Numeracy Higher June 2023 Unit 1 Q11
11. Safety Cones is a company that makes traffic cones.
Its traffic cones have a base in the shape of an octagonal prism, with a hollow cone shape sitting on top of the base.
The bases of the traffic cones are symmetrical and have a circular hole in them to allow the traffic cones to be stacked.
The dimensions of the octagonal base are shown below.
The height of this octagonal base is 5 cm.
Diagram not drawn to scale
The cone that sits on top of the octagonal base is shown below.
It has a height of 60 cm and a circular base area of 800 cm2.
Diagram not drawn to scale
Gary is transporting a stack of 10 of these traffic cones in the back of his van.
Diagram not drawn to scale
Calculate the volume of the space that this stack of 10 traffic cones takes up in the back of Gary’s van.
Give your answer in cm3.
You should assume that the space taken up includes the space underneath the bottom traffic cone and any gaps between the cones. [7]
From the formula list: Area of trapezium \(= \dfrac{1}{2}(a+b)h\); Volume of prism = area of cross-section × length; Volume of cone \(= \dfrac{1}{3}\pi r^2 h\)
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of octagonal prism =) 50 × 20 + 2 × \(\dfrac{1}{2}\) × (50 + 20) × ((50 – 20)/2) or 50 × 20 + 2 × \(\dfrac{1}{2}\) × 70 × 15 (= 1000 + 1050) | M2 | May be embedded within a volume calculation 50 × 20 may be seen as 20×20 + 2×20×((50 – 20)/2) or 20×20 + 2×20×15 2 × \(\dfrac{1}{2}\) × (50 + 20) × ((50 – 20)/2) may be seen as 2 × (20×((50 – 20)/2) + 2×\(\dfrac{1}{2}\)×(50–20)/2×(50–20)/2) or 2 × (20×15 + 2×\(\dfrac{1}{2}\)×15×15) M1 for 50 × 20 + \(\dfrac{1}{2}\) × (50 + 20) × ((50 – 20)/2) or M1 for 50 × 20 + \(\dfrac{1}{2}\) × 70 × 15 OR M1 for ……… + 2 × \(\dfrac{1}{2}\) × (50 + 20) × ((50 – 20)/2) or M1 for ……… + 2 × \(\dfrac{1}{2}\) × 70 × 15 |
| × 5 × 10 = 102 500 (cm3) | m1 A1 | FT from M1 CAO If M0m0A0 awarded, and from an incorrect assumption that the octagon is regular, award SC1 for areas of:
|
| (Volume of a cone =) \(\dfrac{1}{3}\) × 800 × 60 | M1 | Award M1A0 if this is multiplied by 10 |
| = 16 000 (cm3) | A1 | CAO |
| (Total volume = 102 500 + 16 000 =) 118 500 (cm3) | B1 | FT ‘their 102 500’ and ‘their 16 000’ provided at least one M1 mark awarded and is the sum of the volumes of their 10 octagonal bases and 1 (or 10) cone/s |
Alternative method for the first 4 marks
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of octagonal prism =) 502 – 4 × \(\dfrac{1}{2}\) × (50 – 20)/2 × (50 – 20)/2 or 502 – 4 × \(\dfrac{1}{2}\) × 15 × 15 (= 2500 – 450) | M2 | May be embedded within a volume calculation M1 for … – 4 × \(\dfrac{1}{2}\) × (50 – 20)/2 × (50 – 20)/2 or M1 for … – 4 × \(\dfrac{1}{2}\) × 15 × 15 |
| × 5 × 10 = 102 500 (cm3) | m1 A1 | FT from M1 CAO |