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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?

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Your calculation is perfect. You're getting the answer as 56.893449... , this means that the value of car drops down to £5000 after 56th month not in the 56th month.

This implies the answer is during 57th month.

To understand this an analogy may be used. The year 2016 is in 21st century not 20th century.