Ratio & Basic Proportion: WJEC GCSE Higher June 2023 Unit 1 Q12

Higher June 2023 Unit 1 Q12

12. A sphere has a diameter of 6 cm.
A cone has a base radius of 10 cm and a height of 9 cm.

Diagrams not drawn to scale

Find the ratio of the volume of the sphere to the volume of the cone.
Give your answer in its simplest form. [4]

Volume of the sphere : Volume of the cone
= …………………… : ……………………

From the formula list: Volume of sphere \(= \dfrac{4}{3}\pi r^3\); Volume of cone \(= \dfrac{1}{3}\pi r^2 h\)

Mark scheme

Scheme	Marks	Comments
(Volume of sphere =) \(36\pi\) (cm³)	B1	Allow 113·04 (from using \(\pi\) = 3·14).
(Volume of cone =) \(300\pi\) (cm³)	B1	Allow 942 (from using \(\pi\) = 3·14). Accept absence of \(\pi\) (in both expressions). If neither volume has been correctly evaluated, award SC1 for at least one correct volume calculation: \(\frac{4}{3} \times \pi \times 3^3\) OR \(\frac{1}{3} \times \pi \times 10^2 \times 9\) (allowing the use of \(\pi\) = 3·14). Calculations may be seen in stages.
3 : 25	B2	B1 for partially simplified ratio in the form a : b where a, b are integers (with \(\pi\) eliminated and fractions or decimals cleared) or B1 for \(3\pi : 25\pi\). FT ‘their volumes’ if possible (provided at least B1 or SC1 already awarded). An answer of 24 : 25 (from \(288\pi : 300\pi\), using 6 cm for radius of sphere) is awarded B1 FT B2.

Scheme	Marks	Comments
\(\frac{4}{3} \times \pi \times 3^3\) OR \(\frac{1}{3} \times \pi \times 10^2 \times 9\) or equivalent	B1	Calculation for at least one volume (allowing the use of \(\pi\) = 3·14). Calculations may be seen in stages. Accept absence of \(\pi\) (in both expressions).
\(4 \times \pi \times 3^3 \ :\ \pi \times 10^2 \times 9\) or equivalent OR \(\frac{4}{3} \times 3^3 \ :\ \frac{1}{3} \times 10^2 \times 9\) or equivalent	B1	Both calculations correct AND one further step (clearing fractions OR eliminating \(\pi\)). Ratio may be implied.
3 : 25	B2	B1B0 for partially simplified ratio in the form a : b where a, b are integers (with \(\pi\) eliminated and fractions or decimals cleared) or B1B0 for \(3\pi : 25\pi\) or B0B1 for one error in calculation of integers followed by correct simplification e.g. 12:300 = 1:25. FT ‘their volumes’ if possible (provided at least B1 already awarded). An answer of 24 : 25 (from \(288\pi : 300\pi\), using 6 cm for radius of sphere) is awarded B0 B1 FT B2.