Volume word problems worksheet with six exercises on calculating the volume of various geometric shapes.

Let’s solve each problem one by one. We’ll use the formulas for volume of common shapes:

- Cylinder: V = π × r² × h
- Sphere: V = (4/3) × π × r³
- Cube: V = side³
- Box (rectangular prism): V = length × width × height
- Cone: V = (1/3) × π × r² × h

We’ll use π ≈ 3.14 unless told otherwise.

---

Problem 1:
New punch bowl is filled to 10m deep, wide and long → it’s a cube with side = 10 m.
Volume = 10 × 10 × 10 = 1000 m³

---

Problem 2:
Two spheres: radii 250 cm and 750 cm.
V₁ = (4/3)π(250)³ = (4/3)π(15,625,000) = (62,500,000/3)π
V₂ = (4/3)π(750)³ = (4/3)π(421,875,000) = (1,687,500,000/3)π

Ratio V₁ : V₂ = [62,500,000 / 3] : [1,687,500,000 / 3] = 62,500,000 : 1,687,500,000
Divide both by 62,500,000 → 1 : 27
*(Note: Since volume scales with cube of radius, (250/750)³ = (1/3)³ = 1/27 → ratio 1:27)*

Answer: 1:27

---

Problem 3:
Cylinder X: radius = 2 in, height = 2 in → Vₓ = π×2²×2 = π×4×2 = 8π
Cylinder Y: base radius = 4 in, height = 2 in → Vᵧ = π×4²×2 = π×16×2 = 32π
Ratio Vₓ : Vᵧ = 8π : 32π = 8:32 = 1:4

---

Problem 4:
Five metal cubes, each side = 2 cm → volume of one cube = 2³ = 8 cm³
Total volume = 5 × 8 = 40 cm³
Melted into sphere → volume of sphere = 40 cm³
So, (4/3)πr³ = 40
→ r³ = 40 × 3 / (4π) = 120 / (4π) = 30 / π
→ r = ∛(30/π)
But question says “find the volume of the sphere” — that’s just total volume of cubes = 40 cm³

Wait — re-read: “Find the volume of the sphere thus formed.” Yes, since melting doesn’t change volume, answer is 40 cm³

---

Problem 5:
Regular cylinder: radius = 7 cm, height = 2 cm
V = π × 7² × 2 = π × 49 × 2 = 98π ≈ 98 × 3.14 = let’s compute:
98 × 3 = 294, 98 × 0.14 = 13.72 → total 307.72 cm³
But maybe leave as 98π? Problem doesn’t specify. Let’s give numerical value.

Actually, check: 98 × 3.14 =
100×3.14 = 314, minus 2×3.14=6.28 → 314 - 6.28 = 307.72 cm³

---

Problem 6:
Box: 4m × 5m × 6m → V = 4×5×6 = 20×6 = 120 m³

---

Problem 7:
Fifty spheres, each radius 4 m → volume of one sphere = (4/3)π(4)³ = (4/3)π(64) = 256π/3
Total volume = 50 × (256π/3) = (12,800π)/3
This equals volume of cube → s³ = (12,800π)/3
s = ∛[(12,800π)/3]

Compute numerically:
π ≈ 3.14 → 12,800 × 3.14 = let’s calculate:
12,800 × 3 = 38,400
12,800 × 0.14 = 1,792 → total 40,192
Then divide by 3: 40,192 ÷ 3 ≈ 13,397.333
Now cube root of 13,397.333

What’s ∛13,397?
23³ = 12,167
24³ = 13,824 → too big
23.5³ = ? 23.5 × 23.5 = 552.25; ×23.5 ≈ 552.25×20=11,045 + 552.25×3.5≈1,932.875 → total ~12,977.875
Still low
23.7³: 23.7×23.7 = 561.69; ×23.7 ≈ 561.69×20=11,233.8 + 561.69×3.7≈2,078.253 → total ~13,312.053
Close to 13,397
23.8³: 23.8×23.8=566.44; ×23.8 ≈ 566.44×20=11,328.8 + 566.44×3.8≈2,152.472 → total ~13,481.272 → a bit over
So between 23.7 and 23.8 → approx 23.75 m

But maybe we can leave exact or round reasonably. Since problem likely expects calculation:

Alternatively, perhaps they want exact expression? But probably numerical.

Let me recalculate total volume:

One sphere: (4/3)πr³ = (4/3)π(64) = 256π/3
50 spheres: 50 * 256π/3 = 12800π/3
Set equal to s³ → s = ∛(12800π/3)

Using π=3.14: 12800*3.14=40192; /3=13397.333; ∛13397.333≈23.75 m

So approximately 23.75 meters

But let’s see if we can be more precise or if problem expects integer? Probably not. I’ll go with 23.75 m or maybe round to nearest tenth: 23.8 m? Wait 23.7^3=13312, 23.8^3=13481, target 13397, so closer to 23.75.

Actually, 23.75^3 = ? 23.75 * 23.75 = (24-0.25)^2 = 576 - 12 + 0.0625 = 564.0625? Better: 23.75 * 23.75.

23.75 × 23.75:
= (20+3.75)^2 = 400 + 2*20*3.75 + (3.75)^2 = 400 + 150 + 14.0625 = 564.0625
Then ×23.75: 564.0625 × 23.75
First, 564.0625 × 20 = 11,281.25
564.0625 × 3 = 1,692.1875
564.0625 × 0.75 = 423.046875
Sum: 11,281.25 + 1,692.1875 = 12,973.4375 + 423.046875 = 13,396.484375 ≈ 13,396.5 which is very close to 13,397.333

So s ≈ 23.75 m

Answer: 23.75 meters

---

Problem 8:
Original box: 3×2×2 = 12 cm³
Melted into cylinder with radius 1.5 cm → find height h
Volume of cylinder = πr²h = π*(1.5)²*h = π*2.25*h
Set equal to 12: 2.25πh = 12
h = 12 / (2.25π) = 12 / (2.25 * 3.14)
First, 2.25 * 3.14 = 2.25*3 = 6.75, 2.25*0.14=0.315 → total 7.065
h = 12 / 7.065 ≈ ?
12 ÷ 7.065 ≈ 1.698 → approximately 1.70 cm

Exact fraction: h = 12 / (2.25π) = 12 / (9/4 * π) = 12 * 4 / (9π) = 48/(9π) = 16/(3π)
Numerically: 16/(3*3.14) = 16/9.42 ≈ 1.698 → 1.70 cm

---

Problem 9:
Right circular cone-shaped building: height = 3 km, base radius = 14 m
Units mismatch! Height in km, radius in m. Must convert.

Convert height to meters: 3 km = 3000 m
Radius = 14 m
Volume of cone = (1/3)πr²h = (1/3)π(14)²(3000)
= (1/3)π(196)(3000) = π * 196 * 1000 = 196,000π
≈ 196,000 * 3.14 = let’s compute:
200,000 * 3.14 = 628,000
Minus 4,000 * 3.14 = 12,560 → 628,000 - 12,560 = 615,440 m³

Answer: 615,440 cubic meters

---

Problem 10:
Cylinder: radius = 2 inches, height = 7 inches
V = πr²h = π*(2)²*7 = π*4*7 = 28π ≈ 28 * 3.14 = 87.92 in³
Answer: 87.92 cubic inches

---

Now compiling all answers:

1. 1000 m³
2. 1:27
3. 1:4
4. 40 cm³
5. 307.72 cm³ (or 98π cm³, but numerical expected)
6. 120 m³
7. 23.75 m
8. 1.70 cm
9. 615,440 m³
10. 87.92 in³

Double-checking calculations where needed.

For problem 7, I think 23.75 is fine. Some might expect exact form, but context suggests numerical.

For problem 9, units were tricky — converted correctly.

All seem correct.

Final Answer:
1. 1000 m³
2. 1:27
3. 1:4
4. 40 cm³
5. 307.72 cm³
6. 120 m³
7. 23.75 m
8. 1.70 cm
9. 615440 m³
10. 87.92 in³