Foundation November 2024 Unit 1 Q17
17. The shape below has rotational symmetry of order 6.
Diagram not drawn to scale
Find the size of angle \(x\).
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (\(6x\) =) 360 – (42 × 6) (= 108(°)) | M2 | Check diagram for answers. Sight of 108(°) implies M2. Award M1 for sight of 42 × 6 or 252(°). |
| ÷ 6 | m1 | Award m1 only if M2 awarded. |
| (\(x\) =) 18(°) | A1 | CAO. Mark final answer. Unsupported 18 is awarded M2m1A1. Allow an embedded answer but penalise -1 if contradicted by \(x \neq 18\) or equivalent Award SC1 for a final answer of 30(°) or 9·4…(°) (must be clearly using rotational symmetry of order 5 or 7). |
Alternative method 1
| Scheme | Marks | Comments |
|---|---|---|
| 360 ÷ 6 | M1 | Sight of 60(°) implies M1 A1 |
| 60(°) | A1 | |
| (\(x\) =) 60 – 42 or 42 + \(x\) = 60 | m1 | FT ‘their derived 60’– 42, provided M1 awarded and ‘their derived 60’ > 42. Unsupported 18 is awarded M1A1m1A1. |
| 18(°) | A1 | Mark final answer. Allow an embedded answer but penalise -1 if contradicted by \(x \neq 18\) or equivalent If FT leads to a whole number answer, it must be shown as a whole number. Otherwise accept a fraction. Award SC1 for a final answer of 30(°) or 9·4…(°) (must be clearly using rotational symmetry of order 5 or 7). |
Alternative method 2
| Scheme | Marks | Comments |
|---|---|---|
| \(6(42 + x) = 360\) or \(42 + x = 360 \div 6\) | B2 | Award B1 for sight of \(6(42 + x)\) or 360 ÷ 6 |
| \(252 + 6x = 360\) or \(42 + x = 60\) | B1 | FT from \(6x = k\), if B2 B0 previously awarded. |
| (\(x\) =) 18(°) | B1 | Mark final answer. Allow an embedded answer but penalise -1 if contradicted by \(x \neq 18\) or equivalent If FT leads to a whole number answer, it must be shown as a whole number. Otherwise accept a fraction. Award SC1 for a final answer of 30(°) or 9·4…(°) (must be clearly using rotational symmetry of order 5 or 7). |
Corrected by Westie's Workshop
- The board's mark scheme printed “Unsupported 18 is awarded M1A1M1A1”. Changed to “Unsupported 18 is awarded M1A1m1A1” because the third mark in this method is the dependent method mark m1 (as shown in the marks column), and the main method also writes it as m1 (M2m1A1).
Original: see the board's mark scheme PDF (link above).