O-LevelMathematicsMensuration4024/22 · May/June 2024 · Q815 Marks

[Volume of a cone = 1/3 πr²h] [Volume of a sphere = 4/3 πr³] [Curved surface area of cone = πrl] [Surface area of a sphere = 4πr²] A solid is formed by placing a cone on top of a hemisphere. The cone and the hemisphere each have radius r cm. The height of the solid is 10 cm. The curved surface area of the hemisphere is 145 cm². [Figure 8.1]

(a)(i)Show that r = 4.80, correct to 2 decimal places.[3]
(a)(ii)Calculate the volume of the solid.[4]
(a)(iii)Calculate the curved surface area of the cone.[4]
(b)The diagram shows two sectors of circles. Sector A has angle x and radius y. The angle of sector B is 20% smaller than the angle of sector A. The radius of sector B is 20% longer than the radius of sector A. Calculate the area of sector B as a percentage of the area of sector A. [Figure 8.2][4]

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Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[3] r² = 145 / (2π) or 23.1 or 23.06 to 23.09 or r = sqrt(145 / (2π)) / r = 4.803 to 4.804
M1 FT from (their 2) ×π×r² =145 for r² or r
B1 for 4.803 to 4.804
[4] 357 to 358
B1 for height of cone = 10 – 4.8 soi
M1 for 1/3 × π×4.80²×their 5.2 oe
M1 for 2/3 × π×4.80³ oe
[4] 107 or 106.7 to 106.8...
M2 for sqrt(4.80² + (their 5.2)²) soi or M1 for 4.80² + (their 5.2)² soi
M1 for π×4.80×their 7.077
[4] 115.2
M3 for 0.8 × (1.2)² [× 100] oe
OR
B1 for 0.8x oe or 1.2y oe
M2 for (their 0.8x / 360) × π × (their 1.2y)² / (x / 360) × π × y² [×100] oe
or M1 for (their 0.8x / 360) × π × (their 1.2y)² / (x / 360) × π × y²
After 0 scored, SC1 for (x-0.2)/360 × π × (y+0.2)² oe

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