Math 3 Quiz: 3d Solid Volumes
15 questions · exam conditions
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3d Solid VolumesQuestion 1 of 15

A cube has the same volume as a sphere with radius 6 cm. What is the edge length of the cube to the nearest tenth of a centimeter?

7.8 cm
8.4 cm
9.7 cm
9.1 cm
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Math 3 Quiz

Math 3 Quiz: 3d Solid Volumes

Practice 3d Solid Volumes 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 3d Solid Volumes, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

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.

All questions

Question 1

A cube has the same volume as a sphere with radius 6 cm. What is the edge length of the cube to the nearest tenth of a centimeter?

  1. 7.8 cm
  2. 8.4 cm
  3. 9.7 cm
  4. 9.1 cm (correct answer)
Explanation: When you encounter problems involving equal volumes between different shapes, you need to set up an equation using their respective volume formulas and solve for the unknown dimension. Start by finding the sphere's volume using V=43πr3V = \frac{4}{3}\pi r^3. With radius 6 cm: V=43π(6)3=43π(216)=288πV = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi cubic cm. Since the cube has the same volume, and a cube's volume is V=s3V = s^3 where ss is the edge length, you can set up: s3=288πs^3 = 288\pi. Taking the cube root: s=288π3=288×3.141593=904.783≈9.1s = \sqrt[3]{288\pi} = \sqrt[3]{288 \times 3.14159} = \sqrt[3]{904.78} \approx 9.1 cm. Looking at the wrong answers: Choice A (7.8 cm) results from incorrectly using the sphere's surface area formula instead of volume, or making calculation errors early in the process. Choice B (8.4 cm) comes from forgetting the 43\frac{4}{3} coefficient in the sphere volume formula and just using πr3\pi r^3. Choice C (9.7 cm) occurs when students make rounding errors during intermediate steps or use an imprecise value of π. The correct answer is D (9.1 cm). Study tip: For volume equality problems, always write out both volume formulas completely before substituting values. Double-check that you're using volume formulas, not surface area ones, and be careful with π calculations—use at least 3.14159 for accuracy in intermediate steps before final rounding.

Question 2

A cylindrical water tank has a radius of 3 meters and a height of 8 meters. Due to maintenance, the water level is kept at 75% of the tank's capacity. If water is being pumped out at a rate that reduces the volume by 16\frac{1}{6} every hour, what volume of water remains after 2 hours?

  1. 27π27\pi cubic meters
  2. 36π36\pi cubic meters
  3. 45π45\pi cubic meters (correct answer)
  4. 54π54\pi cubic meters
Explanation: First, find the tank's total volume: V=πr2h=π(32)(8)=72πV = \pi r^2 h = \pi(3^2)(8) = 72\pi cubic meters. At 75% capacity: 0.75×72π=54π0.75 \times 72\pi = 54\pi cubic meters initially. Each hour, 16\frac{1}{6} of the current volume is removed, so 56\frac{5}{6} remains. After 2 hours: 54π×(56)2=54π×2536=45π54\pi \times (\frac{5}{6})^2 = 54\pi \times \frac{25}{36} = 45\pi cubic meters. Choice A assumes linear reduction rather than exponential. Choice B uses only one hour of reduction. Choice D is the initial volume.

Question 3

A hollow cylinder has an outer radius of 8 cm, inner radius of 5 cm, and height of 12 cm. If this cylinder is filled with water and then a solid sphere with radius 4 cm is completely submerged in it, what volume of water overflows?

  1. 256π3\frac{256\pi}{3} cubic centimeters
  2. 64π3\frac{64\pi}{3} cubic centimeters
  3. 00 cubic centimeters (correct answer)
  4. 64π64\pi cubic centimeters
Explanation: Hollow cylinder volume: π(82−52)(12)=π(64−25)(12)=468π\pi(8^2 - 5^2)(12) = \pi(64-25)(12) = 468\pi cubic cm. Sphere volume: 43π(43)=256π3≈268.1π\frac{4}{3}\pi(4^3) = \frac{256\pi}{3} \approx 268.1\pi cubic cm. Since the sphere volume (268.1π268.1\pi) is less than the cylinder capacity (468π468\pi), no water overflows. Choice A is the sphere volume. Choice B results from incorrect calculation. Choice D uses wrong sphere volume formula.

Question 4

A sphere has a surface area of 144π144\pi square centimeters. What is the volume of this sphere in cubic centimeters?

  1. 96π96\pi cubic centimeters
  2. 144π144\pi cubic centimeters
  3. 192π192\pi cubic centimeters
  4. 288π288\pi cubic centimeters (correct answer)
Explanation: First, find the radius using surface area formula SA=4πr2SA = 4\pi r^2. So 144π=4πr2144\pi = 4\pi r^2, which gives r2=36r^2 = 36, so r=6r = 6 cm. Then use the volume formula V=43πr3=43π(6)3=43π(216)=288πV = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi cubic centimeters. Choice A results from incorrectly using V=23πr3V = \frac{2}{3}\pi r^3. Choice B assumes the volume equals the surface area numerically. Choice C comes from using V=πr3V = \pi r^3 instead of 43πr3\frac{4}{3}\pi r^3.

Question 5

A rectangular prism has dimensions 6 inches by 8 inches by 10 inches. A smaller rectangular prism with dimensions 2 inches by 3 inches by 4 inches is removed from one corner. What is the volume of the remaining solid?

  1. 456 cubic inches (correct answer)
  2. 460 cubic inches
  3. 464 cubic inches
  4. 480 cubic inches
Explanation: The volume of the large prism is 6×8×10=4806 \times 8 \times 10 = 480 cubic inches. The volume of the removed small prism is 2×3×4=242 \times 3 \times 4 = 24 cubic inches. The remaining volume is 480−24=456480 - 24 = 456 cubic inches. Choice B results from miscalculating the small prism volume as 20 instead of 24. Choice C comes from calculating the small prism volume as 16. Choice D is the original volume without subtracting the removed piece.

Question 6

A cylindrical container has a radius of 6 inches and contains water to a height of 8 inches. If a solid sphere with radius 3 inches is completely submerged in the water, by how many inches will the water level rise?

  1. 12\frac{1}{2} inch
  2. 1 inch (correct answer)
  3. 32\frac{3}{2} inches
  4. 2 inches
Explanation: The volume of the sphere is V=43πr3=43π(3)3=36πV = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3)^3 = 36\pi cubic inches. This volume will displace an equal volume of water in the cylinder. The cross-sectional area of the cylinder is π(6)2=36π\pi(6)^2 = 36\pi square inches. The rise in water level is displaced volumecross-sectional area=36π36π=1\frac{\text{displaced volume}}{\text{cross-sectional area}} = \frac{36\pi}{36\pi} = 1 inch. Choice A uses radius 3 instead of volume calculation. Choice C incorrectly uses 43×1\frac{4}{3} \times 1. Choice D doubles the correct answer.

Question 7

A rectangular prism has dimensions 4 cm × 6 cm × 9 cm. If each dimension is increased by the same factor kk, and the new volume is 1728 cubic cm, what is the value of kk?

  1. 2 (correct answer)
  2. 3
  3. 4
  4. 8
Explanation: Original volume: V1=4×6×9=216V_1 = 4 \times 6 \times 9 = 216 cubic cm. After scaling by factor kk, the new dimensions are 4k4k, 6k6k, and 9k9k. New volume: V2=4k×6k×9k=216k3=1728V_2 = 4k \times 6k \times 9k = 216k^3 = 1728. Solving: k3=1728216=8k^3 = \frac{1728}{216} = 8, so k=2k = 2. Choice B results from incorrectly thinking k3=3k^3 = 3. Choice C comes from 8\sqrt{8} instead of 83\sqrt[3]{8}. Choice D is the value of k3k^3, not kk.

Question 8

A regular tetrahedron has edge length aa. If the volume of this tetrahedron is 18218\sqrt{2} cubic units, what is the volume of a cube with the same edge length aa?

  1. 108108 cubic units
  2. 216216 cubic units (correct answer)
  3. 7272 cubic units
  4. 144144 cubic units
Explanation: This problem tests your ability to work with 3D geometric formulas and use given information to find unknown quantities. When you see a question connecting different geometric shapes with the same dimensions, focus on finding the shared parameter first. Start with the volume formula for a regular tetrahedron: V=a3212V = \frac{a^3\sqrt{2}}{12}, where aa is the edge length. Since the tetrahedron's volume is 18218\sqrt{2}, you can set up the equation: a3212=182\frac{a^3\sqrt{2}}{12} = 18\sqrt{2} Divide both sides by 2\sqrt{2}: a312=18\frac{a^3}{12} = 18 Multiply by 12: a3=216a^3 = 216 Therefore a=6a = 6. Now find the cube's volume using V=a3=63=216V = a^3 = 6^3 = 216 cubic units. Looking at the wrong answers: Choice A (108) represents half the correct volume, likely from forgetting to cube the edge length or making an arithmetic error. Choice C (72) might result from incorrectly using a2⋅6a^2 \cdot 6 instead of a3a^3, mixing up area and volume formulas. Choice D (144) could come from calculation errors when solving for a3a^3 or incorrectly manipulating the tetrahedron formula. The correct answer is B: 216 cubic units. Strategy tip: When problems give you one shape's properties to find another's, always solve for the shared dimension first. Master the standard volume formulas for common 3D shapes, especially cubes (a3a^3) and regular tetrahedra (a3212\frac{a^3\sqrt{2}}{12}), as these appear frequently together in geometry problems.

Question 9

A sphere is inscribed in a cube with edge length 6 cm. What is the ratio of the volume of the sphere to the volume of the cube?

  1. π6\frac{\pi}{6} (correct answer)
  2. π8\frac{\pi}{8}
  3. π4\frac{\pi}{4}
  4. π2\frac{\pi}{2}
Explanation: An inscribed sphere touches all faces of the cube, so its diameter equals the cube's edge length. The sphere's radius is r=3r = 3 cm. Sphere volume: Vs=43πr3=43π(33)=36πV_s = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3^3) = 36\pi cubic cm. Cube volume: Vc=63=216V_c = 6^3 = 216 cubic cm. Ratio: 36π216=π6\frac{36\pi}{216} = \frac{\pi}{6}. Choice B results from using 43πr3\frac{4}{3}\pi r^3 incorrectly as πr3\pi r^3. Choice C comes from confusing diameter with radius. Choice D uses the wrong sphere volume formula.

Question 10

A spherical balloon has a volume of 36π36\pi cubic inches. If the balloon is inflated until its radius doubles, by what factor does the surface area increase?

  1. 2
  2. 4 (correct answer)
  3. 6
  4. 8
Explanation: Original volume: 43πr3=36π\frac{4}{3}\pi r^3 = 36\pi, so r3=27r^3 = 27 and r=3r = 3 inches. Original surface area: 4πr2=4π(32)=36π4\pi r^2 = 4\pi(3^2) = 36\pi square inches. After doubling the radius to 6 inches: New surface area: 4π(62)=144π4\pi(6^2) = 144\pi square inches. Factor of increase: 144π36π=4\frac{144\pi}{36\pi} = 4. Choice A is the radius factor, not surface area. Choice C confuses volume and surface area relationships. Choice D is the volume factor increase.

Question 11

A hemispherical bowl has an inner radius of 15 cm. It is filled with water to a depth of 9 cm from the bottom. What is the volume of water in the bowl?

  1. 324π324\pi cubic centimeters
  2. 486π486\pi cubic centimeters
  3. 729π729\pi cubic centimeters
  4. 972π972\pi cubic centimeters (correct answer)
Explanation: This requires finding the volume of a spherical cap with height h=9h = 9 cm and sphere radius R=15R = 15 cm. The formula is V=πh23(3R−h)=π(92)3(3×15−9)=81π3(45−9)=27π×36=972πV = \frac{\pi h^2}{3}(3R - h) = \frac{\pi (9^2)}{3}(3 \times 15 - 9) = \frac{81\pi}{3}(45 - 9) = 27\pi \times 36 = 972\pi cubic cm. Choice A uses πh2R/3\pi h^2 R/3. Choice B uses πh2(2R)3\frac{\pi h^2(2R)}{3}. Choice C uses πh3\pi h^3.

Question 12

A square pyramid has a base with side length 10 feet and a height of 12 feet. What is the volume of the pyramid in cubic feet?

  1. 120 cubic feet
  2. 360 cubic feet
  3. 400 cubic feet (correct answer)
  4. 1200 cubic feet
Explanation: The volume of a pyramid is V=13BhV = \frac{1}{3}Bh where BB is the base area and hh is the height. The base area is 102=10010^2 = 100 square feet. So V=13(100)(12)=12003=400V = \frac{1}{3}(100)(12) = \frac{1200}{3} = 400 cubic feet. Choice A uses only base area times height divided by 10. Choice B uses 13×10×12×9\frac{1}{3} \times 10 \times 12 \times 9, confusing side length with area. Choice D forgets the 13\frac{1}{3} factor entirely.

Question 13

A cone and a cylinder have the same base radius of 5 meters and the same height of 9 meters. How many times greater is the volume of the cylinder than the volume of the cone?

  1. 2 times greater
  2. 3 times greater (correct answer)
  3. 4 times greater
  4. 9 times greater
Explanation: The volume of a cylinder is Vcylinder=πr2h=π(5)2(9)=225πV_{cylinder} = \pi r^2 h = \pi(5)^2(9) = 225\pi. The volume of a cone is Vcone=13πr2h=13π(5)2(9)=75πV_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(5)^2(9) = 75\pi. The ratio is 225π75π=3\frac{225\pi}{75\pi} = 3. Choice A results from using 12\frac{1}{2} instead of 13\frac{1}{3} in the cone formula. Choice C comes from incorrectly comparing surface areas or using wrong formulas. Choice D uses the height value as the multiplier.

Question 14

Two spheres have radii in the ratio 2:3. If the smaller sphere has a volume of 32π32\pi cubic units, what is the volume of the larger sphere?

  1. 48π48\pi cubic units
  2. 72π72\pi cubic units
  3. 108π108\pi cubic units (correct answer)
  4. 216π216\pi cubic units
Explanation: When radii are in ratio 2:3, volumes are in ratio 23:33=8:272^3:3^3 = 8:27. If the smaller volume is 32π32\pi, then 32π8=4π\frac{32\pi}{8} = 4\pi per unit ratio. The larger volume is 4π×27=108π4\pi \times 27 = 108\pi cubic units. Alternatively, volume scales as the cube of linear dimensions: Vlarge=Vsmall×(32)3=32π×278=108πV_{large} = V_{small} \times (\frac{3}{2})^3 = 32\pi \times \frac{27}{8} = 108\pi. Choice A uses ratio 3:2 instead of (32)3(\frac{3}{2})^3. Choice B uses ratio squared instead of cubed. Choice D uses 32π×27432\pi \times \frac{27}{4}.

Question 15

A cylindrical water tank has a diameter of 8 feet and a height of 12 feet. If the tank is currently filled to 75% of its capacity, how many cubic feet of water are in the tank?

  1. 144π144\pi cubic feet (correct answer)
  2. 192π192\pi cubic feet
  3. 288π288\pi cubic feet
  4. 576π576\pi cubic feet
Explanation: The volume of a cylinder is V=πr2hV = \pi r^2 h. With diameter 8 feet, the radius is 4 feet. The full volume is V=π(4)2(12)=192πV = \pi(4)^2(12) = 192\pi cubic feet. At 75% capacity, the volume is 0.75×192π=144π0.75 \times 192\pi = 144\pi cubic feet. Choice B gives the full tank volume without applying the 75% factor. Choice C incorrectly uses diameter instead of radius: π(8)2(12)×0.75=576π×0.75=432π\pi(8)^2(12) \times 0.75 = 576\pi \times 0.75 = 432\pi, but this doesn't match any calculation. Choice D uses diameter instead of radius for the full volume: π(8)2(12)=768π\pi(8)^2(12) = 768\pi, then incorrectly calculates 768π×0.75=576π768\pi \times 0.75 = 576\pi.