Volume & Surface Areas Questions and Answers

Answer: D

Water touches the base and the lower part of all four walls, but not the tank's top or the walls above the 11 cm line.

\[ \text{base} = 21 \times 13 = 273 \text{ cm}^{2} \]

\[ \text{two long walls} = 2 \times (21 \times 11) = 462 \text{ cm}^{2} \]

\[ \text{two short walls} = 2 \times (13 \times 11) = 286 \text{ cm}^{2} \]

\[ \text{total} = 273 + 462 + 286 = 1021 \text{ cm}^{2} \]

The tank's own height of 18 cm never enters: only the wetted 11 cm counts, and there is no top surface in contact with the water.

Answer: A

Melting conserves volume, so equate the sheet's volume with the bar's.

\[ V_{\text{sheet}} = \pi (25)^{2}(0.65) = \pi (625)(0.65) = 406.25\pi \text{ cm}^{3} \]

\[ V_{\text{bar}} = \pi (4)^{2} L = 16\pi L \]

\[ 16\pi L = 406.25\pi \quad\Longrightarrow\quad L = \frac{406.25}{16} = 25.39 \text{ cm} \]

The \( \pi \) cancels, so no value for it is needed. Both diameters must be halved first - using 50 and 8 directly would give the same ratio, but the thickness would then be mismatched.

Answer: A

A cuboid has three pairs of identical faces.

\[ \text{TSA} = 2(lb + bh + hl) \]

\[ = 2\left(8.9 \times 4.7 + 4.7 \times 3.2 + 3.2 \times 8.9\right) \]

\[ = 2(41.83 + 15.04 + 28.48) = 2(85.35) \]

\[ = 170.7 \text{ cm}^{2} \]

Each of the three products must appear twice; summing them once gives 85.35, exactly half the answer.

Answer: C

Halve the diameter, then rearrange the volume formula for the height.

\[ r = 2.5 \text{ cm} \]

\[ h = \frac{V}{\pi r^{2}} = \frac{650}{3.142 \times 6.25} = \frac{650}{19.64} \]

\[ h = 33.1 \approx 33 \text{ cm} \]

Using the diameter of 5 instead of the radius would divide by four times as much and give about 8.3 cm.

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: C

Only the water is wanted, so use its own height of 11 cm rather than the tank's 18 cm.

\[ V = 21 \times 13 \times 11 = 3003 \text{ cm}^{3} \]

Since \( 1000 \text{ cm}^{3} = 1 \) litre:

\[ V = \frac{3003}{1000} = 3.003 \text{ litres} \]

Using the full 18 cm depth would give 4.914 litres - the tank's capacity, not the water it holds.

Answer: A

The millilitre is defined as one cubic centimetre, so the two units are interchangeable and the number does not change.

\[ 1 \text{ cm}^{3} = 1 \text{ ml} \]

\[ 6800 \text{ cm}^{3} = 6800 \text{ ml} \]

The factor of 1000 belongs to the litre, not the millilitre: 6800 cm³ is 6.8 litres.

Answer: A

Each metre is 100 centimetres, so a cubic metre is \( 100^{3} \) cubic centimetres.

\[ 1 \text{ m}^{3} = 100 \times 100 \times 100 = 1{,}000{,}000 \text{ cm}^{3} \]

\[ 0.8 \text{ m}^{3} = 0.8 \times 1{,}000{,}000 = 800{,}000 \text{ cm}^{3} \]

The conversion factor for a volume is the cube of the length factor - using 100 or 10,000 instead of 1,000,000 produces the smaller distractors.

Answer: B

A centimetre is 10 millimetres, so a cubic centimetre is \( 10^{3} \) cubic millimetres.

\[ 1 \text{ cm}^{3} = 1000 \text{ mm}^{3} \]

\[ 2{,}500{,}000 \text{ mm}^{3} = \frac{2{,}500{,}000}{1000} = 2500 \text{ cm}^{3} \]

Dividing by 10 or by 100 instead - the length and area factors - gives the two nearby distractors.

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.