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: D

The ratio of the volumes is 5 : 3.

The volume of a cylinder is \(\pi r^{2} h\), so the ratio of volumes is the square of the radius ratio times the height ratio:

\[ \begin{aligned} \frac{V_{1}}{V_{2}} &= \left(\frac{5}{3}\right)^{2} \times \frac{3}{5} \\ &= \frac{25}{9} \times \frac{3}{5} = \frac{5}{3} \end{aligned} \]

9 : 25 is the squared radius ratio upside down, and 5 : 5 or 3 : 3 (that is 1 : 1) is what you get by forgetting to square the radii.

Answer: D

The ratio of the curved surface areas is 7 : 3.

The curved surface area of a cylinder is \(2\pi r h\), so the ratio is the product of the radius ratio and the height ratio:

\[ \begin{aligned} \frac{S_{1}}{S_{2}} &= \frac{4}{3} \times \frac{7}{4} \\ &= \frac{28}{12} = \frac{7}{3} \end{aligned} \]

3 : 7 is the same ratio inverted. 5 : 3 and 3 : 5 do not come from these figures; squaring the radius ratio, as for volume, would also be a mistake here.

Answer: C

The number of bricks is 1,00,000.

Divide the volume of the wall by the volume of one brick, using the same units:

\[ \begin{aligned} \text{wall} &= 60 \times 3 \times 5 = 900\ \text{m}^{3} \\ \text{brick} &= 0.30 \times 0.15 \times 0.20 = 0.009\ \text{m}^{3} \\ \text{bricks} &= \frac{900}{0.009} = 100000 \end{aligned} \]

The other options are 25 to 75 percent more or less than this and come from a slip in converting centimetres to metres.

Answer: A

The ratio of the volumes is 1 : 2 : 3.

With a common radius \(r\) and the heights of the cone and cylinder equal to \(r\) (the hemisphere's height is \(r\) anyway):

\[ \begin{aligned} \text{cone} &= \tfrac{1}{3}\pi r^{3} \\ \text{hemisphere} &= \tfrac{2}{3}\pi r^{3} \\ \text{cylinder} &= \pi r^{3} \end{aligned} \]

Multiplying by 3 gives 1 : 2 : 3, Archimedes' classic result. The ratios 1 : 2 : 4, 1 : 2 : 5 and 1 : 2 : 6 overstate the cylinder, whose volume is exactly three times the cone's.