| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Real-world AP: find n satisfying a condition |
| Difficulty | Moderate -0.3 This is a straightforward application of arithmetic and geometric sequences with clear real-world context. Part (i) requires basic arithmetic sequence formulas (nth term and sum), while part (ii) involves a geometric series sum. All techniques are standard P1 content with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part structure and need to interpret the context correctly. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a+(n-1)d = 10+29\times2\) | M1 | Use of \(n\)th term of an AP with \(a=\pm10\), \(d=\pm2\), \(n=30\) or \(29\) |
| \(= 68\) | A1 [2] | Condone \(-68 \rightarrow 68\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}n(20+2(n-1))=2000\) or \(0\) | M1 | Use of \(S_n\) formula for an AP with \(a=\pm10\), \(d=\pm2\) and equated to either \(0\) or \(2000\) |
| \(\rightarrow 2n^2+18n-4000=0\) oe | A1 | Correct 3 term quadratic \(=0\) |
| \((n=)\ 41\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r=1.1\) oe | B1 | e.g. \(\frac{11}{10}\), \(110\%\) |
| Uses \(S_{30}=\frac{10(1.1^{30}-1)}{1.1-1}\) \((=1645)\) | M1 | Use of \(S_n\) formula for a GP, \(a=\pm10\), \(n=30\) |
| Percentage lost \(=\frac{2000-1645}{2000}\times100\) | DM1 | Fully correct method for \(\%\) left with "their \(1645\)" |
| \(= 17.75\) | A1 [4] | allow \(17.7\) or \(17.8\) |
# Question 9(i)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a+(n-1)d = 10+29\times2$ | **M1** | Use of $n$th term of an AP with $a=\pm10$, $d=\pm2$, $n=30$ or $29$ |
| $= 68$ | **A1** [2] | Condone $-68 \rightarrow 68$ |
---
# Question 9(i)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}n(20+2(n-1))=2000$ or $0$ | **M1** | Use of $S_n$ formula for an AP with $a=\pm10$, $d=\pm2$ and equated to either $0$ or $2000$ |
| $\rightarrow 2n^2+18n-4000=0$ oe | **A1** | Correct 3 term quadratic $=0$ |
| $(n=)\ 41$ | **A1** [3] | |
---
# Question 9(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r=1.1$ oe | **B1** | e.g. $\frac{11}{10}$, $110\%$ |
| Uses $S_{30}=\frac{10(1.1^{30}-1)}{1.1-1}$ $(=1645)$ | **M1** | Use of $S_n$ formula for a GP, $a=\pm10$, $n=30$ |
| Percentage lost $=\frac{2000-1645}{2000}\times100$ | **DM1** | Fully correct method for $\%$ left with "their $1645$" |
| $= 17.75$ | **A1** [4] | allow $17.7$ or $17.8$ |
---9 A water tank holds 2000 litres when full. A small hole in the base is gradually getting bigger so that each day a greater amount of water is lost.\\
(i) On the first day after filling, 10 litres of water are lost and this increases by 2 litres each day.
\begin{enumerate}[label=(\alph*)]
\item How many litres will be lost on the 30th day after filling?
\item The tank becomes empty during the $n$th day after filling. Find the value of $n$.\\
(ii) Assume instead that 10 litres of water are lost on the first day and that the amount of water lost increases by $10 \%$ on each succeeding day. Find what percentage of the original 2000 litres is left in the tank at the end of the 30th day after filling.\\[0pt]
[Questions 10 and 11 are printed on the next page.]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q9 [9]}}