Volume & Surface Areas Questions and Answers

Answer: D

8 boxes are needed.

A centimetre cube has a volume of 1 cubic centimetre, so each box holds as many cubes as its volume in cubic centimetres.

\[ \begin{aligned} \text{volume of a box} &= 12 \times 10 \times 7 \\ &= 840\ \text{cm}^{3} \\ \text{boxes} &= 6720 \div 840 = 8 \end{aligned} \]

Check: \(8 \times 840 = 6720\). Seven boxes hold only 5,880 cubes and nine would leave a box partly empty, so 7, 6 and 9 are wrong.

Answer: A

111 dice are left unpacked.

A 1 cm die occupies 1 cubic centimetre, and the internal dimensions are whole centimetres, so each box holds as many dice as its volume in cubic centimetres.

\[ \begin{aligned} \text{cuboid} &= 12 \times 10 \times 8 = 960 \\ \text{cube} &= 9 \times 9 \times 9 = 729 \\ \text{packed} &= 960 + 729 = 1689 \\ \text{unpacked} &= 1800 - 1689 = 111 \end{aligned} \]

The other options come from miscounting one of the two volumes; for example, \(9^{3}\) is often misremembered, so it is worth checking that \(81 \times 9 = 729\).

Answer: A

The volume is 96π m³.

Three times the area of the two bases equals twice the curved surface area:

\[ \begin{aligned} 3 \times 2\pi r^{2} &= 2 \times 2\pi r h \\ 6\pi r^{2} &= 4\pi r h \\ r &= \tfrac{2}{3}h = \tfrac{2}{3} \times 6 = 4\ \text{m} \end{aligned} \]

\[ \begin{aligned} V &= \pi r^{2} h = \pi \times 16 \times 6 \\ &= 96\pi\ \text{m}^{3} \end{aligned} \]

Using the area of only one base gives \(r = 8\) and a volume of \(384\pi\), which is not offered. The other options correspond to radii that do not satisfy the stated condition.

Answer: D

The side of the new cube is 12 cm.

Melting keeps the total volume the same, so add the three volumes and take the cube root.

\[ \begin{aligned} V &= 8^{3} + 6^{3} + 10^{3} \\ &= 512 + 216 + 1000 = 1728 \\ \text{side} &= \sqrt[3]{1728} = 12 \text{ cm} \end{aligned} \]

The common mistake is adding the sides, \(8 + 6 + 10 = 24\) cm, which assumes length is conserved instead of volume. 20 cm and 18 cm are similar guesses that do not cube to 1728.

Answer: D

The capacity is 25.872 litres.

The base area gives the radius:

\[ \begin{aligned} \frac{22}{7} r^2 &= 616 \\ r^2 &= 196,\ r = 14\ \text{cm} \end{aligned} \]

The height is three times the radius, h = 42 cm.

\[ \begin{aligned} V &= \pi r^2 h = 616 \times 42 \\ &= 25872\ \text{cm}^3 \end{aligned} \]

Since 1 litre = 1000 cm³, the capacity is 25.872 litres. 17.248 uses a height of 28 cm (twice the radius), 12.936 uses 21 cm, and 34.496 uses 56 cm.

Answer: B

The capacity is 6.16 litres.

The height is 10 cm, so the 88 cm side wraps round to form the circumference of the base.

\[ \begin{aligned} 2\pi r &= 88 \\ r &= \frac{88 \times 7}{2 \times 22} = 14 \text{ cm} \end{aligned} \]

\[ \begin{aligned} V &= \pi r^{2} h \\ &= \tfrac{22}{7} \times 196 \times 10 \\ &= 6160 \text{ cm}^{3} \end{aligned} \]

Since 1 litre = 1000 cm³, the capacity is 6.16 litres. The other values do not fit a radius of 14 cm and a height of 10 cm; the key step is seeing that the 88 cm edge becomes the circumference, because the 10 cm edge is the height.

Answer: A

The lateral surface area is 676 cm².

The side of the cube is the cube root of its volume, and the lateral surface is the four side faces.

\[ \begin{aligned} a &= \sqrt[3]{2197} = 13 \text{ cm} \\ \text{LSA} &= 4a^{2} = 4 \times 169 = 676 \end{aligned} \]

1014 is the total surface area, \(6a^{2}\), which includes the top and bottom; the question asks for the lateral area only. 845 is \(5a^{2}\), and 576 is \(4 \times 12^{2}\), using the wrong side.

Answer: B

The volume is 0.0385 m³.

Find the radius from the circumference and the height from the curved surface area:

\[ \begin{aligned} 2\pi r &= 110 \Rightarrow r = \frac{110 \times 7}{44} = 17.5\ \text{cm} \\ 2\pi r h &= 4400 \Rightarrow h = \frac{4400}{110} = 40\ \text{cm} \end{aligned} \]

\[ \begin{aligned} V &= \pi r^{2} h = \frac{22}{7} \times 17.5^{2} \times 40 \\ &= 38500\ \text{cm}^{3} = 0.0385\ \text{m}^{3} \end{aligned} \]

1 m³ is 1,000,000 cm³; the other options are close distractors that do not equal 38,500 cm³.

Answer: B

The cost is ₹ 839.10.

The wood is 3 cm thick on all four sides and the base, but the top is open, so the inner length and breadth lose 6 cm and the inner height loses only 3 cm: 142 cm, 110 cm and 80 cm.

\[ \begin{aligned} \text{base} &= 142 \times 110 = 15620\ \text{cm}^{2} \\ \text{walls} &= 2 \times 80 \times (142 + 110) = 40320 \\ \text{total} &= 55940\ \text{cm}^{2} = 5.594\ \text{m}^{2} \\ \text{cost} &= 5.594 \times 150 = 839.10 \end{aligned} \]

Subtracting 6 cm from the height as well, or painting a top that does not exist, gives one of the other figures.

Answer: C

The volume is 168 cubic metres (the stem prints the unit as m²).

Total surface area of a cuboid is \(2(lb + bh + hl)\). With l = 8 and b = 6:

\[ \begin{aligned} 2(48 + 6h + 8h) &= 194 \\ 48 + 14h &= 97 \\ h &= 3.5\ \text{m} \end{aligned} \]

\[ V = l \times b \times h = 8 \times 6 \times 3.5 = 168 \]

A volume is measured in cubic units, so the answer is 168 m³. 112, 126 and 224 correspond to heights of about 2.33, 2.6 and 4.67 m, none of which gives a surface area of 194 m².