OCR H240/02 2022 June — Question 4 5 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
TopicGeometric Sequences and Series
TypeFind year when threshold exceeded
DifficultyModerate -0.3 This is a straightforward geometric series sum problem requiring identification of a=20, r=0.95, and solving 20(1-0.95^n)/(1-0.95)=205 for n using logarithms. While it requires multiple steps and careful arithmetic, it's a standard textbook application with no novel insight needed, making it slightly easier than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1

4 An artist is creating a design for a large painting. The design includes a set of steps of varying heights. In the painting the lowest step has height 20 cm and the height of each other step is \(5 \%\) less than the height of the step immediately below it. In the painting the total height of the steps is 205 cm , correct to the nearest centimetre. Determine the number of steps in the design.

Question 4:
AnswerMarks Guidance
AnswerMark Guidance
\(20+20\times r+20\times r^2+\ldots\) or \(20\times\frac{1-r^n}{1-r}\)M1 Sum of a GP implied. Allow any \(r\), eg \(r=0.05\)
\(20\times\frac{1-0.95^n}{1-0.95}=205\)A1 Correct equation
\(0.95^n=\frac{195}{400}\) or \(\frac{39}{80}\) or \(0.4875\)A1 Allow \(0.487\) or \(0.488\)
\(n=\frac{\ln 0.4875}{\ln 0.95}\) oe or \(n=\log_{0.95}\!\left(\frac{39}{80}\right)\) oeM1 or \(0.95^{14}=0.4875\) or \(0.487\) or \(0.488\) seen. Can be implied by their answer. ft their equation of form \(a^n=b\) (dep M1 gained and \(b>0\))
(Number of steps \(=\)) 14A1 cao. Allow \(n=14\). Allow \(14.0\). Allow \(\approx14\)
# Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $20+20\times r+20\times r^2+\ldots$ or $20\times\frac{1-r^n}{1-r}$ | M1 | Sum of a GP implied. Allow any $r$, eg $r=0.05$ |
| $20\times\frac{1-0.95^n}{1-0.95}=205$ | A1 | Correct equation |
| $0.95^n=\frac{195}{400}$ or $\frac{39}{80}$ or $0.4875$ | A1 | Allow $0.487$ or $0.488$ |
| $n=\frac{\ln 0.4875}{\ln 0.95}$ oe or $n=\log_{0.95}\!\left(\frac{39}{80}\right)$ oe | M1 | or $0.95^{14}=0.4875$ or $0.487$ or $0.488$ seen. Can be implied by their answer. ft their equation of form $a^n=b$ (dep M1 gained and $b>0$) |
| (Number of steps $=$) 14 | A1 | cao. Allow $n=14$. Allow $14.0$. Allow $\approx14$ |

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4 An artist is creating a design for a large painting. The design includes a set of steps of varying heights. In the painting the lowest step has height 20 cm and the height of each other step is $5 \%$ less than the height of the step immediately below it.

In the painting the total height of the steps is 205 cm , correct to the nearest centimetre.

Determine the number of steps in the design.

\hfill \mbox{\textit{OCR H240/02 2022 Q4 [5]}}