Volume & Surface Areas Questions and Answers

Answer: D

A solid hemisphere has two surfaces: the curved dome and the flat circular base.

\[ \text{curved surface} = 2\pi r^{2}, \qquad \text{flat base} = \pi r^{2} \]

\[ \text{total surface area} = 2\pi r^{2} + \pi r^{2} = 3\pi r^{2} \]

The \( 2\pi r^{2} \) option is the curved surface alone - correct for a hollow bowl, but not for a solid hemisphere, which still has its flat face.

Answer: D

The cubes must fit a whole number of times along each edge, so their edge length must divide 6, 12 and 15. To make the fewest cubes, take the largest such edge - the highest common factor.

\[ \text{HCF}(6, 12, 15) = 3 \]

\[ n = \frac{6}{3} \times \frac{12}{3} \times \frac{15}{3} = 2 \times 4 \times 5 = 40 \]

Equivalently \( \tfrac{6 \times 12 \times 15}{3^{3}} = \tfrac{1080}{27} = 40 \). Fewest cubes means largest cubes - the two requirements pull in opposite directions.

Answer: B

With the same base radius and the same height:

\[ V_{\text{cylinder}} = \pi r^{2} h, \qquad V_{\text{cone}} = \frac{1}{3}\pi r^{2} h \]

\[ \frac{V_{\text{cylinder}}}{V_{\text{cone}}} = \frac{\pi r^{2} h}{\tfrac{1}{3}\pi r^{2} h} = 3 \quad\Longrightarrow\quad 3 : 1 \]

Three coneloads of water exactly fill the matching cylinder - the standard classroom demonstration of where the \( \tfrac{1}{3} \) comes from.

Answer: B

The wood occupies the space between the outer box and the hollow inside it, so find both volumes and subtract.

The wood is 4 cm thick on every face, so each external dimension exceeds the internal one by \( 2 \times 4 = 8 \) cm:

\[ 88 \times 56 \times 28 \text{ cm} \]

\[ V_{\text{external}} = 88 \times 56 \times 28 = 137{,}984 \text{ cm}^{3} \]

\[ V_{\text{internal}} = 80 \times 48 \times 20 = 76{,}800 \text{ cm}^{3} \]

\[ V_{\text{wood}} = 137{,}984 - 76{,}800 = 61{,}184 \text{ cm}^{3} \]

Adding 4 cm rather than 8 cm to each dimension is the standard error - the thickness is present on both opposite faces.

Answer: D

The men displace their own volume of water, and that volume spreads over the pool's whole surface.

\[ V_{\text{displaced}} = 150 \times 8 = 1200 \text{ m}^{3} \]

\[ \text{pool area} = 90 \times 40 = 3600 \text{ m}^{2} \]

\[ \text{rise} = \frac{1200}{3600} = \frac{1}{3} \text{ m} = 33.33 \text{ cm} \]

The answer is asked for in centimetres, so the metre figure must be multiplied by 100 - the whole point of the awkward-looking 33.33.

Answer: C

Find the radius from the surface area first, taking \( \pi = \tfrac{22}{7} \).

\[ 4\pi r^{2} = \frac{792}{7} \quad\Longrightarrow\quad r^{2} = \frac{792}{7} \times \frac{7}{4 \times 22} = \frac{792}{88} = 9 \]

\[ r = 3 \text{ cm} \]

\[ V = \frac{4}{3}\pi r^{3} = \frac{4}{3} \times \frac{22}{7} \times 27 = \frac{792}{7} \text{ cm}^{3} \]

The numerical coincidence is real, not a misprint: for \( r = 3 \) the surface area and volume come out to the same number, because \( \tfrac{V}{S} = \tfrac{r}{3} = 1 \).

Answer: D

Find the radius from the volume, then use it for the surface area.

\[ \frac{4}{3}\pi r^{3} = 850 \quad\Longrightarrow\quad r^{3} = \frac{850 \times 3}{4\pi} = \frac{2550}{12.566} = 202.9 \]

\[ r = \sqrt[3]{202.9} = 5.88 \text{ m} \]

\[ S = 4\pi r^{2} = 4 \times 3.1416 \times (5.88)^{2} = 12.566 \times 34.57 \]

\[ S \approx 434 \text{ m}^{2} \]

Only the pair 5.88 m and 434 m² is internally consistent - checking one value against the other is quicker here than computing both from scratch.

Answer: C

Convert the diameter to centimetres first, then halve it.

\[ 45 \text{ mm} = 4.5 \text{ cm} \quad\Longrightarrow\quad r = 2.25 \text{ cm} \]

\[ \text{TSA} = 2\pi r (r + h) = 2 \times 3.142 \times 2.25 \times (2.25 + 12) \]

\[ = 14.139 \times 14.25 = 201.5 \approx 201 \text{ cm}^{2} \]

Both flat ends count, since the whole cylinder is carved; the curved surface alone would come to about 169.7 cm².

Answer: B

With faces \( P = lb \), \( Q = bh \), \( R = hl \), the product of all three is the square of the volume:

\[ PQR = (lbh)^{2} \quad\Longrightarrow\quad V = \sqrt{PQR} \]

\[ V = \sqrt{120 \times 72 \times 60} = \sqrt{518400} = 720 \text{ cm}^{3} \]

Forgetting the square root leaves 518,400; the dimensions themselves work out to 10, 12 and 6 cm, which do reproduce all three given face areas.

Answer: C

Melting conserves volume, so divide the target cylinder's volume by one coin's. Halve both diameters first.

\[ V_{\text{coin}} = \pi (0.75)^{2}(0.2) = 0.1125\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi (2.25)^{2}(10) = 50.625\pi \text{ cm}^{3} \]

\[ n = \frac{50.625\pi}{0.1125\pi} = \frac{50.625}{0.1125} = 450 \]

Using the diameters directly instead of the radii leaves the ratio unchanged only because both are squared - but the thickness is not, so it is safer to convert first.