## Part (a)(i)
$m = m_0k^t$

Using $m = 12$, $m_0 = 24$ and $t = 8$

$12 = 24k^8$ | |

$k^8 = \frac{1}{2}$ or $k = \left(\sqrt[8]{0.5}\right) = 0.917004$ | B1 | Total: 1 | Must see a correct exact expression for $k$ or $k^8$. Accept such as $k = e^{\frac{\ln0.5}{8}}$ or $e^{-0.086643...}$ or $\left(\frac{1}{2}\right)^{1/8}$ or 0.91700404(32....) to at least 8 d.p. AG be convinced

Note that AG so to earn the mark they must show us a correct exact expression for $k$ or $k^8$. Accept such as $k = e^{(\ln0.5)/8}$ or $e^{-0.086643...}$ or $\left(\frac{1}{2}\right)^{1/8}$ or 0.91700404(32 ...) as sufficient evidence but withhold the mark if a clear error has been made – e.g. $k = \sqrt[8]{0.5}$.

Candidates who work with logs must reach an expression such as $\log k = \frac{\log 12 - \log 24}{8}$ first.

## Part (a)(ii)
$1 = m_0(0.917004)^{60}$ | M1 | or $m_0 = (0.917004)^{-60}$ PI by A1 later

$m_0 = 181$ | A1 | Total: 2 | Must be 181 no ISW

NMS scores SC2 for 181 only but sight of greater accuracy (181.0198...) implies M1 if 181 not seen.

## Part (b)
$m = m_0k^t$

$8.106 = 10 \times k^{100}$ | M1 |

$k = \sqrt[100]{0.8106}$ OE | A1 | OE: e.g. $k = e^{\frac{\ln(0.8106)}{100}}$

$\frac{1}{2}m_0 = m_0k^t$

$k^t = \frac{1}{2}$ | |

$t \log k = \log\left(\frac{1}{2}\right)$ | M1 | A linear equation in $t$ from $k^t = \frac{1}{2}$ e.g. $t = \log_k(0.5)$

$t = \frac{\log\left(\frac{1}{2}\right)}{\log k}$ | |

$= 330$ | A1 | Total: 4 | Must be 330 No ISW

For guidance, for first A1, $k = 0.9979$..., PI by later correct work.

The first M1 is for a correct interpretation of the information given so could equally be awarded for an expression involving logs of $k$ such as $\ln 8.106 = \ln10 + 100 \ln k$ then A1 for a correct expression for ink such as $\text{ink} = \frac{\text{ln}8.106 - \text{ln}10}{100}$.

Those who use the value of $k$ from (a) could only score M0 A0 M1 A0.

NMS scores SC4 for 330 only but sight of greater accuracy (330.1006...) implies M1 A1 if 330 not seen.

Total | | 7

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