Higher June 2023 Paper 1 Q4
4 \(ABC\), \(BD\) and \(BE\) are straight lines.
Not drawn accurately
angle \(EBD = 5 \times\) angle \(ABE\)
angle \(DBC = 3 \times\) angle \(ABE\)
Work out the size of angle \(EBD\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – numerical | ||
| 1 and 5 and 3 or 9 (parts) or numbers in the ratio 1 : 5 : 3 or (angle sum on a straight line =) 180 | M1 | oe may be seen in a ratio eg \(\dfrac{1}{5} : 1 : \dfrac{3}{5}\) or \(\dfrac{1}{3} : \dfrac{5}{3} : 1\) numbers can be in any order eg 30, 10, 50 |
| \(180 \div (1 + 5 + 3)\) or 20 or \(180 \div \dfrac{9}{5}\) | M1dep | oe |
| 100 | A1 | |
| Alternative method 2 – algebraic | ||
| \(x\) and \(5x\) and \(3x\) or \(9x\) or (angle sum on a straight line =) 180 | M1 | oe correct terms with any angle as \(x\) any letter, any order may be seen on diagram |
| Correct equation with correct method to solve for one angle | M1dep | eg \(x + 5x + 3x = 180\) and \(180 \div (1 + 5 + 3)\) |
| 100 | A1 | |
Additional guidance
| \(x + 5x + 3x = 360\) or \(360 \div 9\) | M1M0A0 |
| \(\dfrac{1}{5}x + x + \dfrac{3}{5}x = 180\) and \(180 \div \left(\dfrac{1}{5} + 1 + \dfrac{3}{5}\right)\) | M1M1 |
| \(\dfrac{1}{3}x + \dfrac{5}{3}x + x = 180\) and \(180 \div \left(\dfrac{1}{3} + \dfrac{5}{3} + 1\right)\) | M1M1 |
| Angle \(EBD\) marked as 100 on the diagram with answer line blank | M1M1A1 |
| 20 and 100 in working with no or incorrect answer chosen | M1M1A0 |