Volume & Surface Areas Questions and Answers

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

There are 54 cubes with only one face painted.

A 15 cm cube cut into 3 cm cubes has \(15 / 3 = 5\) small cubes along each edge. Cubes with exactly one painted face lie in the middle of each face, away from the edges: a \(3 \times 3\) block on each face.

\[ 6 \times (5 - 2)^{2} = 6 \times 9 = 54 \]

The full breakdown of the 125 small cubes is: 8 corner cubes with three painted faces, \(12 \times 3 = 36\) edge cubes with two, 54 face cubes with one, and \(3^{3} = 27\) inner cubes with none; no cube has four or more painted faces. Check: \(8 + 36 + 54 + 27 = 125\).

36 is the count of edge cubes (two painted faces), not one-face cubes, and 24 and 17 do not match any category.

Answer: D

The larger cube gives 125 smaller cubes.

With 5 small cubes along each edge, the total is:

\[ 5^{3} = 125 \]

Equivalently, the volume ratio is \(15^{3} / 3^{3} = 3375 / 27 = 125\). 25 counts only one layer (\(5^{2}\)), and 75 counts three layers.

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.