Higher June 2024 Paper 1 Q23
23 A solid shape is made by removing a hemisphere, shown shaded, from a cone as shown in the diagram.
Diagram NOT accurately drawn
The radius of the hemisphere is \(2x\) cm
The radius of the base of the cone is \(5x\) cm
The vertical height of the cone is \(6x\) cm
The volume of the solid shape is \(6948\pi\) cm\(^3\)
Work out the total surface area of the solid hemisphere that has been removed from the cone.
Give your answer correct to the nearest integer.
(5)
| Scheme | Marks |
|---|---|
\(\dfrac{1}{3}\pi \times (5x)^2 \times 6x\) oe or \(50\pi x^3\) oe or \(\dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times (2x)^3\) or \(\dfrac{16}{3}\pi x^3\) oe or \(\dfrac{4}{3} \times \pi \times (2x)^3\) or \(\dfrac{32}{3}\pi x^3\) oe | M1 |
\(\dfrac{1}{3}\pi \times (5x)^2 \times 6x - \dfrac{1}{2} \times \dfrac{4}{3} \times \pi \times (2x)^3 = 6948\pi\) oe or \(50\pi x^3 - \dfrac{16}{3}\pi x^3 = 6948\pi\) or \(\dfrac{134}{3}\pi x^3 = 6948\pi\) oe | M1 |
\((x^3 =)\dfrac{6948\pi \times 3}{134\pi}\left(= \dfrac{10422}{67} = 155.(552...)\right)\) oe or \((x =)\sqrt[3]{\dfrac{6948\pi \times 3}{134\pi}}\left(= \sqrt[3]{\dfrac{10422}{67}} = \sqrt[3]{155.(552...)} = 5.37(8...)\right)\) oe | M1 |
| \(3 \times \pi \times (2 \times \text{``}{5.37(8...)}\text{''})^2\) oe or \(12 \times \pi \times \text{``}{5.37(8..)}\text{''}^2\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 1090 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for finding the volume of cone or hemisphere or sphere
NB Ignore missing brackets around \(5x\) and \(2x\) for this mark
M1: for a correct equation for the volume of the shape
NB If not expanded at this stage then must see brackets
M1: for rearranging the correct equation to find the value of \(x^3\) or \(x\)
Accept 5.4 or better
M1: for finding the surface area of the hemisphere
A1: allow 1086 – 1100
Special case for using 6948 without \(\pi\)
SC B3 for \(x^3 = 49.5(138…)\) or
\(x = 3.67(205…)\)
SC B4 for awrt 508