Higher June 2019 Paper 3 Q15
15 Simplify fully \(\quad \dfrac{a^3 b^2}{cd} \times \dfrac{c}{ab^5}\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – answer written as a fraction | ||
| \(a^2\) on numerator | B1 | \(a\) correctly simplified |
| \(b^3\) on denominator or \(b^{-3}\) on numerator | B1 | \(b\) correctly simplified |
| \(c\) cancelled and \(d\) on denominator or \(d^{-1}\) on numerator | B1 | \(d\) correctly simplified |
| Alternative method 2 – answer written only as a product | ||
| \(a^2\) | B1 | \(a\) correctly simplified |
| \(b^{-3}\) | B1 | \(b\) correctly simplified |
| \(d^{-1}\) and \(c\) cancelled | B1 | \(d\) correctly simplified |
Additional guidance
| If answer line is blank, marks can be awarded in the working | |
| Do not award any marks if addition or subtraction is seen in their best attempt | |
| Condone use of capital letters | |
| Penalise use of \(\times\) sign by one mark only if full marks would have been awarded eg \(\ a^2 \times b^{-3} \times d^{-1}\) | B1B1 |
| \(\dfrac{a^2}{db^3}\) or \(\dfrac{a^2 d^{-1}}{b^3}\) or \(\dfrac{a^2 b^{-3}}{d}\) or \(a^2 b^{-3} d^{-1}\) | B1B1B1 |
| \(\dfrac{a^2 b^2}{db^5}\) or \(\dfrac{a^2 b^2 d^{-1}}{b^5}\) or \(a^2 b^2 d^{-1} b^{-5}\) | B1B0B1 |
| \(\dfrac{a^3}{dab^3}\) | B0B1B1 |
| \(\dfrac{a^2 c}{cdb^3}\) | B1B1B0 |
| \(\dfrac{a}{d} \times b^3\) use of \(\times\) sign not penalised as full marks would not be awarded | B0B0B1 |
| \(a^2 + b^{-3} - d^{-1}\) | B0B0B0 |