June 2023 Paper 1 Q7
7 The diagram below shows a cone and a prism.
All measurements are in cm
The cone has base radius \(r\) and perpendicular height \(x\).
The prism has a triangular cross section with base \(y\) and perpendicular height \(y\).
The length of the prism is \(x\).
| Volume of a cone \(= \dfrac{1}{3} \times\) area of base \(\times\) perpendicular height |
The volume of the cone is four times the volume of the prism.
Express \(r\) in terms of \(y\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| (Cone =) \(\dfrac{1}{3}\pi r^2x\) or (Prism =) \(\dfrac{1}{2}y^2x\) | M1 | could appear in M1 dep working height must be \(x\) |
| \(\dfrac{1}{3}\pi r^2x = 2y^2x\) | M1dep | for correctly processing the 4 must be a correct equation oe eg \(\dfrac{\frac{1}{3} \times \pi r^2 \times x}{2} = \dfrac{2y^2 \times x}{2}\) or \(\dfrac{\pi r^2 \times x}{6} = \dfrac{2y^2 \times x}{2}\) \(\dfrac{1}{3}\pi r^2x = 4\left(\dfrac{1}{2}y^2x\right)\) is not enough for M1 here because the 4 has not been processed |
| \(\dfrac{1}{3}\pi r^2 = 2y^2\) or \(r^2 = \dfrac{6y^2}{\pi}\) or \(\dfrac{1}{3}r^2 = \dfrac{2y^2}{\pi}\) or \(\pi r^2 = 6y^2\) | M1 | oe for eliminating \(x\) dependent on both formulae being correct may have been numerical errors in previous M mark allow multiplication signs unprocessed numerical values can be condoned |
| \(r = \sqrt{\dfrac{6}{\pi}}\;y\) or \(r = \sqrt{\dfrac{6y^2}{\pi}}\) | A1 | numbers need to be simplified \(\sqrt{\dfrac{12y^2}{2\pi}}\) or \(\sqrt{\dfrac{2y^2}{\frac{1}{3}\pi}}\) are A0 ensure that root is over the whole fraction (or separate roots over the numerator and denominator) |
Additional guidance
| \(r = \sqrt{\dfrac{6y^2x}{\pi x}}\) | M1 M1 M0 A0 |
| \(\dfrac{4}{3}\pi r^2x = \dfrac{1}{2}y^2x\) (multiplying the wrong side by 4) followed by eg \(\dfrac{4}{3}\pi r^2 = \dfrac{1}{2}y^2\) or \(8r^2 = \dfrac{3y^2}{\pi}\) or \(\dfrac{4}{3}r^2 = \dfrac{y^2}{2\pi}\) or \(\pi r^2 = \dfrac{3}{8}y^2\) | M1 M0 M1 A0 |
| Penalise \(\pm\) on answer line | M3 A0 |
| Either formula incorrect would result in maximum mark of M1 | M1 M0 M0 A0 |
| Possible restarts could use both correct formulae |