Higher November 2019 Paper 2 Q26
26 \(P\), \(Q\) and \(R\) have positive values.
\(P\) is directly proportional to the square of \(Q\).
When \(P = 1.25\), \(Q = 0.5\)
\(Q\) is inversely proportional to \(R\).
When \(Q = 0.5\), \(R = 6\)
Work out the value of \(R\) when \(P = 0.8\) [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(P = kQ^2\) or \(1.25 = k \times 0.5^2\) or \(Q = \dfrac{c}{R}\) or \(0.5 = \dfrac{c}{6}\) | M1 | oe |
| \(k = \dfrac{1.25}{0.5^2}\) or \(k = 5\) or \(P = 5Q^2\) or \(c = 0.5 \times 6\) or \(c = 3\) or \(Q = \dfrac{3}{R}\) | M1 | oe |
| \(P = 5Q^2\) and \(Q = \dfrac{3}{R}\) or \(k = 5\) and \(c = 3\) | A1 | oe |
| \(0.8 =\) their \(5 \times \left(\dfrac{\text{their } 3}{R}\right)^2\) or \({(R =)}\ \sqrt{\dfrac{\text{their } 5 \times (\text{their } 3)^2}{0.8}}\) | M1 | ft their equations of the form \(P = kQ^2\) and \(Q = \dfrac{c}{R}\) oe eg \((Q =)\ \sqrt{\dfrac{0.8}{\text{their } 5}}\) or \(Q = 0.4\) and \((R =)\ \dfrac{\text{their } 3}{\text{their } 0.4}\) |
| 7.5 or \(7\dfrac{1}{2}\) or \(\dfrac{15}{2}\) | A1ft | ft their equations of the form \(P = kQ^2\) and \(Q = \dfrac{c}{R}\) with 3rd M1 scored |
| Alternative method 2 | ||
| \(P = \dfrac{k}{R^2}\) or \(1.25 = \dfrac{k}{6^2}\) | M1 | oe |
| \(k = 1.25 \times 6^2\) | M1dep | oe |
| \(P = \dfrac{45}{R^2}\) or \(k = 45\) | A1 | oe |
| \(0.8 = \dfrac{\text{their } 45}{R^2}\) or \((R =)\ \sqrt{\dfrac{\text{their } 45}{0.8}}\) | M1 | oe ft their equation of the form \(P = \dfrac{k}{R^2}\) |
| 7.5 or \(7\dfrac{1}{2}\) or \(\dfrac{15}{2}\) | A1ft | ft their equation of the form \(P = \dfrac{k}{R^2}\) with 3rd M1 scored |
Additional guidance
| Allow \(k\) and \(c\) to be any letters, including using both as \(k\) in Alt 1 | |
| Alt 1 \(\ kP = Q^2\) leading to \(k = 0.2\) | M1M1 |
| Alt 2 \(\ kP = \dfrac{1}{R^2}\) leading to \(k = \dfrac{1}{45}\) (allow 0.022…) | M1M1A1 |