Higher January 2022 Paper 1 Q15
15 \(A\) is inversely proportional to \(C^2\)
\(A = 40\) when \(C = 1.5\)
Calculate the value of \(C\) when \(A = 1000\)
(3)
| Scheme | Marks |
|---|---|
eg \(40 = \dfrac{k}{1.5^2}\) or \(k = 90\) or \(\dfrac{C^2}{1.5^2} = \dfrac{40}{1000}\;(= 0.04)\) or \((C^2 =)\;1.5^2 \times \dfrac{40}{1000}\;(= 0.09)\) or \(\dfrac{1.5^2}{C^2} = \dfrac{1000}{40}\;(= 25)\) or \((C^2 =)\;1.5^2 \div \dfrac{1000}{40}\;(= 0.09)\) | M1 |
eg \((C =)\;\sqrt{\dfrac{\text{``}{90}\text{''}}{1000}}\) oe or \((C =)\;\sqrt{1.5^2 \times \text{``}{0.04}\text{''}}\) or \((C =)\;\sqrt{1.5^2 \div \text{``}{25}\text{''}}\) or \((C =)\;\sqrt{\text{``}{0.09}\text{''}}\) | M1 |
| 0.3 | A1 |
| (3) | |
| (3 marks) |
Notes
A1: oe, allow ±0.3 oe or −0.3 oe