Surface area and volume — Paper 4 · IGCSE Mathematics (9-1) 0980

10 (a) A solid metal sphere of radius 9 cm is placed into an empty tank. The tank is a cylinder of radius 30 cm and height 18 cm. Water is poured into the tank until it is full. Calculate the number of litres of water poured into the tank. 4 3 [The volume, V, of a sphere with radius r is V = r r . ] 3 … litres [4] (b) A different tank is a cuboid measuring 1.8 m by 1.5 m by 1.2 m. Water flows from a pipe into this empty tank at a rate of 200 cm3 per second. Find the time it takes to fill the tank. Give your answer in hours and minutes. … hours … minutes [4] (c) NOT TO SCALE Area = 159.5 cm2 Area = 295 cm2 17 cm The diagram shows two mathematically similar shapes with areas 295 cm2 and 159.5 cm2. The width of the larger shape is 17 cm. Calculate the width of the smaller shape. … cm [3]

11 marks

Mark scheme: 10(a) 47.8 or 47.84 to 47.85 4 B3 for answer figs 478 or figs 4784 to figs 4785 OR M1 for π × 30 2 × 18 4 3 M1 for × π × 9 3 M1dep for their volume ÷ 1000 soi 10(b) 4 [hours] 30 [ mins] nfww 4 B3 for 16200 or 4.5 or 270 figs 18 × figs 15 × figs 12 or M2 for oe figs 2 or M1 for figs 18 × figs 15 × figs 12 oe 10(c) 12.5 or 12.50… 3 159.5 M2 for 17 × oe 295 159.5 295 or M1 for or seen 295 159.5 159.5 x 2 or for = oe 295 17 2

This question in 0980/42 May/June 2019

4 (a) (i) Calculate the external curved surface area of a cylinder with radius 8 m and height 19 m. … m2 [2] (ii) This surface is painted at a cost of $0.85 per square metre. Calculate the cost of painting this surface. $ … [2] (b) A solid metal sphere with radius 6 cm is melted down and all of the metal is used to make a solid cone with radius 8 cm. Calculate the curved surface area of the cone. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [5] (c) Two cones are mathematically similar. The total surface area of the smaller cone is 80 cm2. The total surface area of the larger cone is 180 cm2. The volume of the smaller cone is 168 cm3. Calculate the volume of the larger cone. … cm3 [3] (d) The diagram shows a pyramid with a P square base ABCD. DB = 8 cm. NOT TO P is vertically above the centre, X, of SCALE the base and PX = 5 cm. D C X A B Calculate the angle between PB and the base ABCD. … [3]

15 marks

Mark scheme: 4(a)(i) 955 or 955.0 to 955.2 2 M1 for 2 × π × 8 × 19 oe 4(a)(ii) 812 or 811.7 to 811.9... 2 FT their (i) × 0.85 M1 for their (i) × 0.85 or their (i) × 85 4(b) 394 or 395 or 394.3 to 394.6... 5 4 3 π× 6 3 M2 for h = or better 1 3 π× 8 2 4 3 1 2 or M1 for × π × 6 = × π × 8 × h oe 3 3 M1 for 82 + their 13.52 or better M1 dep for π × 8 × their slant height dep on use of Pythagoras 4(c) 567 3 3 168  80  2 M2 for =   oe or better V  180  1 1  180  2  80  2 or M1 for   or   oe seen or  80   180  better 4(d) 51.3 or 51.34... 3 5 M2 for tan = oe 4 or M1 for recognition of angle PBX

This question in 0980/41 Oct/Nov 2019

8 (a) C R NOT TO SCALE A B 8 cm P Q 12 cm Triangle ABC is mathematically similar to triangle PQR. The area of triangle ABC is 16 cm2. (i) Calculate the area of triangle PQR. … cm2 [2] (ii) The triangles are the cross-sections of prisms which are also mathematically similar. The volume of the smaller prism is 320 cm3. Calculate the length of the larger prism. … cm [3] (b) A cylinder with radius 6 cm and height h cm has the same volume as a sphere with radius 4.5 cm. Find the value of h. 4 3 [The volume, V, of a sphere with radius r is V = rr . ] 3 h = … [3] (c) A solid metal cube of side 20 cm is melted down and made into 40 solid spheres, each of radius r cm. Find the value of r. 4 3 [The volume, V, of a sphere with radius r is V = rr . ] 3 r = … [3] 7x(d) A solid cylinder has radius x cm and height cm. 2 The surface area of a sphere with radius R cm is equal to the total surface area of the cylinder. Find an expression for R in terms of x. [The surface area, A, of a sphere with radius r is A = 4rr 2 . ] R = … [3]

14 marks

Mark scheme: 8(a)(i) 36 2 2 2  8   12  M1 for   or   oe  12   8  8(a)(ii) 30 3 12 M2 for 320 ÷ 16 × oe 8 or M1 for 320 ÷ 16 8(b) 3.375 cao 3 4 3 π × 4.5 3 M2 for or better π × 6 2 2 4 3 or M1 for π × 6 × h = × π × 4.5 3 8(c) 3.63 or 3.627 to 3.628 3 20 3 M2 for 4 40 × π 3 4 3 3 or M1 for 40 × × π × r = 20 3 8(d) 3x 1 3 B2 for 4 R 2 = 9 x 2 oe or better or 1.5x or x 2 12 2 2 7 x or M1 for 4πR = 2πx + π × 2 x × 2

This question in 0980/42 May/June 2020

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 0980/42 May/June 2022

1 (a) Calculate the volume of (i) a solid cylinder with radius 6 cm and height 14 cm, … cm3 [2] (ii) a solid hemisphere with radius 6 cm. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … cm3 [2] (b) NOT TO SCALE 14 cm 6 cm The cylinder and hemisphere in part (a) are joined to form the solid in the diagram. The solid is made of steel and 1 cm 3 of steel has a mass of 7.85 g. (i) Show that 1 cm 3 of steel has a mass of 0.007 85 kg. [1] (ii) Calculate the total mass of the solid. … kg [2] (c) 2000 cm 3of iron is melted down and some of it is used to make 50 spheres with radius 2 cm. (i) Calculate the percentage of iron that is left over. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 … % [3] (ii) The iron left over is then made into a cube. Calculate the length of an edge of the cube. … cm [1] (d) A solid cone has radius 3R cm and slant height 9R cm. A solid cylinder has radius x cm and height 7x cm. The total surface area of the cone is equal to the total surface area of the cylinder. Given that R = kx , find the value of k. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] k = … [4]

15 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 1580 or 1583 to 1584 2 M1 for π  6 2  14 1(a)(ii) 452 or 452.3 to 452.4... 2 3 M1 for 1 4  π  6  2 3 1(b)(i) 7.85 ÷ 1000 [= 0.00785] M1 1(b)(ii) 16[.0] or 15.95 to 15.99 2 FT {their (a)(i) + their (a)(ii)}  0.00785 evaluated to 3 sig fig or better M1 for (their (a)(i) + their (a)(ii)) × 0.00785 1(c)(i) 16.2 or 16.21 to 16.23 3 4 3 2000 − 50  π 2 3 M2 for 100  2000 4 3 50  π 2 3 or for  100 2000 4 3 50  π 2 3 or M1 for 2000 1(c)(ii) 6.87 or 6.870 to 6.872 1  4 3  FT 3 2000 − their 50  π 2    3  evaluated to 3sf or better 1(d) 2 4 M1 for [π](3 R ) 2 + [π]3 R  9 R oe oe 3 M1 for 2[π]x 2 + 2[π]x  7 x oe M1 for their area of cone = their area of cylinder seen

This question in 0980/41 Oct/Nov 2022

5 (a) NOT TO 15 cm SCALE 8 cm A cone has base diameter 8 cm and perpendicular height 15 cm. (i) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [2] (ii) A label completely covers the curved surface area of the cone. Calculate the area of the label as a percentage of the total surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] … % [5] (b) NOT TO SCALE 0.45 m An empty cylindrical container has radius 0.45 m. 300 litres of water is poured into the container at a rate of 375 ml per second. (i) Find the time taken, in minutes and seconds, for all the water to be poured into the container. … min … s [3] (ii) Calculate the height of the water in the container. … m [3]

13 marks

Mark scheme: 5(a)(i) 251 or 251.3 to 251.4 2 1 2 M1 for  π  4  15 oe 3 5(a)(ii) 79.5 or 79.51… 5 2 2 M3 for π 4 4  15 oe or M2 for 15 2  4 2 oe or M1 for [l2 = ] 42 + 152 oe or π×4× theirl M1 for their curved surfacearea [ 100] their curved surfacearea  π  4 2 oe 5(b)(i) 13 min 20 sec 3 40 B2 for 800 or oe seen 3 or M1 for figs 3 ÷ figs 375 or figs 3 ÷ 22 500 5(b)(ii) 0.472 or 0.4715 to 0.4716… 3 M2 for π  0.452  h  0.3 or π  45 2  h  300000 oe or M1 for π  figs45 2  h  figs3 oe

This question in 0980/42 May/June 2023

8 (a) 3.63.6 cmcm 6.5 cm NOT TO SCALE 5.4 cm The diagram shows a solid formed by joining two hemispheres and a cylinder. The radius of the large hemisphere is 5.4 cm. The radius of the small hemisphere and the radius of the cylinder are both 3.6 cm. The height of the cylinder is 6.5 cm. (i) Show that the volume of the solid is 692 cm 3, correct to the nearest cubic centimetre. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3 [4] (ii) A mathematically similar solid is made of silver. In this solid, the cylinder has radius 0.6 cm. 1 cm 3 of silver has a mass of 10.49 grams. Calculate the total mass of this silver solid. … g [4] (b) A 10 cm NOT TO O 216° SCALE B AOB is a sector of a circle, centre O. AO = 10 cm and the sector angle is 216°. (i) Calculate the length of the arc of this sector. Give your answer as a multiple of r. … cm [2] (ii) A cone is made from this sector by joining OA to OB. Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [4]

14 marks

Mark scheme: 8(a)(i) 2 3 2 3 M3 2 3 2 3 3(3.6) + 3(5.4) + M1 for either 3(3.6) or 3(5.4) (3.6) 2  6.5 M1 for (3.6) 2  6.5 692.1 to 692.2… A1 8(a)(ii) 33.6 or 33.60 to 33.62 4 3  0.6  M3 for  692  10.49 oe    3.6   0.6 3 or M2 for  692 oe    3.6   0.6 3  3.6 3 or M1 for   or   oe  3.6   0.6  If 0 scored, SC1 for their volume  10.49 8(b)(i) 12π final answer 2 216 M1 for  2  10 oe 360 After 0 scored SC1 for final answer 8π or 12π + 20 8(b)(ii) 302 or 301.5 to 301.6… 4 M1 for 2πr = their (b)(i) oe or for 216 2 π 10 = π r 10 oe 360 and M1 for [h =] 10 2 − their 6 2 oe and 1 2 M1 for [V =] (their 6)  (their 8) 3

This question in 0980/41 Oct/Nov 2023

4 (a) NOT TO SCALE 12 cm 1 m The diagram shows a tank in the shape of a half-cylinder of radius 12 cm and length 1 metre. The tank is fixed horizontally and is completely filled with water. (i) Calculate the volume of water in the tank. Give your answer correct to the nearest 10 cm3. … cm3 [3] (ii) NOT TO 6 cm SCALE Water is removed from the tank until the level of water is 6 cm below the top of the tank. The diagram shows the cross-section of the tank. Calculate the volume of water that is now in the tank. … cm3 [5] (b) A rectangular fish tank with length 42 cm and width 35 cm is full of water. A stone lies at the bottom of the tank. When the stone is removed from the tank, the depth of the water decreases by 0.2 cm. The density of the stone is 2.2 g/cm3. Calculate the mass of the stone in grams. [ Density = mass ' volume] … g [3] (c) H G E F 15 cm NOT TO SCALE D C 12 cm A 8 cm B The diagram shows a cuboid, ABCDEFGH. Calculate the angle that AG makes with the base of the cuboid. … [4]

15 marks

Mark scheme: 4(a)(i) 22 620 cao 3 B2 for 7200 or 22 608 to 22 629 1 2 or M1 for   12 [ figs 1] oe 2 4(a)(ii) 8840 or 8850 or 8836 to 8850. 5 6 M1 for cos COM = oe 12 6 or sin AOC = oe 12  theirCOD 2  M1 for    12  oe M  360   1 2  oe M1 for   12  sin  theirCOD    2  M1dep for (their area of sector COD– their area of triangle COD) 100 dep on at least M1M1 oe 4(b) 647 or 646.8 3 m M2 for 2.2  oe 42  35  0.2 or M1 for [vol of stone =] 42×35×0.2 oe If 0 scored SC1 for answer figs 647 or figs 6468 4(c) 46.1 or 46.12 to 46.14 4 15 M3 for tan  oe 8 2  12 2 or M2 for 82 + 122 oe or 82 + 122 + 152 oe or M1 for identifying the angle GAC

This question in 0980/42 May/June 2024

5 (a) Simplify 25x 6 2. … [2] (b) These are the first five terms of a sequence. 1 1 6 36 216 6 Find the nth term of the sequence. … [2] (c) Expand and simplify. ( x + 4)( x - 3)( 3x - 1) … [3] 7 2(d) (i) Show that ( 3x + 5) + = x simplifies to 2x + x - 3 = 0 . x - 2 [4] (ii) Solve by factorisation 2x 2 + x - 3 = 0 . x = … or x = … [3] (e) A solid cylinder has base radius x and height 3x. The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y. 2 75y 2 Show that x = . 8 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]

18 marks

Mark scheme: 5(a) 125x 9 final answer 2 B1 for answer 125 kx or m x 9 or for correct answer seen then spoilt 5(b) 6 n  2 oe final answer 2 B1 for answer of form 6k oe  k  1  or answer of the form   oe  6  or for correct answer seen 5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 5(d)(i) eliminates the fraction correctly M1 eg (3x + 5) (x – 2) + 7 = x (x – 2) 3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2 B1 for 3x2 + 5x –6x –10 [ + 7] oe seen with at least 3 terms correct leading to 2x2 + x – 3 = 0 A1 dep on M1 B2 with no errors or omissions 5(d)(ii)  2 x  3  x  1 M2 or M1 for (2x + a)(x + b) where ab = −3 or 2b + a = [+]1 or for partial factors 2x(x – 1) + 3(x – 1) or x(2x + 3) –[1](2x + 3) −1.5 oe and +1 B1 5(e) 2 M1 [TSA cylinder =] 2x  2x  3 x 2 4(5 y ) 2 M1 [TSA hemisphere=] (5 y )  2 Leading to M1 dep M1M1 2x 2  6x 2  50y 2  25y 2 oe 2 75 y 2 A1 dep on M1M1M1 x  8

This question in 0980/42 May/June 2024

9 (a) O NOT TO 60° 10 cm SCALE 17 cm D C A B OAB is a sector of a circle, centre O, radius 17 cm. OCD is a sector of a circle, centre O, radius 10 cm. OCA and ODB are straight lines and angle AOB = 60° . The perimeter of the shaded shape ABDC can be written in the form ( a r+ b ) cm. Find the value of a and the value of b. a = … b = … [3] (b) NOT TO SCALE The diagram shows a regular hexagon. The area of the hexagon is 127.3 cm2. (i) Show that the length of one side of the hexagon is 7.0 cm , correct to 1 decimal place. [4] (ii) The hexagon is the cross-section of a prism of length 10 cm. 127.3 cm2 NOT TO SCALE 10 cm 7.0 cm (a) Find the volume of the prism. … cm3 [1] (b) Calculate the surface area of the prism. … cm2 [2]

10 marks

Mark scheme: 9(a) [a =] 9 3 B2 for a =9 [b =] 14 OR M2 for 60 60 2  17  2  10  7  7 360 360 oe or M1 for 60 60 2  17 oe or 2  10 oe 360 360 If 0 scored SC1 for b =14 9(b)(i) 60° at centre B1 or interior angle = 120° 1 2 M1 [6]  d  sin60 oe 2 2 127.3 M1 [ d  ] 1 6   sin60 2 6.99[9…] to 7.00[…] A1 Dep on M1M1 9(b)(ii)(a) 1273 1 9(b)(ii)(b) 675 or 674.5 to 674.6 2 M1 for 2 ×127.3 oe or 6 × 7 × 10 oe

This question in 0980/42 May/June 2024

9 H G F E NOT TO 17 cm SCALE D C 8 cm A 10 cm B ABCDEFGH is a solid cuboid. AB = 10 cm, BC = 8 cm and CG = 17 cm. (a) Work out the volume of the cuboid. … cm3 [1] (b) Work out the total surface area of the cuboid. … cm2 [3] (c) Calculate the angle between GA and the base ABCD. … [4] (d) A straight rod PQ is placed inside the cuboid. One end of the rod, P, is placed at the midpoint of AB. The other end of the rod, Q, rests on GH. HQ : QG = 4 : 1 . Q H G F E NOT TO 17 cm SCALE D C 8 cm A P B 10 cm Calculate the length of the rod PQ. … cm [4]

12 marks

Mark scheme: 9(a) 1360 1 9(b) 772 3 M2 for [2 ×] (10 × 8 + 10 × 17 + 8 × 17) oe or M1 for 10 × 8 oe or 10 × 17 oe or 8 × 17 oe 9(c) 53 or 53.0 to 53.01 4 17 M3 for tan [GAC] = oe 10 2 + 8 2 or M2 for 102 + 82 oe or for 102 + 82 + 172 oe or M1 for recognising angle GAC is required 9(d) 19[.0] or 19.02 to 19.03 4 M3 for 32 + 82 + 172 oe OR B1 for QG = 2 soi or HQ = 8 M1 for (5 – 2) 2 + 82 or (5 – 2) 2 + 172

This question in 0980/41 Oct/Nov 2024

6 A solid wooden cone has base radius 4 cm and height 12 cm. The density of the wood is 0.74 g/cm 3. Calculate the mass of the cone. [ Density = Mass ' Volume] … g [3]

3 marks

Mark scheme: 6 149 or 148.7 to 148.8… 3 1 2 M1 for 3π 4  12 oe M1 for 0.74 × their volume

This question in 0980/42 May/June 2025

14 A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm. 256 (a) Show that the volume of the sphere is rcm 3. 3 [1] (b) Calculate the percentage of the cube that is not occupied by the sphere. … % [3] (c) The density of the metal of the sphere is 7.86 g/cm3. Calculate the mass of the sphere. Give your answer in kilograms. [Density = mass ' volume] … kg [2] (d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder. … cm2 [4]

10 marks

Mark scheme: 14(a) 4 3 256 1 π 4 [= π ] 3 3 14(b) 47.6 nfww or 3 50 B2 for 52.4 or 52.35 to 52.37 or π nfww 47.63 to 47.64… nfww 3 OR 3 256 8 − π 3 M2 for 3  100 oe 8 3 256 256 8 − π π 3 3 or M1 for 3 [ 100] oe or 3  100 8 8 oe 14(c) 2.11 or 2.107… 2 256 M1 for π  7.86 3 14(d) 233 or 233.3 to 233.4 4 2 256 M1 for π  3.1  h = π 3 M2dep for 2  π  3.12 + 2  π  3.1  their h or M1dep for 2 π 3.1  theirh or M1 for 2π  3.12

This question in 0980/41 Oct/Nov 2025