Higher June 2018 Paper 6 Q4
4
(a) Show that \(a^5 \times (a^3)^2\) can be expressed as \(a^{11}\). [2]
(b) Write \(\dfrac{1}{125} \times 25^9\) as a power of 5. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(a^5 \times a^6 = a^{5+6} = a^{11}\) or \(a^5 \times a^3 \times a^3 = a^{5+3+3} = a^{11}\) | 2 | B1 for [\((a^3)^2\) =] \(a^6\) or \(a^3 \times a^3\) | \(a^{5+6}\) or \(a^{5+3+3}\) or intent to add indices stated or unambiguously indicated (eg 5 + 6, add indices etc) |
| Alternative: B2 for [\(a^5 \times (a^3)^2 =\)] \(a \times a \times \ldots \times a\) [= \(a^{11}\)] or B1 for [\((a^3)^2\) =] \(a \times a \times a \times a \times a \times a\) | written in full with eleven \(a\)’s. written in full with six \(a\)’s May be implied by \((a \times a \times a \times a \times a \times a)\) seen within an incorrect lengthier product. | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(5^{15}\) | 3 | B1 for \(\left[\dfrac{1}{125} =\right]\ 5^{-3}\) or [125 =] \(5^3\) B1 for \(5^{18}\) | |