Every solution shows the formula, the substitution, and the final volume in cubic units. Use the answers to check your work, and compare the cube-counting method with the multiplication method in problem 5, since both must give the same volume.
Problem 1: Whole-number edges
\(V = \ell wh = 6 \times 4 \times 3 = 24 \times 3 = 72\)
The volume is \(72\text{ ft}^3\).
Problem 2: Box volume
\(V = 7 \times 5 \times 2 = 35 \times 2 = 70\)
The volume is \(70\text{ in}^3\).
Problem 3: Base area times height
\(V = Bh = 24 \times 5 = 120\)
The volume is \(120\text{ in}^3\).
Problem 4: Fractional width
\(V = 4 \times \frac{1}{2} \times 3 = 2 \times 3 = 6\)
The volume is \(6\text{ in}^3\).
Problem 5: Packing with cubes
(a) Along each edge: \(1 : \frac{1}{4} = 4\), \(\frac{3}{4} : \frac{1}{4} = 3\), \(\frac{1}{2} : \frac{1}{4} = 2\). Cubes: \(4 \times 3 \times 2 = 24\).
(b) Four cubes fill 1 inch in each direction, so 64 fit in a 1-inch cube: \(\frac{1}{64}\text{ in}^3\).
(c) \(24 \times \frac{1}{64} = \frac{3}{8}\). Check: \(1 \times \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}\).
Answers: 24 cubes, \(\frac{1}{64}\text{ in}^3\), \(\frac{3}{8}\text{ in}^3\).
Problem 6: Missing height
\(150 = 10 \times 5 \times h = 50h\)
\(h = 150 : 50 = 3\)
The height is \(3\text{ ft}\).
Problem 7: Mixed-number edges
\(V = \frac{5}{2} \times \frac{3}{2} \times 2 = \frac{15}{4} \times 2 = \frac{30}{4} = \frac{15}{2} = 7\frac{1}{2}\)
The volume is \(7\frac{1}{2}\text{ ft}^3\).
Problem 8: Aquarium
Water: \(20 \times 10 \times 9 = 1{,}800\text{ in}^3\).
Full tank: \(20 \times 10 \times 12 = 2{,}400\text{ in}^3\).
To add: \(2{,}400 – 1{,}800 = 600\).
The tank holds \(1{,}800\text{ in}^3\) of water and needs \(600\text{ in}^3\) more.
Problem 9: Storage boxes
Box: \(\frac{5}{2} \times 2 \times \frac{3}{2} = \frac{15}{2} = 7.5\text{ ft}^3\).
Space: \(10 \times 6 \times 3 = 180\text{ ft}^3\).
\(180 : 7.5 = 24\)
Check by rows: \(10 : 2.5 = 4\), \(6 : 2 = 3\), \(3 : 1.5 = 2\), and \(4 \times 3 \times 2 = 24\).
24 boxes fit.
Problem 10: Sandbox
\(V = 6 \times \frac{9}{2} \times \frac{3}{4} = 27 \times \frac{3}{4} = \frac{81}{4} = 20.25\text{ ft}^3\).
Cost: \(20.25 \times \$4 = \$81\).
The sand fills \(20.25\text{ ft}^3\) and costs \(\$81\).
See every chapter on the Grade 6 math page.