Higher November 2024 Paper 4 Q19
19 Two ornaments, A and B, are mathematically similar.
The table shows information about the two ornaments.
| Ornament A | Ornament B | |
|---|---|---|
| Height (m) | \(h\) | 12 |
| Surface area (m2) | 216 | \(A\) |
| Volume (m3) | 240 | 3750 |
Find the value of \(h\) and the value of \(A\).
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\(h\) =] 4.8 [\(A\) =] 1350 with correct working | 6 | “Correct working” requires evidence of at least M2 | |
| M2 for \(\sqrt[3]{\frac{3750}{240}}\) or 2.5 oe or \(\sqrt[3]{\frac{240}{3750}}\) or 0.4 oe or M1 for \(\frac{3750}{240}\) or 15.625 oe or \(\frac{240}{3750}\) or 0.064 oe | |||
| B2 for [\(h\) =] 4.8 or M1 for \(\frac{12}{2.5}\) oe | e.g. 12 × 0.4 | ||
| B2 for [\(A\) =] 1350 or M1 for 216 × (their 2.5)2 oe | e.g. 216 ÷ (their 0.4)2 Note : Use of scale factor = \(\frac{\sqrt[3]{3750}}{\sqrt[3]{240}}\) often leads to accuracy errors so award M2 for \(\frac{\sqrt[3]{3750}}{\sqrt[3]{240}}\) M1 for \(\frac{12}{\textit{their } 2.5}\) oe (penalise accuracy once here at first occurrence) M1 A1 for 216 × (their 2.5) = answer to A involving a rounding error | ||