Higher June 2023 Paper 3 Q20
20
(a) Sunil thinks that \(E\) and \(D\) are linked by the equation \(\quad E = \dfrac{36}{D}\)
The graph shows the values of \(D\) and \(E\) for \(\quad 2 \leqslant D \leqslant 6\)
Choose one point on the graph and state if Sunil’s equation is correct for that point. [1 mark]
(b) \(G\) is directly proportional to the square root of \(H\).
\(G : H = 3 : 2 \quad\) when \(\quad H = 16\)
Work out \(\quad G : H \quad\) when \(\quad H = 100\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Substitutes a correct pair of coordinates and states that the equation is correct | B1 | eg \(18 = \dfrac{36}{2}\) so he is right |
Additional guidance
| Accept ‘Yes’ or a tick or any clear indication that he is correct | |
| Do not accept pairs of values not on the graph | |
| Do not accept a correct answer alongside an incorrect response unless clearly chosen | |
| Do not accept a coordinate with no substitution seen | |
| Pairs with integer \(x\) or \(y\) include \(18 = \dfrac{36}{2}\), \(15 = \dfrac{36}{2.4}\), \(12 = \dfrac{36}{3}\), \(10 = \dfrac{36}{3.6}\) \(9 = \dfrac{36}{4}\), \(8 = \dfrac{36}{4.5}\), \(7.2 = \dfrac{36}{5}\), \(6 = \dfrac{36}{6}\) | |
| Substituting values incorrectly eg \(2 = \dfrac{36}{18}\) or \(4 = \dfrac{36}{9}\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(G \propto \sqrt{H}\) or \(G = k\sqrt{H}\) or \(16 \div 2 \times 3 = k\sqrt{16}\) or \(24 = k\sqrt{16}\) | M1 | oe equation \(k\) may be any letter |
| \(k = \dfrac{\text{their } 24}{\sqrt{16}}\) or \(k = 6\) or \(G =\) their \(6\sqrt{H}\) | M1dep | their 24 must be the result of \(16 \div 2 \times 3\) |
| their \(6 \times\) their \(\sqrt{100}\) or 60 | M1dep | dep on M2 |
| 60 : 100 or 3 : 5 | A1 | oe ratio |
| Alternative method 2 | ||
| \(100 \div 16\) or 6.25 | M1 | |
| \(\sqrt{\text{their } 6.25}\) or 2.5 | M1dep | |
| \(2 \times\) their 2.5 or 5 or \(24 \times\) their 2.5 or 60 | M1dep | dep on M2 |
| 60 : 100 or 3 : 5 | A1 | oe ratio |
Additional guidance
| Ignore an incorrect attempt to simplify a correct ratio eg 60 : 100 followed by 3 : 4 | M1M1M1A1 |
| \(k = 6\) implies M2 unless from incorrect working | |
| \(G \propto k\sqrt{H}\) is M0 unless recovered | |
| \(G = k\sqrt{H} \qquad \sqrt{16} = 4 \qquad G : H = 6 : 4 \qquad 6 = k \times 4 \qquad k = \dfrac{6}{4}\) followed by \(G = 1.5 \times 10 \qquad 150 : 100\) | M1M0M0A0 |
| \(G = 24\) with no correct further work | M0 |