Higher June 2018 Paper 3 Q21
21 The mass of an ornament is \(m\) grams.
The height of the ornament is \(h\) centimetres.
\(m\) is directly proportional to the cube of \(h\).
\(m = 1600\) when \(\;h = 8\)
(a) Work out an equation connecting \(m\) and \(h\). [3 marks]
(b) Work out the mass of an ornament of height 12 centimetres. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(m \propto h^3\) or \(m = \mathrm{k} \times h^3\) or \(1600 = \mathrm{k} \times 8^3\) or \(\mathrm{c} \times m = h^3\) or \(\mathrm{c} \times 1600 = 8^3\) | M1 | oe eg \(h = \mathrm{k}m^{1/3}\) |
| (k =) \(1600 \div 8^3\) or 3.125 or (c =) \(8^3 \div 1600\) or 0.32 | M1dep | oe eg \(\dfrac{1600}{512}\) or \(\dfrac{25}{8}\) \(\dfrac{512}{1600}\) or \(\dfrac{8}{25}\) |
| \(m = 3.125 \times h^3\) or \(0.32 \times m = h^3\) | A1 | oe equation |
Additional guidance
| \(m \propto 3.125 \times h^3\) or \(0.32m \propto h^3\) | M1M1A0 |
| (\(k\) =) 3.125 or (c =) 0.32 | M1M1 |
| \(3.125h^3\) or \(0.32h^3\) | M1M1 |
| Answer | Mark | Comments |
|---|---|---|
| their \(3.125 \times 12^3\) their \(3.125 \times 1728\) or \(1600 \times \left(\dfrac{12}{8}\right)^3\) or \(12^3 \div\) their 0.32 or \(1728 \div 0.32\) or \(1600 \div \left(\dfrac{8}{12}\right)^3\) | M1 | oe |
| 5400 | A1ft | oe ft their 3.125 provided using \(m =\) their \(3.125 \times h^3\) |
Additional guidance
| Must use \(\times\, 12^3\) or \(\times\) 1728 or \(\times \left(\dfrac{12}{8}\right)^3\) for M1 | |
| If in part (a) \(m = \mathrm{k} \times h\) \(1600 = \mathrm{k} \times 8\) \(m = 200h\) and in part (b) \(m = 200 \times 12\), \(m = 2400\) | M0 part (a) M0 part (b) |
| If in part (a) \(m = \mathrm{k} \times h\) \(1600 = \mathrm{k} \times 8\) \(m = 200h\) and in part (b) \(m = 200 \times 12^3\), \(m = 345\,600\) | M0 part (a) M1A1ft part (b) |