Higher June 2019 Paper 2 Q7
7.
(a) The volume of a sphere with a radius of 2·7 cm is equal to the volume of a cuboid.
The base of the cuboid has an area of 14·2 cm2.
Calculate the height of the cuboid. [4]
The base of the cuboid has an area of 14·2 cm2.
Calculate the height of the cuboid. [4]
(b) A piece of paper is in the shape of a circle.
The circumference of the circle is 86 cm.
The paper is cut into 2 semi-circles.
Calculate the perimeter of one of the semi-circles.
Give your answer correct to the nearest \(\frac{1}{10}\) cm. [5]
The circumference of the circle is 86 cm.
The paper is cut into 2 semi-circles.
Calculate the perimeter of one of the semi-circles.
Give your answer correct to the nearest \(\frac{1}{10}\) cm. [5]
| Scheme | Marks | Comments |
|---|---|---|
| (Volume sphere) \(4/3 \times \pi \times 2.7^3\) | M1 | (\(\approx 82.4\ldots\) cm3) |
| (Volume of cuboid) \(14.2 \times \text{height} = 4/3 \times \pi \times 2.7^3\) | M1 | FT for ‘their volume of sphere’ |
| \(\text{Height} = 4/3 \times \pi \times 2.7^3 \div 14.2\) | m1 | FT for ‘their volume of sphere’ |
| Answer in the range 5.8 to 5.81 (cm) | A1 | CAO |
Corrected by Westie's Workshop
- The board's mark scheme printed “(cm³)”. Changed to “(cm)” because the answer is a height, which is a length, so its unit is cm not cm³.
Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| \(86 = \pi \times \text{Diameter}\), or \(86 = 2 \times \pi \times \text{Radius}\) | M1 | |
| \(\text{Diameter} = 86 \div \pi\) or \(\text{Radius} = 86 \div 2\pi\) | m1 | (Diameter = 27.37 to 27.4 cm) (Radius = 13.68 to 13.7 cm) |
| Perimeter semi-circle \(86 \div 2 + \text{Diameter}\), or \(86 \div 2 + 2 \times \text{Radius}\) | M1 | FT ‘their derived diameter’ or ‘their derived 2 × radius’, independent of previous marks |
| 70.4 (cm) or 704 mm | A2 | CAO. For A2 if units are given they must be correct A1 for a correct answer given to the wrong level of accuracy (70.37 to 70.39 or 70(cm)). |
| (9) |