Numeracy Higher November 2023 Unit 2 Q6
6. A solid statue is made from an alloy of copper and tin.
It has a volume of 150 cm\(^3\).
(a)
In the statue, the volume of copper and the volume of tin are in the ratio 22 : 3.
The density of copper is 8·96 g/cm\(^3\).
The density of tin is 7·31 g/cm\(^3\).
Calculate the mass of the statue.
You must show all your working. [4]
(b)
The height of the statue is 12 cm.
A larger statue is mathematically similar to this statue.
It has a height of 21·6 cm.
Calculate the volume of this larger statue. [3]
| Scheme | Marks | Comments |
|---|---|---|
| 150 ÷ (22+3) × 22 OR 150 ÷ (22+3) × 3 | M1 | |
| (Volume of copper =) 132 (cm\(^3\)) AND (Volume of tin =) 18 (cm\(^3\)) | A1 | May be implied in further working |
| (Mass of statue =) 132×8.96 + 18×7.31 (1182.72 + 131.58) | m1 | Allow m1, but A0, for use of rounded or truncated values of 8.96 and 7.31 FT ‘their 132’ and ‘their 18’ |
| (Mass of statue =) 1314(.3) (g) | A1 | ISW FT provided one of their volumes is correct. Accept 1.3(143) kg from sight of 1314(.3) (g) or 1.314(3) (kg) |
Alternative
6(a) Alternative method calculating mass directly:
| Scheme | Marks | Comments |
|---|---|---|
| (Mass of statue =) 150÷(22+3)×22×8.96 + 150÷(22+3)×3×7.31 | M2 | Allow M2, and possible A1 only, for use of rounded or truncated values of 8.96 and 7.31 M1 for 150÷(22+3)×22×8.96 (=1182.72) OR 150÷(22+3)×3×7.31 (=131.58) |
| (Mass of statue =) 1314(.3) (g) | A2 | ISW Accept 1.3(143) kg from sight of 1314(.3) (g) or 1.314(3) (kg) Award A1 for
|
| Scheme | Marks | Comments |
|---|---|---|
| (Volume factor =) \(\left(\dfrac{21.6}{12}\right)^3\) OR \(\left(\dfrac{12}{21.6}\right)^3\) or 1.8\(^3\) OR 0.555…\(^3\) (=5.832) (=0.171…) | B1 | or \(\left(\dfrac{9}{5}\right)^3\) OR \(\left(\dfrac{5}{9}\right)^3\) |
| (Volume of bigger statue =) 150 × \(\left(\dfrac{21.6}{12}\right)^3\) OR 150 ÷ \(\left(\dfrac{12}{21.6}\right)^3\) | M1 | Implies the previous B1 |
| = 874(.8) or 875 (cm\(^3\)) | A1 |