Composite Volume

Volume is $s^3$, so it scales with the cube of the side: tripling $s$ multiplies the volume by $3^3 = 27$. (×9 is how the surface area scales, since $SA$ depends on $s^2$.)

Let the edges be $n, n+1, n+2$. Since $\sqrt[3]{120} \approx 4.9$, the numbers sit around 5: testing gives $4 \times 5 \times 6 = 120$. ($3 \times 4 \times 5 = 60$ and $5 \times 6 \times 7 = 210$ miss; $2 \times 5 \times 12$ is not consecutive.)

Equal volumes: $\pi(4)^2 h_1 = \pi(2)^2 h_2$, so $16h_1 = 4h_2$ and $h_2 = 4h_1$. Halving the radius quarters the base area, so the height must be 4 times as large. (2 times only uses the radius ratio, not its square.)

Cube: $10^3 = 1000$ cm³. Cylinder: $\pi(6)^2(10) = 360\pi \approx 1130$ cm³. Since $1130 > 1000$, the cylinder holds more; its circular base area $36\pi \approx 113$ cm² already exceeds the cube's $100$ cm² base.

The displaced volume equals the cube's volume $6^3 = 216$ cm³. The water rises in a cylinder of base area $\pi(6)^2 = 36\pi$ cm², so the rise is $\dfrac{216}{36\pi} = \dfrac{6}{\pi}$ cm. (You divide the displaced volume by the base area, not by the cube's side.)

Bottom $10 \times 6 \times 3 = 180$ cm³, top $5 \times 6 \times 4 = 120$ cm³, total $180 + 120 = 300$ cm³. (180 cm³ forgot the top block.)

Cube $6^3 = 216$ cm³, cylinder $\pi r^2 h = \pi \times 4 \times 5 = 20\pi$ cm³, total $216 + 20\pi$ cm³. ($216 + 10\pi$ used $2r$; 216 forgot the cylinder.)

Base area $10 \times 8 - 6 \times 4 = 80 - 24 = 56$ cm², so $V = 56 \times 5 = 280$ cm³. (400 cm³ used the full rectangle without the cut.)

The slant height is not the perpendicular height. First use Pythagoras: $h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. Then $V = \dfrac{1}{3}\pi(5)^2(12) = \dfrac{1}{3}\pi(300) = 100\pi$ cm³. ($\tfrac{325}{3}\pi$ used the slant height 13 as the perpendicular height.)

Walls $6 \times 4 \times 3 = 72$ m³. Roof base area $\dfrac{1}{2}(6)(2) = 6$ m², roof volume $6 \times 4 = 24$ m³. Total $72 + 24 = 96$ m³. (72 m³ forgot the roof; 144 m³ forgot the $\dfrac{1}{2}$ or doubled the walls.)

Work back from the surface area: $4\pi r^2 = 36\pi$ gives $r^2 = 9$, so $r = 3$ cm. Then $V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(27) = 36\pi$ cm³. (You must find $r$ from the surface area before using the volume formula; $288\pi$ used $r = 6$.)

$V = \pi(R^2 - r^2)h = \pi(5^2 - 3^2)(20) = \pi(16)(20) = 320\pi$ cm³. ($500\pi$ is the outer cylinder only; $180\pi$ is the inner (empty) cylinder.)

Prism $10 \times 8 \times 5 = 400$ cm³, cylinder $\pi r^2 h = \pi \times 4 \times 5 = 20\pi$ cm³, remaining $400 - 20\pi$ cm³. ($400 + 20\pi$ added; $20\pi - 400$ is the wrong order.)

The full cone has volume $\dfrac{1}{3}\pi(6)^2(8) = 96\pi$ cm³. Cutting halfway up removes a similar cone with half the dimensions (radius 3, height 4): $\dfrac{1}{3}\pi(3)^2(4) = 12\pi$ cm³. The frustum is $96\pi - 12\pi = 84\pi$ cm³. ($96\pi$ forgot to remove the top; $12\pi$ is only the removed piece.)

$V = \dfrac{1}{3}(\text{base area})h$, so $120 = \dfrac{1}{3}(6^2)h = 12h$, giving $h = 10$ cm. ($\tfrac{10}{3}$ cm forgot the $\dfrac{1}{3}$; you must undo the base area and the $\dfrac{1}{3}$ first.)

The cone's slant height is given, so find its perpendicular height with Pythagoras: $h = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$ m. Cylinder $\pi(3)^2(8) = 72\pi$ m³; cone $\dfrac{1}{3}\pi(3)^2(4) = 12\pi$ m³. Total $72\pi + 12\pi = 84\pi$ m³. ($72\pi$ forgot the cone; using the slant $5$ as the cone's height gives $87\pi$.)

Base area $= 200\text{ cm} \times 100\text{ cm} = 20{,}000$ cm². The inflow is $200\text{ L} = 200{,}000$ cm³ per minute, so the level rises $\dfrac{200{,}000}{20{,}000} = 10$ cm each minute. (Not converting metres to cm gives the wrong power of 10.)

The volume is $8 \times 8 \times 27 = 1728$ cm³. A cube of this volume has side $\sqrt[3]{1728} = 12$ cm (since $12^3 = 1728$). (Recasting preserves the volume, so take the cube root of the total.)

For similar solids the volume ratio is the cube of the length ratio: $\left(\dfrac{6}{4}\right)^3 = \dfrac{27}{8}$. So the larger volume is $24\pi \times \dfrac{27}{8} = 81\pi$ cm³. ($36\pi$ scaled by $\tfrac{6}{4}$ once instead of cubing it.)

The cross-sectional area is $\dfrac{1}{2}(6)(8) = 24$ cm², so $V = 24 \times 15 = 360$ cm³. (720 cm³ forgot the $\dfrac{1}{2}$ in the triangle's area.)

Use trigonometry to get both dimensions: the height is $h = 10\sin 30^\circ = 5$ cm and the radius is $r = 10\cos 30^\circ = 5\sqrt{3}$ cm, so $r^2 = 75$. Then $V = \dfrac{1}{3}\pi(75)(5) = 125\pi$ cm³. (Swapping sine and cosine swaps $r$ and $h$.)

$100\pi = \pi r^2 (4)$, so $r^2 = 25$ and $r = 5$ cm. (25 cm gave $r^2$, forgetting the square root.)

Base $4 \times 4 \times 1 = 16$ m³, cylinder $\pi r^2 h = \pi \times 4 \times 5 = 20\pi$ m³, total $16 + 20\pi$ m³. ($16 + 10\pi$ used $2r$; $20\pi$ forgot the base.)

The side cross-section is a trapezium with parallel sides 1 m and 3 m and length 25 m: area $\dfrac{1}{2}(1 + 3)(25) = \dfrac{1}{2}(4)(25) = 50$ m². Volume $= 50 \times 10 = 500$ m³ (× width). (1000 m³ used the sum of depths; 250 m³ used only half the trapezium.)

Curved surface $= 2\pi r h = 2\pi(5)(10) = 100\pi$ cm². Total surface $= 2\pi r^2 + 2\pi r h = 2\pi(25) + 100\pi = 50\pi + 100\pi = 150\pi$ cm². The ratio is $100\pi : 150\pi = 2 : 3$. (The extra $2\pi r^2 = 50\pi$ cm² in the total is the two circular ends.)