Higher November 2021 Paper 1 Q26
26
\(d = 2f\)
\(\dfrac{e - f}{d - e} = \dfrac{1}{4}\)
Work out the ratio \(\quad e : f\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: substitutes \(2f\) for \(d\) | ||
| \(\dfrac{e - f}{2f - e} = \dfrac{1}{4}\) or \(2f - e = 4(e - f)\) | M1 | oe equation in \(e\) and \(f\) |
| \(6f = 5e\) or \(\dfrac{e}{f} = \dfrac{6}{5}\) | M1dep | oe with variables collected eg \(1.5f = 1.25e\) oe with single fractions eg \(\dfrac{f}{5} = \dfrac{e}{6}\) |
| 6 : 5 | A1 | oe ratio |
| Alternative method 2: substitutes \(\dfrac{d}{2}\) for \(f\) | ||
| \(d - e = 4\left(e - \dfrac{d}{2}\right)\) or \(3d = 5e\) | M1 | oe equation in \(d\) and \(e\) |
| \(6f = 5e\) or \(\dfrac{e}{f} = \dfrac{6}{5}\) | M1dep | oe with variables collected eg \(1.5f = 1.25e\) oe with single fractions eg \(\dfrac{f}{5} = \dfrac{e}{6}\) |
| 6 : 5 | A1 | oe ratio |
| Alternative method 3: substitutes \(2f\) for \(d\) and forms simultaneous equations | ||
| \(e - f = 1\) and \(2f - e = 4\) | M1 | oe with rhs in the ratio 1 : 4 eg \(e - f = 2\) and \(2f - e = 8\) |
| \(f = 5\) or \(e = 6\) | M1dep | correct solution for one unknown from their correct simultaneous equations eg \(f = 10\) or \(e = 12\) from above equations |
| 6 : 5 | A1 | oe ratio |
Additional guidance
| 5 : 6 with no method marks awarded | M0M0A0 |