Volume & Surface Areas Questions and Answers

Answer: A

A cone's lateral surface is what the cardboard has to supply, so find the radius first.

\[ r = \sqrt{l^{2} - h^{2}} = \sqrt{25^{2} - 24^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7 \text{ cm} \]

\[ \text{CSA} = \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^{2} \]

\[ \text{sheet area} = 8 \times 550 = 4400 \text{ cm}^{2} \]

A cone rolled from a sheet has no base, so only the curved surface counts - adding \( \pi r^{2} \) would overstate each cone by 154 cm².

Answer: D

Convert the diameter to metres before anything else, since the rate is per square metre.

\[ d = 40 \text{ cm} = 0.4 \text{ m} \quad\Longrightarrow\quad r = 0.2 \text{ m} \]

\[ \text{CSA} = 2\pi r h = 2 \times \frac{22}{7} \times 0.2 \times 7 = 8.8 \text{ m}^{2} \]

\[ \text{cost} = 8.8 \times 20 = \text{Rs. } 176 \]

Only the curved surface is polished, so the two circular ends are excluded; including them would add \( 2\pi r^{2} = 0.251 \) m².

Answer: C

Volume of a cylinder is \( \pi r^{2}h \), so form the ratio with the \( \pi \) cancelling.

\[ \frac{V_{1}}{V_{2}} = \frac{r_{1}^{2}h_{1}}{r_{2}^{2}h_{2}} = \frac{4^{2} \times 21}{7^{2} \times 8} \]

\[ = \frac{16 \times 21}{49 \times 8} = \frac{336}{392} = \frac{6}{7} \]

So the ratio is \( 6 : 7 \). The radii enter squared and the heights only to the first power - treating both alike gives \( 84 : 56 = 3 : 2 \), which is the last distractor.

Answer: B

The surface area of a cube is \( 6a^{2} \), so it scales with the square of the edge.

\[ a' = 0.79a \quad\Longrightarrow\quad \frac{S'}{S} = (0.79)^{2} = 0.6241 \]

\[ \text{decrease} = 1 - 0.6241 = 0.3759 = 37.59\% \]

The drop is not 42% (twice 21%): the correction term \( \left(\tfrac{21}{100}\right)^{2} = 4.41\% \) must be added back, giving \( 42 - 4.41 = 37.59\% \).

Answer: A

Compare volumes, since the box dimensions divide the room's cleanly.

\[ V_{\text{box}} = 4 \times 3 \times 2 = 24 \text{ cm}^{3} \]

\[ n = \frac{4080}{24} = 170 \]

A volume division only gives an upper bound in general, because packing can waste space; here it is achievable, since 24 divides 4080 exactly and the boxes can be laid in aligned layers.

Answer: B

Write the claim as an equation and solve for the constant \( c \).

\[ V = c \times S \times r \]

\[ \frac{4}{3}\pi r^{3} = c \times 4\pi r^{2} \times r = 4c\pi r^{3} \]

\[ \frac{4}{3} = 4c \quad\Longrightarrow\quad c = \frac{1}{3} \]

The same one-third appears in the cone and pyramid formulas, and for the same reason: a solid built of thin pyramids on a surface has volume \( \tfrac{1}{3} \times \text{surface} \times \text{height} \).

Answer: B

Area of a circle is \( \pi r^{2} \), so a 6% rise in the radius acts through the square.

\[ \frac{A'}{A} = \left(\frac{1.06r}{r}\right)^{2} = (1.06)^{2} = 1.1236 \]

\[ \text{increase} = 1.1236 - 1 = 0.1236 = 12.36\% \]

Doubling the percentage to 12% misses the square term \( \left(\tfrac{6}{100}\right)^{2} = 0.36\% \). The exact rule is \( 2x + \tfrac{x^{2}}{100} \).

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: D

A shuttlecock has a rounded cork base and a skirt of feathers that flares outward from it.

The cork nose is a hemisphere. The feather skirt is narrow where it meets the cork and wide at the open end, with both ends circular and neither closing to a point - that is a frustum of a cone.

A full cone would come to a point at the top, and a cylinder would keep the same width all the way up; neither matches the flare.

Answer: B

For a frustum the slant height spans the difference of the radii horizontally and the height vertically.

\[ l = \sqrt{h^{2} + (R - r)^{2}} \]

\[ l = \sqrt{4^{2} + (5-2)^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm} \]

It is the difference of the radii, not their sum, that forms the horizontal leg - using \( 5+2 \) would give \( \sqrt{65} \), the third option.