| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Mixed arithmetic and geometric |
| Difficulty | Standard +0.3 This is a straightforward arithmetic/geometric sequence application with clear scaffolding. Part (a) is direct substitution, (b) involves simple algebraic manipulation to derive a given inequality, (c) is mechanical iteration (no problem-solving), and (d) requires basic real-world reasoning. The question guides students through each step with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Anna \(= 30 + 9 \times 15 = 165\) mins | M1 | Attempt \(u_{10}\) for AP using correct equation; method must be seen |
| Ben \(= 30 \times 1.1^9 = 71\) mins | M1 | Attempt \(u_{10}\) for GP using correct equation; method must be seen |
| \(165 - 71 = 94\) minutes A.G. | A1 | Obtain given answer of 94 minutes; AG so both terms need to be explicitly evaluated; show subtraction or give more accurate value before 94 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Anna: \(u_X = 30 + 15(X-1)\); Ben: \(u_X = 30 \times 1.1^{X-1}\) | B1 | Both \(u_X\) terms correct; condone unknown other than \(X\) |
| \(30 \times 1.1^{X-1} > 30 + 15(X-1)\) → \(30 \times 1.1^{X-1} > 15X + 15\) → \(1.1^{X-1} > 0.5X + 0.5\) → \(X - 1 > \log_{1.1}(0.5X + 0.5)\) | M1 | Link correct expressions and attempt to rearrange; condone incorrect linking sign e.g. not \(>\); must use correct process |
| \(X > \log_{1.1}(0.5X + 0.5) + 1\) A.G. | A1 | Show given answer convincingly; must now be \(>\), with justification if \(=\) used in proof, and with \(X\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 18.9 | B1 | Correct first iterate; allow 19 or 18.8… |
| 25.1, 28.0, 29.0, 29.4, 29.6, 29.6,… | M1 | Use correct iterative process to find at least two further values; allow integer values; could be truncated not rounded |
| \(X = 30\) | A1 | Obtain \(X = 30\); must be an integer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eventually there will not be enough hours in the day for revision | B1 | Comment on long-term behaviour; allow other sensible reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Increasing by 10% will involve decimals of minutes so will no longer be accurate | B1 | Comment on not being able to measure time accurately; allow long-term behaviour if not already given in (i); allow other sensible reason; B0 if referring to reasons that may prevent revision from happening e.g. illness; if a correct reason is given then ISW an incorrect numerical value |
# Question 7:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Anna $= 30 + 9 \times 15 = 165$ mins | M1 | Attempt $u_{10}$ for AP using correct equation; method must be seen |
| Ben $= 30 \times 1.1^9 = 71$ mins | M1 | Attempt $u_{10}$ for GP using correct equation; method must be seen |
| $165 - 71 = 94$ minutes **A.G.** | A1 | Obtain given answer of 94 minutes; AG so both terms need to be explicitly evaluated; show subtraction or give more accurate value before 94 |
**[3 marks]**
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Anna: $u_X = 30 + 15(X-1)$; Ben: $u_X = 30 \times 1.1^{X-1}$ | B1 | Both $u_X$ terms correct; condone unknown other than $X$ |
| $30 \times 1.1^{X-1} > 30 + 15(X-1)$ → $30 \times 1.1^{X-1} > 15X + 15$ → $1.1^{X-1} > 0.5X + 0.5$ → $X - 1 > \log_{1.1}(0.5X + 0.5)$ | M1 | Link correct expressions and attempt to rearrange; condone incorrect linking sign e.g. not $>$; must use correct process |
| $X > \log_{1.1}(0.5X + 0.5) + 1$ **A.G.** | A1 | Show given answer convincingly; must now be $>$, with justification if $=$ used in proof, and with $X$ |
**[3 marks]**
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 18.9 | B1 | Correct first iterate; allow 19 or 18.8… |
| 25.1, 28.0, 29.0, 29.4, 29.6, 29.6,… | M1 | Use correct iterative process to find at least two further values; allow integer values; could be truncated not rounded |
| $X = 30$ | A1 | Obtain $X = 30$; must be an integer |
**[3 marks]**
## Part (d)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eventually there will not be enough hours in the day for revision | B1 | Comment on long-term behaviour; allow other sensible reason |
**[1 mark]**
## Part (d)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Increasing by 10% will involve decimals of minutes so will no longer be accurate | B1 | Comment on not being able to measure time accurately; allow long-term behaviour if not already given in (i); allow other sensible reason; B0 if referring to reasons that may prevent revision from happening e.g. illness; if a correct reason is given then ISW an incorrect numerical value |
**[1 mark]**
---7 Two students, Anna and Ben, are starting a revision programme. They will both revise for 30 minutes on Day 1. Anna will increase her revision time by 15 minutes for every subsequent day. Ben will increase his revision time by $10 \%$ for every subsequent day.
\begin{enumerate}[label=(\alph*)]
\item Verify that on Day 10 Anna does 94 minutes more revision than Ben, correct to the nearest minute.
Let Day $X$ be the first day on which Ben does more revision than Anna.
\item Show that $X$ satisfies the inequality $X > \log _ { 1.1 } ( 0.5 X + 0.5 ) + 1$.
\item Use the iterative formula $x _ { n + 1 } = \log _ { 1.1 } \left( 0.5 x _ { n } + 0.5 \right) + 1$ with $x _ { 1 } = 10$ to find the value of $X$.
You should show the result of each iteration.
\item \begin{enumerate}[label=(\roman*)]
\item Give a reason why Anna's revision programme may not be realistic.
\item Give a different reason why Ben's revision programme may not be realistic.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2020 Q7 [11]}}