Foundation June 2023 Paper 3 Q11
11 Two angles around a point are shown.
Not drawn accurately
The angles are in the ratio \(\quad 2 : 7\)
Show that the larger angle is \(280^\circ\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(360 \div 9\ (= 40)\) and \(40 \times 7 = 280\) or \(360 \div 9\ (= 40)\) and \(40 \times 2\ (= 80)\) and \(80 + 280 = 360\) or \(40 \times 2\ (= 80)\) and \(40 \times 7\ (= 280)\) and \(80 + 280 = 360\) or \(280 \div 7\ (= 40)\) and \(40 \times 9 = 360\) or \(2 : 7 = 80 : 280\) and \(80 + 280 = 360\) or \(360 - 280\ (= 80)\) and \(80 : 280 = 2 : 7\) | B2 | oe B1 \(360 \div 9\) or \(280 \div 7\) or 40 oe or \(\dfrac{2}{9}\) or \(\dfrac{7}{9}\) or \(360 - 280\) or 80 oe |
Additional guidance
| 80 and 280 shown on the diagram is not oe for \(80 + 280 = 360\) | |
| \(360 \div 9 \times 7 = 280\) | B2 |
| \(360 \div 9\) and \(40 \times 2\) and \(2 : 7 = 80 : 280\) | B2 |
| \(360 \div 9 = 40\) and \(2 : 7 = 80 : 280\) (\(40 \times 2\) or \(40 \times 7\) missing) | B1 |
| \(40 \times 7 = 280\) without \(360 \div 9\) eg \(40 \times 7 = 280\) and \(80 + 280 = 360\) (\(360 \div 9 = 40\) or \(40 \times 2\) missing) | B1 |
| \(80 : 280\) and \(80 + 280 = 360\) (\(2 : 7 = 80 : 280\) missing) | B1 |
| \(360 \div 9 = 40\) and \(80 + 280 = 360\) (\(40 \times 2\) or \(40 \times 7\) missing) | B1 |
| \(280 \div 7 = 40\) and \(360 - 280 = 80\) (\(40 \times 2\) or \(40 \times 9\) missing) | B1 |
| \(280 \div 7\) and \(40 \times 2\) and \(80 : 280 = 2 : 7\) (\(80 + 280 = 360\) missing) | B1 |
| \(80 + 280 = 360\) | B1 |