CAIE P1 2018 June — Question 3 5 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2018
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeCompound growth applications
DifficultyModerate -0.8 This is a straightforward application of geometric sequence formulas with clearly stated parameters (a=8000, r=1.02, n=12). Students need only substitute into standard formulas for nth term and sum of GP—no problem-solving insight required, making it easier than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay

3 A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by \(2 \%\) of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt.
  1. Find the amount of salt obtained in the 12th week after the change.
  2. Find the total amount of salt obtained in the first 12 weeks after the change.

Question 3(i):
AnswerMarks Guidance
\(r = 1.02\) or \(\frac{102}{100}\) used in a GP in some wayB1 Can be awarded here for use in \(S_n\) formula
Amount in 12th week \(= 8000\ (their\ r)^{11}\) or \((their\ a\ from\ \frac{8000}{their\ r})(their\ r)^{12}\)M1 Use of \(ar^{n-1}\) with \(a = 8000\) & \(n = 12\) or with \(a = \frac{8000}{1.02}\) and \(n = 13\)
\(= 9950\) (kg) awrtA1 Note: Final answer of either 9943 or 9940 implies M1. Full marks can be awarded for a correct answer from a list of terms
Question 3(ii):
AnswerMarks Guidance
In 12 weeks, total is \(\frac{8000\left((their\ r)^{12}-1\right)}{((their\ r)-1)}\)M1 Use of \(S_n\) with \(a = 8000\) and \(n = 12\) or addition of 12 terms
\(= 107000\) (kg) awrtA1 Correct answer but no working 2/2
## Question 3(i):

$r = 1.02$ or $\frac{102}{100}$ used in a GP in some way | B1 | Can be awarded here for use in $S_n$ formula

Amount in 12th week $= 8000\ (their\ r)^{11}$ or $(their\ a\ from\ \frac{8000}{their\ r})(their\ r)^{12}$ | M1 | Use of $ar^{n-1}$ with $a = 8000$ & $n = 12$ or with $a = \frac{8000}{1.02}$ and $n = 13$

$= 9950$ (kg) awrt | A1 | Note: Final answer of either 9943 or 9940 implies M1. Full marks can be awarded for a correct answer from a list of terms

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## Question 3(ii):

In 12 weeks, total is $\frac{8000\left((their\ r)^{12}-1\right)}{((their\ r)-1)}$ | M1 | Use of $S_n$ with $a = 8000$ and $n = 12$ or addition of 12 terms

$= 107000$ (kg) awrt | A1 | Correct answer but no working 2/2

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3 A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by $2 \%$ of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt.\\
(i) Find the amount of salt obtained in the 12th week after the change.\\

(ii) Find the total amount of salt obtained in the first 12 weeks after the change.\\

\hfill \mbox{\textit{CAIE P1 2018 Q3 [5]}}