| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a standard log-linear modelling question requiring routine application of logarithm laws to linearize an exponential model, then reverse the process to find constants. Part (a) is pure recall, parts (b)-(c) are mechanical calculations with clear methods, and part (d) requires a generic criticism of extrapolation. Slightly easier than average due to its highly procedural nature with no problem-solving insight required. |
| Spec | 1.02z Models in context: use functions in modelling1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\log_{10}S = \log_{10}(ab^t)\) → \(\log_{10}S = \log_{10}a + \log_{10}b^t\) | M1 | Attempt to show reduction to linear form; introduce logs on both sides, correctly split to sum of two terms |
| \(\log_{10}S = t\log_{10}b + \log_{10}a\) | A1 | Obtain correct equation; condone no base, any bases seen must be 10; A0 for \(\log_{10}bt\) unless previously seen as \(t\log_{10}b\) |
| which is of the form \(Y = mX + c\) | A1 | Link to equation of straight line; base 10 must now be explicit throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\log_{10}a = 0.583 \Rightarrow a = 10^{0.583} = 3.8\) | B1 | Obtain \(a = 3.8\) or better; must clearly be value for \(a\) |
| \(\log_{10}b = 0.146 \Rightarrow b = 10^{0.146} = 1.4\) | B1 | Obtain \(b = 1.4\) or better; must clearly be value for \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3.8 \times 1.4^t = 200\), \(1.4^t = 52.63\) | M1 | Link model to 200 and attempt to solve for \(t\); must use correct solution method; allow M1 if \(S = 200{,}000{,}000\); allow if \(a\) and \(b\) transposed |
| \(t = 11.8\) | A1 | Obtain \(t = 11.8\) or better; condone 11.7 as truncated value |
| so year is 2027 | A1FT | FT their value for \(t\); answer in context; FT on \(2015 +\) integer number of years, rounding up their \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Unlikely that sales will continue at same rate / Finite market | B1 | Any sensible reason e.g. pattern not necessarily continuing or market being limited by number of customers; allow 'extrapolation unreliable'; reason needed not just 'other external factors' |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10}S = \log_{10}(ab^t)$ → $\log_{10}S = \log_{10}a + \log_{10}b^t$ | M1 | Attempt to show reduction to linear form; introduce logs on both sides, correctly split to sum of two terms |
| $\log_{10}S = t\log_{10}b + \log_{10}a$ | A1 | Obtain correct equation; condone no base, any bases seen must be 10; A0 for $\log_{10}bt$ unless previously seen as $t\log_{10}b$ |
| which is of the form $Y = mX + c$ | A1 | Link to equation of straight line; base 10 must now be explicit throughout |
**[3 marks]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\log_{10}a = 0.583 \Rightarrow a = 10^{0.583} = 3.8$ | B1 | Obtain $a = 3.8$ or better; must clearly be value for $a$ |
| $\log_{10}b = 0.146 \Rightarrow b = 10^{0.146} = 1.4$ | B1 | Obtain $b = 1.4$ or better; must clearly be value for $b$ |
**[2 marks]**
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3.8 \times 1.4^t = 200$, $1.4^t = 52.63$ | M1 | Link model to 200 and attempt to solve for $t$; must use correct solution method; allow M1 if $S = 200{,}000{,}000$; allow if $a$ and $b$ transposed |
| $t = 11.8$ | A1 | Obtain $t = 11.8$ or better; condone 11.7 as truncated value |
| so year is 2027 | A1FT | FT their value for $t$; answer in context; FT on $2015 +$ integer number of years, rounding up their $t$ |
**[3 marks]**
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Unlikely that sales will continue at same rate / Finite market | B1 | Any sensible reason e.g. pattern not necessarily continuing or market being limited by number of customers; allow 'extrapolation unreliable'; reason needed not just 'other external factors' |
**[1 mark]**
---6 A mobile phone company records their annual sales on $31 ^ { \text {st } }$ December every year.\\
Paul thinks that the annual sales, $S$ million, can be modelled by the equation $S = a b ^ { t }$, where $a$ and $b$ are both positive constants and $t$ is the number of years since $31 ^ { \text {st } }$ December 2015.
Paul tests his theory by using the annual sales figures from $31 ^ { \text {st } }$ December 2015 to $31 { } ^ { \text {st } }$ December 2019. He plots these results on a graph, with $t$ on the horizontal axis and $\log _ { 10 } S$ on the vertical axis.
\begin{enumerate}[label=(\alph*)]
\item Explain why, if Paul's model is correct, the results should lie on a straight line of best fit on his graph.
The results lie on a straight line of best fit which has a gradient of 0.146 and an intercept on the vertical axis of 0.583 .
\item Use these values to obtain estimates for $a$ and $b$, correct to 2 significant figures.
\item Use this model to predict the year in which, on the $31 { } ^ { \text {st } }$ December, the annual sales would first be recorded as greater than 200 million.
\item Give a reason why this prediction may not be reliable.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2020 Q6 [9]}}