Math 3 Quiz: Multi Step Volume Problems
12 questions · exam conditions
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Multi Step Volume ProblemsQuestion 1 of 12

A rectangular prism-shaped container is being designed to hold exactly 500 cubic inches of liquid. The length must be twice the width, and the height must be 4 inches less than the width. What should the width be, rounded to the nearest tenth of an inch?

6.8 inches
7.2 inches
7.6 inches
8.1 inches
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Math 3 Quiz

Math 3 Quiz: Multi Step Volume Problems

Practice Multi Step Volume Problems in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Multi Step Volume Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A rectangular prism-shaped container is being designed to hold exactly 500 cubic inches of liquid. The length must be twice the width, and the height must be 4 inches less than the width. What should the width be, rounded to the nearest tenth of an inch?

  1. 6.8 inches
  2. 7.2 inches
  3. 7.6 inches (correct answer)
  4. 8.1 inches
Explanation: Let width = ww, length = 2w2w, height = w−4w-4. Volume equation: w×2w×(w−4)=500w \times 2w \times (w-4) = 500, so 2w2(w−4)=5002w^2(w-4) = 500, giving 2w3−8w2=5002w^3 - 8w^2 = 500, or w3−4w2−250=0w^3 - 4w^2 - 250 = 0. Solving numerically gives w≈7.6w \approx 7.6 inches. Choice A uses incorrect constraint relationships. Choice B solves w3−4w2=250w^3 - 4w^2 = 250 instead. Choice D adds 4 to width instead of subtracting for height.

Question 2

A cylindrical water tank has a radius of 4 feet and a height of 10 feet. Water is being pumped out at a rate of 15 cubic feet per minute. If the tank starts completely full, how many minutes will it take for the water level to drop to exactly 3 feet from the bottom?

  1. About 23.5 minutes (correct answer)
  2. About 28.3 minutes
  3. About 33.5 minutes
  4. About 35.2 minutes
Explanation: The tank volume is πr2h=π(4)2(10)=160π\pi r^2 h = \pi(4)^2(10) = 160\pi cubic feet. When water level is 3 feet from bottom, remaining volume is π(4)2(3)=48π\pi(4)^2(3) = 48\pi cubic feet. Volume to be removed is 160π−48π=112π160\pi - 48\pi = 112\pi cubic feet. At 15 cubic feet per minute: 112π15≈23.5\frac{112\pi}{15} \approx 23.5 minutes. Choice B uses the wrong height calculation. Choice C incorrectly calculates total tank volume. Choice D uses diameter instead of radius.

Question 3

A cylindrical oil drum with radius 1.5 feet and height 4 feet is lying on its side. Oil fills the drum to a depth of 2 feet (measured vertically from the bottom). What is the volume of oil in the drum?

  1. About 14.1 cubic feet (correct answer)
  2. About 16.8 cubic feet
  3. About 18.5 cubic feet
  4. About 21.2 cubic feet
Explanation: This requires finding the area of a circular segment. The circle has radius 1.5 feet, and oil depth is 2 feet, so the oil surface is 0.5 feet above the center. The central angle θ satisfies cos⁡(θ/2)=0.5/1.5=1/3\cos(\theta/2) = 0.5/1.5 = 1/3, so θ/2=arccos⁡(1/3)≈1.23\theta/2 = \arccos(1/3) \approx 1.23 radians, giving θ≈2.46\theta \approx 2.46 radians. Segment area is 12r2(θ−sin⁡θ)≈3.53\frac{1}{2}r^2(\theta - \sin\theta) \approx 3.53 square feet. Volume is 3.53×4≈14.13.53 \times 4 \approx 14.1 cubic feet. Choice B assumes semicircular cross-section. Choice C uses wrong depth measurement. Choice D calculates full cylinder volume incorrectly.

Question 4

A rectangular swimming pool is 20 feet long, 12 feet wide, and 6 feet deep. The pool must be filled to 80% capacity to comply with safety regulations. If water costs $0.003 per cubic foot, what is the total cost to fill the pool to the required level?

  1. $3.46 (correct answer)
  2. $4.32
  3. $5.18
  4. $6.05
Explanation: Pool volume is 20×12×6=144020 \times 12 \times 6 = 1440 cubic feet. At 80% capacity: 1440×0.8=11521440 \times 0.8 = 1152 cubic feet. Cost is 1152×0.003=3.456≈$3.461152 \times 0.003 = 3.456 \approx \$3.46. Choice B uses 100% capacity instead of 80%. Choice C miscalculates the volume as using height of 4.8 feet. Choice D uses the wrong percentage (85% instead of 80%).

Question 5

A conical funnel has a top diameter of 12 inches and height of 8 inches. Sand flows through it at a rate of 2 cubic inches per second. When the sand level in the funnel is 3 inches deep, how fast is the sand level dropping?

  1. About 0.071 inches per second
  2. About 0.089 inches per second
  3. About 0.112 inches per second
  4. About 0.141 inches per second (correct answer)
Explanation: The funnel radius is 6 inches. By similar triangles, when sand depth is hh, radius is r=6h8=3h4r = \frac{6h}{8} = \frac{3h}{4}. Sand volume is V=13πr2h=13π(3h4)2h=3πh316V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{3h}{4}\right)^2 h = \frac{3\pi h^3}{16}. Taking derivative: dVdt=9πh216⋅dhdt\frac{dV}{dt} = \frac{9\pi h^2}{16} \cdot \frac{dh}{dt}. Given dVdt=−2\frac{dV}{dt} = -2 and h=3h = 3: −2=9π(9)16⋅dhdt=81π16⋅dhdt-2 = \frac{9\pi(9)}{16} \cdot \frac{dh}{dt} = \frac{81\pi}{16} \cdot \frac{dh}{dt}. So dhdt=−3281π≈−0.141\frac{dh}{dt} = \frac{-32}{81\pi} \approx -0.141 inches per second. Choice A uses wrong similar triangle ratio. Choice B miscalculates the volume formula. Choice C uses diameter instead of radius in calculations.

Question 6

A company manufactures cylindrical cans. To reduce material costs, they want to design a can that holds 355 mL (355 cubic cm) while minimizing the total surface area. If the height must be at least 8 cm for labeling requirements, what radius should they use?

  1. 2.8 cm
  2. 3.3 cm (correct answer)
  3. 3.8 cm
  4. 4.2 cm
Explanation: For fixed volume V=πr2h=355V = \pi r^2 h = 355, we have h=355πr2h = \frac{355}{\pi r^2}. Surface area is S=2πr2+2πrh=2πr2+2πr⋅355πr2=2πr2+710rS = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r \cdot \frac{355}{\pi r^2} = 2\pi r^2 + \frac{710}{r}. Taking the derivative: dSdr=4πr−710r2=0\frac{dS}{dr} = 4\pi r - \frac{710}{r^2} = 0. Solving: 4πr3=7104\pi r^3 = 710, so r3=7104π≈56.5r^3 = \frac{710}{4\pi} \approx 56.5, giving r≈3.8r \approx 3.8. But this gives h=355π(3.8)2≈7.8<8h = \frac{355}{\pi(3.8)^2} \approx 7.8 < 8. Since h≥8h \geq 8, we use h=8h = 8: 355=πr2(8)355 = \pi r^2(8), so r2=3558πr^2 = \frac{355}{8\pi} and r≈3.3r \approx 3.3 cm. Choice A uses wrong constraint. Choice C ignores height constraint. Choice D miscalculates the volume relationship.

Question 7

A cone-shaped grain silo has a base radius of 8 meters and height of 15 meters. Due to structural limitations, the silo can only be filled to 75% of its volume. If grain costs $240 per cubic meter, what is the maximum value of grain that can be stored in the silo?

  1. $60,319 (correct answer)
  2. $72,382
  3. $80,425
  4. $96,510
Explanation: Cone volume is 13πr2h=13π(8)2(15)=13π(64)(15)=320π\frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(8)^2(15) = \frac{1}{3}\pi(64)(15) = 320\pi cubic meters. At 75% capacity: 320π×0.75=240π≈753.98320\pi \times 0.75 = 240\pi \approx 753.98 cubic meters. Cost is 753.98×240≈$60,319753.98 \times 240 \approx \$60,319. Choice B uses 90% capacity. Choice C forgets the 1/3 factor for cone volume. Choice D uses 100% capacity with wrong volume formula.

Question 8

A company manufactures cone-shaped ice cream containers. Due to a machine calibration error, all containers are produced with heights that are 15% less than specified, while the radius remains correct. If a correctly manufactured container should hold 200 cubic cm of ice cream, how much ice cream will the defective containers actually hold?

  1. 170.0 cubic cm (correct answer)
  2. 154.4 cubic cm
  3. 185.0 cubic cm
  4. 162.5 cubic cm
Explanation: Let the correct radius be rr and correct height be hh. The correct volume is 13πr2h=200\frac{1}{3}\pi r^2 h = 200 cubic cm. The defective containers have the same radius rr but height 0.85h0.85h (15% reduction). The defective volume is 13πr2(0.85h)=0.85×13πr2h=0.85×200=170\frac{1}{3}\pi r^2(0.85h) = 0.85 \times \frac{1}{3}\pi r^2 h = 0.85 \times 200 = 170 cubic cm. Choice B (154.4) might result from incorrectly applying the 15% reduction to both radius and height: (0.85)2×200=144.5≈154.4(0.85)^2 \times 200 = 144.5 \approx 154.4. Choice C (185.0) might come from misunderstanding the percentage (using 92.5% instead of 85%). Choice D (162.5) could result from calculation errors in applying the percentage reduction.

Question 9

A rectangular swimming pool is 25 meters long, 12 meters wide, and 2 meters deep. The pool needs to be filled to 80% capacity, but the water source can only provide 15 cubic meters per hour. Additionally, evaporation causes a loss of 0.5 cubic meters per hour. What is the net time required to fill the pool to the desired level?

  1. 33.1 hours
  2. 41.4 hours (correct answer)
  3. 48.0 hours
  4. 52.8 hours
Explanation: First, find the volume needed: Pool volume = 25×12×2=60025 \times 12 \times 2 = 600 cubic meters. At 80% capacity: 600×0.8=480600 \times 0.8 = 480 cubic meters. Net fill rate = inflow rate - evaporation rate = 15−0.5=14.515 - 0.5 = 14.5 cubic meters per hour. Time required = 48014.5≈33.1\frac{480}{14.5} \approx 33.1 hours. Choice A incorrectly ignores evaporation (48015=32\frac{480}{15} = 32 hours). Choice C uses full capacity instead of 80% (60012.5=48\frac{600}{12.5} = 48 hours). Choice D compounds both errors.

Question 10

A triangular prism has a right triangular base with legs measuring 5 feet and 12 feet. The prism is 8 feet long. This prism must fit inside a cylindrical shipping container. What is the minimum radius of the cylinder needed to accommodate the prism if the prism's length aligns with the cylinder's height?

  1. 6.5 feet (correct answer)
  2. 7.2 feet
  3. 8.1 feet
  4. 9.0 feet
Explanation: The triangular base has legs of 5 ft and 12 ft, so the hypotenuse is 52+122=25+144=169=13\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 feet. To fit the triangular cross-section in a circular cross-section, we need the radius of the circumscribed circle around the right triangle. For a right triangle, the circumradius equals half the hypotenuse length: R=132=6.5R = \frac{13}{2} = 6.5 feet. Choice B (7.2) might result from incorrectly using the formula for an equilateral triangle. Choice C (8.1) might come from incorrectly using the semi-perimeter. Choice D (9.0) might result from using the longest side as the diameter instead of recognizing the circumradius relationship.

Question 11

A cylindrical grain silo with radius 12 feet and height 40 feet has a conical roof with height 8 feet. The silo is filled to 85% of its total storage capacity. How many cubic feet of grain are stored?

  1. About 15,240 cubic feet
  2. About 17,320 cubic feet
  3. About 19,580 cubic feet (correct answer)
  4. About 21,840 cubic feet
Explanation: Cylindrical volume: π(12)2(40)=5760π\pi(12)^2(40) = 5760\pi cubic feet. Conical roof volume: 13π(12)2(8)=1152π3=384π\frac{1}{3}\pi(12)^2(8) = \frac{1152\pi}{3} = 384\pi cubic feet. Total capacity: 5760π+384π=6144π5760\pi + 384\pi = 6144\pi cubic feet. At 85% capacity: 6144π×0.85=5222.4π≈19,5806144\pi \times 0.85 = 5222.4\pi \approx 19,580 cubic feet. Choice A uses only cylindrical volume at 85%. Choice B forgets the conical roof entirely. Choice D uses 100% capacity instead of 85%.

Question 12

A hemispherical dome has an interior radius of 20 feet. The dome is filled with water to a depth of 8 feet from the bottom. What is the volume of water in the dome?

  1. About 1,675 cubic feet (correct answer)
  2. About 2,144 cubic feet
  3. About 2,513 cubic feet
  4. About 2,890 cubic feet
Explanation: For a spherical cap of height hh in a sphere of radius RR: V=πh23(3R−h)V = \frac{\pi h^2}{3}(3R - h). Here h=8h = 8 and R=20R = 20: V=π(8)23(3(20)−8)=64π3(60−8)=64π3(52)=3328π3≈1,675V = \frac{\pi(8)^2}{3}(3(20) - 8) = \frac{64\pi}{3}(60 - 8) = \frac{64\pi}{3}(52) = \frac{3328\pi}{3} \approx 1,675 cubic feet. Choice B uses wrong radius value. Choice C applies full hemisphere formula incorrectly. Choice D miscalculates the cap height relationship.