Volume & Surface Areas Questions and Answers

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: C

Take the original dimensions as \( 1k, 2k, 3k \).

\[ V_{\text{old}} = 1k \times 2k \times 3k = 6k^{3} \]

An increase of 100% doubles a length, and 200% triples it:

\[ 1k \to 2k, \qquad 2k \to 6k, \qquad 3k \to 9k \]

\[ V_{\text{new}} = 2k \times 6k \times 9k = 108k^{3} \]

\[ V_{\text{new}} - V_{\text{old}} = 102k^{3} = 17 \times 6k^{3} \]

So the old box falls short by 17 times its own volume. Note that a 200% increase means the length becomes three times the original, not two - that single slip turns the answer into 7.

Answer: A

Same material means the same density, so weight is proportional to volume.

\[ V_{\text{solid}} = \frac{4}{3}\pi (3)^{3} = 36\pi \text{ cm}^{3} \;\longrightarrow\; 6 \text{ kg} \]

The hollow sphere has outer radius 9 cm and inner radius 6 cm:

\[ V_{\text{hollow}} = \frac{4}{3}\pi\left(9^{3} - 6^{3}\right) = \frac{4}{3}\pi (729 - 216) = \frac{4}{3}\pi (513) = 684\pi \text{ cm}^{3} \]

\[ \text{weight} = 6 \times \frac{684\pi}{36\pi} = 6 \times 19 = 114 \text{ kg} \]

The cavity must be subtracted; treating the object as a solid 18 cm sphere would give 162 kg.

Answer: A

The volume of a cuboid is the product of its three dimensions.

\[ V = l \times b \times h = 7.8 \times 3.7 \times 2.7 \]

\[ 7.8 \times 3.7 = 28.86 \]

\[ 28.86 \times 2.7 = 77.922 \text{ cm}^{3} \]

Three decimal places appear in the answer because the factors contribute one each - a quick check that the decimal point has landed correctly.

Answer: B

Find the wall area from the cost, then use the perimeter.

\[ \text{area of 4 walls} = \frac{24000}{20} = 1200 \text{ m}^{2} \]

The four walls form a band of height \( h \) around the floor's perimeter:

\[ \text{area} = \text{perimeter} \times h \quad\Longrightarrow\quad 1200 = 150h \]

\[ h = 8 \text{ m} \]

Writing the wall area as \( 2(l+b)h \) is the same thing, since the perimeter \( 2(l+b) \) is already given as 150 m - the individual length and breadth are never needed.

Answer: B

For the sphere to be enclosed snugly, it must touch the curved wall all the way round, so its radius equals the cylinder's radius \( r \).

\[ \text{diameter of sphere} = 2r \]

The condition \( h > 2r \) simply guarantees the cylinder is tall enough to contain that sphere; the height itself never limits the fit, so it does not appear in the answer.

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

A closed cylinder has a curved surface plus two circular ends.

\[ \text{TSA} = 2\pi r h + 2\pi r^{2} = 2\pi r (r + h) \]

With \( r = 9 \) cm and \( h = 28 \) cm:

\[ = 2 \times 3.142 \times 9 \times (9 + 28) = 2 \times 3.142 \times 9 \times 37 \]

\[ = 56.556 \times 37 = 2092.572 \text{ cm}^{2} \]

The curved surface alone is \( 2\pi r h = 1583.6 \) cm²; the two ends add the remaining 509 cm².

Answer: A

For cones of equal height the volume \( \tfrac{1}{3}\pi r^{2}h \) varies as \( r^{2} \), so the height cancels from the ratio.

\[ \frac{V_{1}}{V_{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{3^{2}}{5^{2}} = \frac{9}{25} \]

The ratio is \( 9 : 25 \). Carrying the radii through unsquared would give \( 3 : 5 \), which is the shape of the second distractor.

Answer: C

Work in centimetres throughout: the cylinder is \( r = 100 \) cm, \( h = 500 \) cm.

\[ V_{\text{cyl}} = \pi (100)^{2}(500) = 5{,}000{,}000\pi \text{ cm}^{3} \]

\[ V_{\text{cone}} = \frac{1}{3}\pi (50)^{2}(50) = \frac{125000\pi}{3} \text{ cm}^{3} \]

\[ n = \frac{5{,}000{,}000\pi}{125000\pi/3} = \frac{5{,}000{,}000 \times 3}{125000} = 120 \]

Mixing metres and centimetres is the trap here - the units must be made consistent before any division.