| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | 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 standard arithmetic and geometric series formulas with minimal problem-solving required. Students simply identify the first term and common difference/ratio, apply memorized sum formulas, and solve a basic quadratic equation. The context is clear and all necessary information is explicitly given. |
| Spec | 1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks |
|---|---|
| 5(i) | x 2+( x−1 )( −/+0.02 ) or 1.01x−0.01x2 or 0.99x+0.01x2 oe |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | Allow ‒ or + 0.02. Allow n used |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 5(ii) | Equate to 13 then either simplify to a 3-term quadratic equation or |
| Answer | Marks | Guidance |
|---|---|---|
| quadratic | M1 | Expect n2 ‒ 101n + 1300 (=0) or 0.99x+0.01x2 =13. Allow x used |
| 16 | A1 | Ignore 85.8 or 86 |
| Answer | Marks |
|---|---|
| 5(iii) | a(1−rn) |
| Answer | Marks |
|---|---|
| 1−r | M1 |
| (=) 10.1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| ∞ 1−r | M1 | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| ∞ | A1 | Conclusion required – 'Shown' is insufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | x 2+( x−1 )( −/+0.02 ) or 1.01x−0.01x2 or 0.99x+0.01x2 oe
2 | B1 | Allow ‒ or + 0.02. Allow n used
1
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | Equate to 13 then either simplify to a 3-term quadratic equation or
find at least 1 solution (need not be correct) to an unsimplified
quadratic | M1 | Expect n2 ‒ 101n + 1300 (=0) or 0.99x+0.01x2 =13. Allow x used
16 | A1 | Ignore 85.8 or 86
2
--- 5(iii) ---
5(iii) | a(1−rn)
Use of with a = 1, r = 0.92, n = 20 soi
1−r | M1
(=) 10.1 | A1
( =) a
Use of S with a = 1, r = 0.92
∞ 1−r | M1 | ( )
1 (1−0.92n)
OR =13 → 0.92n =−0.04 oe
1−0.92
S = 12.5 so never reaches target or < 13
∞ | A1 | Conclusion required – 'Shown' is insufficient
No solution so never reaches target or < 13
4
Question | Answer | Marks | GuidanceTwo heavyweight boxers decide that they would be more successful if they competed in a lower weight class. For each boxer this would require a total weight loss of 13 kg. At the end of week 1 they have each recorded a weight loss of 1 kg and they both find that in each of the following weeks their weight loss is slightly less than the week before.
Boxer A's weight loss in week 2 is 0.98 kg. It is given that his weekly weight loss follows an arithmetic progression.
\begin{enumerate}[label=(\roman*)]
\item Write down an expression for his total weight loss after $x$ weeks. [1]
\item He reaches his 13 kg target during week $n$. Use your answer to part (i) to find the value of $n$. [2]
\end{enumerate}
Boxer B's weight loss in week 2 is 0.92 kg and it is given that his weekly weight loss follows a geometric progression.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item Calculate his total weight loss after 20 weeks and show that he can never reach his target. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2019 Q5 [7]}}