Numeracy Higher November 2016 Unit 2 Q9
9. A metal round-headed nail can be thought of as a cone sitting on top of a cylinder, which sits on top of a hemisphere.
A company produces round-headed nails of different sizes, but made of the same metal.
Each nail has the following dimensions:
- height of cone = \(9r\),
- height of cylinder = \(15r\),
- radius of the hemisphere = \(12r\),
where \(r\) is the radius of the cylinder and the base radius of the cone.

Diagram not drawn to scale
A metal cuboid of volume 18 000 mm3 is melted down, and re-cast to form round-headed nails of size A, where \(r = 0{\cdot}4\) mm.
From the formula list: Volume of sphere \(= \dfrac{4}{3}\pi r^3\); Volume of cone \(= \dfrac{1}{3}\pi r^2 h\)
| STATEMENT | ||
|---|---|---|
| A nail double the height of a size A nail will have a total height of 28·8 mm. | TRUE | FALSE |
| A nail double the height of a size A nail will be 8 times the weight of a size A nail. | TRUE | FALSE |
| A nail 3 times the height of a size A nail will have a total surface area 6 times that of a size A nail. | TRUE | FALSE |
| When \(r = 0{\cdot}8\) mm, the number of nails that could be produced from the same metal cuboid will be double the number of size A nails. | TRUE | FALSE |
| Scheme | Marks | Comments |
|---|---|---|
| Volume of hemisphere = ⅔ × π × 4·83 OR ⅔ × π × (12r)3 | B1 | Allow for sight of 231·5 to 231·7 |
| (Tot vol=) ⅔ π×4·83 + π×0·42×6 + ⅓π ×0·42×3·6 ( 231·… + 3·0… + 0·6…) | M2 | M1 for summing 3 terms with any 2 correct, OR M1 for use of r in a correct method |
| = 235 to 235·3 (mm3) | A1 | OR 74·88π. FT from M1 |
| No. of nails = 18 000 ÷ 235(·3) or equivalent | M1 | Allow M1 for 18 000 ÷ ‘their 235(·3)’ provided previous M1 at least awarded |
| = 76 | A1 | Needs to be a whole number, and be the result of rounding down their answer |
| Scheme | Marks | Comments |
|---|---|---|
| True True False False | B2 | B1 for 3 correct |