| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find n satisfying a condition |
| Difficulty | Moderate -0.8 This is a straightforward application of arithmetic sequences requiring only direct formula substitution (finding a₁₀) and solving a quadratic inequality (sum exceeding £500,000). The context is familiar, the methods are standard textbook exercises, and part (c) requires only basic commentary rather than mathematical reasoning. Easier than average A-level questions. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Identify AP with \(a = 16000\) and \(d = 1200\) | B1 | Seen or implied. Could be implied by use in \(u_n\), even if \(n \neq 10\), or by use in \(S_n\) |
| \(u_{10} = 16000 + 9 \times 1200 = £26{,}800\) | B1 | Obtain £26,800. Units required. Answer only of 26,800 would be B1B0, as AP implied but no units |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_N = 0.5N(32000 + (N-1)1200)\) | M1 | Attempt \(S_N\) of AP, with their \(a\) and \(d\). FT their \(a\) and \(d\) from part (a) |
| \(600N^2 + 15400N - 500000 = 0\) | A1 | Equate sum of AP to 500000 and rearrange to any correct 3 term quadratic. Allow \(=\), or any inequality sign |
| \(N = 18.8\) (and possibly \(N = -44.4\)) | A1 | At least correct positive root. BC. Condone 18.7 (i.e. truncated not rounded). Sight of 19 implies A1, even if 18.8 never seen. If other root given then must be correct |
| 19 years | B1FT | Conclude with 19 (years). FT on their positive non-integer root being rounded up. Allow just 19, rather than 19 years. B0 for \(\geq 19\), or equivalent in words. The FT is just on their incorrect root (which must have come from an attempt at the sum of an AP, but could follow M0), and not any other aspect of the question. Must see a non-integer value first to get the B1FT |
| Answer | Marks | Guidance |
|---|---|---|
| 'not realistic' or 'unlikely to be realistic' along with a reason such as: Sam is unlikely to stay in the same role for that long / Unlikely that salary will increase by same amount / may be an upper limit on annual salary e.g. 'pay cap', 'level off', 'plateau' / Could reach a point where the company cannot afford to pay that salary / Unlikely that salary will increase by same amount as likely to be a percentage increase | B1 | Identify that model is not (very) realistic, with a sensible, specific reason which refers why the increase is unlikely to remain constant. Could use other language such as 'model may not hold', or 'model may not be valid'. Could also identify that salary might increase by more e.g. Sam might get a promotion, be paid a bonus or be paid more due to inflation. B0 for statements that do not pertain to this model, e.g. death, retirement or changing companies (allow B1 BOD for changing jobs as this could refer to a different role within the same company). If a correct reason has been given, then ignore other incorrect or irrelevant reasons, unless directly contradictory |
## Question 4:
### Part (a)
Identify AP with $a = 16000$ and $d = 1200$ | **B1** | Seen or implied. Could be implied by use in $u_n$, even if $n \neq 10$, or by use in $S_n$
$u_{10} = 16000 + 9 \times 1200 = £26{,}800$ | **B1** | Obtain £26,800. Units required. Answer only of 26,800 would be B1B0, as AP implied but no units
### Part (b)
$S_N = 0.5N(32000 + (N-1)1200)$ | **M1** | Attempt $S_N$ of AP, with their $a$ and $d$. FT their $a$ and $d$ from part **(a)**
$600N^2 + 15400N - 500000 = 0$ | **A1** | Equate sum of AP to 500000 and rearrange to any correct 3 term quadratic. Allow $=$, or any inequality sign
$N = 18.8$ (and possibly $N = -44.4$) | **A1** | At least correct positive root. **BC**. Condone 18.7 (i.e. truncated not rounded). Sight of 19 implies A1, even if 18.8 never seen. If other root given then must be correct
19 years | **B1FT** | Conclude with 19 (years). FT on their positive non-integer root being rounded up. Allow just 19, rather than 19 years. B0 for $\geq 19$, or equivalent in words. The FT is just on their incorrect root (which must have come from an attempt at the sum of an AP, but could follow M0), and not any other aspect of the question. Must see a non-integer value first to get the B1FT
### Part (c)
'not realistic' or 'unlikely to be realistic' along with a reason such as: Sam is unlikely to stay in the same role for that long / Unlikely that salary will increase by same amount / may be an upper limit on annual salary e.g. 'pay cap', 'level off', 'plateau' / Could reach a point where the company cannot afford to pay that salary / Unlikely that salary will increase by same amount as likely to be a percentage increase | **B1** | Identify that model is not (very) realistic, with a sensible, specific reason which refers why the increase is unlikely to remain constant. Could use other language such as 'model may not hold', or 'model may not be valid'. Could also identify that salary might increase by more e.g. Sam might get a promotion, be paid a bonus or be paid more due to inflation. B0 for statements that do not pertain to this model, e.g. death, retirement or changing companies (allow B1 BOD for changing jobs as this could refer to a different role within the same company). If a correct reason has been given, then ignore other incorrect or irrelevant reasons, unless directly contradictory4 Sam starts a job with an annual salary of $\pounds 16000$. It is promised that the salary will go up by the same amount every year. In the second year Sam is paid $\pounds 17200$.
\begin{enumerate}[label=(\alph*)]
\item Find Sam's salary in the tenth year.
\item Find the number of complete years needed for Sam's total salary to first exceed $\pounds 500000$.
\item Comment on how realistic this model may be in the long term.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2019 Q4 [7]}}