What is the volume of the sphere shown, in terms of \pi?
Hint
The sphere volume formula on the formula sheet is V = \tfrac{4}{3}\pi r^3. Cube the radius first.
Step-by-step solution
Formula (on the formula sheet): V = \tfrac{4}{3}\pi r^3.
The radius is 6 cm. Cube it: 6^3 = 6 \times 6 \times 6 = 216.
Multiply by \tfrac{4}{3}: \tfrac{4}{3} \times 216 = \tfrac{864}{3} = 288.
So V = 288\pi.
Answer: 288\pi cm³ (about 904.8\text{ cm}^3)
Check: Divide by 3 first to keep numbers small: 216 \div 3 = 72, and 72 \times 4 = 288 ✓.
Tip: Volume is cubic, so the radius is cubed (r^3). Surface area is square, so the radius is squared (r^2). Matching the exponent to the units keeps the two formulas straight.
Question 2 of 36
Fill in
A ball has the diameter shown.
What is the surface area of the ball, rounded to the nearest square inch?
in²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The figure shows the diameter. The formula SA = 4\pi r^2 needs the radius.
Step-by-step solution
Find the radius: r = 10 \div 2 = 5 in.
Formula (on the formula sheet): SA = 4\pi r^2.
Square the radius: 5^2 = 25.
Multiply: SA = 4\pi(25) = 100\pi.
Evaluate: 100\pi \approx 314.16, which rounds to 314.
Answer: about 314 in²
Check: The surface area of a sphere is exactly 4 times the area of its middle circle. That circle has area \pi(5^2) = 25\pi \approx 78.5, and 4 \times 78.5 = 314 ✓.
Tip: Using the diameter (10) instead of the radius gives 400\pi \approx 1{,}257 — four times too big. Always halve the diameter first.
Question 3 of 36
A rubber ball has a diameter of 6 inches. What is the volume of the ball, in terms of \pi?
Hint
Find the radius first. Then use V = \tfrac{4}{3}\pi r^3.
Step-by-step solution
Radius: r = 6 \div 2 = 3 in.
Formula: V = \tfrac{4}{3}\pi r^3.
Cube the radius: 3^3 = 3 \times 3 \times 3 = 27.
Multiply by \tfrac{4}{3}: 27 \div 3 = 9, and 9 \times 4 = 36.
So V = 36\pi.
Answer: 36\pi in³ (about 113.1\text{ in}^3)
Check: A sphere fits inside a cube whose edge equals the diameter. That cube holds 6^3 = 216\text{ in}^3. The ball should hold a little more than half of it: 36\pi \approx 113, and 113 \div 216 \approx 0.52 ✓.
Tip: "Diameter" in a word problem is a signal: divide by 2 before you touch the formula.
Question 4 of 36
A gumball is a sphere with a radius of 1.5 cm. What is the volume of the gumball, in terms of \pi?
Hint
Use V = \tfrac{4}{3}\pi r^3. Cubing means multiplying the radius by itself three times.
Step-by-step solution
Formula (on the formula sheet): V = \tfrac{4}{3}\pi r^3.
Multiply by \tfrac{4}{3}: \tfrac{4}{3} \times 3.375 = \tfrac{13.5}{3} = 4.5.
So V = 4.5\pi.
Answer: 4.5\pi cm³ (about 14.1 cm³)
Check: Work backwards: 4.5\pi \div \tfrac{4}{3}\pi = 4.5 \times \tfrac{3}{4} = 3.375, and \sqrt[3]{3.375} = 1.5 ✓.
Tip: With decimals, cube on the calculator with the ^ key (1.5^3). A common slip is 1.5^2 (2.25), which gives 3\pi — always count three factors.
Question 5 of 36
Fill in
A bowling ball has a diameter of 8.5 inches. What is the volume of the ball? Round to the nearest cubic inch. (Ignore the finger holes.)
in³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The sphere formula uses the radius. What is half of 8.5?
Step-by-step solution
Radius: r = 8.5 \div 2 = 4.25 in.
Formula (on the formula sheet): V = \tfrac{4}{3}\pi r^3.
Cube the radius: 4.25^3 \approx 76.77.
Multiply: \tfrac{4}{3}\pi(76.77) \approx 321.56.
Round to the nearest whole number: 322.
Answer: about 322 in³
Check: Estimate with r \approx 4: \tfrac{4}{3}(3)(64) = 256; with r \approx 4.5: \tfrac{4}{3}(3)(91) \approx 364. Our answer, 322, is between them ✓.
Tip: On the TI-30XS, type 4÷3×π×4.25^3enter in one line. With \pi \approx 3.14 you get about 321.4, which rounds to 321; that answer is also accepted.
Question 6 of 36
The tank of a city water tower is a sphere with a radius of 15 ft. A crew will paint the entire outside of the tank.
About how many square feet will they paint? Round to the nearest square foot.
Hint
Painting covers the outside, so you need surface area: 4\pi r^2.
Step-by-step solution
Paint covers a surface, so use surface area: SA = 4\pi r^2 (on the formula sheet).
Square the radius: 15^2 = 225.
Multiply: 4 \times 225 = 900, so SA = 900\pi.
Use the \pi key: 900\pi \approx 2{,}827.43, which rounds to 2{,}827.
Answer: about 2,827 ft²
Check: A sphere's surface area is exactly 4 times the area of its "equator" circle: \pi(15^2) \approx 706.9, and 4 \times 706.9 \approx 2{,}827.6 ✓ (tiny difference from rounding).
Tip: Ask "is it covering or filling?" Paint, wrapping and leather cover → square units (4\pi r^2). Water, air and sand fill → cubic units (\tfrac{4}{3}\pi r^3).
Question 7 of 36
A classroom globe has a diameter of 16 inches. About how many square inches of printed map cover the entire globe? Round to the nearest square inch.
Hint
"Cover" means surface area. Use SA = 4\pi r^2 with the radius, not the diameter.
Step-by-step solution
The map covers the outside of the globe, so find the surface area.
Evaluate: 256\pi \approx 804.25, which rounds to 804.
Answer: about 804 in²
Check: Estimate with \pi \approx 3: 4 \times 3 \times 64 = 768. The answer should be a bit more than 768, and 804 is ✓.
Tip: Words like cover, paint, wrap, coat, material signal surface area (square units). Words like hold, fill, contain, capacity signal volume (cubic units).
Question 8 of 36
Fill in
A sphere has a volume of 972\pi cubic centimeters. What is the radius of the sphere?
cm
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Set \tfrac{4}{3}\pi r^3 = 972\pi. The \pi cancels. Then undo the \tfrac{4}{3} and take a cube root.
Step-by-step solution
Formula: \tfrac{4}{3}\pi r^3 = 972\pi.
Divide both sides by \pi: \tfrac{4}{3}r^3 = 972.
Multiply both sides by \tfrac{3}{4}: r^3 = 972 \times \tfrac{3}{4} = \tfrac{2916}{4} = 729.
Cube root: r = \sqrt[3]{729} = 9, because 9 \times 9 \times 9 = 729.
Tip: Undo the steps in reverse order: divide by \pi, multiply by \tfrac{3}{4}, then cube root. Know the small cubes: 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 9^3 = 729, 10^3 = 1000.
Question 9 of 36
Drop-down
The surface area of a ball is 196\pi square inches. Complete the statements.
The radius of the ball is inches, and its diameter is inches.
Choose an answer from each menu.
Hint
Set 4\pi r^2 = 196\pi. Divide by 4\pi, then take a square root.
Step-by-step solution
Formula: 4\pi r^2 = 196\pi.
Divide both sides by 4\pi: r^2 = \dfrac{196}{4} = 49.
Square root: r = \sqrt{49} = 7 in.
Diameter is twice the radius: d = 2 \times 7 = 14 in.
Answer: radius 7 in, diameter 14 in
Check:4\pi(7^2) = 4\pi(49) = 196\pi ✓.
Tip: Working backwards from surface area ends with a square root; from volume it ends with a cube root. Read the last line of the question to see whether it wants the radius or the diameter.
Question 10 of 36
A mixing bowl is shaped like a hemisphere (half of a sphere) with a radius of 4 inches.
How many cubic inches can the bowl hold? Round to the nearest tenth.
Hint
A hemisphere is half a sphere. Find the volume of the whole sphere, then take half.
Step-by-step solution
Full sphere: V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi(4^3) = \tfrac{4}{3}\pi(64) = \tfrac{256}{3}\pi.
Half of it: \tfrac{1}{2} \times \tfrac{256}{3}\pi = \tfrac{128}{3}\pi.
Check: The full sphere is \tfrac{256}{3}\pi \approx 268.1; half of 268.1 is about 134.0 ✓.
Tip: Hemisphere volume = \tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3. This shortcut is not on the formula sheet, but you can always build it from the sphere formula.
Question 11 of 36
Select all
A sphere has a radius of 3 units. Select all of the statements that are true.
Select all that apply.
Hint
Compute \tfrac{4}{3}\pi r^3 and 4\pi r^2 with r = 3. Then think about what happens to r^3 when r doubles.
Step-by-step solution
Volume: \tfrac{4}{3}\pi(3^3) = \tfrac{4}{3}\pi(27) = 36\pi. Statement A is true.
Surface area: 4\pi(3^2) = 4\pi(9) = 36\pi. Statement B is true (for r = 3 the two numbers happen to match, though the units differ).
12\pi comes from 4\pi(3) without squaring. Statement C is false.
108\pi comes from 4\pi(27) without dividing by 3. Statement D is false.
Doubling the radius to 6 gives \tfrac{4}{3}\pi(216) = 288\pi, which is 288 \div 36 = 8 times as large, not 2 times. Statement E is false.
Diameter = 2 \times 3 = 6. Statement F is true.
Answer: A, B, and F
Check:36\pi \approx 113.1 for both the volume (cubic units) and surface area (square units) ✓.
Tip: When the radius is multiplied by k, surface area is multiplied by k^2 and volume by k^3.
Question 12 of 36
A candy maker has 1,000 cubic centimeters of melted chocolate. She pours it into spherical molds, each with a radius of 1 cm. What is the greatest number of complete chocolate balls she can make?
Hint
Find the volume of one ball, then divide. Can she sell part of a ball?
Step-by-step solution
Volume of one ball: V = \tfrac{4}{3}\pi(1^3) = \tfrac{4}{3}\pi \approx 4.18879\text{ cm}^3.
Number of balls: 1{,}000 \div 4.18879 \approx 238.7.
Tip: In "how many complete items" problems, round down even if the decimal is .7 or more. In "how many containers are needed" problems, round up.
Question 13 of 36
A clear plastic hamster ball is a sphere with a radius of 7 inches. What is the volume of the space inside the ball, to the nearest cubic inch?
Hint
Volume of a sphere is \tfrac{4}{3}\pi r^3. What is 7^3?
Step-by-step solution
Formula (on the formula sheet): V = \tfrac{4}{3}\pi r^3.
Cube the radius: 7^3 = 7 \times 7 \times 7 = 343.
Multiply by \tfrac{4}{3}: \tfrac{4}{3}(343) = \tfrac{1{,}372}{3} \approx 457.33, so V \approx 457.33\pi.
Use the \pi key: 457.33\pi \approx 1{,}436.76, which rounds to 1{,}437.
Answer: about 1,437 in³
Check: A cube that just holds the ball has edges of 14 in and volume 14^3 = 2{,}744. A sphere fills a little more than half of its box (\approx 52\%): 0.52 \times 2{,}744 \approx 1{,}427 — close to 1,437 ✓.
Tip: Enter \tfrac{4}{3} as 4 ÷ 3 or with the n/d key, then × π × 7 ^ 3. Doing it in one line keeps full precision.
Question 14 of 36
A round marker buoy floating in a harbor is a sphere with a diameter of 24 inches. What is its volume, in terms of \pi?
Hint
Find the radius first. Then cube it and multiply by \tfrac{4}{3}.
Check: Divide first to keep numbers small: 1{,}728 \div 3 = 576, and 576 \times 4 = 2{,}304 ✓.
Tip: Using the diameter in a sphere formula makes the volume 2^3 = 8 times too large (8 \times 2{,}304 = 18{,}432). If an answer choice is exactly 8 times another, that is a sign of the radius–diameter trap.
Question 15 of 36
Select all
A science teacher shows the class a solid steel ball bearing whose diameter measures 8 cm across. Which of these facts about the ball are correct? Select all that apply.
Select all that apply.
Hint
Start with the radius. Then compute 4\pi r^2 and \tfrac{4}{3}\pi r^3 and compare with each statement.
Step-by-step solution
Radius: r = 8 \div 2 = 4 cm → statement A is true.
Surface area: 4\pi r^2 = 4\pi(16) = 64\pi cm² → statement B is true.
Volume: \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi(64) = \tfrac{256}{3}\pi \approx 268.1 cm³ → statement C is true.
Statement D, 16\pi, is only \pi r^2, the area of one flat circle → false.
Statement E, about 2,144.7 cm³, comes from cubing the diameter: \tfrac{4}{3}\pi(8^3) \approx 2{,}144.7 → false.
Answer: A, B, and C
Check:\tfrac{256}{3} \approx 85.33, and 85.33 \times 3.1416 \approx 268.1 ✓.
Tip: In select-all items, judge every statement on its own. Write down r, SA and V once, then compare — do not stop after finding one true statement.
Question 16 of 36
A hotel lobby has decorative balls with a diameter of 2 ft. Covering a ball in gold leaf costs $12 per square foot.
To the nearest dollar, how much does it cost to cover one ball?
Hint
Gold leaf covers the outside, so find the surface area. Remember to use the radius, not the diameter.
Tip: Prices "per square foot" go with surface area; prices "per cubic foot" go with volume. Matching the unit in the price tells you which formula to use.
Question 17 of 36
A ball with a radius of 4 in fits snugly inside a cylinder-shaped case. The ball touches the top, the bottom and the curved side of the case.
What is the volume of the case, in terms of \pi?
Hint
If the ball touches the top and bottom, how tall is the case compared with the ball's radius?
Step-by-step solution
The ball touches the curved side, so the case's radius equals the ball's radius: r = 4 in.
The ball touches the top and bottom, so the case's height equals the ball's diameter: h = 2 \times 4 = 8 in.
Cylinder volume (on the formula sheet): V = \pi r^2 h = \pi(4^2)(8) = \pi(16)(8) = 128\pi.
Answer: 128\pi in³ (about 402.1 in³)
Check: The ball itself is \tfrac{4}{3}\pi(64) = \tfrac{256}{3}\pi \approx 85.3\pi. A snug ball always fills \tfrac{2}{3} of its cylinder: \tfrac{2}{3} \times 128\pi = 85.3\pi ✓.
Tip: "Fits snugly" means the container's dimensions come from the ball: width and height both equal the diameter 2r.
Question 18 of 36
Fill in
A spherical storage tank holds 288\pi cubic feet of propane. What is the diameter of the tank?
ft
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Set \tfrac{4}{3}\pi r^3 = 288\pi. Cancel \pi, undo the \tfrac{4}{3}, and use a cube root. Then think about the diameter.
Step-by-step solution
Formula: \tfrac{4}{3}\pi r^3 = 288\pi.
Divide both sides by \pi: \tfrac{4}{3}r^3 = 288.
Multiply both sides by \tfrac{3}{4}: r^3 = 288 \times \tfrac{3}{4} = 216.
Cube root: r = \sqrt[3]{216} = 6 ft (because 6 \times 6 \times 6 = 216).
Tip: On the TI-30XS, a cube root is 32nd^ (ⁿ√) 216. Know the small cubes by heart — 8, 27, 64, 125, 216, 343, 512, 729, 1000 — they appear often.
Question 19 of 36
A spherical party balloon has a surface area of 324\pi square inches. What is the radius of the balloon?
Hint
Set 4\pi r^2 = 324\pi and solve. The last step is a root — which one goes with r^2?
Step-by-step solution
Formula: 4\pi r^2 = 324\pi.
Divide both sides by 4\pi: r^2 = 324 \div 4 = 81.
Square root: r = \sqrt{81} = 9 in.
Answer: 9 in
Check:4\pi(9^2) = 4\pi(81) = 324\pi ✓.
Tip: Surface area has r^2, so you finish with a square root. Volume has r^3, so you finish with a cube root. Using the wrong formula (volume instead of area) is what produces the 6.24 trap.
Question 20 of 36
Fill in
The dome of a planetarium is a hemisphere (half of a sphere) with a radius of 10 m. The outside of the dome — the curved part only, not the floor — will be covered with new roofing.
How many square meters of roofing are needed? Round to the nearest square meter.
m²
Type your answer in the box. Fractions like 3/4 are fine.
Hint
The curved part of a hemisphere is half of a sphere's surface area, 4\pi r^2.
Step-by-step solution
A full sphere has surface area 4\pi r^2.
The dome is half a sphere and has no floor, so its curved area is \tfrac{1}{2}(4\pi r^2) = 2\pi r^2.
Substitute r = 10: 2\pi(10^2) = 2\pi(100) = 200\pi.
Use the \pi key: 200\pi \approx 628.32, which rounds to 628.
Answer: about 628 m²
Check: A full sphere of radius 10 would be 400\pi \approx 1{,}256.6 m². Half of that is 628.3 ✓.
Tip: For a solid hemisphere (like a paperweight), add the flat circle \pi r^2 to get 3\pi r^2. For a dome or bowl, use only the curved 2\pi r^2.
Question 21 of 36
Leo found the volume of a marble with a diameter of 4 cm. Here is his work.
V = \tfrac{4}{3}\pi r^3
r = 4 \div 2 = 2
r^3 = 2 \times 3 = 6
V = \tfrac{4}{3}\pi(6) = 8\pi\text{ cm}^3
In which step did Leo make the first mistake?
Hint
What does the small 3 in r^3 tell you to do?
Step-by-step solution
Step 1 is the correct sphere formula from the formula sheet.
Step 2 is correct: the radius is half of the 4 cm diameter, r = 2.
Step 3 is wrong. r^3 means r \times r \times r, not r \times 3. So 2^3 = 2 \times 2 \times 2 = 8.
Correct finish: V = \tfrac{4}{3}\pi(8) = \tfrac{32}{3}\pi \approx 33.5\text{ cm}^3.
Answer: Step 3
Check:\tfrac{32}{3} \approx 10.67, and 10.67 \times 3.1416 \approx 33.5 ✓. Leo's 8\pi \approx 25.1 is too small because 6 < 8.
Tip: An exponent counts how many times a number is multiplied by itself. With small numbers, 2^3 = 8 and 2 \times 3 = 6 are close enough to confuse. A bigger number shows the difference: 5^3 = 125, but 5 \times 3 = 15.
Question 22 of 36
A cook has 2,000 cm³ of meatball mix. Each meatball is a sphere with a radius of 2.5 cm.
What is the greatest number of complete meatballs the cook can make?
Hint
Find the volume of one meatball. Then divide. Can the cook make part of a meatball count as a whole one?
Step-by-step solution
Volume of one meatball: V = \tfrac{4}{3}\pi(2.5^3) = \tfrac{4}{3}\pi(15.625) \approx 65.45\text{ cm}^3.
Number of meatballs: 2{,}000 \div 65.45 \approx 30.56.
Only complete meatballs count, so round down: 30.
Answer: 30 meatballs
Check:30 \times 65.45 = 1{,}963.5 \le 2{,}000 ✓, but 31 \times 65.45 = 2{,}029 > 2{,}000, so a 31st meatball is not possible ✓.
Tip: "How many complete…" or "how many can be filled" → round down. "How many containers are needed" → round up. Read the question before you round.
Question 23 of 36
Fill in
A ball fits exactly inside a cube-shaped gift box, touching all six faces. Each edge of the box is 10 inches.
How many cubic inches of empty space are inside the box around the ball? Round to the nearest tenth.
in³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Empty space = volume of the box − volume of the ball. If the ball fits exactly, what is its diameter?
Step-by-step solution
The ball touches opposite faces, so its diameter equals the edge: d = 10 in, and r = 5 in.
Volume of the box: 10^3 = 1{,}000\text{ in}^3.
Volume of the ball: \tfrac{4}{3}\pi(5^3) = \tfrac{4}{3}\pi(125) = \tfrac{500}{3}\pi \approx 523.60\text{ in}^3.
Empty space: 1{,}000 - 523.60 = 476.40.
Round to the nearest tenth: 476.4.
Answer: about 476.4 in³
Check: The ball fills 523.6 \div 1{,}000 \approx 52\% of the box — a sphere always fills a little more than half of the cube it fits in ✓.
Tip: "Fits exactly" or "inscribed" means a length is shared: here the sphere's diameter is the cube's edge. Look for that shared length in every composite problem. (With \pi \approx 3.14, the answer is 476.7, also accepted.)
Question 24 of 36
Drag and drop
A beach ball has a radius 3 times the radius of a toy ball. Drag the correct tile into each box. Two tiles are not used.
Beach ball's diameter compared with the toy ball's
Beach ball's surface area compared with the toy ball's
Beach ball's volume compared with the toy ball's
Tap a tile, then tap a box to place it.
Hint
Lengths scale by k, areas by k^2, volumes by k^3. Here k = 3.
Step-by-step solution
The scale factor for lengths is k = 3.
Diameter is a length (d = 2r), so it grows by the same factor: ×3.
Surface area 4\pi r^2 depends on r^2: 3^2 = 9 → ×9.
Tip: Two-step sphere problems often go "area → radius → volume." Find the radius exactly first; it is the key that unlocks every other measurement.
Question 26 of 36
The inside of a snow igloo is shaped like a hemisphere with a radius of 2.5 m.
What is the volume of air inside the igloo, to the nearest tenth of a cubic meter?
Hint
A hemisphere is half of a sphere. Find the sphere's volume, then take half.
Step-by-step solution
Full sphere: V = \tfrac{4}{3}\pi r^3.
Hemisphere: half of that, V = \tfrac{1}{2} \cdot \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3.
Cube the radius: 2.5^3 = 15.625.
V = \tfrac{2}{3}\pi(15.625) \approx \tfrac{2}{3}(49.087) \approx 32.72.
Round to the nearest tenth: 32.7.
Answer: about 32.7 m³
Check: The full sphere would be \tfrac{4}{3}\pi(15.625) \approx 65.4\text{ m}^3, and half of 65.4 is 32.7 ✓.
Tip: Hemisphere volume = \tfrac{2}{3}\pi r^3. Hemisphere curved area = 2\pi r^2. Both are simply "half of the sphere."
Question 27 of 36
Fill in
A cylinder-shaped can holds 3 tennis balls stacked on top of each other. Each ball has a diameter of 2.6 in. The can has a radius of 1.3 in and a height of 7.8 in, so the balls fit snugly.
How many cubic inches of empty space are inside the can around the balls? Round to the nearest tenth.
in³
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Empty space = volume of the can − volume of all 3 balls. Each ball's radius is half of 2.6 in.
Step-by-step solution
Can (cylinder): V = \pi r^2 h = \pi(1.3^2)(7.8) = \pi(1.69)(7.8) = 13.182\pi \approx 41.41\text{ in}^3.
One ball: r = 2.6 \div 2 = 1.3, so V = \tfrac{4}{3}\pi(1.3^3) = \tfrac{4}{3}\pi(2.197) \approx 2.929\pi.
Three balls: 3 \times 2.929\pi = 8.788\pi \approx 27.61\text{ in}^3.
Convert: each cubic foot holds 7.48 gallons, so multiply: 113.10 \times 7.48 \approx 845.97.
Round to the nearest gallon: 846.
Answer: about 846 gallons
Check: Estimate: 36 \times 3 = 108 ft³ (using \pi \approx 3) and 108 \times 7.5 = 810. A little more than 810 makes sense because \pi > 3 ✓.
Tip: Gallons are smaller than cubic feet, so the number of gallons must be bigger than the number of cubic feet. If your answer got smaller, you divided by mistake.
Question 29 of 36
A toy company makes a new bouncy ball whose volume is 8 times the volume of its original ball. Both balls are spheres.
The radius of the new ball is how many times the radius of the original ball?
Hint
Volume grows by the cube of the radius factor. Which number cubed is 8?
Step-by-step solution
Sphere volume is \tfrac{4}{3}\pi r^3, so if the radius is multiplied by k, the volume is multiplied by k^3.
We know the volume factor: k^3 = 8.
Undo the cube with a cube root: k = \sqrt[3]{8} = 2, because 2 \times 2 \times 2 = 8.
Answer: 2 times
Check: Original r = 1: V = \tfrac{4}{3}\pi. New r = 2: V = \tfrac{4}{3}\pi(8) = \tfrac{32}{3}\pi, which is 8 times \tfrac{4}{3}\pi ✓.
Tip: Going forward (radius → volume) you cube the factor. Going backward (volume → radius) you take the cube root. For areas, use square and square root.
Question 30 of 36
A leather company covers softballs that have a diameter of 9 cm. Which expression gives the amount of leather needed to cover one ball?
Hint
Covering means surface area. Check two things in each expression: the formula and the number used for r.
Step-by-step solution
Covering a ball uses its surface area: SA = 4\pi r^2 (on the formula sheet).
The radius is half of the diameter: r = 9 \div 2 = 4.5 cm.
Check:4\pi(4.5)^2 = 81\pi, and 81\pi also equals \pi d^2 = \pi(9)^2 — a handy second form of the same formula ✓.
Tip: In "which expression" items, you do not need to calculate everything. Eliminate choices with the wrong formula (volume, circle area) and the wrong measurement (diameter in place of radius).
Question 31 of 36
Kai found the volume of a hemisphere-shaped soup bowl with a radius of 6 cm. Here is the work.
Whole sphere: \tfrac{4}{3}\pi(6^3) = 288\pi
Hemisphere: 288\pi \div 2 = 144\pi
144\pi \approx 452.4
The bowl holds about 452.4\text{ cm}^2.
In which step did Kai make the first mistake?
Hint
Check the numbers in each step, and also check the units of the final answer. What units does volume have?
Step-by-step solution
Step 1 is correct: 6^3 = 216, and \tfrac{4}{3}(216) = 288, so 288\pi.
Step 2 is correct: a hemisphere is half of a sphere, 144\pi.
Step 3 is correct: 144 \times 3.14159 \approx 452.4.
Step 4 is wrong. Volume is measured in cubic units. The answer should be about 452.4\text{ cm}^3, not cm².
Answer: Step 4
Check: Area uses two lengths multiplied (cm × cm = cm²), and volume uses three (cm × cm × cm = cm³). Since r is cubed here, the answer must be in cm³ ✓.
Tip: On the GED, answer choices sometimes differ only by units. Always ask: "Is this a length, an area or a volume?"
Question 32 of 36
A decorative glass ball has a volume of about 523.6 cubic inches. What is the diameter of the ball, to the nearest inch?
Hint
Solve \tfrac{4}{3}\pi r^3 = 523.6 for r^3, take the cube root, and remember what the question asks for.
Tip: A quick way on the calculator: r^3 = \dfrac{3V}{4\pi}. Then use the cube root (3, then 2nd^, then the number). Always re-read the question at the end: radius or diameter?
Question 33 of 36
Drag and drop
A weather balloon is a sphere. Drag the correct tile into each box to show how its measurements change. Two tiles are not used.
Surface area when the radius is doubled
Volume when the radius is doubled
Volume when the radius is tripled
Tap a tile, then tap a box to place it.
Hint
Surface area depends on r^2 and volume depends on r^3. Try r = 1 and r = 2.
Step-by-step solution
Surface area is 4\pi r^2 and volume is \tfrac{4}{3}\pi r^3. The constants (4\pi and \tfrac{4}{3}\pi) stay the same, so only r^2 and r^3 matter.
Radius doubled, surface area: r^2 goes from 1^2 = 1 to 2^2 = 4 → ×4.
Radius doubled, volume: r^3 goes from 1^3 = 1 to 2^3 = 8 → ×8.
Radius tripled, volume: r^3 goes from 1 to 3^3 = 27 → ×27.
Answer: ×4, ×8, ×27 (the ×2 and ×9 tiles are not used)
Check: With real numbers, r = 3: V = 36\pi. With r = 9: V = \tfrac{4}{3}\pi(729) = 972\pi, and 972 \div 36 = 27 ✓.
Tip: Scale factor k on lengths → k^2 on areas → k^3 on volumes. This works for every solid, not just spheres.
Question 34 of 36
An art student has a cylinder of clay with a radius of 3 cm and a height of 16 cm. She rolls all of the clay into small balls, each with a radius of 1 cm, without wasting any.
How many balls can she make?
Hint
The total volume stays the same. Find the cylinder's volume and one ball's volume, then divide. Keeping \pi in both makes it cancel.
Step-by-step solution
Cylinder: V = \pi r^2 h = \pi(3^2)(16) = 144\pi\text{ cm}^3.
One ball: V = \tfrac{4}{3}\pi(1^3) = \tfrac{4}{3}\pi\text{ cm}^3.
Check:28 \times 500 = 14{,}000 cm³, very close to the ball's 14,137 cm³ ✓.
Tip: In rate problems, write the units: \dfrac{\text{cm}^3}{\text{cm}^3\text{/s}} = \text{s}. Also halve the diameter before cubing — doing it after (or never) multiplies the error by 8.
