Intermediate November 2017 Unit 1 Q17
17. A cylinder just fits inside a hollow cube with sides of length \(m\) cm.
Diagram not drawn to scale
The radius of the cylinder is \(\dfrac{m}{2}\) cm.
The height of the cylinder is \(m\) cm.
The ratio of the volume of the cube to the volume of the cylinder is given by
volume of cube : volume of cylinder
\(= k : \pi\),
where \(k\) is a number.
Find the value of \(k\).
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of cube =) m3 OR m × m × m OR m2 × m | B1 | For sight of m3 or equivalent. |
| (Volume of cylinder =) \(\dfrac{\pi m^3}{4}\) OR \(\dfrac{\pi \times m \times m \times m}{4}\) OR \(\dfrac{\pi \times m^2 \times m}{4}\) | B2 | For sight of πm3/4 or equivalent. B1 for \(\pi \times \left(\dfrac{m}{2}\right)^2 \times m\). Also allow this B1 if brackets are missing. \(m^3 : \dfrac{\pi m^3}{4}\) OR \(4m^3 : \pi m^3\) OR \(1 : \dfrac{\pi}{4}\) all imply B1B2. |
| k = 4 | B1 | Allow B1 if left as 4 : π . F.T only for πm3 / 2 (giving k = 2 or 2 : π) Note : If a value is used for m then mark as above and penalise −1 from total mark gained. |
Corrected by Westie's Workshop
- The board's mark scheme printed “(Volume of cylinder =) = πm³/4”. Changed to “(Volume of cylinder =) πm³/4” because the equals sign is printed twice.
Original: see the board's mark scheme PDF (link above).