CAIE P1 2005 November — Question 6 8 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2005
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeCompound growth applications
DifficultyModerate -0.8 This is a straightforward application of geometric progression formulas to a compound growth context. Part (i) requires simple GP term calculation, part (ii) uses the standard GP sum formula, and part (iii) involves equating GP and AP sums then solving for D. All steps are routine with no problem-solving insight required beyond recognizing the GP/AP structures.
Spec1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum

6 A small trading company made a profit of \(\\) 250000\( in the year 2000. The company considered two different plans, plan \)A\( and plan \)B$, for increasing its profits. Under plan \(A\), the annual profit would increase each year by \(5 \%\) of its value in the preceding year. Find, for plan \(A\),
  1. the profit for the year 2008,
  2. the total profit for the 10 years 2000 to 2009 inclusive. Under plan \(B\), the annual profit would increase each year by a constant amount \(\\) D\(.
  3. Find the value of \)D$ for which the total profit for the 10 years 2000 to 2009 inclusive would be the same for both plans.

(i) GP with \(a = 250000\)
\(r = 1.05\)
Year 2008 is the 9th term
AnswerMarks Guidance
\(ar^8 = 250000 \times 1.05^8 = 369000\)B1 M1 A1√ For any use of \(r=1.05\) (25000 + 0.05 × 25000 gets B1) Use of \(ar^{n-1}\) with \(r=8\) or 9 Answer rounding to 369 000. ft on r.
(ii) \(S_{10} = 250000(1.05^{10}-1)-0.05 = 3140000\)M1 A1 Use of correct \(S_n\) formula – for 10 only Co – must round to the correct answer (adds 10 numbers correctly M1 A1)
(iii) AP
AnswerMarks Guidance
\(S_{10} = 5(500000 + 9D) = \text{answer to (ii)} \rightarrow D = 14300\)M1 DM1 A1 Correct \(S_n\) formula. Forming + soln Co.
(i) GP with $a = 250000$
$r = 1.05$
Year 2008 is the 9th term
$ar^8 = 250000 \times 1.05^8 = 369000$ | B1 M1 A1√ | For any use of $r=1.05$ (25000 + 0.05 × 25000 gets B1) Use of $ar^{n-1}$ with $r=8$ or 9 Answer rounding to 369 000. ft on r. | [3]

(ii) $S_{10} = 250000(1.05^{10}-1)-0.05 = 3140000$ | M1 A1 | Use of correct $S_n$ formula – for 10 only Co – must round to the correct answer (adds 10 numbers correctly M1 A1) | [2]

(iii) AP
$S_{10} = 5(500000 + 9D) = \text{answer to (ii)} \rightarrow D = 14300$ | M1 DM1 A1 | Correct $S_n$ formula. Forming + soln Co. | [3]
6 A small trading company made a profit of $\$ 250000$ in the year 2000. The company considered two different plans, plan $A$ and plan $B$, for increasing its profits.

Under plan $A$, the annual profit would increase each year by $5 \%$ of its value in the preceding year. Find, for plan $A$,\\
(i) the profit for the year 2008,\\
(ii) the total profit for the 10 years 2000 to 2009 inclusive.

Under plan $B$, the annual profit would increase each year by a constant amount $\$ D$.\\
(iii) Find the value of $D$ for which the total profit for the 10 years 2000 to 2009 inclusive would be the same for both plans.

\hfill \mbox{\textit{CAIE P1 2005 Q6 [8]}}