Higher November 2020 Paper 6 Q9
9 \(x\) is directly proportional to \(y\).
\(y\) is directly proportional to \(z\).
When \(x = 10\), \(y = 60\).
When \(y = 8\), \(z = 1.6\).
Find a formula for \(z\) in terms of \(x\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(z = 1.2x\) or \(z = \frac{6x}{5}\) | 4 | B3 for a correct equation involving just \(x\) and \(z\) but not in required form OR B1 for \(y = 6x\) oe B1 for \(y = 5z\) oe M1 for a correct equation involving just \(x\) and \(z\) using their two equations OR B1 for \(y = 60\) when \(z = 12\) B1 for \(x : [y :]\ z\) is 10 : [60 :]12 M1 for a correct equation involving just \(x\) and \(z\) using their triple ratio or their two ratios with a common \(y\) value If 0 scored SC1 for \(y = kx\) and \(k = 6\) found oe or for \(y = kz\) and \(k = 5\) found oe | Condone \(\propto\) for = in B1 marks and SC1 but not at B3 or full marks e.g. \(5z = 6x\), or \(x = \frac{5z}{6}\) Their two equations of the form \(y = ax\) oe and \(y = bz\) oe Allow B2 for other triple ratios of the form \(5k : 30k : 6k\) or two correct ratios with a common \(y\) value |