Higher November 2018 Paper 3 Q24
24
| Volume of a sphere \(= \dfrac{4}{3}\pi r^3 \quad\) where \(r\) is the radius |
| Volume of a cone \(= \dfrac{1}{3}\pi r^2 h \quad\) where \(r\) is the radius and \(h\) is the perpendicular height |
A sphere has radius \(2x\) cm
A cone has
radius \(3x\) cm
perpendicular height \(h\) cm
The sphere and the cone have the same volume.
Work out radius of cone : perpendicular height of cone
Give your answer in the form \(\quad a : b \quad\) where \(a\) and \(b\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{4}{3}\pi(2x)^3\) or \(\dfrac{1}{3}\pi(3x)^2h\) | M1 | oe |
| \(\dfrac{4}{3}\pi(2x)^3 = \dfrac{1}{3}\pi(3x)^2h\) or \(\dfrac{4}{3}\pi 8x^3 = \dfrac{1}{3}\pi 9x^2h\) | M1dep | oe Sets up equation |
| \(32x = 9h\) or \(x = \dfrac{9}{32}h\) or \(h = \dfrac{32}{9}x\) or \(\dfrac{32}{3}r = 9h\) or \(r = \dfrac{27}{32}h\) or \(h = \dfrac{32}{27}r\) or \(27h = 32r\) or \(\dfrac{27}{32}h : h\) or \(3x : \dfrac{32}{9}x\) or \(\dfrac{27}{32} : 1\) or \(3 : \dfrac{32}{9}\) or 0.84… : 1 or 3 : 3.55… | M1dep | oe linear equation or ratio |
| 27 : 32 | A1 |
Additional guidance
| 32 : 27 | M1M1M1A0 |
| Note \(\dfrac{4}{3}\pi(2)^3 = [33.49, 33.52]\) \(\dfrac{1}{3}\pi(3)^2h = [9.42h, 9.43h]\) |