Numeracy Higher November 2016 Unit 1 Q10
10. The shaded part of the diagram below shows the top surface of an engine part.
Diagram not drawn to scale
The measurements taken by a motor engineer are:
- reflex angle \(B\hat{O}C = 240^\circ\),
- AO = OD = 6 cm,
- BO = OC = 3 cm.
The length of the major arc AD is to be sealed by attaching a flexible anti-rust strip.
Each flexible anti-rust strip is of length 35 cm.
What length of the anti-rust strip will be left over after sealing the length of the major arc AD?
Give your answer in terms of \(\pi\), in its simplest form. [3]
Length of anti-rust strip left over = …………… cm
The top surface of the engine part is to be painted.
The paint costs 15p per cm2.
Calculate the cost of the paint to be used.
Give your answer in terms of \(\pi\), in its simplest form. [4]
Using \(\pi = 3\), calculate how much it costs to paint the top surface of 20 engine parts.
Give your answer in pounds. [1]
Paint cost is £ ……………
| Scheme | Marks | Comments |
|---|---|---|
| \(35 - \dfrac{240}{360} \times \pi \times 2 \times 6\) or equivalent | M2 | M1 for sight of \(\dfrac{240}{360} \times \pi \times 2 \times 6\) or equivalent |
| 35 - 8π (cm) | A1 | Ignore attempts to substitute a value for π into this expression Do not ignore further incorrect simplification of the expression If no marks allow SC1 for an answer of 35 - 4π (cm) from sight of \(\dfrac{240}{360} \times \pi \times 6\) or equivalent |
| Scheme | Marks | Comments |
|---|---|---|
| (i) Sight of \(\dfrac{240}{360} \times \pi \times 6^2\) OR \(\dfrac{240}{360} \times \pi \times 3^2\) | B1 | Or equivalent expressions |
| Area of region \(\dfrac{240}{360} \times \pi \times (6^2 - 3^2)\) or equivalent | M1 | Accept shown in stages |
| = 18π (cm2) | A1 | |
| Cost of paint is 270π (p) or £ 2.7π | A1 | FT 15 × ‘their 18π’ provided their answer is a multiple of π, and in its simplest form If units are given they must be correct |
| (ii) (£)162 | B1 | FT ‘their 2.7’ or ‘their 270’ provided M1 previously awarded |