Volume & Surface Areas Questions and Answers

Answer: C

The level rises by 7 cm in 2 hours.

The volume needed is the tank's base area times the rise in level, 7 cm or 0.07 m:

\[ 100 \times 44 \times 0.07 = 308\ \text{m}^{3} \]

The pipe has radius 7 cm, 0.07 m, and water moves 10 km, 10 000 m, in an hour, so the flow per hour is:

\[ \begin{aligned} \pi r^{2} \times 10000 &= \tfrac{22}{7} \times 0.0049 \times 10000 \\ &= 154\ \text{m}^{3} \end{aligned} \]

Time = \(308 \div 154 = 2\) hours. Using the diameter instead of the radius would give a flow four times larger and a time of half an hour.

Answer: B

The value of x is 15.

The cost is ₹ 20 per 100 square cm, so the area painted is:

\[ \begin{aligned} \text{Area} &= \frac{1032}{20} \times 100 = 5160 \\ 2(40 \times 36 + 36x + 40x) &= 5160 \\ 1440 + 76x &= 2580 \\ x &= \frac{1140}{76} = 15\ \text{cm} \end{aligned} \]

A check: 2(1440 + 540 + 600) = 5160 square cm, which costs 51.6 × 20 = ₹ 1,032. The other values give total surface areas of 5920 (x = 20), 6528 (x = 24) and 4704 (x = 12), none costing ₹ 1,032.

Answer: A

The curved surface area is 2640 square cm.

The length is 1.2 m = 120 cm. From the volume:

\[ \begin{aligned} \frac{22}{7} r^{2} \times 120 &= 4620 \\ r^{2} &= \frac{4620 \times 7}{22 \times 120} = 12{\cdot}25 \\ r &= 3{\cdot}5\ \text{cm} \\ \text{CSA} &= 2 \times \frac{22}{7} \times 3{\cdot}5 \times 120 = 2640 \end{aligned} \]

The common error is to leave the length as 1.2 or 12 instead of converting it to 120 cm, which gives a wrong radius; the other options do not follow from a radius of 3.5 cm.

Answer: B

40 planks can be made.

Put all lengths in centimetres. The block is 800 cm × 70 cm × 40 cm and each plank is 200 cm × 35 cm × 8 cm.

\[ \begin{aligned} \text{planks} &= \frac{800 \times 70 \times 40}{200 \times 35 \times 8} \\ &= \frac{2240000}{56000} = 40 \end{aligned} \]

The dimensions also fit exactly: 800/200 = 4 along the length, 70/35 = 2 across the width and 40/8 = 5 through the thickness, and 4 × 2 × 5 = 40, so no wood is wasted.

Forgetting to convert metres to centimetres is the usual source of the other answers.

Answer: A

The new total surface area will be thrice the original.

The big cube makes \((12/4)^3 = 27\) small cubes.

\[ \begin{aligned} \text{original} &= 6 \times 12^2 = 864 \text{ cm}^2 \\ \text{new} &= 27 \times 6 \times 4^2 = 2592 \text{ cm}^2 \\ 2592 \div 864 &= 3 \end{aligned} \]

Cutting creates new faces, so the total area increases, not decreases. In general, cutting into cubes of 1/n the side multiplies the surface area by n, here 3.

The volume stays the same, 1728 cm³, because no material is lost; only the exposed surface grows.

Answer: D

The difference is 10000 cm².

The volume stays the same when the box is recast:

\[ \begin{aligned} V &= 200 \times 50 \times 100 = 10^{6}\ \text{cm}^{3} \\ \text{side} &= \sqrt[3]{10^{6}} = 100\ \text{cm} \\ S_{\text{cuboid}} &= 2(10000 + 5000 + 20000) \\ &= 70000 \\ S_{\text{cube}} &= 6 \times 100^{2} = 60000 \end{aligned} \]

The difference is 70000 - 60000 = 10000 cm². The cube has the smaller surface area, as expected: for a fixed volume, a cube encloses it with less surface than an elongated box.

Answer: C

The maximum number of boxes is 4800.

Place each box with its 70 cm (0.7 m) side along the 56 m length, its 2.8 m side along the 42 m breadth and its 2.5 m side along the 10 m height:

\[ \begin{aligned} 56 \div 0.7 &= 80 \\ 42 \div 2.8 &= 15 \\ 10 \div 2.5 &= 4 \\ 80 \times 15 \times 4 &= 4800 \end{aligned} \]

Every division is exact, so the boxes fill the godown completely. The volume check agrees: \(56 \times 42 \times 10 = 23520\ \text{m}^{3}\) and each box is \(2.8 \times 2.5 \times 0.7 = 4.9\ \text{m}^{3}\), and 23520 ÷ 4.9 = 4800. Other orientations leave gaps and hold fewer boxes.

Answer: C

The volume is 4.4352 m³.

From the circumference, find the radius, then use \(V = \pi r^{2} h\) with h = 2 m = 200 cm:

\[ \begin{aligned} 2\pi r &= 528 \\ r &= 528 \times \frac{7}{44} = 84\ \text{cm} \\ V &= \frac{22}{7} \times 84^{2} \times 200 \\ &= 4435200\ \text{cm}^{3} \end{aligned} \]

Since 1 m³ = 1,000,000 cm³, V = 4.4352 m³. 2.2176 m³ is half the answer, as if the height were taken as 1 m; 6.6528 m³ is one and a half times it.

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 \(81 \times 9 = 729\).