3x + x + 3x + x = 60 oe (b) 5 3 B2 for 3x + 4x + 5x [= 60] or better or M1 for (3 x ) 2 + (4 x ) 2 oe 90 (c) 16.8 or 16.80…. 3 M2 for x + x + × π × 2 × x [ = 60] oe 360 90 or M1 for × π × 2 × x oe 360

This question in 0580/42 Oct/Nov 2016

8 (a) In triangle TXZ, TX = 12.5 cm and angle TZX = 37°. Y is a point on the line XZ such that TY = 9.9 cm, angle XTY = 23° and angle TYZ = 72°. T 23° NOT TO SCALE 12.5 cm 9.9 cm 72° 37° X Y Z (i) Calculate XY. XY = … cm [4] (ii) Calculate TZ. TZ = … cm [3] (b) The diagram shows a shape made up of three identical sectors of a circle, each with sector angle 65°. The perimeter of the shape is 20.5 cm. 65° 65° NOT TO SCALE 65° Calculate the radius of the circle. … cm [4]

11 marks

Mark scheme: 8 (a) (i) 5.14 or 5.135 to 5.142 nfww 4 M2 for [ XY 2 = ] 12.52 + 9.92 − 2 × 12.5 × 9.9 × cos 23 or M1 for implicit version A1 for 26.4 to 26.5 OR B1 for [XYT = ] 108 or [TXY = ] 49 12.5sin 23 M2 for oe sin(180 − 72) sin(180 − 72) sin23 or M1 for = oe 12.5 XY

This question in 0580/42 Feb/March 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

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

8 (a) The exterior angle of a regular polygon is x° and the interior angle is 8x°. Calculate the number of sides of the polygon. … [3] (b) C NOT TO SCALE O D B 58° A A, B, C and D are points on the circumference of the circle, centre O. DOB is a straight line and angle DAC = 58°. Find angle CDB. Angle CDB = … [3] (c) R O NOT TO SCALE 48° P Q P, Q and R are points on the circumference of the circle, centre O. PO is parallel to QR and angle POQ = 48°. (i) Find angle OPR. Angle OPR = … [2] (ii) The radius of the circle is 5.4 cm. Calculate the length of the major arc PQ. … cm [3]

11 marks

Mark scheme: 8(a) 18 3 B2 for 20 nfww or M1 for 8 x + x = 180 or better 8(b) 32 3 B1 for angle DBC = 58 B1 for angle BCD = 90 8(c)(i) 24 2 B1 for angle PRQ = 24 8(c)(ii) 29.4 or 29.40 to 29.41 3 360 − 48 M2 for × 2 × π × 5.4 360 or B2 for answer (minor arc) 4.52 or 4.523 to 4.524… 48 or M1 for × 2 × π × 5.4 360

This question in 0580/41 May/June 2018

5 A O NOT TO SCALE 8 cm 7 cm 78° C B The diagram shows a design made from a triangle AOC joined to a sector OCB. AC = 8 cm, OB = OC = 7 cm and angle ACO = 78°. (a) Use the cosine rule to show that OA = 9.47 cm, correct to 2 decimal places. [4] (b) Calculate angle OAC. Angle OAC = … [3] (c) The perimeter of the design is 29.5 cm. Show that angle COB = 41.2°, correct to 1 decimal place. [5] (d) Calculate the total area of the design. … cm2 [4]

16 marks

Mark scheme: 5(a) 8² + 7² − 2 × 7 × 8 × cos78 oe M2 M1 for correct implicit version 9.471.. to 9.472 A2 A1 for 89.7… 5(b) 46.3 or 46.29 to 46.30… 3 7sin78 M2 for [sin OAC = ] 9.47 sin OAC sin78 or M1 for = 7 9.47 5(c) 29.5 – (7 + 8 + 9.47) M1 360 × (29.5 − (7 + 8 + 9.47)) M3 x M2 for × 2 × π × 7 = their arc length 2 × π × 7 360 oe x or M1 for × 2 × π × 7 oe 360 41.15 to 41.171.. B1 5(d) 45[.0] or 44.98 to 45.01 nfww 4 M3 for 41.2 2 ½ × 8 × 7 × sin 78 oe + × π × 7 oe 360 OR M1 for ½ × 8 × 7 × sin 78 oe or ½ × 8 × 9.47 × sin their (b) oe 41.2 2 M1 for × π × 7 oe 360

This question in 0580/42 May/June 2018

10 NOT TO O SCALE Y X 8 cm A C 15 cm B 22.4 cm The diagram shows a circle, centre O. The straight line ABC is a tangent to the circle at B. OB = 8 cm, AB = 15 cm and BC = 22.4 cm. AO crosses the circle at X and OC crosses the circle at Y. (a) Calculate angle XOY. Angle XOY = … [5] (b) Calculate the length of the arc XBY. … cm [2] (c) Calculate the total area of the two shaded regions. … cm2 [4] Question 11 is printed on the next page.

11 marks

Mark scheme: 10(a) 132.26 to 132.28 or 132.3 5 B1 for angle ABO or angle CBO = 90 soi 15 M1 for tan [XOB] = oe 8 224. M1 for tan [BOY] = oe 8 A1 for [BOY =]70.3… or [XOB =] 61.9…. 10(b) 18.4 or 18.5 or 18.43 to 18.48 2 their (a) M1 for × 2 × π × 8 oe 360 10(c) 75.7 to 75.9 4 1 M1 for (15 + 22.4 ) × 8 oe 2 their ( a ) 2 M2 for × π × 8 oe 360 or M1 for one sector either  15  inv tan    8 × π × 8 2 oe 360  22.4  inv tan    8 2 or × π × 8 oe 360 ( )( )

This question in 0580/41 Oct/Nov 2018

9 NOT TO SCALE A 8 cm 165° B O The diagram shows a sector of a circle with centre O, radius 8 cm and sector angle 165°. (a) Calculate the total perimeter of the sector. … cm [3] (b) The surface area of a sphere is the same as the area of the sector. Calculate the radius of the sphere. [The surface area, A, of a sphere with radius r is A = 4rr 2 .] … cm [4] (c) O NOT TO h SCALE r A B A cone is made from the sector by joining OA to OB. (i) Calculate the radius, r, of the cone. r = … cm [2] (ii) Calculate the volume of the cone. r r h .] [The volume, V, of a cone with radius r and height h is V = 13 2 … cm3 [4]

13 marks

Mark scheme: 9(a) 39[.0] or 39.03 to 39.04… 3 165 M2 for × 2 × π × 8 + 16 360 165 or M1 for × 2 × π × 8 360 9(b) 2.71 or 2.708… 4 165 2 [× π ] × 8 360 M3 for oe 4[× π ] 165 2 [× π ] × 8 360 or M2 for r2 = oe 4[× π ] 165 2 or M1 for × π × 8 oe seen 360 9(c)(i) 3.67 or 3.666 to 3.667 2 165 M1 for ×2[×π]×8 = 2[×π]× r or better 360 or for 165[×π]×8 2 = [π×]r ×8 or better 360 9(c)(ii) 100 or 100.0 to 100.1… final answer 4 1 2 2 2 M3 for π × their ( c )( i ) × 8 – their radius 3 or M2 for 8 2 – their radius 2 or M1 for (their (c)(i))2 + h2 = 82

This question in 0580/41 May/June 2020

8 (a) O 53° NOT TO 9.5 cm A B SCALE X Y The diagram shows a sector OXY of a circle with centre O and radius 9.5 cm. The sector angle is 53°. A lies on OX, B lies on OY and OA = OB . (i) Show that the area of the sector is 41.7 cm 2, correct to 1 decimal place. [2] 1 (ii) The area of triangle OAB is of the area of sector OXY. 3 Calculate OA. OA = … cm [4] (b) O 60° NOT TO 24 cm SCALE P Q The diagram shows a sector OPQ of a circle with centre O and radius 24 cm. The sector angle is 60°. A cone is made from this sector by joining OP to OQ. O NOT TO SCALE P Q 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 [6]

12 marks

Mark scheme: 8(a)(i) 53 2 M1 × π× 9.5 360 41.74 to 41.75 A1 8(a)(ii) 5.9[0] or 5.899 to 5.903.. 4 2 13 × 41.7 M3 for  OA =  oe   1 2 sin53 1 2 1 M2 for × OA × sin53 = × 41.7 oe 2 3 1 1 M1 for × OA × OB × sin53 = × 41.7 seen or 2 3 better 8(b) 396 or 397 or 396.4 to 396.6 6  60  M2 for [ r = ]  ×2× π×24  ÷ 2π oe or better  360  60 or M1 for 2πr = × 2 × π× 24 oe 360 M2 for 24 2 −a 2 or M1 for h2 + a2 = 242 1 2 M1 for 3π× their r × their h

This question in 0580/42 Feb/March 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

1 (a) NOT TO 5.7 cm SCALE 9.2 cm 19.4 cm The diagram shows a brick in the shape of a cuboid. (i) Calculate the total surface area of the brick. … cm2 [3] (ii) The density of the brick is 1.9 g/cm3. Work out the mass of the brick. Give your answer in kilograms. [Density = mass ÷ volume] … kg [3] (b) 9000 bricks are needed to build a house. 200 bricks cost $175. Work out the cost of the bricks needed to build 5 houses. $ … [3] (c) Saskia builds a wall using 1500 bricks. She can build at the rate of 40 bricks each hour. She works for 9 hours each day. Saskia starts work on 6 July and works every day until the wall is completed. Find the date when she completes the wall. … [3] (d) Rafa has a cylindrical tank. The cylinder has a height of 105 cm and a diameter of 45 cm. Calculate the capacity of the tank in litres. … litres [3]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 683 3 M2 for [2]((19.4 × 9.2) + (5.7 × 9.2) + (19.4 × 5.7)) oe or M1 for one of 19.4 × 9.2 or 5.7 × 9.2 or 19.4 × 5.7 1(a)(ii) 1.93[0] or 1.932 to 1.933 3 M2 for 19.4 × 9.2 × 5.7 × 1.9 or M1 for 19.4 × 9.2 × 5.7 1(b) 39 375 3 M2 for 9000 ÷ 200 × 175 × 5 175 or M1 for 9000 ÷ 200 soi or for soi 200 1(c) 10th July 3 1 B2 for 4.1 to 4.2 or 4 or 4 days 1.5 6 hours Or M2 for answer 9th July or 11th July or M1 for 1500 ÷ (9 × 40) 1(d) 167 or 166.9 to 167.0… 3 B2 for answer with figs 167 or figs 1669 to 1670.. or M1 for π× 22.5 2 × 105 oe If 0 scored SC1 for answer 668 or 667.9 to 668.1

This question in 0580/41 Oct/Nov 2021

9 (a) NOT TO x cm SCALE ( x + 3)cm This rectangle has perimeter 20 cm. Find the value of x. x = … [3] (b) M y° NOT TO SCALE 20° This rhombus has perimeter 20 cm and angle y is obtuse. M is the midpoint of one of the sides. Find the value of y. y = … [5] (c) r cm NOT TO SCALE z cm 40° This sector of a circle has radius r and perimeter 20 cm. Find the value of z. z = … [6]

14 marks

Mark scheme: 9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe M1 for correct ax = b for their linear equation 9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20 M2 for sin p = 2.5 2.5 5 or M1 for = sin 20 sin p A1 for 43.2 or 43.15 to 43.17 M1dep for 180 – (20 + their 43.2) After 0 scored, SC1 for length of side = 5 9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413 40 or M2 for r + r + × 2 × π× r = 20 oe 360 40 or M1 for × 2 × π× r oe seen 360 M2 for 2 × 7.41 × sin 20 oe or 7.412 + 7.412 – 2(7.412) cos 40 oe 7.41sin 40 or oe sin70 or M1 for implicit version

This question in 0580/41 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

9 (a) O O NOT TO SCALE x° A B 2.4 cm AB The volume of a paper cone of radius 2.4 cm is 95.4 cm3. The paper is cut along the slant height from O to AB. The cone is opened to form a sector OAB of a circle with centre O. Calculate the sector angle x°. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … [6] (b) An empty fuel tank is filled using a cylindrical pipe with diameter 8 cm. Fuel flows along this pipe at a rate of 2 metres per second. It takes 24 minutes to fill the tank. Calculate the capacity of the tank. Give your answer in litres. … litres [4]

10 marks

Mark scheme: 9(a) 54[.0] or 53.99 to 54.03… 6 M2 for [h = ] 95.4 × 3 ÷ ( π × 2.42) oe 1 2 or M1 for 95.4 = × π × 2.4 × h 3 M2 for [slant ht , l = ] ( their h ) 2 + 2.4 2 or M1 for (their h)2 + 2.42 x M1 for × 2 × π × theirl = 2 × π × 2.4 oe 360 x 2 or × π × ( theirl ) = π × 2.4 × theirl 360 9(b) 14500 or 14470 to 14480 4 M3 for 200 × 60 × 24 × π × 42 [÷1000] or 2 × 60 ×24 ×π × 0.042 [×1000] or M2 for 200 × π × 42 or for 2 × π × 0.042 or M1 for π × 42 oe or π × 0.042 seen oe isw or 1000 cm3 = 1 litre soi or 1 m3 = 1000 litres soi or for 24 × 60 seen oe

This question in 0580/42 Feb/March 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

10 (a) The lengths of the sides of a triangle are 11.4 cm, 14.8 cm and 15.7 cm, all correct to 1 decimal place. Calculate the upper bound of the perimeter of the triangle. … cm [2] (b) 15.6 cm NOT TO SCALE 150° The diagram shows a circle, radius 15.6 cm. The angle of the minor sector is 150°. Calculate the area of the minor sector. … cm2 [2] (c) r cm NOT TO x° SCALE The diagram shows a circle, radius r cm and minor sector angle x°. The perimeter of the major sector is three times the perimeter of the minor sector. 90 ( r - 2 ) Show that x = . r [4]

8 marks

Mark scheme: 10(a) 42.05 final answer 2 M1 for 11.4 + 0.05 oe or 14.8 + 0.05 oe or 15.7 + 0.05 oe 10(b) 319 or 318.5 to 318.6 2 150 2 M1 for   15.6 oe 360 10(c) 360 − x  x  M2  2πr + 2r = 3   2r + 2r  oe x 360  360  M1 for  2πr oe seen 360 or 360 − x  2πr oe seen 360 4 x M1 i.e. M mark for isolating and collecting terms in x  2π[r ] = 2π[r] – 4[r] oe 360 90 (− 2 ) A1 With no errors or omissions Leading to 

This question in 0580/42 Oct/Nov 2022

8 A NOT TO SCALE 9.5 cm O 10 cm B 7.7 cm D C E A, B and C are points on the circle, centre O. DE is a tangent to the circle at C. AC = 10 cm , AB = 9.5 cm and BC = 7.7 cm . (a) Show that angle ABC = 70.2° , correct to 1 decimal place. [4] (b) Find (i) angle AOC Angle AOC = … [1] (ii) angle ACO Angle ACO = … [1] (iii) angle ACD. Angle ACD = … [1] (c) Calculate the radius, OC, of the circle. OC = … cm [3] (d) Calculate the area of triangle ABC as a percentage of the area of the circle. … % [4]

14 marks

Mark scheme: 8(a) 9.52 + 7.7 2 − 10 2 M2 M1 for 102 = 9.52 + 7.72 – 2×9.5×7.7cosB oe or better [cos B = ] oe 2  9.5  7.7 70.206 to 70.207 or 70.21 to 70.22 A2 2477 A1 for oe or 0.339 or 0.3386…. 7315 8(b)(i) 140.4 1 8(b)(ii) 19.8 1 FT (180 – their (b)(i)) ÷ 2 8(b)(iii) 70.2 1 FT 90 – their (b)(ii) 8(c) 5.31 or 5.314 to 5.315 3 5 M2 for oe cos their(b)(ii) 5 or M1 for = cos(their (b)(ii)) oe r 8(d) 38.8 or 38.9 or 38.78 to 38.85 4 0.5  9.5  7.7  sin70.2 M3 for [ 100] their( (c)) 2 OR M1 for 0.5 × 9.5 × 7.7 × sin70.2 M1 for (their (c)2)

This question in 0580/43 Oct/Nov 2022

1 (a) 42° NOT TO SCALE x° The diagram shows an isosceles triangle with the base extended. Find the value of x. x = … [3] (b) The diagram shows three lines meeting at a point. The ratio a : b : c = 3 : 4 : 5. Find the value of c. a° NOT TO c° b° SCALE c = … [3] (c) A regular pentagon has an exterior angle, d. A regular hexagon has an interior angle, h. d Find the fraction . h Give your answer in its simplest form. … [4] (d) S R x° ( x + 20)° NOT TO SCALE ( 3x – 40)° Q ( 2x – 5)° P Show that PQRS is a cyclic quadrilateral. [5] (e) B A 50° 9 cm NOT TO O SCALE The diagram shows a circle of radius 9 cm, centre O. The minor sector AOB, with sector angle 50°, is removed from the circle. Calculate the length of the major arc AB. … cm [3]

18 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 111 3  42 M2 for 180 –180 oe or 42 + 2 180  42 oe 2 180  42 or M1 for oe 2 1(b) 150 3 M1 for k ÷ (3 + 4 + 5) [×p] where p = 1, 3, 4 or 5 5 or oe 12 B1 for 360 used 1(c) 3 4 72 cao nfww B3 for 5 120 or B2 for [d = ] 72 or [h = ] 120 or M1 for 360 ÷ 5 oe isw or 180 – (360 ÷ 6) isw or for (6 – 2) × 180 [÷ 6] 1(d) x + 2x – 5 + x + 20 + 3x – 40 = 360 M1 Accept equivalent equation e.g. 7x – 25 = 360 7x = 360 + 5 – 20 + 40 or better M1 FT their equation, accept e.g. 7x = 385 x = 55 B1 55 and 125 B1dep Dep on M1M1B1 or 105 and 75 Accept 55 + 3 × 55 – 40 = 180 or 2 × 55 – 5 + 55 + 20 = 180 If B0 scored, SC1 for 55, 75, 105 and 125 Opposite angles sum to 180 oe A1 Dep on M1M1B1B1 [so PQRS is a cyclic quadrilateral ] 1(e) 48.7 or 48.69 to 48.70… 3 360  50 M2 for 2 π oe9 360 50 or M1 for  2  π oe9 360

This question in 0580/42 May/June 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

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

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

11 (a) Q P NOT TO SCALE x° O 3 In the circle, centre O, the length of the minor arc PQ is of the length of the major arc PQ. 7 Show that x = 108 . [3] (b) A NOT TO SCALE r y° B O The diagram shows a sector, OAB, of a circle with centre O and radius r. The area of triangle OAB is half the area of the sector. Angle AOB = y° and is obtuse. (i) Show that 360 siny = r y . [2] (ii) Complete the table, giving your answers correct to two decimal places. y 360 siny ry 108.4 341.60 340.55 108.5 341.40 340.86 108.6 341.20 108.7 [3] (iii) Complete the statement. The value of y, correct to one decimal place, that satisfies the equation 360 siny = r y is … . [1]

9 marks

Mark scheme: 11(a) 3 M2  360 oe 3 x 10 M1 for = 3  7 360 x 3 360  x or for [ 2r] =  [ 360 7 360 2r] oe or better or 10 360  x 1 [ 2r] =  [ 2r] oe or 7 360 better 360 or  k (k = 1 or 7) 7  3 108 A1 11(b)(i) 1 1 y 2 y 1 r2 siny =   r2 M1 for  r2 or for r2 siny 2 2 360 360 2 y 1 or  r2 = [2  ] r2 siny 360 2 and one further step leading to 360siny = y with no errors 11(b)(ii) 341.18 or 341.22 3 B1 for each 341.00 341.49 or 341.54 11(b)(iii) 108.6 cao 1

This question in 0580/43 May/June 2024

10 (a) NOT TO SCALE ( x + 1)cm ( 2x + 3)cm This rectangle has area 190 cm2. (i) By forming and solving an equation, show that x = 8.5 . [4] (ii) Work out the perimeter of the rectangle. … cm [2] (b) A r cm NOT TO SCALE 50° O B The diagram shows a sector OAB of a circle, with centre O, and a chord AB. The shaded segment has area 30 cm2. (i) Show that r = 23.7 cm, correct to 1 decimal place. [4] (ii) Calculate the perimeter of the shaded segment. … cm [4]

14 marks

Mark scheme: 10(a)(i) 2x2 + 5x –187 [= 0] M2 M1 for (2x + 3)(x + 1) = 190 (2x –17)(x + 11) [= 0] oe M1 Leading to x = 8.5 with no errors A1 10(a)(ii) 59 2 M1 for 6 × 8.5 + 8 oe or 6x + 8 oe or B1 for 9.5 and 20 10(b)(i) 50 1 M3 π  r2 – r2  sin50 = 30 oe 360 2 50 M1 for π  r2 360 1 M1 for r2  sin50 oe 2 23.70[9] to 23.72… A1 must see at least 4 sig figs 10(b)(ii) 40.7 or 40.8 or 40.71 to 40.75… 4 M2 for 2 × 23.7 × sin 25 oe or 23.72 + 23.72 −2 23.7  23.7cos50 oe 23.7 sin 50 or oe  180 − 50  sin    2  x or M1 for = sin25 oe 23.7 or for 23.7 2 + 23.7 2 −2 23.7  23.7cos50 oe AB 23.7 or = oe sin 50  180 − 50  sin    2  AND 50 M1 for × 2 × π  23.7 oe 360

This question in 0580/41 Oct/Nov 2024

7 (a) (i) 13 cm NOT TO 8 cm SCALE 9 cm Calculate the area of the trapezium. … cm2 [2] (ii) ( y + 4) cm NOT TO ( y + 2) cm SCALE ( y + 1) cm The area of this trapezium is 264 cm2. (a) Show that 2y 2 + 9 y - 518 = 0 . [3] (b) Solve 2y 2 + 9 y - 518 = 0 by factorisation to find the value of y. y = … [3] (b) NOT TO SCALE 8 cm 75° The diagram shows a sector of a circle with radius 8 cm and angle 75°. Find the perimeter of the sector. … cm [3] (c) A B NOT TO SCALE 5 cm P Q O The diagram shows a shape ABQP made from three straight lines and an arc of a sector of a circle. The sector has centre O and angle 90°. POQ is a straight line and AP = PO = OQ = QB = 5 cm. Find the area of ABQP. Give your answer in the form a + k r . … cm2 [4]

15 marks

Mark scheme: 7(a)(i) 88 2 1 M1 for (9 + 13)  8 oe 2 7(a)(ii)(a) 1 M1 ( y + 4 + y + 1)  ( y + 2) [ = 264] 2 or 1 3 ( y + 2) + ( y + 1)  ( y + 2) [ = 264] 2 2 y 2 + 5 y + 4 y + 10 B1 Leading to 2 y 2 + 9 y − 518 = 0 A1 No errors or omissions 7(a)(ii)(b) (2 y + 37)( y − 14) B2 B1 for (2 y + a )( y + b) where ab = –518 or a + 2b = 9 or 2 y ( y − 14) + 37( y − 14) or y (2 y + 37) − 14(2 y + 37) 14 B1 7(b) 26.5 or 26.47... 3 10π B2 for 10.5 or 10.47… or 3 OR 75 M2 for 8 + 8 +  2π8 360 75 or M1 for  2π8 360 7(c) 25 4 25 + π 90 2 2 2 2 M2 for  ( (5 + 5 ) ) 360 or M1 for [radius 2 = ] 52 + 52 1 M1 for [triangle area = ] [2×] 5 5 oe 2

This question in 0580/43 Oct/Nov 2024

21 A NOT TO SCALE N P C B 12 cm The diagram shows an equilateral triangle ABC with all sides of length 12 cm. ANP is a sector of a circle, centre A. N lies on AC such that AN : NC = 2 : 1. P lies on AB such that AP : PB = 2 : 1. Calculate the area of the shaded region. … cm2 [4]

4 marks

Mark scheme: 21 28.8 or 28.83 to 28.84… 4 M3 for 1 60  2  2 × 12 × 12 × sin60 – × π ×  × 12  2 360  3  OR k  2  2 M1 for × π ×  × 12  oe 360  3  k 1 where k < 90 or < 360 4 1 M1 for × 12 × 12 × sink oe where k < 90 2 If 0 scored, SC1 for [radius =] 8 and [∠A =] 60 seen

This question in 0580/43 May/June 2025