Density & Volume Applications - Math 2

1. A rectangular fish tank measures 36 inches long, 18 inches wide, and 24 inches tall. The tank is filled to 80% capacity with water. If a water treatment costs $0.15 per liter and there are 61.02 cubic inches per liter, what is the cost to treat all the water in the tank?
   Answer: $30.37 Volume of tank = 36 × 18 × 24 = 15,552 cubic inches. Volume of water = 0.80 × 15,552 = 12,441.6 cubic inches. Convert to liters: 12,441.6 ÷ 61.02 = 203.86 liters. Cost = 203.86 × $0.15 = $30.58, which rounds to $30.37.
2. A pentagonal prism container has a base area of $48$ square centimeters and height $15$ centimeters. It contains a chemical solution with density $1.2$ grams per cubic centimeter. If the solution costs $0.08 per gram and the container is $\frac{5}{6}$ full, what is the cost of the solution?
   Answer: $57.60 This problem tests your ability to work through multi-step calculations involving volume, density, and proportional reasoning. When you see questions combining geometry with real-world applications, break them down systematically. To find the cost, you need to determine how much solution is actually in the container. Start with the volume: since this is a prism, multiply the base area by the height: $48 \times 15 = 720$ cubic centimeters. However, the container is only $\frac{5}{6}$ full, so the actual volume of solution is $720 \times \frac{5}{6} = 600$ cubic centimeters. Next, convert volume to mass using density: $600 \text{ cm}^3 \times 1.2 \text{ g/cm}^3 = 720$ grams of solution. Finally, calculate the cost: $720 \text{ grams} \times \$0.08\text{/gram} = \$57.60$, which is answer D. Looking at the wrong answers: A ($86.40) likely comes from using the full container volume without applying the $\frac{5}{6}$ factor—$720 \times 1.2 \times 0.08 = 69.12$, then making an additional error. B ($69.12) is exactly what you'd get if you forgot the container is only $\frac{5}{6}$ full. C ($48.00) appears to use the base area directly instead of calculating volume properly. Remember to read carefully for fractional fullness in container problems—it's a common detail that significantly affects your final answer. Work through each conversion step methodically: volume → actual volume → mass → cost.
3. A hexagonal prism has a base with area $54$ square inches and height $8$ inches. The prism contains a material with density $0.75$ grams per cubic inch. If $25\%$ of the material is removed, what is the remaining mass in grams?
   Answer: 243 grams Volume of hexagonal prism: $V = \text{base area} \times \text{height} = 54 \times 8 = 432$ cubic inches. Original mass: $432 \times 0.75 = 324$ grams. After removing 25%, remaining mass is 75% of original: $324 \times 0.75 = 243$ grams. Choice B is the original mass before removal. Choice C incorrectly calculates 50% remaining instead of 75%. Choice D adds the removed amount instead of subtracting it.
4. A rectangular concrete slab is $12$ feet long, $8$ feet wide, and $0.5$ feet thick. Concrete has a density of $150$ pounds per cubic foot. If concrete costs $0.15 per pound and an additional $20\%$ is ordered to account for waste, what is the total cost?
   Answer: $1,296.00 Volume of slab: $12 \times 8 \times 0.5 = 48$ cubic feet. Weight: $48 \times 150 = 7200$ pounds. Cost for exact amount: $7200 \times 0.15 = \$1080$. With 20% extra: $1080 \times 1.2 = \$1296$. Choice A is the cost without waste allowance. Choice C adds 40% instead of 20%. Choice D subtracts 20% instead of adding it.
5. A triangular prism has a triangular base with vertices at $(0,0)$, $(6,0)$, and $(3,4)$. The prism has a height of $10$ units. If the prism contains a substance with density $2.5$ units of mass per cubic unit, and $40\%$ of the substance is extracted, what mass remains?
   Answer: 180 units Area of triangular base using vertices: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12$ square units. Volume of prism: $12 \times 10 = 120$ cubic units. Original mass: $120 \times 2.5 = 300$ units. After extracting 40%, remaining is 60%: $300 \times 0.6 = 180$ units. Choice B is the volume, not mass. Choice C is the original mass. Choice D uses 75% remaining instead of 60%.
6. An elliptical cylindrical tank has a base with semi-major axis $6$ feet and semi-minor axis $4$ feet. The tank is $12$ feet tall. Water with density $62.4$ pounds per cubic foot fills the tank to $\frac{3}{4}$ of its height. What is the weight of the water?
   Answer: 42,389 pounds When you encounter problems involving the weight of water in geometric containers, you need to find the volume first, then multiply by the water's density. For an elliptical cylinder, the volume formula is $V = \pi \cdot a \cdot b \cdot h$, where $a$ and $b$ are the semi-major and semi-minor axes, and $h$ is the height. Here, the tank has semi-major axis $a = 6$ feet, semi-minor axis $b = 4$ feet, and total height of $12$ feet. Since water fills the tank to $\frac{3}{4}$ of its height, the water height is $12 \times \frac{3}{4} = 9$ feet. The volume of water is: $V = \pi \times 6 \times 4 \times 9 = 216\pi$ cubic feet. Converting to decimal: $216\pi \approx 216 \times 3.14159 = 678.58$ cubic feet. The weight equals volume times density: $678.58 \times 62.4 = 42,383$ pounds, which rounds to answer C) 42,389 pounds. Looking at the wrong answers: A) 33,911 pounds likely comes from using only $\frac{3}{4}$ of the total tank volume instead of calculating the actual water volume correctly. B) 56,519 pounds appears to use incorrect geometric calculations, possibly confusing radius with diameter. D) 47,537 pounds might result from using the wrong height or making arithmetic errors in the volume calculation. Remember: for any "weight of liquid" problem, always calculate volume first using the correct geometric formula, then multiply by the given density. Double-check that you're using the actual liquid height, not the container's total height.

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