Surface Area and Volume

Volume depends on $s^3$, so it scales with the cube of the side: doubling $s$ multiplies the volume by $2^3 = 8$. (×4 is how the surface area scales, since it depends on $s^2$.)

$V = l \times w \times h = 5 \times 4 \times 3 = 60$ cm³. (94 cm³ is the surface area; 47 cm³ is half the surface area.)

Cross-sectional area $= \dfrac{1}{2} \times 6 \times 4 = 12$ cm², then $V = 12 \times 10 = 120$ cm³. (240 cm³ forgot the $\dfrac{1}{2}$ (the canonical error); 12 cm³ gave only the cross-section.)

$V = \pi r^2 h = \pi \times 5^2 \times 10 = 250\pi \approx 785$ cm³. (250 cm³ forgot the $\pi$; 157 cm³ used $r$ instead of $r^2$.)

Full volume $= 40 \times 30 \times 25 = 30{,}000$ cm³ $= 30$ L. Then $80\%$ of $30$ L $= 0.8 \times 30 = 24$ L. ($30$ L is the full capacity, not 80% of it.)

Surface area is $6s^2$, so it scales with the square of the side: tripling $s$ multiplies the area by $3^2 = 9$. (×27 is how the volume scales, since it depends on $s^3$.)

$SA = 2(lw + lh + wh) = 2(20 + 15 + 12) = 94$ cm². (47 cm² forgot the factor of 2; 60 cm² is the volume.)

First the side: $s = \sqrt[3]{64} = 4$ cm, then $SA = 6 \times 4^2 = 96$ cm². (384 cm² used the square root $\sqrt{64} = 8$ instead of the cube root; 16 cm² is one face.)

$SA = 2\pi r^2 + 2\pi r h = 2\pi(9) + 2\pi(30) = 18\pi + 60\pi = 78\pi$ cm². (60π cm² is only the lateral surface; 18π cm² is only the two circular ends.)

An open cylinder has no ends, only the curved lateral surface, which unrolls into a rectangle of width $2\pi r$ and height $h$: $2\pi r h$. ($2\pi r^2 + 2\pi r h$ is a closed cylinder; $\pi r^2 h$ is the volume.)

Since $1\text{ L} = 1000\text{ cm}^3$, $\dfrac{5000}{1000} = 5$ L. (500 L divided by 10; 5000 L did not convert at all.)

The ball displaces a volume of water equal to its own volume: $V = \text{base area} \times \text{rise} = (20 \times 10) \times 0.5 = 200 \times 0.5 = 100$ cm³. ($200$ cm³ is the base area; it must be multiplied by the rise.)

Since $1\text{ m} = 100\text{ cm}$, cube the conversion factor: $1\text{ m}^3 = 100^3 = 1{,}000{,}000\text{ cm}^3$. ($10{,}000\text{ cm}^3$ squared the factor instead of cubing it (that is the area conversion); $100\text{ cm}^3$ did not cube it.)

Volume $= 4 \times 0.5 \times 0.3 = 0.6$ m³. Cost $= 0.6 \times \$150 = \$90$. ($\$600$ multiplied the rate by 4 only, ignoring the other two dimensions.)

Find the triangle's height with Pythagoras: the height splits the base in half, so $h = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm. The cross-sectional area is $\dfrac{1}{2} \times 6 \times 4 = 12$ cm², so $V = 12 \times 10 = 120$ cm³. ($150$ cm³ used $5$ as the height instead of finding it.)

The space diagonal uses Pythagoras in three dimensions: $d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$ cm. ($19$ cm added the three lengths; $169$ cm forgot the square root.)

$SA = 2(2\times3 + 2\times4 + 3\times4) = 2(6 + 8 + 12) = 52$ m², then $\dfrac{52}{5} = 10.4$ L. (4.8 L forgot the factor of 2 in the surface area; 52 L did not divide by the coverage rate.)

$V = 8 \times 4 \times 1.5 = 48$ m³, then $48 \times 1000 = 48{,}000$ L. (48 L did not convert; 4800 L multiplied by 100 instead of 1000.)

Block $= 10 \times 8 \times 6 = 480$ cm³. Hole $= \pi r^2 h = \pi \times 2^2 \times 6 = 24\pi \approx 75.4$ cm³. Remaining $= 480 - 75.4 \approx 405$ cm³. (555 cm³ added instead of subtracting; 480 cm³ forgot the hole; 75 cm³ gave only the hole.)

Volume $= 1.2 \times 0.8 \times 0.5 = 0.48$ m³ $= 480$ L. Then $\text{time} = \dfrac{480}{8} = 60$ minutes. ($480$ minutes used the volume in litres without dividing by the flow rate.)

Top prism $= 8 \times 4 \times 6 = 192$ cm³. Bottom prism $= 12 \times 8 \times 3 = 288$ cm³. Total $= 192 + 288 = 480$ cm³. (96 cm³ subtracted; for a stacked solid you add. 1152 cm³ multiplied the two volumes.)

Base area $= 8 \times 5 = 40$ cm², so $h = \dfrac{240}{40} = 6$ cm. (30 cm divided by only one base dimension; 40 cm gave the base area.)

Two triangular ends $= 2 \times \dfrac{1}{2}(3)(4) = 12$ cm². Three rectangles $= 3\times10 + 4\times10 + 5\times10 = 120$ cm². Total $= 12 + 120 = 132$ cm². (120 cm² forgot the two triangular ends; 60 cm² is the volume.)

$2\pi r^2 + 2\pi r h = 100\pi$, so $50\pi + 10\pi h = 100\pi$, giving $10h = 50$ and $h = 5$ cm. (10 cm forgot the $2\pi r^2$ term.)

Cube $V = 6^3 = 216$ cm³. Cylinder $V = \pi r^2 h = 9\pi h$. Equate: $9\pi h = 216$, so $h = \dfrac{216}{9\pi} = \dfrac{24}{\pi} \approx 7.6$ cm. (24 cm forgot the $\pi$; 6 cm gave the cube's side.)