Volume Word Problems Worksheets With Answers - Fill and Sign ... - Free Printable

Let's solve each problem step by step from the worksheet titled "Volume Word Problems 1 Pack - Worksheet 1". We'll use the formulas for volume:

- Cylinder: $ V = \pi r^2 h $
- Sphere: $ V = \frac{4}{3} \pi r^3 $
- Cube: $ V = s^3 $
- Rectangular Prism (Box): $ V = l \times w \times h $

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1. How much soil is needed to fill a pit that is 10m deep and 8m wide and 17m long?

This is a rectangular prism.

$$
V = \text{length} \times \text{width} \times \text{depth} = 17 \times 8 \times 10 = 1,360 \, \text{m}^3
$$

✔ Answer: 1,360 m³ of soil

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2. Two spheres with radius of 250 cm and 750 cm are used in a respiratory system. Find the ratio of their volumes.

Use sphere volume formula: $ V = \frac{4}{3} \pi r^3 $

Let’s compute the ratio:
$$
\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi (250)^3}{\frac{4}{3} \pi (750)^3} = \left(\frac{250}{750}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}
$$

✔ Answer: The ratio of their volumes is $ \frac{1}{27} $

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3. Cylinder A has a base radius of 7 cm and a height of 3 cm. Cylinder B has a base radius of 4 cm and a height of 7 cm. Find the ratio of their volumes.

Use cylinder volume: $ V = \pi r^2 h $

- Volume of A: $ \pi (7)^2 (3) = \pi \cdot 49 \cdot 3 = 147\pi $
- Volume of B: $ \pi (4)^2 (7) = \pi \cdot 16 \cdot 7 = 112\pi $

Ratio: $ \frac{147\pi}{112\pi} = \frac{147}{112} $

Simplify: divide numerator and denominator by 7 → $ \frac{21}{16} $

✔ Answer: Ratio of volumes is $ \frac{21}{16} $

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4. Five metal cubes with sides of 3 cm are melted and cast into a spherical ball. Find the volume of the sphere that is formed.

First, find total volume of cubes:

- Volume of one cube: $ 3^3 = 27 \, \text{cm}^3 $
- Total volume: $ 5 \times 27 = 135 \, \text{cm}^3 $

This becomes the volume of the sphere.

So, $ V_{\text{sphere}} = 135 \, \text{cm}^3 $

✔ Answer: 135 cm³

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5. What is the volume of a regular cylinder whose base has a radius of 7 cm and a height of 2 cm?

$$
V = \pi r^2 h = \pi (7)^2 (2) = \pi \cdot 49 \cdot 2 = 98\pi \, \text{cm}^3
$$

If we use $ \pi \approx 3.14 $, then:

$ 98 \times 3.14 = 307.72 \, \text{cm}^3 $

But unless asked, leave it as $ 98\pi $

✔ Answer: $ 98\pi \, \text{cm}^3 $ or approximately $ 307.72 \, \text{cm}^3 $

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6. Find the volume of a box with the dimensions 4 cm × 3 cm × 6 cm.

$$
V = 4 \times 3 \times 6 = 72 \, \text{cm}^3
$$

✔ Answer: 72 cm³

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7. Fifty metal spheres with the radii of 4 cm are melted and this melted solution is poured into a box with base dimensions of 5 m × 6 m. Find the height of the cube filled with solution.

First, find volume of one sphere:

$$
V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3
$$

Total volume of 50 spheres:

$$
50 \times \frac{256}{3} \pi = \frac{12,800}{3} \pi \, \text{cm}^3 \approx 13,404.1 \, \text{cm}^3
$$

Now, convert to meters because the box is in meters.

Since $ 1 \, \text{m} = 100 \, \text{cm} $, so $ 1 \, \text{m}^3 = 1,000,000 \, \text{cm}^3 $

Convert volume to m³:

$$
\frac{12,800}{3} \pi \div 1,000,000 = \frac{12,800\pi}{3,000,000} = \frac{128\pi}{30,000} \, \text{m}^3
$$

Now, base area of box: $ 5 \times 6 = 30 \, \text{m}^2 $

Height $ h = \frac{V}{A} = \frac{128\pi / 30,000}{30} = \frac{128\pi}{900,000} \, \text{m} $

Approximate:

$ \pi \approx 3.14 $, so:

$ \frac{128 \times 3.14}{900,000} \approx \frac{401.92}{900,000} \approx 0.0004466 \, \text{m} $

Convert to cm: $ 0.0004466 \times 100 = 0.04466 \, \text{cm} $

Wait — this seems too small. Let's double-check units.

We made a mistake in unit conversion.

Better approach: Work in cm throughout.

Box base: $ 5 \, \text{m} = 500 \, \text{cm}, \quad 6 \, \text{m} = 600 \, \text{cm} $

Base area = $ 500 \times 600 = 300,000 \, \text{cm}^2 $

Total volume of melted spheres: $ 50 \times \frac{4}{3} \pi (4)^3 = 50 \times \frac{4}{3} \pi \times 64 = 50 \times \frac{256}{3} \pi = \frac{12,800}{3} \pi \, \text{cm}^3 $

Now, height $ h = \frac{V}{A} = \frac{12,800\pi / 3}{300,000} = \frac{12,800\pi}{900,000} = \frac{128\pi}{9,000} \, \text{cm} $

Now approximate:

$ \frac{128 \times 3.14}{9,000} = \frac{401.92}{9,000} \approx 0.04466 \, \text{cm} $

Still very small. But let's check:

Each sphere volume: $ \frac{4}{3} \pi (64) \approx \frac{4}{3} \times 3.14 \times 64 \approx 268.08 \, \text{cm}^3 $

50 spheres: $ 50 \times 268.08 = 13,404 \, \text{cm}^3 $

Box base area: $ 500 \times 600 = 300,000 \, \text{cm}^2 $

Height: $ \frac{13,404}{300,000} \approx 0.0447 \, \text{cm} $

Yes, that's correct. It's only about 0.045 cm, which is 0.45 mm — very shallow layer.

✔ Answer: Height ≈ 0.0447 cm (or 0.447 mm)

But perhaps the question meant "a box" with base 5 m × 6 m and unknown height — but yes, that's what we did.

Alternatively, maybe it's a typo — but based on given info, this is correct.

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8. A cubical box with side dimension 3 × 3 × 3 is melted into another cube whose width is 15. Find the length and height of cube formed if l=h.

Original cube: $ 3 \times 3 \times 3 = 27 \, \text{units}^3 $

New cube has same volume: $ 27 $

It's a cube with width = 15, and length = height = $ x $

But wait — if width is 15, and it's a cube, then all sides must be equal. But 15 ≠ 3.

So likely, it's a rectangular prism where width = 15, and length = height = $ x $, and volume = 27.

So:

$$
V = \text{length} \times \text{width} \times \text{height} = x \times 15 \times x = 15x^2
$$

Set equal to 27:

$$
15x^2 = 27 \Rightarrow x^2 = \frac{27}{15} = 1.8 \Rightarrow x = \sqrt{1.8} \approx 1.34
$$

But the problem says “cube” — contradiction.

Wait: “melted into another cube whose width is 15” — but a cube can't have width 15 unless all sides are 15.

But original volume is 27, new cube volume would be $ 15^3 = 3375 $ — impossible.

So likely a misstatement.

Re-read: “A cubical box with side dimension 3 × 3 × 3 is melted into another cube whose width is 15.”

That can’t be. Unless it's not a cube, but a rectangular prism.

But it says “cube”.

Possibility: typo — perhaps “whose side is 15”? But then volume is $ 15^3 = 3375 $, but original is 27 — no.

Or: “another cube” — but that can't have width 15 unless it's 15×15×15.

So probably the new shape is a rectangular prism with width 15, and length = height = $ x $, and volume = 27.

So:

$$
V = x \times 15 \times x = 15x^2 = 27 \Rightarrow x^2 = \frac{27}{15} = 1.8 \Rightarrow x = \sqrt{1.8} \approx 1.3416
$$

So length and height are both $ \sqrt{1.8} $ or $ \frac{3\sqrt{2}}{5} $? Let's simplify:

$ \sqrt{1.8} = \sqrt{\frac{18}{10}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \approx 1.34 $

✔ Answer: Length and height ≈ 1.34 units

But since it says “cube”, maybe it's a typo and should be “rectangular prism”.

Assuming that, we proceed.

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9. Find the volume of a right circular cone-shaped building with a height of 3m and a radius of base 14 cm.

Units mismatch! Height is in meters, radius in cm.

Convert radius to meters: $ 14 \, \text{cm} = 0.14 \, \text{m} $

Cone volume: $ V = \frac{1}{3} \pi r^2 h $

$$
V = \frac{1}{3} \pi (0.14)^2 (3) = \frac{1}{3} \pi (0.0196)(3) = \pi (0.0196) = 0.0196\pi \, \text{m}^3
$$

Approximately: $ 0.0196 \times 3.14 \approx 0.0615 \, \text{m}^3 $

✔ Answer: $ 0.0196\pi \, \text{m}^3 $ or ~0.0615 m³

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10. A cylinder has a radius of 2 inches and a height of 7 inches. What is the volume of this cylinder?

$$
V = \pi r^2 h = \pi (2)^2 (7) = \pi \cdot 4 \cdot 7 = 28\pi \, \text{in}^3
$$

✔ Answer: $ 28\pi \, \text{in}^3 $ or approximately 87.96 in³

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✔ Final Answers Summary:

1. 1,360 m³
2. 1:27
3. 21:16
4. 135 cm³
5. 98π cm³ (~307.72 cm³)
6. 72 cm³
7. ≈ 0.0447 cm (very thin layer)
8. Length and height ≈ 1.34 units (assuming rectangular prism, not cube)
9. 0.0196π m³ (~0.0615 m³)
10. 28π in³ (~87.96 in³)

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