Volume & Surface Areas Questions and Answers

Answer: B

The wood occupies the space between the outer box and the hollow inside it, so find both volumes and subtract.

The wood is 4 cm thick on every face, so each external dimension exceeds the internal one by \( 2 \times 4 = 8 \) cm:

\[ 88 \times 56 \times 28 \text{ cm} \]

\[ V_{\text{external}} = 88 \times 56 \times 28 = 137{,}984 \text{ cm}^{3} \]

\[ V_{\text{internal}} = 80 \times 48 \times 20 = 76{,}800 \text{ cm}^{3} \]

\[ V_{\text{wood}} = 137{,}984 - 76{,}800 = 61{,}184 \text{ cm}^{3} \]

Adding 4 cm rather than 8 cm to each dimension is the standard error - the thickness is present on both opposite faces.

Answer: D

The men displace their own volume of water, and that volume spreads over the pool's whole surface.

\[ V_{\text{displaced}} = 150 \times 8 = 1200 \text{ m}^{3} \]

\[ \text{pool area} = 90 \times 40 = 3600 \text{ m}^{2} \]

\[ \text{rise} = \frac{1200}{3600} = \frac{1}{3} \text{ m} = 33.33 \text{ cm} \]

The answer is asked for in centimetres, so the metre figure must be multiplied by 100 - the whole point of the awkward-looking 33.33.

Answer: C

Find the radius from the surface area first, taking \( \pi = \tfrac{22}{7} \).

\[ 4\pi r^{2} = \frac{792}{7} \quad\Longrightarrow\quad r^{2} = \frac{792}{7} \times \frac{7}{4 \times 22} = \frac{792}{88} = 9 \]

\[ r = 3 \text{ cm} \]

\[ V = \frac{4}{3}\pi r^{3} = \frac{4}{3} \times \frac{22}{7} \times 27 = \frac{792}{7} \text{ cm}^{3} \]

The numerical coincidence is real, not a misprint: for \( r = 3 \) the surface area and volume come out to the same number, because \( \tfrac{V}{S} = \tfrac{r}{3} = 1 \).

Answer: D

Find the radius from the volume, then use it for the surface area.

\[ \frac{4}{3}\pi r^{3} = 850 \quad\Longrightarrow\quad r^{3} = \frac{850 \times 3}{4\pi} = \frac{2550}{12.566} = 202.9 \]

\[ r = \sqrt[3]{202.9} = 5.88 \text{ m} \]

\[ S = 4\pi r^{2} = 4 \times 3.1416 \times (5.88)^{2} = 12.566 \times 34.57 \]

\[ S \approx 434 \text{ m}^{2} \]

Only the pair 5.88 m and 434 m² is internally consistent - checking one value against the other is quicker here than computing both from scratch.

Answer: C

Convert the diameter to centimetres first, then halve it.

\[ 45 \text{ mm} = 4.5 \text{ cm} \quad\Longrightarrow\quad r = 2.25 \text{ cm} \]

\[ \text{TSA} = 2\pi r (r + h) = 2 \times 3.142 \times 2.25 \times (2.25 + 12) \]

\[ = 14.139 \times 14.25 = 201.5 \approx 201 \text{ cm}^{2} \]

Both flat ends count, since the whole cylinder is carved; the curved surface alone would come to about 169.7 cm².

Answer: B

With faces \( P = lb \), \( Q = bh \), \( R = hl \), the product of all three is the square of the volume:

\[ PQR = (lbh)^{2} \quad\Longrightarrow\quad V = \sqrt{PQR} \]

\[ V = \sqrt{120 \times 72 \times 60} = \sqrt{518400} = 720 \text{ cm}^{3} \]

Forgetting the square root leaves 518,400; the dimensions themselves work out to 10, 12 and 6 cm, which do reproduce all three given face areas.

Answer: C

Melting conserves volume, so divide the target cylinder's volume by one coin's. Halve both diameters first.

\[ V_{\text{coin}} = \pi (0.75)^{2}(0.2) = 0.1125\pi \text{ cm}^{3} \]

\[ V_{\text{cylinder}} = \pi (2.25)^{2}(10) = 50.625\pi \text{ cm}^{3} \]

\[ n = \frac{50.625\pi}{0.1125\pi} = \frac{50.625}{0.1125} = 450 \]

Using the diameters directly instead of the radii leaves the ratio unchanged only because both are squared - but the thickness is not, so it is safer to convert first.

Answer: A

A surahi is a water pot with a rounded, bulbous body and a long straight neck.

The body is a sphere and the neck is a cylinder.

A hemisphere would leave the body flat on one side, and a cone would taper the neck rather than keeping it a uniform tube.

Answer: B

The largest cone that fits inside a cube has its base inscribed in one face and its apex at the centre of the opposite face.

The base circle is limited by the face's width, so its diameter equals the cube's edge:

\[ d = 4.2 \text{ cm} \quad\Longrightarrow\quad r = \frac{4.2}{2} = 2.1 \text{ cm} \]

The cone's height is the full 4.2 cm edge, but the question asks only for the radius. The 4.2 option confuses the diameter with the radius.

Answer: A

Divide the granary's volume by the space one bag needs.

\[ V = 12 \times 6 \times 5 = 360 \text{ m}^{3} \]

\[ n = \frac{360}{0.48} = 750 \]

The division by a decimal is where slips creep in: \( \tfrac{360}{0.48} = \tfrac{36000}{48} = 750 \), which is cleaner to compute after multiplying both numbers by 100.