Higher June 2023 Paper 2R Q24
24 The surface area of sphere A is nine times the surface area of sphere B
The difference between the volume of sphere A and the volume of sphere B is \(117\pi\) cm3
Find the radius of the smaller sphere.
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
| eg \(4\pi R^2 = 9 \times 4\pi r^2\) oe or | M1 |
\(R = 3r\) oe or 1:3 or 3:1 or 3 or \(\dfrac{1}{3}\) | M1 |
eg \(\dfrac{4}{3}\pi(3r)^3 - \dfrac{4}{3}\pi r^3 = 117\pi\) oe or \(\dfrac{4}{3}\pi r^3 - \dfrac{4}{3}\pi\left(\dfrac{1}{3}r\right)^3 = 117\pi\) or \(27 \times \dfrac{4}{3}\pi r^3 - \dfrac{4}{3}\pi r^3 = 117\pi\) oe or \(\dfrac{4}{3}\pi r^3 - \dfrac{1}{27} \times \dfrac{4}{3}\pi r^3 = 117\pi\) oe or oe | M1 |
\((r =)\;\sqrt[3]{\dfrac{117 \times 3}{104}}\left(= \sqrt[3]{\dfrac{27}{8}}\right)\) or \((R =)\;\sqrt[3]{\dfrac{117 \times 81}{104}}\left(= \sqrt[3]{\dfrac{729}{8}} = \dfrac{9}{2}\right)\) | M1 |
Working required Answer: \(\dfrac{3}{2}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: (a correct scale factor of 3 or \(R = 3r\) oe implies the first M1)
M1: for a correct equation based on volumes with only one variable eg \(R\) or \(r\) or \(x\)
(M3 for
\(26 \times \dfrac{4}{3}\pi r^3 = 117\pi\) oe or
\(26 \times (Vol)_B = 117\pi\) or
\(\dfrac{26}{27} \times \dfrac{4}{3}\pi r^3 = 117\pi\) oe or
\(\dfrac{26}{27} \times (Vol)_A = 117\pi\)
M1: dep on previous M mark
A1: oe dep on M2
M2 for (vol SF =) 27 or \(\dfrac{1}{27}\) or \(3^3\) or \(\dfrac{1}{3^3}\)