Surface area and volume — Paper 4 · IGCSE Mathematics - International 0607

5 8 cm NOT TO SCALE 16 cm The diagram shows a solid sphere of radius 4 cm inside a hollow cone of radius 8 cm and height 16 cm. The sphere touches the interior of the cone. (a) Calculate the volume of the cone that is not occupied by the sphere. … cm3 [3] (b) Calculate the curved surface area of the cone. … cm2 [3] (c) 8 cm NOT TO SCALE O 4cm 16 cm V The centre, O, of the sphere is directly above the vertex, V, of the cone. Calculate the length OV. OV = … cm [4]

10 marks

Mark scheme: 5(a) 804 or 804.2 to 804.4 3 1 2 M1 for 3π× 8 × 16 4 3 M1 for π× 4 3 5(b) 450 or 449.5 to 449.6… 3 2 2 M2 for π× 8 × 8 + 16 or M1 for 8 2 + 16 2 or π× 8 ×their l 5(c) 8.94 or 8.944… 4 P is point of contact between slant edge and circle. B2 for PV = 8 nfww 8 16 or M1 for = oe 4 PV M1 for OV 2 = 4 2 + PV 2 OR B2 for l = 320 oe or M1 for l 2 = 82 + 16 2 8 l M1 for = soi 4 OV OR x is semi-vertical angle of cone 8 M1 for tan x = oe 16 4 M2 for sin x 4 or M1 for = sin x OV

This question in 0607/43 May/June 2017

6 V NOT TO SCALE 8 cm B C √72 cm P A D √72 cm The diagram shows a pyramid with a square base ABCD of side 72 cm. The diagonals of the base, AC and BD, meet at P. The vertex, V, is vertically above P and VP = 8 cm. (a) Find the volume of the pyramid. Give the units of your answer. … … [3] (b) Find the length AC. AC = … cm [2] (c) Find the length DV. DV = … cm [3] (d) Find angle VDP. Angle VDP = … [2] (e) X is the midpoint of the side CD. (i) Find the length VX. VX = … cm [3] (ii) Find angle VXP. Angle VXP = … [2] (f) The pyramid is cut parallel to ABCD to form a smaller pyramid VEFGH. The volume of VEFGH is 24 cm3. Find the vertical height of this pyramid. … cm [3]

18 marks

Mark scheme: 6(a) 192 2 1 2 M1 for × 72 × 8 oe ( ) 3 cm3 1 6(b) 12 2 M1 for ( 72) 2 + ( 72) 2 oe 6(c) 10 3 2 2 M2 for 8 + ( 0.5 their (b) ) or M1 for [PD oe =] 0.5 × their (b) (d) 53.1 or 53.13 2 8 8 M1 for tan = or sin = 0.5 × their (b) their (c) 0.5 × their (b) or cos = their (c) 6(e) (i) 82 or 9.06 or 9.055... 3 M2 for 8 2 + (0.5 × 72) 2 or (their (c))2 – (0.5 × 72) 2 or M1 for (0.5 × 72) 2 6(e)(ii) 62.1 or 62[.0] or 62.00 to 62.10 2 8 M1 for tan = oe 0.5 × 72 6(f) 4 cao 3 24 their (a) M2 for 3 or 3 their (a) 24 1 soi by 2 or 2 24 their (a) or M1 for or their (a) 24 1 soi by 8 or 8

This question in 0607/43 Oct/Nov 2017

7 30 cm 10 cm The diagram shows a hollow metal hemisphere. The outside diameter of the hemisphere is 30 cm and the inside diameter is 10 cm. (a) Find the volume of metal used to make the hemisphere. … cm3 [3] (b) Find the total surface area of the hemisphere. … cm2 [5]

8 marks

Mark scheme: 7(a) 6810 or 6806 to 6808 3 1 4 3 3 M2 for × π(15 − 5 ) 2 3 1 4 3 1 4 3 or M1 for either [ ×] π × 15 or [ ×] π × 5 2 3 2 3 7(b) 2200 or 2199… 5 M4 for 2 × π× 5 2 + 2 × π× 15 2 + π× (15 2 − 5 2 ) or M1 for each term

This question in 0607/43 Oct/Nov 2017

6 (a) NOT TO SCALE 12 cm 40 cm (i) The rectangle can be made into a hollow cylinder with height 40 cm. (a) Show that the radius of this cylinder is 1.910 cm, correct to 3 decimal places. [2] (b) Calculate the volume of this cylinder. … cm3 [2] (ii) The rectangle can also be made into a hollow cylinder with height 12 cm. Calculate the difference between the volumes of this cylinder and the cylinder in part (i). Give your answer correct to the nearest 10 cm3. … cm3 [4] (b) A model of a car is mathematically similar to the actual car. The volume of the model is 75 cubic centimetres and the volume of the actual car is 4.8 cubic metres. The scale is model : actual = 1 : n . Find the value of n. n = … [4]

12 marks

Mark scheme: 6(a)(i)(a) 2πr = 12 oe M1 1.9096 to 1.9099 A1 6(a)(i)(b) 458 or 457.9 to 458.5 2 M1 for π × 1.91[0]2 × 40 6(a)(ii) 1070 4 B3 for volume of other cylinder 1530 or 1527 to 1529. …  40  2 or M2 for π × × 12    2π  or M1 for 40 ÷ ( 2 π) oe 6(b) 40 4 3 4.8 × 100 M3 for 3 oe 75 figs 48 figs 75 or M1 for 3 oe or 3 oe figs 75 figs 48 and M1 for 4.8 × 1003 or 75 ÷ 1003 oe

This question in 0607/41 May/June 2018

7 D NOT TO SCALE A 105° E F 0.6 m 0.8 m 1.2 m B C ABCDEF is a solid triangular prism. (a) Calculate the volume of the prism. … m3 [3] (b) Calculate the total surface area of the prism. … m2 [5] (c) ABCDEF is made of metal and has a mass of 2170 kg. It is melted down and made into prisms similar to ABCDEF. Each of these prisms has a mass of 2.17 kg. Calculate the total surface area of each of these smaller prisms. … m2 [3]

11 marks

Mark scheme: 7(a) 0.278 or 0.2781 to 0.2782 3 M2 for 0.5 × 0.6 × 0.8 × sin105 × 1.2 oe or M1 for 0.5 × 0.6 × 0.8 × sin105 7(b) 3.48 to 3.49 5 2 2 M2 for 0.6 + 0.8 − 2 × 0.6 × 0.8 × cos105 or M1 for 0.6 2 + 0.8 2 − 2 × 0.6 × 0.8 × cos105 A1 for 1.12 or 1.117... M1 for their 1.117 × 1.2+ 2 × their area of ABC + 0.6 × 0.12 + 0.8 × 1.2 7(c) 0.0348 to 0.0349 3 FT their (b) ÷ 100 2170  2170  2 M2 for their (b) ÷ 3 ×  3  oe   2.17  2.17  2170 2.17 3 2 or M1 for or or k ( ) 2.17 2170 implied by their (b) ÷ 1000

This question in 0607/43 May/June 2018

11 NOT TO SCALE 8 m 1.8 m The diagram shows a polythene structure in which a farmer grows vegetables. The structure consists of a prism with a quarter of a sphere at one end. The cross-section of the prism is a semicircle. The semicircle has a radius of 1.8 m and the length of the prism is 8 m. (a) Calculate the volume of the structure. … m3 [3] (b) The curved surface of the prism and the two ends of the structure are made of polythene. Calculate the area of the polythene. … m2 [4]

7 marks

Mark scheme: 11(a) 46.8 or 46.82 to 46.83 3 1 M1 for × π × 1.82 × 8 oe 2 1 4 M1 for × × π × 1.83 oe 4 3 11(b) 60.5 or 60.49 to 60.51... 4 1 M1 for × π × 1.82 oe 2 1 M1 for × 2 × π × 1.8 × 8 oe 2 1 M1 for × 4 × π × 1.82 oe 4

This question in 0607/42 Oct/Nov 2018

4 NOT TO SCALE 3 cm l cm The diagram shows a solid made from a cylinder and two hemispheres. The radius of the cylinder and each hemisphere is 3 cm. The total volume of the solid is 144r cm3. (a) The length of the cylinder is l cm. Find the value of l. l = … [3] (b) The solid is made of steel. 1 cm3 of steel has a mass of 7.8 g. Calculate the mass of the solid. Give your answer in kilograms. … kg [2] (c) The solid is melted down and made into 20 cubes each of side length 2.8 cm. Calculate the volume of steel not used for the cubes as a percentage of the 144r cm3. … % [3] (d) A solid that is mathematically similar to the original solid has a volume of 18r cm3. Find the radius of the new cylinder. … cm [3]

11 marks

Mark scheme: 4(a) 12 cao final answer 3 B2 for 11.98 to 12.02 2 2 3 or M1 for π × 3 × l + 2 × × π × 3 [ = 144π] 3 oe 4(b) 3.53 or 3.528 to 3.529... 2 M1 for 144 π × 8.7 soi by figs 353 or 3528 to 3529 4(c) 2.95 or 2.96 or 2.950 to 2.963... 3 144 π − 20 × 8.2 3 M2 for [× 100 ] 144 π 20 × 8.2 3 or × 100 oe 144 π 3 20 × 8.2 3 or M1 for 144 π − 20 × 8.2 or oe 144 π 4(d) 1.5 oe cao final answer 3 B2 for 1.498 to 1.502 18π or M2 for 3 × 3 oe 144 π 18π 144 π or M1 for 3 or 3 oe or better 144 π 18π  3 3 144π or for = oe    x  18π

This question in 0607/42 May/June 2019

9 NOT TO SCALE h cm r cm 18 cm 26 cm 24 cm r cm 14 cm The diagram shows three solids, a prism, a sphere and a cone. The radius of the sphere is equal to the base radius of the cone. The volume of each solid is the same. (a) Show that the volume of the prism is 7392 cm3. [3] (b) A similar prism has a volume of 924 cm3. The length of the original prism is 24 cm. Find the length of this similar prism. … cm [3] (c) Find the value of r. r = … [2] (d) Find the value of h. h = … [2] (e) When exact values of h and r are used, h = 4 r. Find, in terms of r, an exact expression for the curved surface area of the cone. Give your answer in its simplest form. … [3]

13 marks

Mark scheme: 9(a) [(14 × 18) + 0.5 × 14 × 8] × 24 oe M3 i.e. area × length or 18 × 14 × 24 + 0.5 × 14 × 8 × 24 oe volume + volume leading to 7392 M2 for 14 × 18 + 0.5 × 14 × 8 or M1 for 14 × 18 or 0.5 × 14 × 8 or 0.5 × (18 + 26) × 7 9(b) 12 cao 3 7392 M2 for 24 ÷ 3 oe 924 7392 or M1 for 3 soi 924 9(c) 12.1 or 2.08… 2 3 3 7392 M1 for r = × oe 4 π 9(d) 48.2 or 48.3 or 48.4 2 3 × 7392 M1 for h = or 48.20 to 48.37… 2 π × (their12.1) 9(e) πr 2 17 final answer 3 M2 for πr r 2 + (4 r ) 2 or M1 for l 2 = r 2 + (4 r ) 2 If 0 scored , SC1 for πr 2 5

This question in 0607/42 Oct/Nov 2019

7 NOT TO SCALE 6 cm 4 cm The diagram shows a child’s toy made of a cone joined to a hemisphere. The cone and the hemisphere each have a radius of 4 cm. The perpendicular height of the cone is 6 cm. (a) (i) Find the volume of the hemisphere. … cm3 [2] (ii) Find the volume of the cone. … cm3 [2] (iii) Each cubic centimetre of the hemisphere has a mass of 7.85 g. Each cubic centimetre of the cone has a mass of 0.65 g. Find the total mass of the toy. … g [2] (b) Find the total surface area of the toy. … cm2 [5] (c) The height of the cone on a similar toy is 9 cm. Find the total surface area of this toy. … cm2 [2]

13 marks

Mark scheme: 7(a)(i) 134 or 134.0 to 134.1 2 1 4 M1 for × × π × 43 oe 2 3 7(a)(ii) 101 or 100.5... 2 1 M1 for × π × 42 × 6 oe 3 7(a)(iii) 1120 or 1117 to 1118 2 2 M1 for × π × 43 × 7.85 soi or 3 their (i) × 7.85 1 or for × π × 42 × 6 × 0.65 soi or 3 their (ii) × 0.65 7(b) 191 or 191.1 to 191.2 5 M1 for 62 + 42 A1 for 7.21 or 7.211... or 52 M1 for π × 4 × their 7.21 M1 for 2 × π × 42 7(c) 430 or 429.7 to 430.2 2 FT their (b)  9  2 M1 for their (b) ×   oe  6 

This question in 0607/43 Oct/Nov 2019

11 (a) C NOT TO SCALE 11 cm 60° A B 12 cm Calculate the shortest distance from B to AC. … cm [7] (b) V NOT TO SCALE h cm S R M 6 cm P 8 cm Q The diagram shows a pyramid on a rectangular base PQRS. The diagonals of the base meet at M and V is vertically above M. PQ = 8 cm, QR = 6 cm and VM = h cm. The volume of the pyramid is 112 cm3. (i) Show that h = 7. [2] (ii) Calculate the length of VR. VR = … cm [3] (iii) K is the mid-point of PS and L is the mid-point of QR. Calculate angle KVL. Angle KVL = … [3]

15 marks

This question in 0607/41 May/June 2020

4 A piece of metal is in the shape of a cuboid. The cuboid has length 18 cm, width 12 cm and height 12 cm. A cylinder is removed from the cuboid. The cylinder has length 18 cm and radius 4 cm. NOT TO SCALE 18 cm 12 cm 4 cm 12 cm (a) (i) Find the volume of the metal remaining after the cylinder has been removed. … cm3 [3] (ii) Write your answer to part (i) in standard form. … cm3 [1] (b) Find the total surface area of the metal remaining after the cylinder has been removed. … cm2 [4] (c) The cylinder removed is melted and formed into 16 identical spheres. (i) Calculate the volume of one sphere. … cm3 [1] (ii) Calculate the radius of one sphere. … cm [2]

11 marks

Mark scheme: 4(a)(i) 1690 or 1687 or 1687.1 to 1687.3 3 M2 for 18 × 12 2 − 18 × π × 4 2 or M1 for either term correct 4(a)(ii) 1.69[0] × 10 3 1 FT their (i) 4(b) 1500 or 1503 to 1504 4 M1 for [4 ×]18 × 12 M1 for [2 ×](12 2 − π × 4 2 ) M1 for π × 2 × 4 × 18 4(c)(i) 56.5 or 56.6 or 56.54 to 56.56 1 4(c)(ii) 2.38 or 2.380 to 2.382 2 3 × their ( c ) M1 for 4 × π

This question in 0607/42 May/June 2020

7 The diagram shows a radio in the shape of a prism. This diagram shows the base of the radio. E F A D G B C H I ABC is an equilateral triangle. The circles have their centres at A, B and C and each has a radius of 5 cm. DE, FG and HI are tangents to the circles. (a) Show that AB = 8.66 cm, correct to 3 significant figures. [3] (b) Calculate the area of the base of the radio. … cm2 [4] (c) The height of the radio is 12 cm. Calculate the volume of the radio. … cm3 [1]

8 marks

Mark scheme: 7(a) 2 × 5 × cos 30 M2 x or M1 for = cos 30 oe 5 8.660... A1 7(b) 241 or 240.9... to 241.2... 4 M1 for 3 × 8.66 × 5 120 M1 for 3 × × π × 52 360 M1 for × 8.662 × sin 60 7(c) 2890 to 2895 1 FT 12 × their (b)

This question in 0607/43 May/June 2020

9 5 cm NOT TO SCALE 24 cm 12 cm 16 cm The diagram shows a solid made from a cuboid and a solid hemisphere. The cuboid measures 12 cm by 16 cm by 24 cm. The hemisphere has radius 5 cm. (a) Find (i) the volume of the solid, … cm3 [3] (ii) the volume of a similar solid where the radius of the hemisphere is 3 cm. … cm3 [2] (b) Find (i) the total surface area of the original solid, … cm2 [3] (ii) the total surface area of a similar solid where the radius of the hemisphere is 6 cm. … cm2 [2]

10 marks

Mark scheme: 9(a)(i) 4870 or 4869 to 4870 3 M1 for 24 × 16 × 12 1 4 3 M1 for × × π × 5 2 3 9(a)(ii) 1050 or 1051 to 1052 nfww 2 3  3  M1 for   oe  5  1 4 3 or × × π × 3 + 14.4 × 9.6 × 7.2 2 3 9(b)(i) 1810 or 1806 to 1807 3 M1 for 24 × 16 × 2 + 24 × 12 ×2 + 16 × 12 × 2 [–π × 52] M1 for 0.5 × 4 × π × 5 2 9(b)(ii) 2600 or 2610 or 2600 to 2606. ... 2 2  6  nfww M1 for   oe soi  5  or 0.5 × 4 × π × 62 + 28.8 × 19.2 × 2 + 28.8 × 14.4 × 2 + 19.2 × 14.4 × 2 – π × 62

This question in 0607/41 Oct/Nov 2020

6 11 cm NOT TO SCALE 3 cm The diagram shows a solid made from a cylinder and two hemispheres. The cylinder has radius 3 cm and length 11 cm. Each hemisphere has radius 3 cm. (a) Find the volume of the solid. Give your answer in terms of r. … cm3 [3] (b) The solid is melted down and all the metal is used to make a cylinder of length 15 cm. (i) Use your answer to part (a) to find the radius of this cylinder. … cm [2] (ii) A rectangular tank contains water. The base of the tank measures 20 cm by 10 cm. The cylinder is placed in the tank so that it is completely covered by water. No water overflows the tank. Calculate the increase in the depth of the water in the tank. … cm [2]

7 marks

Mark scheme: 6(a) 135π 3 M1 for π × 3 2 × 11 4 3 M1 for π × 3 3 6(b)(i) 3 2 M1 for π × r 2 × 15 = their (a) 6(b)(ii) 2.12 or 2.120 to 2.121 2 M1 for 20 × 10 × h = their (a)

This question in 0607/43 Oct/Nov 2020

5 (a) B 49 mm NOT TO 91 mm SCALE A C Calculate the length of AC. AC = … mm [2] (b) 305° NOT TO O SCALE B 16 cm A The diagram shows a circle with centre O and radius 16 cm. Calculate the length of the major arc AB. … cm [2] (c) NOT TO SCALE 12 cm The diagram shows a prism with length 12 cm. The cross-section of the prism is a quarter of a circle. The radius of the circle is 6 cm. Calculate the volume of the prism. … cm3 [2] (d) C NOT TO (2x + 4) cm SCALE B D (x + 1) cm A E (x – 3) cm Shape ABCDE is made by joining rectangle ABDE and triangle BCD. The perpendicular height of triangle BCD is (2x + 4) cm. The total area of ABCDE is 11 cm2. (i) Show that 2x 2 - 3x - 20 = 0 . [3] (ii) Factorise 2x 2 - 3x - 20 . … [2] (iii) Use your answer to part (ii) to solve the equation 2x 2 - 3x - 20 = 0 . x = … or x = … [1] (iv) Find the perpendicular height of triangle BCD. … cm [1]

13 marks

Mark scheme: 5(a) 103 or 103.3 to 103.4 2 M1 for 492 + 912 oe 5(b) 85.2 or 85.17 to 85.18 2 305 M1 for × π × 2 × 16 360 5(c) 339 or 339.2 to 339.3… 2 1 2 M1 for × π × 6 × 12 4 5(d)(i) 1 M1 (x – 3)(x + 1) + (x – 3)(2x + 4) 2 [=11] x2 – 3x + x – 3 B1 one correct expansion seen 1 or (2x2 – 6x + 4x – 12)  2 or x2 – 3x + 2x – 6 At least one more line of A1 no errors or omissions working leading to 2x2 – 3x – 20 = 0 5(d)(ii) (2x + 5)(x – 4) 2 M1 for (2x + a)(x + b) where ab = –20 or a + 2b = –3 or 2x(x – 4) + 5(x – 4) or x(2x + 5) – 4(2x + 5) 5(d)(iii) 4 , –2.5 1 Strict FT their factors Dep on factors in part (ii) 5(d)(iv) 12 1 FT 2 × (their positive root (d)(iii)) + 4

This question in 0607/42 Feb/March 2021

10 B C NOT TO A SCALE h r Cone A has radius r and perpendicular height h. Cone B is mathematically similar to cone A. Solid C is formed by removing cone A from cone B. The ratio height of cone A : height of cone B = 2 : 3. (a) Find the ratio volume of cone A : volume of solid C. … : … [3] (b) Cone A has radius 4 cm and height 10 cm. Calculate the total surface area of solid C. … cm2 [8] Question 11 is printed on the next page.

11 marks

Mark scheme: 10(a) 8 : 19 oe 3 M1 for [Vol A : Vol B =] 23 : 33 oe M1 for [Vol C =] 27k – 8k k any variable OR 1  3r ×2 3h M1 for π   3  2  2 1 2 1  19  2 M1 for [VA : VC =] πr h :   πr h 3 3  8  10(b) 503 or 502.6 to 502.8 8 3 3 M1 for × 4 oe or 10 2 2× 3 32 or × their l oe if their l is from Pythagoras or 2 2 2 M2 for 4 2 + 10 2 or (their R ) 2 + (their H ) 2 or M1 for 4 2 + 10 2 or (their R)2 + (theirH)2 M1 for π× 4 × 116 3 32 M1 for π× 6 × 116 or 2 × π× 4 116 2 2 M2 for CSAa + CSAb + π × (their R)2 – π × 42 oe or M1 for for CSAa + CSAb or π × (their R)2 – π × 42 oe

This question in 0607/42 Feb/March 2021

9 5 cm NOT TO SCALE 12 cm The diagram shows a cup in the shape of a cone. (a) Calculate the curved surface area of the cup. … cm2 [3] (b) The cup is filled with water. A metal sphere of radius r cm is lowered into the cup. The top of the sphere is level with the surface of the water. NOT TO SCALE r cm (i) Use similar triangles to show that r = 3.33 cm correct to 3 significant figures. [3] (ii) Calculate the volume of the water in the cup. … cm3 [3]

9 marks

Mark scheme: 9(a) 204 or 204.2... 3 2 2 M2 for π× 5 × 5 + 12 ( ) or M1 for 52 + 122 (implied by 13) 9(b)(i) r 5 M1 r 5 = oe = 12 − r their13 13 − 5 12 r(their 13) = 5(12 – r) M1 M1 dep on first M1 for 12 r = 5(13 − 5) 1 10 A1  Completion to r = 3.3 or 3 or or 3 3 3.333... with no errors 9(b)(ii) 159 or 159.0 to 159.5 3 1 2 M1 for × π× 5 × 12 3 4 3 M1 for × π× 3.33 3

This question in 0607/42 May/June 2021

4 12 cm NOT TO SCALE 20 cm 5 cm The diagram shows a solid made by joining a cone and a hemisphere to a cylinder. The radius of each of the three shapes is 5 cm. The height of the cylinder is 20 cm and the height of the cone is 12 cm. (a) Calculate the total surface area of the solid. … cm2 [5] 2050 r(b) The total volume of the solid is cm3 . 3 It is melted down and made into spheres of radius 1.2 cm. (i) Find the greatest number of spheres that can be made. … [3] 2050 r (ii) Work out the percentage of the cm3 that remains after the spheres have been made. 3 … % [3]

11 marks

Mark scheme: 4(a) 990 or 989.6 to 989.7... 5 2 2 M2 for π × 5 × 5 + 12 or M1 for 5 2 + 12 2 implied by 13 M1 for 2π × 5 × 20 1 2 M1 for × 4π × 5 2 4(b)(i) 296 3 2050π  4 3  M2 for ÷  π × 1.2  implied by final 3  3  answer 296.5 to 296.6 4 3 or M1 for π × 1.2 implied by 7.24 or 7.238 to 3 7.239... 4(b)(ii) 0.197 or 3 M2 for 0.1972 to 0.1975  2050π 4 3  2050π  − their 296 × × π × 1.2  ÷ oe  3 3  3  4 3  2050π or for  their 296 × × π × 1.2  ÷ × 100 oe  3  3 2050π 4 3 or M1 for − their 296 × × π × 1.2 3 3  4 3  2050π or for  their 296 × × π × 1.2  ÷  3  3

This question in 0607/43 May/June 2021

10 NOT TO 4 cm SCALE 16 cm 12 cm The diagram shows a solid made from a cylinder, a hemisphere and a cone, each with radius 4 cm. The cylinder has length 16 cm. The slant height of the cone is 12 cm. (a) Find the volume of the solid. … cm3 [5] (b) Show that the total surface area of the solid is 208 r cm2. [4] (c) A mathematically similar solid has a total surface area of 468 r cm2. Find the radius of the cylinder in this solid. … cm [3]

12 marks

Mark scheme: 10(a) 1130 or 1127 to 1128 5 M1 for π × 16 × 4 2 1 4 3 M1 for × × π × 4 2 3 M1 for 12 2 − 4 2 or better 1 2 M1 for × π × 4 ×their h 3 10(b) 2 × π × 16 × 4 M1 1 2 M1 × 4 × π × 4 oe 2 π × 12 × 4 M1 32π + 128π + 48π [=208π] B1 10(c) 6 3 468 M2 for × 4 oe 208 468 208 or M1 for or 208 468 or  4 =2 208π oe    r  468π

This question in 0607/43 Oct/Nov 2021

6 (a) NOT TO C SCALE 5 cm B 30° O A The diagram shows a circle, centre O, with radius 5 cm. BA and BC are tangents to the circle at A and C. Angle ABC = 30° . Calculate the area of the shaded minor segment. … cm2 [4] (b) O NOT TO height O SCALE 40° 12 cm D E D E The circle, centre O, has radius 12 cm. Angle DOE = 40° . The minor sector DOE is removed. The major sector is formed into a cone by joining OD to OE. Calculate the height of the cone. … cm [5]

9 marks

Mark scheme: 6(a) 26.5 or 26.47 to 26.48 4 B1 for AOC = 150 or BOC = 75 soi their150  5 2 M1 for or 360 their 75  5 2  2 360 1 M1 for 5 5 sin[their150] or 2 1 5 5 sin[their 75  2] 2 6(b) 5.5[0] or 5.49 to 5.51... 5 (360  40) M2 for 2 12 = 2r 360 (360  40) 2 oe or  12 = πr12 oe 360 (360  40) or M1 for 2 12 or 360 (360  40) 2  12 360 4 If minor sector SC1 for radius = oe 3 AND M2 for 12 2 their ( radius ) 2 oe dependent on at least M1 or SC1 or M1 for h 2  their ( radius ) 2  12 2 oe dependent on at least M1 or SC1

This question in 0607/41 May/June 2022

7 Abbi makes wooden boards in three sizes, small, medium and large. They are all cuboids. The medium board has height 2 cm, width 23 cm and length 50 cm. (a) Calculate the volume of the medium board. … cm3 [2] (b) The small board is mathematically similar to the large board. The small board has a volume of 287.5 cm 3and a height of 1.15 cm. The large board has a volume of 18400 cm 3. (i) Find the height of the large board. … cm [3] (ii) Is the medium board mathematically similar to the large board? Explain how you decide. … because … … [3]

8 marks

Mark scheme: 7(a) 2300 2 M1 for 2 × 23 × 50 oe 7(b)(i) 4.6 3 18400 M2 for 1.15  3 oe 287.5 h 3 18400 18400 or M1 for  or 3 1.153 287.5 287.5 287.5 or 3 seen 18400 7(b)(ii) 2 : their (b)(i) or their (a) : 18400 soi M1 FT their figures showing comparison of length ratio or M1 volume ratio not similar A1 Dep on M1M1

This question in 0607/41 May/June 2022

11 NOT TO SCALE 2.1 m 110° 110° 0.9 m The diagram shows the symmetrical cross-section of a ditch containing water. The angle between the base and each side of the ditch is 110°. The width of the base is 0.9 m and the depth of the water is 2.1 m. The ditch is 100 m long. (a) Calculate the volume of water in the ditch. … m3 [4] (b) On a different day, the ditch contains 300 m 3of water. Water is pumped out of the ditch at a rate of 4.2 litres per second. Calculate the time taken to empty the ditch completely. Give your answer in hours and minutes, correct to the nearest minute. … h … min [4]

8 marks

Mark scheme: 11(a) 349 or 350 or 349.4 to 349.5… 4 M3 for [0.9  2.1  2  12 (2.1  2.1tan 20)] × 100 oe 1 or (0.9  (2  2.1tan 20  0.9))  2.1 × 2 100 oe or M2 for area of cross section 0.9  2.1  2[ 12 (2.1  2.1tan 20)] oe 1 or (0.9  (2  2.1tan 20  0.9))  2.1 2 oe or M1 for tan 20 x 2.1 If 0 scored SC1 for triangle marked/drawn with 20 or 70 11(b) 19 h 50 (or 51) min 4 B3 for 19.8 or 19.84… 300  1000 or M2 for oe 4.2  60  60 or M1 for 300  1000 or 4.2  60  60 or for their volume divided by their rate

This question in 0607/41 May/June 2022

11 (a) A pyramid has a square base with sides of length 9 cm and vertical height h cm. Find an expression, in terms of h, for the volume of the pyramid. … cm3 [1] (b) A NOT TO 10 cm SCALE B C a cm h cm D E 9 cm ADE is an isosceles triangle. BC is parallel to DE, BC = a cm and DE = 9 cm. The vertical height of triangle ADE is h cm and the vertical height of triangle ABC is 10 cm. 90 Show that a = h [1] (c) A square-based pyramid with base of side 9 cm and vertical height h cm contains some water. When the pyramid is placed on level ground the surface of the water is 10 cm below the vertex of the pyramid (see Diagram 1). When the pyramid stands vertically on its vertex, the surface of the water is 1 cm below the base of the pyramid (see Diagram 2). 9 cm 10 cm 1 cm b cm a cm h cm h cm 9 cm Diagram 1 Diagram 2 (i) Use Diagram 1 to find an expression, in terms of a and h, for the volume of the water. … cm3 [1] (ii) Use Diagram 2 to find an expression, in terms of b and h, for the volume of the water. … cm3 [1] (iii) Show that h 3 - 1000 = ( h - 1) 3 . [3] (iv) The equation h 3 - 1000 = ( h - 1) 3 simplifies to h 2 - h - 333 = 0 . Use a graphical method to find the value of h. h = … [2] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) (1/3)  92  h or 181h or 27h oe 1 3 11(b) 10 a 9 a  90  1  ,  a  oe   h 9 h 10  h  11(c)(i) 1 2 1 2 1 FT their (a) V   9 h  a  10 oe isw 3 3 11(c)(ii) 1 2 1 V   b ( h  1) oe isw 3 2 2 M211(c)(iii) h 1 1 90 1 h     1  1 b M1 for oe 9 2 h   10  9 2  h  1     h 9 3 3 3 h h     oe 3 A1 No errors or omissions seen 1000  h  1 3 3 h   , h  1000  ( h  1) h 2 h 2 11(c)(iv) 18.8 or 18.75 to 18.76 cao 2 B1 for 18.8 or 18.75 to 18.76 and negative root as final answers M1 for [quadratic/cubic]sketch(es)

This question in 0607/42 May/June 2022

6 V NOT TO SCALE 12 cm B 12 cm A O D C VABC is a pyramid with a triangular base. All the edges have length 12 cm. O is vertically below V. 2 D is the mid‑point of AC and BO = BD . 3 (a) Show that BO = 6. 928 cm , correct to 3 decimal places. [4] (b) Calculate the volume of the pyramid. … cm3 [4]

8 marks

Mark scheme: 6(a) 2 2 2 M3 M2 for 122 – 62 12  6 oe or M1 for attempt at Pythagoras e.g. BD2 + 62 = 122 3 6.9282... A1 6(b) 204 or 203.6 to 203.7 4 2 2 M1 for 12  6.928 M1 for 0.5 × 12 × 12 × sin60 or 0.5 × their BD × 12 1 M1 for × their 62.35 × their 9.798 dependent on 3 use of Pythagoras and not BD or BO.

This question in 0607/43 May/June 2022

10 15 cm NOT TO SCALE A 5 cm 8 cm B C 11 cm Triangle ABC is the cross-section of a prism of length 15 cm. AB = 5 cm , AC = 8 cm and BC = 11 cm . (a) Show that the area of triangle ABC = 18.33 cm 2 correct to 2 decimal places. [4] (b) Find the volume of the prism. … cm3 [1] (c) Find the total surface area of the prism. … cm2 [2] (d) A mathematically similar prism has a volume of 500 cm 3. Calculate the total surface area of this similar prism. Give your answer correct to 2 significant figures. … cm2 [3]

10 marks

Mark scheme: 10(a) 5 2 + 8 2 − 112 M2 M1 for 112 = 52 + 82 – 2 × 5 × 8 × cosA [cos A =] oe 2 5 8 5 2 + 112 − 8 2 or [cos B =] oe or 82 = 52 + 112 – 2 × 5 × 11 × cosB 2 5 11 112 + 8 2 − 5 2 or [cos C =] oe or 52 = 112 + 82 – 2 × 11 × 8 × cosC 2  11  8 0.5 × 5 × 8 × sin(their A) oe M1 or 0.5 × 5 × 11 × sin(their B) oe or 0.5 × 11 × 8 × sin(their C) oe 18.330... A1 Dep on no errors seen and on M2 and M1 awarded 10(b) 275 or 274.9... 1 10(c) 397 or 396.6 to 396.7 2 M1 for 8 × 15 + 11 × 15 + 5 × 15 + 2 × 18.33 10(d) 590 cao 3 2  500  3 M2 for  ( their (c))   oe  their (b)  1  500  3 or M1 for   oe soi  their(b)   their (c) 3  their (b)  2 or = oe      A   500 

This question in 0607/42 Oct/Nov 2022

2 (a) Calculate the volume of each shape. (i) A cuboid with a square base of side 5 cm and height 3 cm. … cm3 [2] (ii) A sphere with radius 4 cm. … cm3 [2] (b) A cylinder has volume 120 cm3 and height 6 cm. Calculate its radius. … cm [2] (c) A cone has volume 120 cm3 and height 6 cm. Calculate the length of its sloping edge. … cm [3]

9 marks

Mark scheme: 2(a)(i) 75 2 M1 for 52[3] 2(a)(ii) 268 or 268.0 to 268.1... 2 4 M1 for    43 3 2(b) 2.52 or 2.522 to 2.523... 2 M1 for   r2  6  120 or better 2(c) 7.42 or 7.422 to 7.423 3 1 M1 for    r2  6  120 oe or 3 better M1 for 62 (their r)2 or better

This question in 0607/41 May/June 2023

11 E 16 cm C D NOT TO SCALE 34 cm A 20 cm B The diagram shows a cuboid with base ABCD. AB = 20 cm , BC = 34 cm and CE = 16 cm . Water is poured into the cuboid to a height of 8 cm. (a) Find the volume of water in the cuboid. … cm3 [2] (b) A sphere of radius 4 cm is placed so that it rests on the base of the cuboid. The water level is now q cm above the base of the cuboid. Find the value of q. … [4] (c) The sphere is removed from the cuboid. 15 identical cubes of side x cm are placed so that they rest on the base of the cuboid. (i) Find the maximum value of x. x = … [3] (ii) The water level is now p cm above the base of the cuboid. Find the maximum value of p. p = … [3]

12 marks

Mark scheme: 11(a) 5440 2 M1 for 20  34  8 or 20  34  16 11(b) 8.39 or 8.394… 4 B3 for 0.394 or 0.394… OR 4 3  4  their a  3 () M3 for q  680 or M2 for 4 3  20  34  q   4 their ( a )  3 4 3 or M1 for  4 3 11(c)(i) 6.67 or 6.666 to 6.667 3 20 34 M2 for and 3 5 20 34 or M1 for or 3 5 or recognition of 3 by 5 11(c)(ii) 14.5 or 14.53 to 14.55 3 B2 for 6.53 to 6.55 OR M2 for 3 15   their (c)(i)  their (a) p  20  34 or M1 for 3 20  34 p  15   their (c)(i)    their (a) 

This question in 0607/43 May/June 2023

8 P NOT TO SCALE 140° O 9 cm Q The diagram shows the sector of a circle with radius 9 cm and sector angle 140°. (a) Calculate the length of the arc PQ. … cm [2] (b) Calculate the area of the sector. … cm2 [2] (c) The sector is the cross-section of a solid of length 20 cm. Calculate the total surface area of the solid. … cm2 [4] (d) Another solid is mathematically similar to the solid in part (c). The radius of the sector in this solid is 10 cm. Calculate the total surface area of this solid. … cm2 [2]

10 marks

Mark scheme: 8(a) 22[.0] or 21.99... 2 140 M1 for 2 π  9 oe 360 8(b) 99[.0] or 98.96 to 98.97... 2 140 2 M1 for π 9 oe 360 8(c) 998 or 997.7 to 998.0 4 M1 for their (a) × 20 M1 for their (b) × 2 M1 for [2 ×] 9 × 20 8(d) 1230 or 1231 to 1232...nfww 2 2  10  FT their (c)    9   10  2  9  2 M1 for   or    9   10 

This question in 0607/41 Oct/Nov 2023

6 V NOT TO SCALE 12 cm B C O M A 10 cm D VABCD is a square-based pyramid. V is vertically above the centre of the base O. AD = 10 cm and VO = 12 cm . (a) (i) Calculate the volume of the pyramid. … cm3 [2] (ii) M is the mid-point of CD. Show that VM = 13 cm . [2] (b) V Q R P S Q 8 cm R NOT TO SCALE S P B C A 10 cm D A pyramid VPQRS is cut from the larger pyramid so that the face PQRS is parallel to the face ABCD. QR = 8 cm . (i) Calculate the volume of the remaining solid, ABCDPQRS. … cm3 [4] (ii) Calculate the total surface area of the remaining solid. … cm2 [4]

12 marks

Mark scheme: 6(a)(i) 400 2 1 M1 for × 102 × 12 3 6(a)(ii) 2 M2 2  10 12 2  10  M1 for [VM =]   + 122  + 12  2   2  leading to 13 6(b)(i) 195.2 or 195 4 B3 for 204.8 OR B2 for 9.6 or awrt 9.6 or M1 for [height of small pyramid] = 8 12  oe 10 1 M1 for × 82 × (their 9.6) 3 OR   8  3  M3 for their 400   1 −    oe    10     8 3 or M2 for their 400   oe  10   8  3 or M1 for   oe  10  6(b)(ii) 257.6 or 258 4 2 8 B1 for 13  or 13  10 10 M1 for 1 8 × (their 2.6) + 2 × × 1×(their 2.6) oe 2 M1 for 82 and 102 soi

This question in 0607/43 Oct/Nov 2023

9 NOT TO SCALE 60 cm 40 cm The diagram shows a solid cone with base radius 40 cm and slant height 60 cm. (a) Find the volume of the cone. … cm3 [3] (b) Show that the total surface area of the cone is 4000 r cm 2 . [2] (c) A mathematically similar cone has a surface area of 1000 r cm 2 . Show that the radius of this cone is 20 cm. [2] (d) A cone with radius 20 cm is removed from the top of the cone with radius 40 cm to leave a solid. Calculate the surface area of the remaining solid. … cm2 [3]

10 marks

Mark scheme: 9(a) 74 900 or 74 929 to 74 941.1 3 1 2 2 2 M2 for  π  40  60  40 oe 3 or M1 for 602 – 402 [=2000] oe 9(b) π × 402 + π × 40 × 60 M2 M1 for π × 402 or π × 40 × 60 = 4000π with no errors 9(c) Ratio areas = 4000π : 1000π M1 implies Ratio sides = 2 : 1 oe [r=] 40 × 0.5 = 20 oe A1 ALTERNATIVE (M2) 1000π 4000π M1 for or oe 1000π 4000π 1000π 40  oe 4000π 2  40 4000π  oe = 20 with no errors or    x  1000π 9(d) 11 900 or 11 930 to 11 940 or 3800π 3 M2 for  40 2  20 2  60  40  30  20 or M1 for π  60  40  π  30  20 If 0 scored, SC1 for 3400π or 10 700 or 10 680 to 10 681.4…

This question in 0607/41 May/June 2024

1 (a) The volume of a triangular prism is 476 cm 3. The base of the triangle is 8 cm and the perpendicular height is 7 cm. Calculate the length of the prism. … cm [3] (b) The volume of a solid steel cube is 8000 cm 3. (i) The mass of 1 cm 3 of the steel is 7.86 g. Calculate the mass of the cube. Give your answer in kilograms. … kg [1] (ii) Calculate the total surface area of the cube. … cm2 [3] (iii) The steel cube is melted down and made into spheres with radius 3.5 cm. Calculate the number of these spheres that are made. … [3]

10 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 17 3 1 M1 for 8 7 2 M1 for 476 ÷ their area of triangle 1(b)(i) 62.88 1 1(b)(ii) 2400 3 3 2 M2 for 6 × 8000 oe   or M1 for 3 8000 oe 1(b)(iii) 44 cao 3  4 3  M2 for 8000  π3.5   3  4 3 or M1 for π3.5 3

This question in 0607/42 May/June 2024

8 NOT TO SCALE 0.8 cm 0.7 cm 0.7 cm The diagram shows a square-based pyramid. The side of the base of the pyramid is 0.7 cm. The length of each sloping edge is 0.8 cm. (a) Show that the perpendicular height of the pyramid is 0.628 cm, correct to 3 significant figures. [4] (b) 4.9 cm NOT TO SCALE 6 cm End view The diagram shows a kitchen tool made from wood. The tool is formed from a cuboid, a cylinder and 49 of the square-based pyramids from part (a). The cylinder has a radius of 1.2 cm and length 25 cm. The cuboid measures 4.9 cm by 4.9 cm by 6 cm. The mass of 1 cm3 of the wood is 0.63 grams. Calculate the total mass of the tool. … g [5]

9 marks

Mark scheme: 8(a) h2 = 0.82 – 0.352 – 0.352 oe M3 M2 for 0.82 = 0.352 + 0.352 + h2 oe or M1 for 0.352 + 0.352 or 0.72 + 0.72 oe 0.6284 to 0.6285 A1 8(b) 165 or 165.0 to 165.2 5 B4 for 262 or 262.1 to 262.2… OR M1 for 6 × 4.9 × 4.9 M1 for π × 1.22 × 25 1 M1 for 3 0.7  0.7  0.628 [× 49] M1 for at least 1 of their volumes × 0.63

This question in 0607/43 May/June 2024

10 In this question all lengths are in centimetres. NOT TO SCALE h r r A solid cone has radius r and vertical height h. A solid hemisphere also has radius r. The curved surface area of the cone is the same as the curved surface area of the hemisphere. (a) Show that h = r 3 . [4] (b) The cone is placed directly on top of the hemisphere. 1 3 Show that the volume of this solid is rr ( 2 + 3 ) . 3 NOT TO SCALE [2] (c) A larger solid is mathematically similar to the solid in part (b). The larger solid has volume 243r r 3 ( 2 + 3) . (i) Find, in terms of r, the radius of the hemisphere of the larger solid. … [2] (ii) The surface area of the larger solid is 5000 cm 2. Find the volume of this solid. … cm3 [4] Question 11 is printed on the next page.

12 marks

Mark scheme: 10(a) rl = 2  r 2 B1 2 2 M1 l = r + h 2 2 M1 2 r = r + h 4 r 2 = r 2 + h 2 h 2 = 3r 2 A1 No errors or omissions  h = r 3    10(b) 2 3 1 2 M1 r or r  r 3 seen 3 3 2 3 1 2 A1  r +  r  r 3 3 3  1 3  V = r 2 + 3  ( )   3  10(c)(i) 9r 2 1  1  3 M1 for  243   , implied by 9 seen  3  10(c)(ii) 31000 or 31018… 4 3 5000 M3 for V = 243 2 + 3 or ( ) 324 3 5000 V = 27  2 + 3 ( ) 4 5000 5000 M2 for r = or R = 324  4  M1 for 4  (9 r ) 2 = 5000 or 4 R 2 = 5000

This question in 0607/41 Oct/Nov 2024

10 (a) NOT TO SCALE A metal sphere has a volume of 9203 cm3. The sphere is inside a cube and touches each face of the cube. (i) Find the volume of the cube. … cm3 [4] (ii) The sphere is melted and poured into the cube. Find the depth of the metal. … cm [2] (b) D 15 cm 8 cm C H NOT TO A 4 cm SCALE B G E F ABCDEFGH is a cuboid. (i) Calculate the length DF. … cm [4] (ii) Calculate the angle that the diagonal DF makes with the base EFGH. … [2]

12 marks

Mark scheme: 10(a)(i) 17600 or 17570 to 17580 4 3  9203 M3 for [cube side] = 3  2 4π 3  9203 or M2 for [ r ] = 3 4π 3 3  9203 or M1 for r = 4π 10(a)(ii) 13.6 or 13.61… 2 M1 FT for h  (their cube side) 2 = 9203 or better 10(b)(i) 17.5 or 17.46… 4 2 2 2 M3 for (4 + 15 + 8 ) or M2 for (152 + 82 ) , (42 + 152 ) , (4 2 + 82 ) or 15 2 + 8 2 + 4 2 or M1 for 15 2 + 8 2 , 4 2 + 15 2 , 4 2 + 82 10(b)(ii) 13.2 or 13.23 to 13.24… 2 4 M1 FT for sin[ ] = oe or other trig (their 17.5)

This question in 0607/43 Oct/Nov 2024

12 A container is in the shape of a cuboid. The cuboid measures 41 cm by 32 cm by 25 cm. (a) Show that the capacity of the container is 32.8 litres. [2] (b) At 10 40 the container is empty. Water flows into the container at a rate of 2 litres per hour until the container is full. (i) At 12 10 there are x litres of water in the container. Find x as a percentage of 32.8 . … % [3] (ii) Find the time when the container is full. … [3]

8 marks

Mark scheme: 12(a) 41  32  25 M2 M1 for 41 × 32 × 25 or for dividing by 1000 seen 1000 = 32.8 with no errors 12(b)(i) 9.15 or 9.146… 3 B1 for 1.5 oe or 3 seen 2  their1.5[ 100] M1 for 32.8 12(b)(ii) [0]3 04 3 32.8 M1 for 2 M1 FT for correct conversion of their length of time to hours and minutes 16 hours 24 min

This question in 0607/42 Feb/March 2025

6 8 cm NOT TO SCALE A 5.7 cm 72° B N 3.6 cm C Triangle ABC is the cross-section of a prism. AC = 5.7 cm, NC = 3.6 cm and the length of the prism is 8 cm. Angle ABN = 72° and angle ANC = 90°. (a) AN is the perpendicular height of the triangle. Show that AN = 4.42 cm correct to 3 significant figures. [2] (b) Calculate the volume of the prism. … cm3 [5]

7 marks

Mark scheme: 6(a) 3.62 + x 2 = 5.7 2 or better M1 4.419… A1 6(b) 89[.0] or 89.1 or 89.01 to 89.11 5 4.42 M2 for oe tan72 4.42 or M1 for tan72 = oe BN 1 M2 for  ( their BN + 3.6 )  4.42  8 oe 2 1 or M1 for  ( their BN + 3.6 )  4.42 oe 2 If 0 scored, SC1 for answer 63.6 or 63.63 to 63.65

This question in 0607/41 May/June 2025

10 5 cm NOT TO SCALE 12 cm 15 cm The diagram shows a solid cylinder and a solid cone. The radius of the cylinder is 5 cm and its height is 12 cm. The height of the cone is 15 cm. The volume of the cylinder is equal to the volume of the cone. (a) Show that the volume of the cylinder is 300rcm 3. [1] (b) Find the total surface area of the cone. … cm2 [5]

6 marks

Mark scheme: 10(a)  25  12 1 10(b) 599 or 599.1 to 599.4 5 B4 for 410.7 to 410.9 seen OR B2 for r = 7.745 to 7.746 1 2 or M1 for  π  r  15 = 300  π 3 M1 for 152 + ( theirr ) 2 M1 for π × (their r)×their l

This question in 0607/41 May/June 2025

7 22 cm NOT TO SCALE x cm 5 cm x cm The diagram shows a wooden cuboid with a cylinder removed. The cuboid has length 22 cm, width x cm and height x cm. The cylinder has length 22 cm and diameter 5 cm. The volume of wood remaining after the cylinder has been removed is 976 cm3. Find the value of x. x = … [3]

3 marks

Mark scheme: 7 8 [.00] or 7.999 to 8.000... 3 2 976 + 22  π  2.5 2 M2 for  x =  oe   22 or M1 for 976 = 22  x 2 − 22  π  2.5 2 oe If 0 scored SC1 for answer 11.1 or 11.08 to 11.09

This question in 0607/42 May/June 2025

9 The volume of a sphere is 96.5 cm3 . Calculate the radius of the sphere. … cm [2]

2 marks

Mark scheme: 9 2.85 or 2.845... 2 4 M1 for 96.5  π or better 3

This question in 0607/43 May/June 2025

13 NOT TO SCALE 6 cm 15 cm A solid is formed from a cone and a hemisphere. The cone and the hemisphere each have a radius of 6 cm. The cone has a height of 15 cm. Find the volume of the solid. … cm3 [3]

3 marks

Mark scheme: 13 1020 or 1017 to 1018. ... 3 1 2 M1 for 3π 6  15 or better 1 4 3 M1 for  π 6 or better 2 3

This question in 0607/43 May/June 2025

17 A solid cone has base radius r and vertical height 3r. The total surface area of the cone is 209.22 cm2. (a) Find r. r = … cm [4] (b) A mathematically similar cone has a total surface area of 1882.98 cm2. Find the radius of this cone. … cm [3]

7 marks

Mark scheme: 17(a) 4 4 209.22 M3 for  r =  oe π(1 + 10) or M2 for 209.22 = πr 2 + πr r 2 + (3r ) 2 oe or M1 for l 2 = r 2 + (3r ) 2 oe 17(b) 12 3 1882.98 M2 for x =  their r oe 209.22 1882.98  x  2 or M1 for = oe   209.22  their r  1882.98 209.22 or oe or oe 209.22 1882.98

This question in 0607/42 Oct/Nov 2025