4MA0 Higher June 2015 Paper 3R Q19
19 The diagram shows a circular pond, of radius \(r\) metres, surrounded by a circular path.
The circular path has a constant width of 1.5 metres.
Diagram NOT accurately drawn
The area of the path is \(\dfrac{1}{10}\) the area of the pond.
Show your working clearly.
Give your answer correct to 3 significant figures. (5)
| Scheme | Marks |
|---|---|
| \(\pi(r + 1.5)^2 - \pi r^2\) (= 0.1 × \(\pi r^2\)) | M1 |
| \(r^2 + 3r + 2.25 - r^2 = 0.1r^2\) | M1 |
| \(2r^2 - 60r - 45 = 0\) | A1 |
| (3) |
Notes
M1: Correct expression for area of path (may be seen as part of an equation)
M1: ind. \(r^2 + 3r + 2.25\) or \(r^2 + 3r + 1.5^2\)
(i.e. correct expansion of brackets with or without \(\pi\))
A1: Correct algebraic steps to \(2r^2 - 60r - 45 = 0\)
| Scheme | Marks |
|---|---|
| \(\dfrac{--60 \pm \sqrt{(-60)^2 - 4 \times 2 \times -45}}{2 \times 2}\) | M1 |
| \(\dfrac{60 \pm \sqrt{3600 + 360}}{4}\) | M1 |
| 30.7(32 ..) or \(\dfrac{30 \pm 3\sqrt{110}}{2}\) | A1 |
| (Area =) \(\pi\) × “30.732..”\(^2\) (= 2967.12..) | M1 |
| 2970 | A1 |
| (5) |
Notes
M1: Condone 1 sign error; condone missing brackets around −60 ; accept \(60^2\); some evaluation may be seen
NB: allow + instead of \(\pm\)
M1: for \(\sqrt{3600 + 360}\) or \(\sqrt{3960}\)
A1: dep on at least M1 awarded
NB: Ignore −0.73…
M1: ind (ft for \(r\) (at least 3 sf)) do not award for substitution of \(r = 1.5\)
A1: for 2960 – 2970