A car depreciates in value according to the model $$V=Ak^t$$ where £$V$ is the value of the car $t$ months from when it was new. Its value when new was £$12499$ and $36$ months later its value was £$7000$.
Find the value of $A$ \begin{align} Ak^0 & = 12499 \\ \Rightarrow A & = 12499 \\ \end{align} Find the value of $k$ \begin{align} 12499k^{36} & = 7000 \\ \Rightarrow k^{36} & = \frac{7000}{12499} \\ \Rightarrow k & = \left(\frac{7000}{12499}\right)^{\frac{1}{36}} \\ \Rightarrow k & = 0.984025... \\ \end{align}
The value of this car first dropped below £$5000$ $\color{red}{during}$ the $n$th month from new. Find the value of $n$.
\begin{align} 12499\cdot0.984025^n & = 5000 \\ \Rightarrow 0.984025^n & = \frac{5000}{12499} \\ \Rightarrow n\log 0.984025 & = \log \frac{5000}{12499} \\ \Rightarrow n & = \frac{\log \frac{5000}{12499}}{\log 0.984025} \\ \Rightarrow n & = 56.893449... \\ \end{align} So therefore since $56<n<57$ it was $\color{red}{during}$ the $56$th month that the value of this car first dropped below £$5000$.
The problem is that the Mark Scheme is only accepting the answer $n=57$.
How can it be during the $57$th month if we're beginning at $0$ months?