Volume & Surface Areas Questions and Answers

Answer: D

The minimum volume is \(8r^{3}\).

The smallest cube that can hold a sphere of radius \(r\) is the one the sphere just touches on all six faces, so the cube’s edge equals the sphere’s diameter, \(2r\). Its volume is

\[ (2r)^{3} = 8r^{3} \]

The options \(r^{3}\) and \(r^{3}/8\) come from taking the edge as \(r\) or \(r/2\), which are too small to hold the sphere: the sphere’s own volume, \(\tfrac{4}{3}\pi r^{3} \approx 4.19r^{3}\), is already larger than either. \(2r^{3}\) is also smaller than the sphere itself, so it cannot be a box that encloses it.

Answer: A

The ratio is \(1/\pi\).

Joining the two 54 cm edges makes a tube 54 cm long whose circumference is the 4 cm edge, so its radius is \(r = 4/2\pi = 2/\pi\) cm:

\[ \begin{aligned} V_{tube} &= \pi r^{2} h \\ &= \pi \cdot \frac{4}{\pi^{2}} \cdot 54 = \frac{216}{\pi} \end{aligned} \]

The sheet's area is \(54 \times 4 = 216\ \text{cm}^{2}\). A cube with that surface area has \(6a^{2} = 216\), so \(a = 6\) cm and \(V_{cube} = 216\ \text{cm}^{3}\). The ratio is \(\frac{216/\pi}{216} = \frac{1}{\pi}\).

The trap is to roll the sheet the other way (circumference 54 cm, length 4 cm), which gives a far larger tube; the stem joins the longer edges, so the circumference is the short side. The other ratios do not follow from either reading.

Answer: D

The surface area is 96 square units.

Sixty-four unit blocks make a \(4 \times 4 \times 4\) cube with surface area \(6 \times 4^{2} = 96\).

Removing a corner block takes away 3 unit squares of the outer surface (the block's three exposed faces) but exposes 3 new unit squares (the faces of the three neighbouring blocks that it was touching). The net change at each corner is zero, so after removing all eight corners the area is still 96.

72 comes from subtracting the 3 removed squares at each corner (\(96 - 24 = 72\)) and forgetting the newly exposed faces; 64 is the number of blocks, not an area, and 56 has no consistent derivation.

Answer: C

The ratio of the volumes is 15:16.

A cone holds one third of the cylinder on the same base and height: \(V = \tfrac{1}{3}\pi r^2 h\) for the cone and \(V = \pi r^2 h\) for the cylinder.

\[ \begin{aligned} \text{cone} &= \tfrac{1}{3}\pi \times 3^2 \times 5 = 15\pi \\ \text{cylinder} &= \pi \times 2^2 \times 4 = 16\pi \end{aligned} \]

So the ratio is \(15\pi : 16\pi = 15 : 16\).

45:16 is the answer if the one third in the cone formula is forgotten. 15:8 halves the cylinder by mistake, and 1:3 is the cone-to-cylinder ratio only when both share the same base and height, which these do not.

Answer: B

A hexagonal pyramid has 7 vertices, 12 edges and 7 faces.

The base is a hexagon with 6 vertices and 6 edges, and the apex adds 1 vertex joined to each base vertex by 6 slant edges. The faces are the base and 6 triangular sides:

\[ \begin{aligned} V &= 6 + 1 = 7 \\ E &= 6 + 6 = 12 \\ F &= 1 + 6 = 7 \end{aligned} \]

Euler's formula checks it: \(V - E + F = 7 - 12 + 7 = 2\). The other options fail this check or miscount the base (6 faces leaves out the base, 6 vertices leaves out the apex).

Answer: B

The embankment is 5 meter high.

The earth dug out is a cylinder of radius 7 m and depth 15 m. It is spread as a ring from radius 7 m to radius 7 + 7 = 14 m, so its volume equals the area of that ring times the height \(h\):

\[ \begin{aligned} \pi (7^{2})(15) &= \pi (14^{2} - 7^{2}) h \\ 735 &= 147h \\ h &= 5 \text{ m} \end{aligned} \]

Using the full circle of radius 14 m instead of the ring would give \(735/196 = 3.75\) m, which is not offered; the well itself is not covered by the embankment.

Answer: D

The correct relation is 3h = 2d.

Set the cylinder's volume equal to the sphere's, with the same radius \(r\):

\[ \begin{aligned} \pi r^{2} h &= \tfrac{4}{3}\pi r^{3} \\ h &= \tfrac{4}{3} r = \tfrac{2}{3} d \end{aligned} \]

since \(d = 2r\). Multiplying by 3 gives \(3h = 2d\). Option C reverses the ratio, and \(h = d\) or \(2h = d\) would make the cylinder too tall or too short to match the sphere's volume.

Answer: B

The volume is 48 cm³.

Let the edges be \(x\), \(2x\) and \(3x\). The surface area is twice the sum of the three face areas:

\[ \begin{aligned} 2(2x^{2} + 6x^{2} + 3x^{2}) &= 22x^{2} = 88 \\ x^{2} = 4 &\Rightarrow x = 2 \end{aligned} \]

The edges are 2, 4 and 6 cm, so the volume is \(2 \times 4 \times 6 = 48\) cm³. 24 cm³ is half of it, as if the surface area had been taken without the factor 2.

Answer: A

The side of the new cube is 12 cm.

Melting and recasting keeps the total volume:

\[ \begin{aligned} 6^{3} + 8^{3} + 10^{3} &= 216 + 512 + 1000 \\ &= 1728 = 12^{3} \end{aligned} \]

Adding the edges (6 + 8 + 10 = 24) is the trap that gives the 24 option; volumes add, edges do not. 11 and 13 cm give volumes of 1331 and 2197 cm³.

Answer: D

The key's answer is 900 cubic centimetre.

The volume of a cylinder is \(\pi r^{2} h\), so it varies as the square of the radius when the height is unchanged. The key reads 'increased by two times of the original radius' as an increase of 2r, making the new radius 3r:

\[ \begin{aligned} V' &= 100 \times 3^{2} \\ &= 900\ \text{cm}^{3} \end{aligned} \]

If the radius were simply doubled, the volume would be \(100 \times 2^{2} = 400\) cm³, which is not offered. 200 and 600 come from treating the volume as proportional to the radius rather than its square.