June 2022 Paper 2 Q4
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. [5]
| Scheme | Marks | AO |
|---|---|---|
| \(20 + 20 \times r + 20 \times r^2 + \ldots\) or \(20 \times \dfrac{1 - r^n}{1 - r}\) | M1 | 3.1b |
| \(20 \times \dfrac{1 - 0.95^n}{1 - 0.95} = 205\) | A1 | 1.1 |
| \(0.95^n = \dfrac{195}{400}\) or \(\dfrac{39}{80}\) or 0.4875 | A1 | 1.1 |
| \(n = \dfrac{\ln 0.4875}{\ln 0.95}\) oe or \(n = \log_{0.95}\left(\dfrac{39}{80}\right)\) oe | M1 | 2.1 |
| (Number of steps =) 14 | A1 | 1.1 |
| [5] |
Notes
M1: Sum of a GP implied. Allow any \(r\), eg \(r = 0.05\)
A1: Correct equation
A1: Allow 0.487 or 0.488
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 \(b \gt 0\))
A1: cao. Allow \(n = 14\). Allow 14.0. Allow \(\approx 14\)
Alternative method
| Scheme | Marks |
|---|---|
| Sum of GP implied | M1 |
| \(20 + 20 \times r + 20 \times r^2 + \ldots\) | M1 |
| \(20 + 20 \times 0.95 + 20 \times 0.95^2 + \ldots + 20 \times 0.95^{13}\) | A1 |
| \(= 205\) (3 sf) | A1 |
| Number of steps = 14 | A1 |
M1: Attempt add \(\geqslant 10\) terms. Allow any value of \(r\) for this mark
A1: Correct 14 terms added
NB Unsupported correct answer: SC B3
Alternative (incorrect) methods using \(r = 1.05\), or \(\frac{1}{0.95}\) or \(\frac{20}{19}\)
(For info’ only: \(r = \dfrac{1}{0.95}\) or \(\dfrac{20}{19}\) comes from misinterpreting “lowest” to mean “shortest”)
| Scheme | Marks |
|---|---|
| \(20 + 20 \times r + 20 \times r^2 + \ldots\) or \(20 \times \dfrac{1 - r^n}{1 - r}\) | M1 |
| \(20 \times \dfrac{1 - \left(\frac{1}{0.95}\right)^n}{1 - \frac{1}{0.95}} = 205\) or \(20 \times \dfrac{1 - \left(\frac{20}{19}\right)^n}{1 - \frac{20}{19}} = 205\) | A1 |
| \(\left(\dfrac{20}{19}\right)^n = \dfrac{117}{76}\) or \(1.05^n = 1.51\) or 1.54 | A1 |
| \(n = \dfrac{\ln\frac{117}{76}}{\ln\frac{20}{19}}\) or \(\ln_{1.053} 1.539\) or \(\ln_{1.05} 1.51\) | M1 |
| Number of steps = 8 or 9 | A0 |
M1: Allow any value of \(r\) for this mark
A1: oe using 1.05. Correct equation
M1: oe, eg \(\dfrac{\ln 1.539}{\ln 1.053}\) or \(\dfrac{\ln 1.51}{\ln 1.05}\) ft their “\(\frac{117}{76}\)”
Can be implied by their answer