Volume & Surface Areas Questions and Answers

Answer: C

The pipe weighs 3.696 kg.

The external radius is 4 cm and, with a 1 cm wall, the internal radius is 3 cm. The volume of iron is the difference of the two cylinders:

\[ \begin{aligned} V &= \pi(4^{2} - 3^{2}) \times 21 \\ &= \tfrac{22}{7} \times 7 \times 21 = 462\ \text{cm}^{3} \\ \text{mass} &= 462 \times 8 = 3696\ \text{g} \end{aligned} \]

3696 g is 3.696 kg. 36.96 kg misplaces the decimal point. A solid rod of the same size would weigh \(\tfrac{22}{7} \times 16 \times 21 \times 8\) g, about 8.4 kg, so forgetting the hollow core does not give the other options either.

Answer: A

The volume is 125 cm³.

A cube has 6 equal square faces, so each face has area \(150 / 6 = 25\) cm² and the edge is 5 cm:

\[ V = 5^{3} = 125\ \text{cm}^{3} \]

216 and 343 are the cubes of 6 and 7, the volumes of cubes with surface areas 216 and 294 cm². 164 is not a perfect cube, so it cannot be the volume of a cube with a whole-number edge.

Answer: C

The longest pole is \(10\sqrt{2}\) m.

The longest straight pole in a box-shaped hall runs along the space diagonal, from one bottom corner to the opposite top corner:

\[ \begin{aligned} d &= \sqrt{10^{2} + 8^{2} + 6^{2}} \\ &= \sqrt{100 + 64 + 36} = \sqrt{200} \\ &= 10\sqrt{2} \approx 14.1\ \text{m} \end{aligned} \]

\(8\sqrt{2}\) m (about 11.3 m) and 11 m are shorter than this. 9 m is shorter even than the 10 m length of the floor, so it cannot be the longest pole.

Answer: A

The volume is 729 cm³.

A cube has six square faces, so its surface area is \(6a^{2}\):

\[ \begin{aligned} 6a^{2} &= 486 \\ a^{2} &= 81 \Rightarrow a = 9\ \text{cm} \\ V &= a^{3} = 9^{3} = 729\ \text{cm}^{3} \end{aligned} \]

625 is \(25^{2}\), 216 is \(6^{3}\) and 743 is not a cube at all; none comes from an edge of 9 cm.

Answer: B

The edge of the new cube is 9 cm.

No metal is lost, so the new volume is the sum of the three volumes:

\[ \begin{aligned} V &= 8^{3} + 6^{3} + 1^{3} \\ &= 512 + 216 + 1 = 729 \\ a &= \sqrt[3]{729} = 9\ \text{cm} \end{aligned} \]

Adding the edges (15 cm) is the trap; volumes add, edges do not. 8 cm is the largest original cube and must be exceeded, while 10 cm and 7 cm give volumes of 1000 and 343, not 729.

Answer: C

He needs about 71 buckets.

Fill the drum to 80% of 150 cm, that is 120 cm. Each bucket delivers only 90% of its volume.

\[ \begin{aligned} V_{drum} &= \pi(80)^{2}(120) = 768000\pi \\ V_{bucket} &= 0.9 \times \pi(20)^{2}(30) = 10800\pi \\ \frac{768000}{10800} &\approx 71.1 \end{aligned} \]

So about 71 buckets. Ignoring the spillage gives 64, and filling the drum completely gives about 89; 73 and 74 come from rounding slips.

Answer: A

Melting conserves volume, so divide the cylinder's volume by one coin's.

\[ V_{\text{coin}} = \pi (0.75)^{2}(0.2) = \pi (0.5625)(0.2) = 0.1125\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi (3)^{2}(8) = 72\pi \text{ cm}^{3} \]

\[ n = \frac{72\pi}{0.1125\pi} = \frac{72}{0.1125} = 640 \]

The \( \pi \) cancels, so no approximation is needed anywhere in the calculation.

Answer: C

The unsharpened body of the pencil is a straight cylinder of uniform circular cross-section. Sharpening tapers one end to a point while keeping that cross-section circular, which is exactly a cone.

So the shape is a cone mounted on a cylinder.

A hemisphere would end in a rounded dome rather than a point, and a frustum would be cut off flat instead of tapering to the graphite tip.

Answer: B

A cube of volume 27 m³ has edge \( \sqrt[3]{27} = 3 \) m, so it cuts into \( 3 \times 3 \times 3 \) unit cubes.

Only the cubes strictly inside - not touching any face - escape the paint. Removing one layer from each of the six faces leaves a cube of side

\[ 3 - 2 = 1 \]

\[ \text{unpainted cubes} = 1^{3} = 1 \]

The general result for an \( n \times n \times n \) cube is \( (n-2)^{3} \), which is why a 3-cube yields just the single central cube.

Answer: B

Rearrange the cuboid volume formula for the missing dimension.

\[ V = l \times b \times h \quad\Longrightarrow\quad h = \frac{V}{l \times b} \]

\[ h = \frac{450}{9 \times 8} = \frac{450}{72} = 6.25 \text{ cm} \]

Checking: \( 9 \times 8 \times 6.25 = 450 \) cm³, as required.