Middle School Math Quiz: Apply Volume Formulas
20 questions · exam conditions
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Apply Volume FormulasQuestion 1 of 20

A science class uses a cylindrical container to store sand. The container has radius 3 cm3\text{ cm} and height 10 cm10\text{ cm}. What is the volume of the container? (Give your answer in terms of π\pi or as an approximation.)

900π cm3 (≈2826 cm3)900\pi\text{ cm}^3\ (\approx 2826\text{ cm}^3)
90π cm3 (≈283 cm3)90\pi\text{ cm}^3\ (\approx 283\text{ cm}^3)
60π cm3 (≈188 cm3)60\pi\text{ cm}^3\ (\approx 188\text{ cm}^3)
30π cm3 (≈94 cm3)30\pi\text{ cm}^3\ (\approx 94\text{ cm}^3)
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Middle School Math Quiz

Middle School Math Quiz: Apply Volume Formulas

Practice Apply Volume Formulas in Middle School Math 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 Apply Volume Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

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 science class uses a cylindrical container to store sand. The container has radius 3 cm3\text{ cm} and height 10 cm10\text{ cm}. What is the volume of the container? (Give your answer in terms of π\pi or as an approximation.)

  1. 900π cm3 (≈2826 cm3)900\pi\text{ cm}^3\ (\approx 2826\text{ cm}^3)
  2. 90π cm3 (≈283 cm3)90\pi\text{ cm}^3\ (\approx 283\text{ cm}^3) (correct answer)
  3. 60π cm3 (≈188 cm3)60\pi\text{ cm}^3\ (\approx 188\text{ cm}^3)
  4. 30π cm3 (≈94 cm3)30\pi\text{ cm}^3\ (\approx 94\text{ cm}^3)
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=3 cm and h=10 cm, V=π(3²)(10)=π(9)(10)=90π≈282.7 cm³. Common errors include using the cone formula by mistake (adding 1/3), treating radius as diameter, incorrect squaring (like 3²=6), or forgetting to cube the units (cm instead of cm³). Steps: (1) identify shape as cylinder (circular base and constant height), (2) gather dimensions (r=3 cm, h=10 cm), (3) select formula πr²h, (4) substitute values, (5) calculate 910π=90π, (6) add cubic units cm³. Always double-check arithmetic and formula selection to avoid mistakes like confusing cylinder with cone.

Question 2

A model planet is a sphere with diameter 10 ft10\text{ ft}. What is its volume? (Give your answer in terms of π\pi or as an approximation.)

  1. 2503π ft3 (≈262 ft3)\frac{250}{3}\pi\text{ ft}^3\ (\approx 262\text{ ft}^3)
  2. 40003π ft3 (≈4189 ft3)\frac{4000}{3}\pi\text{ ft}^3\ (\approx 4189\text{ ft}^3)
  3. 500π ft3 (≈1570 ft3)500\pi\text{ ft}^3\ (\approx 1570\text{ ft}^3)
  4. 5003π ft3 (≈524 ft3)\frac{500}{3}\pi\text{ ft}^3\ (\approx 524\text{ ft}^3) (correct answer)
Explanation: This question tests applying volume formulas: cylinder V=πr2hV=πr^2 h, cone V=13πr2hV=\frac{1}{3}πr^2 h, sphere V=43πr3V=\frac{4}{3}πr^3, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr2πr^2 times height hh giving V=πr2hV=πr^2 h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=13πr2hV=\frac{1}{3}πr^2 h (tapers to point reducing volume to one-third), sphere is V=43πr3V=\frac{4}{3}πr^3 (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2r=d/2), calculate (exponents first, multiply, approximate π≈3.14π≈3.14 or leave exact). For this sphere with d=10 ft so r=5 ft, V=43π(53)=43π(125)=5003π≈523.6 ft3V=\frac{4}{3}π(5^3)=\frac{4}{3}π(125)=\frac{500}{3}π≈523.6 \text{ ft}^3. Common errors include using diameter as radius (r=10r=10), forgetting to halve diameter, wrong exponent (r2r^2), calculation mistakes (53=1255^3=125 not 25), or incorrect units. Steps: (1) identify shape as sphere, (2) gather dimensions (r=5 ft from d=10), (3) select formula 43πr3\frac{4}{3}πr^3, (4) substitute values, (5) calculate (43∗125π=5003π\frac{4}{3}*125π=\frac{500}{3}π), (6) add cubic units ft³. Always convert diameter to radius to avoid doubling errors.

Question 3

A spherical ornament has volume 288π cm3288\pi\text{ cm}^3. What is the radius of the sphere?

  1. r=3 cmr=3\text{ cm}
  2. r=12 cmr=12\text{ cm}
  3. r=6 cmr=6\text{ cm} (correct answer)
  4. r=9 cmr=9\text{ cm}
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). To find r for sphere V=288π, set (4/3)πr³=288π, divide by π, multiply by 3/4, r³=216, r=6 cm. Common errors include using wrong formula (like πr²h), incorrect solving (forgetting cube root), or arithmetic mistakes (2161/3216^{1/3}=6). Steps: (1) identify sphere, (2) use V=(4/3)πr³, (3) set equal to 288π, (4) solve r³=288*(3/4)=216, (5) r=∛216=6, (6) check units cm. Solving backwards requires isolating r and taking cube root accurately.

Question 4

A spherical ornament has diameter 10 ft10\text{ ft}. What is its volume, rounded to the nearest cubic foot? (Use π≈3.14\pi\approx 3.14.)

  1. ≈524 ft3\approx 524\text{ ft}^3 (correct answer)
  2. ≈314 ft3\approx 314\text{ ft}^3
  3. ≈419 ft3\approx 419\text{ ft}^3
  4. ≈1047 ft3\approx 1047\text{ ft}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as sphere, use corresponding formula with given dimensions (convert diameter to radius: r=10/2=5 ft), calculate (exponents first, multiply, approximate π≈3.14, round). Example: sphere r=5 ft using V=(4/3)π(125)≈(4/3)3.14125≈523.33, rounded to 524 ft³. Correct formula selection and calculation yield ≈524 ft³ for this sphere. Error like using diameter as radius (r=10, V=(4/3)π1000≈4186.67), wrong exponent, arithmetic error, or incorrect rounding. Steps: (1) identify shape (sphere), (2) gather dimensions (r=5), (3) select formula ((4/3)πr³), (4) substitute, (5) calculate (5³=125, 1254/3≈166.67, *3.14≈523.33, round to 524), (6) units (cubic: ft³).

Question 5

A cone-shaped paper cup has a diameter of 8 cm and a height of 12 cm. If the cup is filled to exactly 34\frac{3}{4} of its height with water, what fraction of the total cup volume is occupied by water?

  1. The water occupies exactly 2764\frac{27}{64} of the total cup volume (correct answer)
  2. The water occupies exactly 916\frac{9}{16} of the total cup volume
  3. The water occupies exactly 34\frac{3}{4} of the total cup volume
  4. The water occupies exactly 12\frac{1}{2} of the total cup volume
Explanation: This involves similar cones. The water forms a smaller cone with height 34×12=9\frac{3}{4} \times 12 = 9 cm. By similar triangles, if the original radius is 4 cm, the water cone's radius is 912×4=3\frac{9}{12} \times 4 = 3 cm. Volume scales as the cube of the linear scale factor. The scale factor is 34\frac{3}{4}, so the volume fraction is (34)3=2764(\frac{3}{4})^3 = \frac{27}{64}. Choice C incorrectly assumes volume scales linearly with height. Choice B uses (34)2(\frac{3}{4})^2. Choice D is 12\frac{1}{2}, which doesn't relate to the given fraction.

Question 6

A science club is building a cylindrical container with radius 3 cm3\text{ cm} and height 10 cm10\text{ cm}. What is the volume of the cylinder? Give your answer in terms of π\pi and in cubic centimeters.

  1. 180π cm3180\pi\text{ cm}^3
  2. 283 cm3283\text{ cm}^3
  3. 90π cm390\pi\text{ cm}^3 (correct answer)
  4. 30π cm330\pi\text{ cm}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=3 cm and h=10 cm, V=π(3²)(10)=π(9)(10)=90π cm³, matching choice B. Common errors include using diameter instead of radius, wrong exponent like r³ for cylinder, arithmetic mistakes such as 3²=6, or forgetting cubic units. Steps: (1) identify cylinder, (2) gather r=3 cm, h=10 cm, (3) select V=πr²h, (4) substitute values, (5) calculate 90π, (6) add cm³. Remember, leaving in terms of π keeps it exact, as required here.

Question 7

A cylindrical water tank has radius 2 m2\text{ m} and height 5 m5\text{ m}. About how much water can it hold? (Give your answer in terms of π\pi and in cubic meters.)

  1. 20π m320\pi\text{ m}^3 (correct answer)
  2. 203π m3\tfrac{20}{3}\pi\text{ m}^3
  3. 40π m340\pi\text{ m}^3
  4. 10π m310\pi\text{ m}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with r=2 m and h=5 m, V=π(2²)(5)=π(4)(5)=20π m³, matching choice B. Common errors include using cone formula for 20/3 π, doubling radius incorrectly, arithmetic like 2²=2, or non-cubic units. Steps: (1) identify cylinder, (2) gather r=2 m, h=5 m, (3) select V=πr²h, (4) substitute values, (5) calculate 20π, (6) add m³. This represents the tank's capacity in exact terms with π.

Question 8

Which formula correctly gives the volume of a cylinder with radius rr and height hh?

  1. V=43πr3V=\tfrac{4}{3}\pi r^3
  2. V=2πrhV=2\pi r h
  3. V=13πr2hV=\tfrac{1}{3}\pi r^2 h
  4. V=πr2hV=\pi r^2 h (correct answer)
Explanation: This question tests applying volume formulas: cylinder V=πr2hV=\pi r^2 h, cone V=13πr2hV=\frac{1}{3} \pi r^2 h, sphere V=43πr3V=\frac{4}{3} \pi r^3, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr2\pi r^2 times height hh giving V=πr2hV=\pi r^2 h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=13πr2hV=\frac{1}{3} \pi r^2 h (tapers to point reducing volume to one-third), sphere is V=43πr3V=\frac{4}{3} \pi r^3 (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2r=d/2), calculate (exponents first, multiply, approximate π≈3.14\pi \approx 3.14 or leave exact). The correct cylinder formula is V=πr2hV=\pi r^2 h, matching choice B, representing base area times height. Common errors include adding 1/3 like cone, using r3r^3 like sphere, or surface area like 2πrh2 \pi r h. Steps: (1) identify cylinder by uniform height, (2) recall πr2h\pi r^2 h, (3) select from options, (4) avoid extras like fractions unless tapered. Formulas derive from geometry, so visualize shapes.

Question 9

A playground ball is a sphere with radius 6 m6\text{ m}. What is the volume of the ball? (Give the answer in terms of π\pi.)

  1. 288π m3288\pi\text{ m}^3 (correct answer)
  2. 432π m3432\pi\text{ m}^3
  3. 144π m3144\pi\text{ m}^3
  4. 216π m3216\pi\text{ m}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as sphere, use corresponding formula with given dimensions (radius given), calculate (exponents first, multiply, leave exact in terms of π). Example: sphere r=6 m using V=(4/3)π(6³)=(4/3)π(216)=288π m³. Correct formula selection and calculation yield 288π m³ for this sphere. Error like wrong formula (using r² instead of r³, like (4/3)π36=48π), arithmetic error (6³=216 mistaken as 36), or forgetting the 4/3 factor. Steps: (1) identify shape (sphere, perfect roundness), (2) gather dimensions (r=6), (3) select formula ((4/3)πr³), (4) substitute, (5) calculate (6³=216, 2164/3=288, times π), (6) units (cubic: m³).

Question 10

A snow cone is shaped like a cone with radius 4 in4\text{ in} and height 9 in9\text{ in}. What is the volume of the cone? (Give the answer in terms of π\pi.)

  1. 48π in348\pi\text{ in}^3 (correct answer)
  2. 144π in3144\pi\text{ in}^3
  3. 96π in396\pi\text{ in}^3
  4. 16π in316\pi\text{ in}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cone, use corresponding formula with given dimensions (radius given, no need to convert), calculate (exponents first, multiply, leave exact in terms of π). Example: cone r=4 in, h=9 in using V=(1/3)π(4²)(9)=(1/3)π(16)(9)=48π in³. Correct formula selection and calculation yield 48π in³ for this cone. Error like wrong formula (cylinder π169=144π, missing 1/3), using diameter as radius (if misread), wrong exponent (r instead of r²), arithmetic error (169=128), or units not cubed. Steps: (1) identify shape (cone with circular base and point), (2) gather dimensions (r=4, h=9), (3) select formula ((1/3)πr²h), (4) substitute, (5) calculate (169=144, 144/3=48, times π), (6) units (cubic: in³).

Question 11

A school has a cylindrical water tank with radius 2 m2\text{ m} and height 5 m5\text{ m}. About how much water can it hold?

  1. 10π m3 (≈31 m3)10\pi\text{ m}^3\ (\approx 31\text{ m}^3)
  2. 20π m3 (≈63 m3)20\pi\text{ m}^3\ (\approx 63\text{ m}^3) (correct answer)
  3. 40π m3 (≈126 m3)40\pi\text{ m}^3\ (\approx 126\text{ m}^3)
  4. 203π m3 (≈21 m3)\frac{20}{3}\pi\text{ m}^3\ (\approx 21\text{ m}^3)
Explanation: This question tests applying the correct volume formula for a cylinder, V=πr2hV=\pi r^2 h. With r=2r=2 m and h=5h=5 m, V=π(22)(5)=π(4)(5)=20π≈62.8V=\pi(2^2)(5)=\pi(4)(5)=20\pi \approx 62.8 m^3. Common errors include using the cone formula (one third of this value), using the diameter instead of the radius, or making an exponent mistake. Squaring the radius before multiplying by the height, then by pi, gives the correct volume of 20π20\pi cubic meters, about 63 cubic meters.

Question 12

A conical tent has a base diameter of 14 feet and a slant height of 25 feet. The tent manufacturer needs to know the volume to determine ventilation requirements. What is the volume of the tent?

  1. The tent has a volume of approximately 1,078 cubic feet for ventilation calculations
  2. The tent has a volume of approximately 1,232 cubic feet for ventilation calculations (correct answer)
  3. The tent has a volume of approximately 896 cubic feet for ventilation calculations
  4. The tent has a volume of approximately 1,155 cubic feet for ventilation calculations
Explanation: First, find the vertical height using the Pythagorean theorem. Radius = 7 feet, slant height = 25 feet. h2+72=252h^2 + 7^2 = 25^2, so h2+49=625h^2 + 49 = 625, giving h2=576h^2 = 576 and h=24h = 24 feet. Volume: V=13πr2h=13π(72)(24)=13π(49)(24)=1,176π3=392π≈1,232V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(7^2)(24) = \frac{1}{3}\pi(49)(24) = \frac{1,176\pi}{3} = 392\pi \approx 1,232 cubic feet. Choice A might result from using slant height instead of vertical height. Choice C could come from calculation errors. Choice D might result from rounding errors or using approximate values for π.

Question 13

A classroom has a cylindrical container for storing rice. The container has radius 3 cm3\text{ cm} and height 10 cm10\text{ cm}. What is the volume of the container? (Give the answer in terms of π\pi.)

  1. 60π cm360\pi\text{ cm}^3
  2. 90π cm390\pi\text{ cm}^3 (correct answer)
  3. 900π cm3900\pi\text{ cm}^3
  4. 30π cm330\pi\text{ cm}^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use corresponding formula with given dimensions (radius given, no need to convert), calculate (exponents first, multiply, leave exact in terms of π). Example: cylinder r=3 cm, h=10 cm using V=π(3²)(10)=π(9)(10)=90π cm³. Correct formula selection and calculation yield 90π cm³ for this cylinder. Error like not squaring radius (π310=30π), using wrong formula like for cone ((1/3)π910=30π), arithmetic error (3²=6), or units not cubed (cm vs cm³). Steps: (1) identify shape (cylinder with circular base and height), (2) gather dimensions (r=3, h=10), (3) select formula (πr²h), (4) substitute (plug values: r=3, h=10), (5) calculate (exponents: 3²=9, multiply: π×9×10=90π), (6) units (cubic: cm³).

Question 14

Which shape has the greater volume?

Shape 1: a cylinder with radius 3 cm3\text{ cm} and height 4 cm4\text{ cm}

Shape 2: a sphere with radius 3 cm3\text{ cm}

  1. Not enough information to compare.
  2. The sphere has greater volume.
  3. They have equal volume. (correct answer)
  4. The cylinder has greater volume.
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For comparison, cylinder V=π(3²)(4)=36π, sphere V=(4/3)π(3³)=36π, so equal volumes. Common errors include wrong formulas (cylinder as (1/3)πr²h), exponent mix-ups (sphere r²), or not calculating both. Steps: (1) identify shapes, (2) gather dimensions, (3) select formulas, (4) substitute and calculate each (36π both), (5) compare numerically, (6) conclude equality. Comparing requires computing both volumes fully to see they match.

Question 15

A cylindrical pencil holder has diameter 10 cm10\text{ cm} and height 8 cm8\text{ cm}. What is its volume? (Give the answer in terms of π\pi.)

  1. 100π cm3100\pi\text{ cm}^3
  2. 400π cm3400\pi\text{ cm}^3
  3. 800π cm3800\pi\text{ cm}^3
  4. 200π cm3200\pi\text{ cm}^3 (correct answer)
Explanation: This question tests applying volume formulas: cylinder V=πr2hV=πr^2 h, cone V=13πr2hV=\frac{1}{3}πr^2 h, sphere V=43πr3V=\frac{4}{3}πr^3, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr2πr^2 times height hh giving V=πr2hV=πr^2 h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=13πr2hV=\frac{1}{3}πr^2 h (tapers to point reducing volume to one-third), sphere is V=43πr3V=\frac{4}{3}πr^3 (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use formula with dimensions (convert diameter to radius: r=10/2=5r=10/2=5 cm), calculate (exponents first, multiply, leave exact). Example: cylinder r=5r=5 cm, h=8h=8 cm using V=π(25)(8)=200πV=π(25)(8)=200π cm³. Correct formula selection and calculation yield 200π200π cm³. Error like using diameter as radius (r=10r=10, π∗100∗8=800ππ*100*8=800π), wrong formula, or arithmetic (52=205^2=20 mistaken). Steps: (1) identify cylinder, (2) gather r=5r=5, h=8h=8, (3) select πr2hπr^2 h, (4) substitute, (5) calculate (25∗8=20025*8=200, times ππ), (6) units cm³.

Question 16

A science class uses a cylindrical water tank with radius 2 m2\text{ m} and height 5 m5\text{ m}. About how much water can it hold? (Use π≈3.14\pi\approx 3.14 and round to the nearest cubic meter.)

  1. ≈31 m3\approx 31\text{ m}^3
  2. ≈126 m3\approx 126\text{ m}^3
  3. ≈20 m3\approx 20\text{ m}^3
  4. ≈63 m3\approx 63\text{ m}^3 (correct answer)
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape as cylinder, use corresponding formula with given dimensions (radius given), calculate (exponents first, multiply, approximate π≈3.14, round). Example: cylinder r=2 m, h=5 m using V=π(4)(5)=20π≈203.14=62.8, rounded to 63 m³. Correct formula selection and calculation yield ≈63 m³ for this cylinder. Error like not squaring radius (π25=10π≈31), using cone formula ((1/3)π45≈20.93), or wrong rounding. Steps: (1) identify shape (cylinder), (2) gather dimensions (r=2, h=5), (3) select formula (πr²h), (4) substitute, (5) calculate (2²=4, 45=20, 20*3.14=62.8, round to 63), (6) units (cubic: m³).

Question 17

Which formula correctly gives the volume of a cone with radius rr and height hh?

  1. V=πr2hV=\pi r^2h
  2. V=13πr2hV=\tfrac{1}{3}\pi r^2h (correct answer)
  3. V=πr3hV=\pi r^3h
  4. V=43πr3V=\tfrac{4}{3}\pi r^3
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). The correct cone formula is V=(1/3)πr²h, matching choice C, as it accounts for the tapering volume. Common errors include confusing with cylinder (no 1/3), sphere (r³), or made-up like πr³h. Steps: (1) identify cone by its point and base, (2) recall need for 1/3 factor, (3) select (1/3)πr²h, (4) distinguish from others. Memorize formulas to avoid mixing cone and cylinder.

Question 18

Which formula gives the volume of a cylinder with radius rr and height hh?

  1. V=43πr3V=\frac{4}{3}\pi r^3
  2. V=πr2hV=\pi r^2h (correct answer)
  3. V=13πr2hV=\frac{1}{3}\pi r^2h
  4. V=πr3hV=\pi r^3h
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For a cylinder, the correct formula is V=πr²h, based on base area times height. Common errors include selecting cone formula (with 1/3), sphere (with r³), or invented ones like πr³h. Steps: (1) identify shape as cylinder, (2) recall need for r and h, (3) choose πr²h from options, (4) verify against geometry (no 1/3, no r³), (5) distinguish from similar shapes, (6) memorize for application. Understanding derivations helps select the right formula without confusion.

Question 19

A clear tube is a cylinder with diameter 12 cm12\text{ cm} and height 7 cm7\text{ cm}. What is the volume of the tube? (Give your answer in terms of π\pi or as an approximation.)

  1. 252π cm3 (≈791 cm3)252\pi\text{ cm}^3\ (\approx 791\text{ cm}^3) (correct answer)
  2. 168π cm3 (≈528 cm3)168\pi\text{ cm}^3\ (\approx 528\text{ cm}^3)
  3. 84π cm3 (≈264 cm3)84\pi\text{ cm}^3\ (\approx 264\text{ cm}^3)
  4. 504π cm3 (≈1583 cm3)504\pi\text{ cm}^3\ (\approx 1583\text{ cm}^3)
Explanation: This question tests applying volume formulas: cylinder V=πr²h, cone V=(1/3)πr²h, sphere V=(4/3)πr³, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr² times height h giving V=πr²h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=(1/3)πr²h (tapers to point reducing volume to one-third), sphere is V=(4/3)πr³ (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2), calculate (exponents first, multiply, approximate π≈3.14 or leave exact). For this cylinder with d=12 cm so r=6 cm, h=7 cm, V=π(6²)(7)=π(36)(7)=252π≈791.7 cm³. Common errors include using diameter as radius (r=12), adding 1/3 like cone, exponent mistakes (6²=36 not 72), arithmetic errors (367=242), or wrong units. Steps: (1) identify cylinder, (2) gather r=6 cm from d=12, h=7 cm, (3) select πr²h, (4) substitute, (5) calculate 367*π=252π, (6) add cubic units cm³. Converting diameter correctly is key for accurate radius.

Question 20

A sphere has volume 288π m3288\pi\text{ m}^3. What is the radius of the sphere?

  1. r=6 mr=6\text{ m} (correct answer)
  2. r=3 mr=3\text{ m}
  3. r=12 mr=12\text{ m}
  4. r=9 mr=9\text{ m}
Explanation: This question tests applying volume formulas: cylinder V=πr2hV=\pi r^2 h, cone V=13πr2hV=\frac{1}{3} \pi r^2 h, sphere V=43πr3V=\frac{4}{3} \pi r^3, selecting correct formula for shape and calculating accurately. Each shape has specific formula based on geometry: cylinder is base area πr2\pi r^2 times height h giving V=πr2hV=\pi r^2 h (circular cross-section throughout height), cone is 1/3 of cylinder with same base and height giving V=13πr2hV=\frac{1}{3} \pi r^2 h (tapers to point reducing volume to one-third), sphere is V=43πr3V=\frac{4}{3} \pi r^3 (radius cubed, no height—symmetric). Apply: identify shape, use corresponding formula with given dimensions (convert diameter to radius if needed: r=d/2r=d/2), calculate (exponents first, multiply, approximate π≈3.14\pi\approx3.14 or leave exact). For V=288πV=288\pi, (4/3)πr3=288π(4/3)\pi r^3=288\pi so r3=288×(3/4)=216r^3=288\times(3/4)=216, r=6 mr=6 \text{ m} (63=2166^3=216), matching B. Common errors: ignoring 4/3, getting r=9 (wrong inverse), or confusing with other formulas. Steps: (1) identify sphere, (2) set (4/3)πr3=288π(4/3)\pi r^3=288\pi, (3) solve r3=216r^3=216, (4) cube root to r=6, (5) check units m. Reverse calculations test formula understanding.