Higher June 2018 Paper 1 Q16
16 A frustum is made by removing a small cone from a large cone.
The cones are mathematically similar.
Diagram NOT accurately drawn
The large cone has base radius \(r\) cm and height \(h\) cm.
Given that\[\frac{\text{volume of frustum}}{\text{volume of large cone}} = \frac{98}{125}\]find an expression, in terms of \(h\), for the height of the frustum.
(4)
| Scheme | Marks |
|---|---|
| \(1 - \dfrac{98}{125} \left(= \dfrac{27}{125}\right)\) or 0.216 or 125 – 98 (=27) | M1 |
| \(\sqrt[3]{\text{``}{\dfrac{27}{125}}\text{''}} \left(= \dfrac{3}{5}\right)\) or \(\sqrt[3]{\text{``}{\dfrac{125}{27}}\text{''}} \left(= \dfrac{5}{3}\right)\) | M1 |
| \(1 - \text{``}{\dfrac{3}{5}}\text{''}\) or \(h - \text{``}{\dfrac{3}{5}}\text{''}h\) oe | M1 |
| \(\dfrac{2}{5}h\) oe | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for the length scale factor
may be seen as a ratio E.g. 3 : 5
A1: for \(\dfrac{2}{5}h\) oe (may not be simplified)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{3}\pi r^2 h - \dfrac{1}{3}\pi (kr)^2 kh = \dfrac{98}{125} \times \dfrac{1}{3}\pi r^2 h\) oe | M1 |
| \(k = \dfrac{3}{5}\) | M1 |
| \(1 - \text{``}{\dfrac{3}{5}}\text{''}\) or \(h - \text{``}{\dfrac{3}{5}}\text{''}h\) oe | M1 |
| \(\dfrac{2}{5}h\) oe | A1 |
Notes
M1: sets up an equation using scale factor
M1: for the length scale factor
A1: for \(\dfrac{2}{5}h\) oe (may not be simplified)