Math 6 · 6.G.A.2

Volume of a box with fraction edge lengths

A box is 2\tfrac{1}{2} inches long, so whole unit cubes no longer fit inside it. How do you find how much it holds?

You need

Volume of a boxGrade 5 Multiplying fractionsGrade 5

Where you are 9 in Volume and capacity

Measurement12

Volume and capacity9

▴ 4 above Volume with unit cubes Volume of a box Surface area, and how it compares with volume

Volume of a box with fraction edge lengths

Volume of a solid made of boxes

When you reach for it

A box with whole-number edges can be filled with whole layers of unit cubes, and counting them gives the rule for the volume of a box: length times width times height. A box 4 by 3 by 2 has 4 \times 3 = 12 unit cubes in each layer and 2 layers, so it holds 12 \times 2 = 24 unit cubes and its volume is 24 cubic units.

Real boxes rarely measure whole units. A fudge box may be 1\tfrac{1}{2} inches deep, and a terrarium 1\tfrac{1}{4} feet wide. Unit cubes no longer fit exactly, so the counting that made V = lwh true seems to break down.

Reach for this lesson whenever a box has an edge that is a fraction or a mixed number. It shows two things. First, such a box can still be packed exactly, with smaller cubes whose edge is a unit fraction such as \tfrac{1}{2} or \tfrac{1}{4}, and the count gives the volume. Second, that count always agrees with length times width times height worked out by multiplying fractions, so the rule you already know keeps working, and so does its shorter form V = Bh. Once that is settled, a missing edge can be found from the volume by division.

The procedure

Part 1: packing with smaller cubes

What a small cube is worth. Cut a unit cube, a cube with edge 1, in half along its length, its width and its height. You get 2 \times 2 \times 2 = 8 small cubes, each with edge \tfrac{1}{2}. Eight equal pieces make one cubic unit, so each piece is \tfrac{1}{8} of a cubic unit.

Figure 1Halving every edge makes 2 \times 2 \times 2 = 8 small cubes; quartering it makes 4 \times 4 \times 4 = 64.

Cutting each edge into four instead gives 4 \times 4 \times 4 = 64 cubes of edge \tfrac{1}{4}, each worth \tfrac{1}{64} of a cubic unit. In general, cubes of edge \tfrac{1}{n} fit n along each edge of the unit cube, so n \times n \times n of them fill it, and each is worth \tfrac{1}{n \times n \times n} of a cubic unit. A cube of edge \tfrac{1}{2} is not half a cubic unit: it is half as long, half as wide and half as tall, so it is only an eighth.

Notice that \tfrac{1}{2} \times \tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{8} and \tfrac{1}{4} \times \tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{1}{64}. For one small cube, multiplying its three edges already gives its volume.

Packing a box. Take a box 2\tfrac{1}{2} units long, 1\tfrac{1}{2} units deep and 1 unit tall. Unit cubes do not fill it, but cubes of edge \tfrac{1}{2} do, because every edge is a whole number of halves: 2\tfrac{1}{2} = \tfrac{5}{2} is five halves, 1\tfrac{1}{2} = \tfrac{3}{2} is three halves, and 1 = \tfrac{2}{2} is two halves.

Figure 2The box holds 5 \times 3 \times 2 = 30 cubes of edge \tfrac{1}{2}, each worth \tfrac{1}{8} of a cubic unit.

So the box holds 5 small cubes along its length, 3 along its depth and 2 up its height, 5 \times 3 \times 2 = 30 in all. Each is worth \tfrac{1}{8} of a cubic unit, so

V = 30 \times \frac{1}{8} = \frac{30}{8} = 3\frac{3}{4} \text{ cubic units}

Now multiply the three edges as fractions instead:

\frac{5}{2} \times \frac{3}{2} \times \frac{2}{2} = \frac{5 \times 3 \times 2}{2 \times 2 \times 2} = \frac{30}{8}

The same 30 and the same 8 appear, and that is no accident. Fractions multiply numerator by numerator and denominator by denominator. The numerators 5, 3 and 2 are the counts along the edges, so their product is the number of small cubes. The denominators multiply to 2 \times 2 \times 2 = 8, the number of small cubes in one unit cube. Dividing the count by 8 is exactly what the fraction product does.

Why it always works. Write the three edges over one denominator n, as \tfrac{a}{n}, \tfrac{b}{n} and \tfrac{c}{n}, where a, b, c and n are whole numbers. Cubes of edge \tfrac{1}{n} fit a along the length, b along the width and c up the height, so the box holds a \times b \times c of them, each worth \tfrac{1}{n \times n \times n} of a cubic unit. The volume is

\frac{a \times b \times c}{n \times n \times n} = \frac{a}{n} \times \frac{b}{n} \times \frac{c}{n}

which is length times width times height. So for every box,

V = l \times w \times h

whether its edges are whole numbers, fractions or mixed numbers. When the edges have different denominators, rewrite them over a common denominator first; that chooses a small cube that fits every edge exactly, as in Example 3.

Part 2: base times height, and a missing edge

Base times height. The bottom face of a box is a rectangle l by w, with area B = l \times w square units. A layer of the box 1 unit thick, sitting on that face, holds B cubic units, and the box is h such layers tall, so

V = B \times h, \qquad\text{written}\qquad V = Bh

A fractional height changes nothing: a layer \tfrac{1}{2} unit thick holds half of B, and a layer h units thick holds B \times h. This is the same rule as V = lwh with the first two edges grouped, since l \times w \times h = (l \times w) \times h by the associative property of multiplication. It splits the work into two steps: the area of the base, then times the height. When the base area is given directly, only one multiplication is left.

Any face may be the base. By the commutative property and the associative property, the three edges can be multiplied in any order. Standing a box on a different face changes which product is called B, but not V. Pick the order that makes the arithmetic easiest, often the one that lets a factor cancel.

Mixed numbers. Write each mixed number as an improper fraction before multiplying, as in Multiplying mixed numbers: 2\tfrac{1}{2} \times 1\tfrac{1}{2} is \tfrac{5}{2} \times \tfrac{3}{2} = \tfrac{15}{4}. Multiplying the whole parts and the fraction parts separately leaves out the cross pieces, so it gives the wrong answer. Turn the result back into a mixed number at the end if that reads more easily.

A missing edge. V = Bh is a multiplication fact, so it can be read backwards. If the volume and the base are known, the height is the number that B must be multiplied by to give V:

h = V \div B

Any other edge works the same way: divide the volume by the product of the other two edges. The division is division of fractions, done by multiplying by the reciprocal. Multiplying back checks it.

Units. All three edges must be in the same unit. Inches times inches times inches gives cubic inches, written \text{in}^3, so a box of 9\ \text{in}^3 holds nine cubic inches.

Worked examples

Examples 1 to 3 go with Part 1, and Examples 4 to 7 with Part 2.

Example 1

Fudge cubes in a box

Svea, who runs a sweet shop, cuts fudge into cubes with edges of \tfrac{1}{2} inch. She packs them into a box that is 3 inches long, 2 inches wide and 1\tfrac{1}{2} inches deep inside. How many cubes fill the box exactly? What is the volume of the box?

Solution

Count along each edge. Every inch holds two half-inch cubes, so 3 inches holds 6 cubes, 2 inches holds 4, and 1\tfrac{1}{2} inches holds 3. The box takes

6 \times 4 \times 3 = 72 \text{ cubes}

Eight half-inch cubes make one cubic inch, so the volume is 72 \div 8 = 9 cubic inches.

Check with the rule: 3 \times 2 \times \tfrac{3}{2} = 6 \times \tfrac{3}{2} = 9 cubic inches.

Example 2

Quarter cubes

A box measures \tfrac{3}{4} cm by \tfrac{1}{2} cm by 1\tfrac{1}{4} cm. How many cubes with edge \tfrac{1}{4} cm fill it? Use the count to find its volume, then check by multiplying the edges.

Solution

Write each edge in quarters: \tfrac{3}{4} is three quarters, \tfrac{1}{2} = \tfrac{2}{4} is two quarters, and 1\tfrac{1}{4} = \tfrac{5}{4} is five quarters. So the edges hold 3, 2 and 5 small cubes, and the box holds

3 \times 2 \times 5 = 30 \text{ cubes}

A cubic centimeter holds 4 \times 4 \times 4 = 64 quarter cubes, so each is worth \tfrac{1}{64} of a cubic centimeter, and

V = 30 \times \frac{1}{64} = \frac{30}{64} = \frac{15}{32} \text{ cubic centimeters}

Check: \tfrac{3}{4} \times \tfrac{1}{2} \times \tfrac{5}{4} = \tfrac{3 \times 1 \times 5}{4 \times 2 \times 4} = \tfrac{15}{32}. Less than half a cubic centimeter is sensible, since two of the edges are well under 1 cm.

Example 3

Edges with different denominators

A box measures 1\tfrac{1}{2} feet by \tfrac{2}{3} foot by \tfrac{1}{2} foot. Find a size of small cube that packs it exactly, and use it to find the volume.

Solution

Half-foot cubes fit the first and last edges but not the \tfrac{2}{3}, and cubes of edge \tfrac{1}{3} fit the \tfrac{2}{3} but not the halves. The cube needed has an edge that goes a whole number of times into halves and into thirds, and sixths do. Write all three edges over the common denominator 6:

1\tfrac{1}{2} = \frac{9}{6}, \qquad \frac{2}{3} = \frac{4}{6}, \qquad \frac{1}{2} = \frac{3}{6}

Cubes of edge \tfrac{1}{6} foot fit 9, 4 and 3 along the edges, so the box holds 9 \times 4 \times 3 = 108 of them. One cubic foot holds 6 \times 6 \times 6 = 216, so

V = \frac{108}{216} = \frac{1}{2} \text{ cubic foot}

Check: \tfrac{3}{2} \times \tfrac{2}{3} \times \tfrac{1}{2}. The first two factors multiply to 1, so the volume is \tfrac{1}{2} cubic foot. Multiplying was quicker; the count is what shows it is right.

Example 4

A moving van

Adelmo hires a moving van. Its cargo space is 10\tfrac{1}{2} feet long, 6 feet wide and 6\tfrac{2}{3} feet high. Find the volume of the cargo space with V = Bh.

Solution

The floor is the base:

B = 10\tfrac{1}{2} \times 6 = \frac{21}{2} \times 6 = 63 \text{ square feet}

So each foot of height holds 63 cubic feet. The height is 6\tfrac{2}{3} = \tfrac{20}{3} feet:

V = 63 \times \frac{20}{3} = 21 \times 20 = 420 \text{ cubic feet}

Dividing 63 by 3 before multiplying kept the numbers small. Check with another face as the base, the front wall: 6 \times \tfrac{20}{3} = 40 square feet, and 40 \times \tfrac{21}{2} = 420 cubic feet.

Example 5

Fudge cut from a tray

Berit pours fudge into a tray 10 inches long and 7\tfrac{1}{2} inches wide, and it sets \tfrac{3}{4} inch deep. She cuts it into pieces 1\tfrac{1}{4} inches by 1\tfrac{1}{4} inches, each the full depth. What volume of fudge is there, and how many pieces does she get?

Figure 3The fudge in Example 5, cut into 8 \times 6 = 48 pieces.

Solution

The base is 10 \times 7\tfrac{1}{2} = 10 \times \tfrac{15}{2} = 75 square inches, so

V = 75 \times \frac{3}{4} = \frac{225}{4} = 56\tfrac{1}{4} \text{ cubic inches}

For the pieces, count along the edges, using division to ask how many 1\tfrac{1}{4} inch lengths fit: 10 \div \tfrac{5}{4} = 10 \times \tfrac{4}{5} = 8, and \tfrac{15}{2} \div \tfrac{5}{4} = \tfrac{15}{2} \times \tfrac{4}{5} = 6. The cuts make 8 \times 6 = 48 pieces.

Check with volume. Each piece is \tfrac{5}{4} \times \tfrac{5}{4} \times \tfrac{3}{4} = \tfrac{75}{64} cubic inch, and \tfrac{225}{4} \div \tfrac{75}{64} = \tfrac{225}{4} \times \tfrac{64}{75} = 48. The two ways agree because the pieces fit every edge exactly, with nothing left over.

Example 6

How deep is the sand?

Gerd’s terrarium has a base 2 feet by 1\tfrac{1}{4} feet and is 1\tfrac{1}{2} feet tall. He pours in \tfrac{5}{8} cubic foot of sand and levels it. How deep is the sand? How much space is left above it?

Solution

The level sand is a box standing on the terrarium’s base, so V = Bh with the height unknown. The base is B = 2 \times \tfrac{5}{4} = \tfrac{5}{2} square feet, and the height is the volume divided by the base:

h = \frac{5}{8} \div \frac{5}{2} = \frac{5}{8} \times \frac{2}{5} = \frac{10}{40} = \frac{1}{4} \text{ foot}

Check: \tfrac{5}{2} \times \tfrac{1}{4} = \tfrac{5}{8}. The sand is \tfrac{1}{4} foot, or 3 inches, deep.

The space above it is a box with the same base and height 1\tfrac{1}{2} - \tfrac{1}{4} = 1\tfrac{1}{4} feet, so it holds \tfrac{5}{2} \times \tfrac{5}{4} = \tfrac{25}{8} = 3\tfrac{1}{8} cubic feet. The whole terrarium holds \tfrac{5}{2} \times \tfrac{3}{2} = \tfrac{15}{4} = 3\tfrac{3}{4} cubic feet, and 3\tfrac{3}{4} - \tfrac{5}{8} = 3\tfrac{1}{8} agrees.

Example 7

When dividing volumes over-counts

Evren loads Adelmo’s van from Example 4 with moving boxes that are cubes of edge 1\tfrac{1}{2} feet, standing in straight rows. How many fit? Compare with the cargo volume divided by the volume of one box.

Solution

Each box is \tfrac{3}{2} \times \tfrac{3}{2} \times \tfrac{3}{2} = \tfrac{27}{8} cubic feet, and

420 \div \frac{27}{8} = 420 \times \frac{8}{27} = \frac{3{,}360}{27} = 124\tfrac{4}{9}

That suggests 124 boxes. Count along the edges instead. Along the length, \tfrac{21}{2} \div \tfrac{3}{2} = \tfrac{21}{2} \times \tfrac{2}{3} = 7 boxes. Across the width, 6 \div \tfrac{3}{2} = 4. Up the height, \tfrac{20}{3} \div \tfrac{3}{2} = \tfrac{20}{3} \times \tfrac{2}{3} = 4\tfrac{4}{9}, so only 4 layers fit, leaving a gap 6\tfrac{2}{3} - 6 = \tfrac{2}{3} foot high at the top. The van holds

7 \times 4 \times 4 = 112 \text{ boxes}

The quotient of volumes over-counts because the gap at the top is real space, 63 \times \tfrac{2}{3} = 42 cubic feet of it, yet too thin for a single box. Volume says how much space there is. Whether rigid boxes can use it depends on how many fit along each edge. In Example 5 they fit exactly, so both ways gave 48.

Boundaries

Packing needs small cubes that fit every edge exactly. Each edge must be a whole number of small-cube edges, so choose the edge \tfrac{1}{n} with n a common denominator of all three edges. The rule V = lwh needs no packing once Part 1 has shown it holds for fraction edges.

A small cube’s volume is not its edge. A cube of edge \tfrac{1}{n} is \tfrac{1}{n \times n \times n} of a cubic unit, because it shrinks in all three directions at once.

One unit for all three edges. Inches times feet times feet is no named unit. Convert first: a box 6 inches by 2 feet by 1 foot is \tfrac{1}{2} foot by 2 feet by 1 foot, which is 1 cubic foot.

Boxes only. V = lwh needs every face of the solid to be a rectangle. V = Bh also holds for other prisms, with B the area of the end face, but then B is not l \times w; that is Volume of prisms and cylinders.

The height needs the base. h = V \div B needs B known and not zero. The volume alone does not fix the edges: a box \tfrac{1}{2} by 2 by 3 and a box 1 by 1 by 3 both hold 3 cubic units.

Space is not the same as how many solid objects fit. Dividing a container’s volume by an object’s volume gives the most that could fit, as in Example 7; rigid objects may fit fewer. Fudge poured warm, sand and water take the shape of the space, and boxes do not.

Where this goes next

V = Bh is the rule for every prism, not only boxes, and Volume of prisms and cylinders uses it with triangles, trapezoids and other shapes as the base. A solid built from several boxes is measured by adding the boxes, in Volume of a solid made of boxes. Volume is also what density divides a mass by, and real objects seldom have whole-number edges. The outside of a box, the paper it would take to cover it, is a different measure, its surface area, found by unfolding the box into a net.

Exercises

Part 1: packing with small cubes

How many cubes with edge \tfrac{1}{3} foot fill a cube with edge 1 foot? What is the volume of each small cube?
How many cubes with edge \tfrac{1}{2} cm fill a cube with edge 2 cm? Use the count to find the volume of the large cube.
A box is 1\tfrac{1}{2} units by 1 unit by 1 unit. How many cubes of edge \tfrac{1}{2} unit fill it, and what is its volume?
A box is \tfrac{1}{4} foot by 1 foot by 1\tfrac{1}{2} feet. How many cubes of edge \tfrac{1}{4} foot fill it, and what is its volume?
A box is 3\tfrac{1}{2} feet by 1 foot by \tfrac{1}{2} foot. Find its volume by packing it with half-foot cubes, and check by multiplying the edges.
Svea’s narrow fudge box is 4 inches long, 1\tfrac{1}{2} inches wide and 1 inch deep. How many half-inch fudge cubes fill it, and what is its volume?
A box is 1\tfrac{3}{4} inches by 2 inches by \tfrac{1}{4} inch. How many cubes of edge \tfrac{1}{4} inch fill it, and what is its volume?
A box is \tfrac{1}{3} m by \tfrac{3}{4} m by \tfrac{1}{2} m. Find a size of small cube that packs it exactly, how many of them it takes, and the volume of the box.

Part 2: base times height, and missing edges

Find the volume of a box 3\tfrac{1}{2} cm by 2 cm by 1\tfrac{1}{2} cm.
A box has a base of area 5\tfrac{1}{3} square feet and a height of \tfrac{3}{4} foot. Find its volume.
Find the volume of a cube whose edges are 2\tfrac{1}{2} inches long.
A terrarium is 2\tfrac{1}{2} feet long, 1\tfrac{1}{5} feet wide and 1\tfrac{1}{3} feet tall. How much space is inside it?
Berit sets fudge \tfrac{5}{8} inch deep in a tray 12 inches by 8 inches and cuts it into pieces 1\tfrac{1}{2} inches by 1 inch, each the full depth. Find the volume of the fudge and the number of pieces, and check that they agree.
A box holds 15 cubic inches. Its base is 4 inches by 2\tfrac{1}{2} inches. How tall is it?
A box holds 9\tfrac{3}{8} cubic feet. It is 2\tfrac{1}{2} feet long and 1\tfrac{1}{2} feet tall. How wide is it?
Gerd spreads gravel \tfrac{1}{3} foot deep over the floor of a terrarium 3 feet by 1\tfrac{1}{2} feet. Gravel is sold in bags of \tfrac{2}{5} cubic foot. How many bags must he buy?
Box A is 2\tfrac{1}{2} inches by 2\tfrac{1}{2} inches by 2 inches. Box B is 3\tfrac{1}{3} inches by 1\tfrac{1}{2} inches by 2\tfrac{1}{2} inches. Which holds more?
A fudge box has a base 2\tfrac{1}{2} inches by 2 inches, and exactly 80 half-inch fudge cubes fill it in full layers. How tall is it inside?
A van’s cargo space is 7\tfrac{1}{2} feet by 5 feet by 4 feet. How many cube-shaped boxes of edge 1\tfrac{1}{4} feet fit, standing in straight rows? Compare with the cargo volume divided by the volume of one box.

Reasoning

Explain why 64 cubes of edge \tfrac{1}{4} fill a unit cube, and why each has volume \tfrac{1}{64} of a cubic unit rather than \tfrac{1}{4}.
A box is \tfrac{5}{6} by \tfrac{1}{2} by \tfrac{2}{3} units. Pack it with cubes of edge \tfrac{1}{6}, find the volume from the count, and explain why the count matches \tfrac{5}{6} \times \tfrac{1}{2} \times \tfrac{2}{3}.
Explain why the height of a box is V \div B, and why the volume alone cannot tell you the height.
Can a box whose edges are all shorter than 1 unit hold 1 cubic unit or more? Explain.

Answers

27 cubes, each \tfrac{1}{27} cubic foot. Three fit along each edge, so 3 \times 3 \times 3 = 27 equal cubes make one cubic foot.
64 cubes, so the volume is 64 \times \tfrac{1}{8} = 8 cubic centimeters. Four fit along each edge: 4 \times 4 \times 4 = 64. Check: 2 \times 2 \times 2 = 8.
12 cubes, volume 1\tfrac{1}{2} cubic units. The edges hold 3, 2 and 2 cubes, and \tfrac{12}{8} = 1\tfrac{1}{2}. Check: \tfrac{3}{2} \times 1 \times 1 = \tfrac{3}{2}.
24 cubes, volume \tfrac{3}{8} cubic foot. The edges hold 1, 4 and 6 cubes, and \tfrac{24}{64} = \tfrac{3}{8}. Check: \tfrac{1}{4} \times 1 \times \tfrac{3}{2} = \tfrac{3}{8}.
1\tfrac{3}{4} cubic feet. The edges hold 7, 2 and 1 cubes, 14 in all, and \tfrac{14}{8} = \tfrac{7}{4}. Check: \tfrac{7}{2} \times 1 \times \tfrac{1}{2} = \tfrac{7}{4}.
48 cubes, volume 6 cubic inches. The edges hold 8, 3 and 2 cubes, and 48 \div 8 = 6. Check: 4 \times \tfrac{3}{2} \times 1 = 6.
56 cubes, volume \tfrac{7}{8} cubic inch. In quarters the edges are \tfrac{7}{4}, \tfrac{8}{4} and \tfrac{1}{4}, so they hold 7, 8 and 1 cubes, and \tfrac{56}{64} = \tfrac{7}{8}. Check: \tfrac{7}{4} \times 2 \times \tfrac{1}{4} = \tfrac{7}{8}.
Cubes of edge \tfrac{1}{12} m; 216 of them; volume \tfrac{1}{8} cubic meter. Over 12 the edges are \tfrac{4}{12}, \tfrac{9}{12} and \tfrac{6}{12}, so 4 \times 9 \times 6 = 216 cubes. A cubic meter holds 12 \times 12 \times 12 = 1{,}728 of them, and \tfrac{216}{1{,}728} = \tfrac{1}{8}. Check: \tfrac{1}{3} \times \tfrac{3}{4} \times \tfrac{1}{2} = \tfrac{3}{24} = \tfrac{1}{8}.
10\tfrac{1}{2} cubic centimeters. \tfrac{7}{2} \times 2 \times \tfrac{3}{2} = 7 \times \tfrac{3}{2} = \tfrac{21}{2}.
4 cubic feet. \tfrac{16}{3} \times \tfrac{3}{4} = \tfrac{48}{12} = 4.
15\tfrac{5}{8} cubic inches. \tfrac{5}{2} \times \tfrac{5}{2} \times \tfrac{5}{2} = \tfrac{125}{8}.
4 cubic feet. The base is \tfrac{5}{2} \times \tfrac{6}{5} = 3 square feet, and 3 \times \tfrac{4}{3} = 4.
60 cubic inches of fudge in 64 pieces. The base is 96 square inches, and 96 \times \tfrac{5}{8} = 60. Along the edges 12 \div 1\tfrac{1}{2} = 8 and 8 \div 1 = 8, so 64 pieces. Each is \tfrac{3}{2} \times 1 \times \tfrac{5}{8} = \tfrac{15}{16} cubic inch, and 64 \times \tfrac{15}{16} = 60.
1\tfrac{1}{2} inches. B = 4 \times \tfrac{5}{2} = 10 square inches, and h = 15 \div 10 = \tfrac{3}{2}.
2\tfrac{1}{2} feet. The two known edges multiply to \tfrac{5}{2} \times \tfrac{3}{2} = \tfrac{15}{4}, and \tfrac{75}{8} \div \tfrac{15}{4} = \tfrac{75}{8} \times \tfrac{4}{15} = \tfrac{5}{2}.
4 bags. The gravel is 3 \times \tfrac{3}{2} \times \tfrac{1}{3} = \tfrac{3}{2} cubic feet, and \tfrac{3}{2} \div \tfrac{2}{5} = \tfrac{15}{4} = 3\tfrac{3}{4}, so 3 bags fall short.
Neither: each holds 12\tfrac{1}{2} cubic inches. A is \tfrac{25}{4} \times 2 = \tfrac{25}{2}. B is \tfrac{10}{3} \times \tfrac{3}{2} = 5 square inches of base times \tfrac{5}{2}, which is also \tfrac{25}{2}.
2 inches. One layer is 5 \times 4 = 20 cubes, so 80 cubes make 4 layers, each \tfrac{1}{2} inch thick. Check: 80 \div 8 = 10 cubic inches, the base is 5 square inches, and 10 \div 5 = 2.
72 boxes. Along the edges 7\tfrac{1}{2} \div 1\tfrac{1}{4} = 6, 5 \div 1\tfrac{1}{4} = 4 and 4 \div 1\tfrac{1}{4} = 3\tfrac{1}{5}, so 3 layers and 6 \times 4 \times 3 = 72. The cargo volume is 150 cubic feet and one box is \tfrac{125}{64}, so the quotient is 150 \times \tfrac{64}{125} = 76\tfrac{4}{5}, more than fit, because a gap \tfrac{1}{4} foot high is left at the top.
Four quarter cubes fit along each edge, so 4 \times 4 = 16 cover the floor in one layer \tfrac{1}{4} thick, and 4 such layers fill the cube: 64. They are equal and together make 1 cubic unit, so each is \tfrac{1}{64}. A quarter cube is a quarter as long, a quarter as wide and a quarter as tall as the unit cube, so it is far less than a quarter of it.
60 cubes and \tfrac{5}{18} cubic unit. With \tfrac{1}{2} = \tfrac{3}{6} and \tfrac{2}{3} = \tfrac{4}{6}, the edges hold 5, 3 and 4 cubes, so 5 \times 3 \times 4 = 60. A unit cube holds 216, and \tfrac{60}{216} = \tfrac{5}{18}. Written as \tfrac{5}{6} \times \tfrac{3}{6} \times \tfrac{4}{6}, the product has numerator 5 \times 3 \times 4, the cube count, and denominator 6 \times 6 \times 6, the cubes in one unit, so it is the same fraction.
V = B \times h says the height is the number the base area must be multiplied by to give the volume, and that number is V \div B. Without the base the height is not fixed: a box of 6 cubic units could have base 6 and height 1, or base 4 and height 1\tfrac{1}{2}.
No. Multiplying by a number between 0 and 1 gives less than you started with. So l \times w is less than l, which is less than 1, and (l \times w) \times h is smaller still. In packing terms, such a box fits inside a unit cube, so it cannot hold as much as the whole cube.