Anvi pours cups of seed into the backyard feeder every day. How much seed does she pour in days, and what if the number of days is a mixed number too?

Grade 5 · 5.NF.B.4 · 5.NF.B.6

Multiplying mixed numbers

Anvi pours \displaystyle 2\frac{1}{4} cups of seed into the backyard feeder every day. How much seed does she pour in 3 days, and what if the number of days is a mixed number too?

You need

A fraction times a fractionGrade 5 Mixed numbers and improper fractions: rewriting and comparingGrade 4

Where you are 4 in Multiplying fractions

Fractions, ratio and percent3

Fractions15

Multiplying fractions4

A fraction times a whole number A fraction times a fraction Multiplying as scaling

Multiplying mixed numbers

When you reach for it

Anvi, Duc and Funmi keep bird feeders in their backyards. Anvi pours \displaystyle 2\frac{1}{4} cups of seed into her feeder every day. In 3 days she pours 3 groups of \displaystyle 2\frac{1}{4} cups, which is \displaystyle 3 \times 2\frac{1}{4} cups. Duc’s seed bags weigh \displaystyle 12\frac{1}{2} pounds, and Funmi’s ground tray is a rectangle whose sides are mixed numbers. Each of these asks for a product where at least one factor is a mixed number: a whole number and a fraction written side by side.

A mixed number is a sum. \displaystyle 2\frac{1}{4} means \displaystyle 2 + \frac{1}{4}, two wholes and one fourth more. The space between the 2 and the \displaystyle \frac{1}{4} is a hidden plus sign, never a times sign. It can also be written as one fraction, an improper fraction, with a top bigger than its bottom: \displaystyle 2\frac{1}{4} = \frac{9}{4}, because 2 wholes are 8 fourths and one more fourth makes 9 (Mixed numbers and improper fractions: rewriting and comparing).

Those two ways of seeing a mixed number give two ways to multiply it. Seen as a sum, you multiply each part and add, which is the distributive property. Seen as one fraction, you use the rule you already have, tops times tops and bottoms times bottoms (A fraction times a fraction). This lesson takes three days. Day 1 multiplies a mixed number by a whole number or a fraction, by splitting it into parts. Day 2 multiplies two mixed numbers, by rewriting both as fractions, and checks the answer with a four-part area model. Day 3 estimates a product before working it out and uses the skill in problems about seed, recipes and areas.

The procedure

Day 1: split the mixed number into its parts

Take \displaystyle 3 \times 2\frac{1}{4}, Anvi’s seed for 3 days. Since \displaystyle 2\frac{1}{4} is \displaystyle 2 + \frac{1}{4}, three of them are three 2s and three \displaystyle \frac{1}{4}s:

\begin{aligned} 3 \times 2\frac{1}{4} &= 3 \times \left(2 + \frac{1}{4}\right) \\ &= 3 \times 2 + 3 \times \frac{1}{4} \end{aligned}

That step is the distributive property: multiplying a sum is the same as multiplying each part and adding the results. Now each part is a product you know: 3 \times 2 = 6, and \displaystyle 3 \times \frac{1}{4} = \frac{3}{4} (A fraction times a whole number). So

3 \times 2\frac{1}{4} = 6 + \frac{3}{4} = 6\frac{3}{4}

Anvi pours \displaystyle 6\frac{3}{4} cups in 3 days.

An area model shows why both parts count. Draw a rectangle 3 units tall and \displaystyle 2\frac{1}{4} units wide. Its area is \displaystyle 3 \times 2\frac{1}{4} square units. Draw a line down the rectangle where the width 2 ends. The left part is 3 by 2, which holds 6 unit squares. The right part is a thin strip 3 by \displaystyle \frac{1}{4}. Each row of the strip is a quarter of a unit square, and there are 3 rows, so the strip has area \displaystyle \frac{3}{4}.

Figure 1The rectangle is 3 by \displaystyle 2\frac{1}{4}. Its two parts have areas 3 \times 2 = 6 and \displaystyle 3 \times \frac{1}{4} = \frac{3}{4}, so \displaystyle 3 \times 2\frac{1}{4} = 6\frac{3}{4}.

The whole rectangle is both parts together, \displaystyle 6 + \frac{3}{4}. Leaving out the strip would leave out real area: the sides of the rectangle are 3 and \displaystyle 2\frac{1}{4}, and every bit of the width gets multiplied by the height.

The other road gives the same answer. Rewrite \displaystyle 2\frac{1}{4} as \displaystyle \frac{9}{4} and multiply:

3 \times \frac{9}{4} = \frac{27}{4} = 6\frac{3}{4}

since 27 fourths is 6 wholes (24 fourths) and 3 fourths more. For a whole number times a mixed number, splitting is usually quicker, because each part is small and easy.

The same split works when the other factor is a fraction. To find \displaystyle \frac{1}{2} \times 4\frac{2}{3}, take half of each part: half of 4 is 2, and half of \displaystyle \frac{2}{3} is \displaystyle \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}. So \displaystyle \frac{1}{2} \times 4\frac{2}{3} = 2 + \frac{1}{3} = 2\frac{1}{3}. The order of the factors does not matter, by the commutative property, so \displaystyle 4\frac{2}{3} \times \frac{1}{2} is the same \displaystyle 2\frac{1}{3}.

Sometimes the fraction part grows past 1. Then the answer is tidied by moving the extra whole into the whole-number part, the same regrouping as in Adding mixed numbers. Example 1 does this.

Day 2: two mixed numbers

Take \displaystyle 1\frac{3}{4} \times 2\frac{2}{5}. Now both factors are sums, and the area model shows what that means. Draw a rectangle \displaystyle 2\frac{2}{5} wide and \displaystyle 1\frac{3}{4} tall. Split the width into 2 and \displaystyle \frac{2}{5}, and split the height into 1 and \displaystyle \frac{3}{4}. One line down and one line across cut the rectangle into four parts.

Figure 2Splitting both sides cuts the rectangle into four parts, and the product is all four added together.

Each part is a width times a height:

Part	Width	Height	Area
top left	2	1	2
top right	\displaystyle \frac{2}{5}	1	\displaystyle \frac{2}{5}
bottom left	2	\displaystyle \frac{3}{4}	\displaystyle \frac{6}{4} = \frac{3}{2}
bottom right	\displaystyle \frac{2}{5}	\displaystyle \frac{3}{4}	\displaystyle \frac{6}{20} = \frac{3}{10}

The product is the whole rectangle, so it is all four parts added. Over a common bottom of 10:

\begin{aligned} &2 + \frac{2}{5} + \frac{3}{2} + \frac{3}{10} \\ &= \frac{20}{10} + \frac{4}{10} + \frac{15}{10} + \frac{3}{10} \\ &= \frac{42}{10} \end{aligned}

and \displaystyle \frac{42}{10} = 4\frac{2}{10} = 4\frac{1}{5}.

Notice that the wholes times each other (2 \times 1) and the fractions times each other (\displaystyle \frac{2}{5} \times \frac{3}{4}) give only two of the four parts, the top left and the bottom right corners. The other two parts, a whole times a fraction, are real area too. The picture makes sure none is skipped: the product is the area of the whole rectangle.

Four parts and a sum over a common bottom is a lot of work. The other road is shorter. Rewrite each mixed number as an improper fraction:

\begin{aligned} 1\frac{3}{4} &= \frac{4}{4} + \frac{3}{4} = \frac{7}{4} \\ 2\frac{2}{5} &= \frac{10}{5} + \frac{2}{5} = \frac{12}{5} \end{aligned}

These are the same numbers with new names, so the product does not change:

1\frac{3}{4} \times 2\frac{2}{5} = \frac{7}{4} \times \frac{12}{5}

Now the rule for fractions takes over. Before multiplying, look for a top and a bottom that share a factor, as in A fraction times a fraction. The top 12 and the bottom 4 share 4: 12 \div 4 = 3 and 4 \div 4 = 1. So

\frac{7}{4} \times \frac{12}{5} = \frac{7}{1} \times \frac{3}{5} = \frac{21}{5} = 4\frac{1}{5}

because 21 fifths is 4 wholes (20 fifths) and 1 fifth more. Both roads give \displaystyle 4\frac{1}{5}, which is a good check on each.

Simplifying waits until both factors are improper fractions. A mixed number is a sum, and dividing a top and a bottom by a common factor works only when everything on top is multiplied together and everything on the bottom is multiplied together. In \displaystyle \frac{7}{4} \times \frac{12}{5} that is true, so the 12 and the 4 may be paired. In \displaystyle 1\frac{3}{4} the 1 is added, not multiplied, so its parts cannot be paired with anything yet.

For two mixed numbers, the improper-fraction road is the one to reach for. The four-part area model stays useful as a picture of what the product means and as a check.

Day 3: estimate first, then use it

A mixed number sits between two whole numbers. \displaystyle 3\frac{1}{3} is between 3 and 4, and \displaystyle 1\frac{1}{5} is between 1 and 2. So \displaystyle 3\frac{1}{3} \times 1\frac{1}{5} is between 3 \times 1 = 3 and 4 \times 2 = 8. Making a factor bigger makes the product bigger, so the product of the smaller wholes is too small and the product of the larger wholes is too big.

That range comes before any fraction work, and it catches an answer of the wrong size. Here is the exact product, with both factors rewritten:

3\frac{1}{3} \times 1\frac{1}{5} = \frac{10}{3} \times \frac{6}{5}

The top 10 and the bottom 5 share 5, and the top 6 and the bottom 3 share 3:

\frac{10}{3} \times \frac{6}{5} = \frac{2}{1} \times \frac{2}{1} = 4

The product is exactly 4, inside the range from 3 to 8 and near its low end, which makes sense: \displaystyle 3\frac{1}{3} is close to 3 and \displaystyle 1\frac{1}{5} is close to 1, so the product should sit a little above 3 \times 1.

A product of mixed numbers is often an area. If Funmi spreads seed on a patch of yard \displaystyle 3\frac{1}{3} yards long and \displaystyle 1\frac{1}{5} yards wide, the patch covers 4 square yards, because the area of a rectangle is its length times its width even when the sides are not whole numbers (Area of rectangles and squares). It is also often a recipe scaled up or down, or a rate times an amount: cups per day times days, or pounds per bag times bags. In each case, check that the units make sense together before multiplying. Yards times yards gives square yards. Cups per day times days gives cups.

Worked examples

Examples 1 and 2 go with Day 1, Examples 3 and 4 with Day 2, and Examples 5 to 7 with Day 3.

Example 1

Four days of seed

Duc’s feeder empties every day, and he refills it with \displaystyle 1\frac{3}{8} cups of seed. How much seed does he use in 4 days?

Solution

Four days of \displaystyle 1\frac{3}{8} cups is \displaystyle 4 \times 1\frac{3}{8} cups. Split the mixed number into \displaystyle 1 + \frac{3}{8} and multiply each part by 4, by the distributive property:

4 \times 1\frac{3}{8} = 4 \times 1 + 4 \times \frac{3}{8} = 4 + \frac{12}{8}

The fraction part, \displaystyle \frac{12}{8}, is more than one whole: 8 eighths make 1, and 4 eighths are left. And \displaystyle \frac{4}{8} = \frac{1}{2}. So \displaystyle \frac{12}{8} = 1\frac{1}{2}, and

4 + 1\frac{1}{2} = 5\frac{1}{2}

Duc uses \displaystyle 5\frac{1}{2} cups in 4 days. Check with the other road: \displaystyle 4 \times \frac{11}{8} = \frac{44}{8} = \frac{11}{2} = 5\frac{1}{2}.

Example 2

A smaller feeder

Funmi’s big feeder holds \displaystyle 3\frac{3}{4} cups of seed. Her small feeder holds \displaystyle \frac{2}{3} as much. How much does the small feeder hold?

Solution

Two thirds as much as \displaystyle 3\frac{3}{4} cups is \displaystyle \frac{2}{3} \times 3\frac{3}{4} cups. Split the mixed number and take two thirds of each part:

\begin{aligned} \frac{2}{3} \times 3\frac{3}{4} &= \frac{2}{3} \times \left(3 + \frac{3}{4}\right) \\ &= \frac{2}{3} \times 3 + \frac{2}{3} \times \frac{3}{4} \end{aligned}

Two thirds of 3 is 2, since one third of 3 is 1. For the second part, the top 3 and the bottom 3 share 3, so \displaystyle \frac{2}{3} \times \frac{3}{4} = \frac{2}{1} \times \frac{1}{4}, which is \displaystyle \frac{2}{4} = \frac{1}{2}. Adding the parts:

2 + \frac{1}{2} = 2\frac{1}{2}

The small feeder holds \displaystyle 2\frac{1}{2} cups. That is less than \displaystyle 3\frac{3}{4}, as it should be, because \displaystyle \frac{2}{3} of an amount is less than the whole amount.

In an area model, this is a rectangle \displaystyle \frac{2}{3} tall and \displaystyle 3\frac{3}{4} wide, cut into a part \displaystyle \frac{2}{3} by 3 and a strip \displaystyle \frac{2}{3} by \displaystyle \frac{3}{4}. The two parts have areas 2 and \displaystyle \frac{1}{2}.

Example 3

A feeder platform, two ways

Duc builds a flat feeder platform \displaystyle 2\frac{1}{3} feet long and \displaystyle 1\frac{1}{2} feet wide. What is its area? Find it with a four-part area model, then with improper fractions.

Solution

The area is \displaystyle 2\frac{1}{3} \times 1\frac{1}{2} square feet.

Four parts. Split the length into 2 and \displaystyle \frac{1}{3} and the width into 1 and \displaystyle \frac{1}{2}. The four parts are

\begin{aligned} 2 \times 1 &= 2, & 2 \times \frac{1}{2} &= 1, \\ \frac{1}{3} \times 1 &= \frac{1}{3}, & \frac{1}{3} \times \frac{1}{2} &= \frac{1}{6} \end{aligned}

Add them. The wholes give 2 + 1 = 3. The fractions give \displaystyle \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}. The total is \displaystyle 3\frac{1}{2}.

Improper fractions. \displaystyle 2\frac{1}{3} = \frac{7}{3} and \displaystyle 1\frac{1}{2} = \frac{3}{2}. The top 3 and the bottom 3 share 3:

\frac{7}{3} \times \frac{3}{2} = \frac{7}{1} \times \frac{1}{2} = \frac{7}{2} = 3\frac{1}{2}

The platform has area \displaystyle 3\frac{1}{2} square feet. Both roads agree.

Example 4

Two mixed numbers, simplified first

Find \displaystyle 4\frac{1}{2} \times 2\frac{2}{3}.

Solution

First a range: the product is between 4 \times 2 = 8 and 5 \times 3 = 15.

Rewrite both factors. \displaystyle 4\frac{1}{2} is 8 halves and 1 half, \displaystyle \frac{9}{2}. \displaystyle 2\frac{2}{3} is 6 thirds and 2 thirds, \displaystyle \frac{8}{3}. So the product is \displaystyle \frac{9}{2} \times \frac{8}{3}.

Look for a top and a bottom with a common factor. The top 9 and the bottom 3 share 3: they become 3 and 1. The top 8 and the bottom 2 share 2: they become 4 and 1. So

\frac{9}{2} \times \frac{8}{3} = \frac{3}{1} \times \frac{4}{1} = 12

\displaystyle 4\frac{1}{2} \times 2\frac{2}{3} = 12, a whole number, and inside the range from 8 to 15. Without simplifying first, the product would be \displaystyle \frac{72}{6}, which also equals 12, but with bigger numbers along the way.

Example 5

A seed-cake recipe

Anvi makes seed cakes to hang in the trees. One batch uses \displaystyle 2\frac{3}{4} cups of seed. She makes \displaystyle 1\frac{1}{2} batches. Estimate how much seed she needs, then find it exactly.

Solution

One and a half batches need \displaystyle 1\frac{1}{2} times as much seed: \displaystyle 1\frac{1}{2} \times 2\frac{3}{4} cups.

Estimate. \displaystyle 2\frac{3}{4} is close to 3, and \displaystyle 1\frac{1}{2} batches of 3 cups is \displaystyle 3 + 1\frac{1}{2} = 4\frac{1}{2} cups. Since \displaystyle 2\frac{3}{4} is a little less than 3, the answer should be a little less than \displaystyle 4\frac{1}{2} cups.

Exact. \displaystyle 1\frac{1}{2} = \frac{3}{2} and \displaystyle 2\frac{3}{4} = \frac{11}{4}. No top shares a factor with a bottom (3 and 11 against 2 and 4), so multiply straight through:

\frac{3}{2} \times \frac{11}{4} = \frac{3 \times 11}{2 \times 4} = \frac{33}{8}

33 eighths is 4 wholes (32 eighths) and 1 eighth: \displaystyle \frac{33}{8} = 4\frac{1}{8}.

Anvi needs \displaystyle 4\frac{1}{8} cups of seed, a little less than the estimate of \displaystyle 4\frac{1}{2}, as expected.

Example 6

Seed for the ground tray

Funmi’s ground tray is \displaystyle 1\frac{2}{3} feet long and \displaystyle 1\frac{4}{5} feet wide. She scatters \displaystyle 1\frac{1}{4} cups of seed on each square foot. How much seed covers the tray?

Solution

Two products, one after the other. First the area of the tray, then the seed.

Area. \displaystyle 1\frac{2}{3} \times 1\frac{4}{5} = \frac{5}{3} \times \frac{9}{5}. The top 5 and the bottom 5 share 5; the top 9 and the bottom 3 share 3:

\frac{5}{3} \times \frac{9}{5} = \frac{1}{1} \times \frac{3}{1} = 3

The tray covers 3 square feet.

Seed. Each square foot gets \displaystyle 1\frac{1}{4} cups, so 3 square feet get \displaystyle 3 \times 1\frac{1}{4} cups. Split the mixed number, as on Day 1:

3 \times 1\frac{1}{4} = 3 + \frac{3}{4} = 3\frac{3}{4}

Funmi uses \displaystyle 3\frac{3}{4} cups. The units work out: feet times feet is square feet, and cups per square foot times square feet is cups.

Example 7

A big bag, two ways

Each of Duc’s seed bags weighs \displaystyle 12\frac{1}{2} pounds. Over the winter his feeders went through \displaystyle 3\frac{1}{5} bags. How many pounds of seed was that?

Solution

\displaystyle 3\frac{1}{5} bags of \displaystyle 12\frac{1}{2} pounds is \displaystyle 12\frac{1}{2} \times 3\frac{1}{5} pounds. The range: between 12 \times 3 = 36 and 13 \times 4 = 52 pounds.

Improper fractions. \displaystyle 12\frac{1}{2} = \frac{25}{2} and \displaystyle 3\frac{1}{5} = \frac{16}{5}. The top 25 and the bottom 5 share 5; the top 16 and the bottom 2 share 2:

\frac{25}{2} \times \frac{16}{5} = \frac{5}{1} \times \frac{8}{1} = 40

Four parts, as a check. Split \displaystyle 12\frac{1}{2} into \displaystyle 12 + \frac{1}{2} and \displaystyle 3\frac{1}{5} into \displaystyle 3 + \frac{1}{5}. The parts are 12 \times 3 = 36, \displaystyle 12 \times \frac{1}{5} = \frac{12}{5}, \displaystyle \frac{1}{2} \times 3 = \frac{3}{2}, and \displaystyle \frac{1}{2} \times \frac{1}{5} = \frac{1}{10}. Over a bottom of 10:

\begin{aligned} 36 + \frac{12}{5} + \frac{3}{2} + \frac{1}{10} &= 36 + \frac{24}{10} + \frac{15}{10} + \frac{1}{10} \\ &= 36 + \frac{40}{10} = 36 + 4 = 40 \end{aligned}

Duc used 40 pounds of seed, inside the range from 36 to 52. The three small parts add to exactly 4 pounds: the fifth of a bag holds \displaystyle 2\frac{1}{2} pounds, and the 3 full bags hold \displaystyle \frac{1}{2} pound each beyond their 12 pounds, another \displaystyle 1\frac{1}{2} pounds. With numbers this size, the improper-fraction road took one line, while the four parts needed a common bottom.

Boundaries

The fractions here have counting numbers on the bottom, never 0, because a fraction with 0 on the bottom names no amount. The mixed numbers are amounts like cups and feet, so they are positive.

A mixed number is a sum, so it is not ready for the fraction rule as written. Either split it and multiply every part (the distributive property), or rewrite it as one improper fraction first. Dividing a top and a bottom by a common factor is allowed only after the rewrite, when the product is all tops multiplied over all bottoms multiplied.

When both factors are mixed numbers, the area model has four parts, and the product is the sum of all four, including the two where a whole meets a fraction.

The range from “smaller wholes times smaller wholes” to “larger wholes times larger wholes” works because both factors are positive: a bigger positive factor gives a bigger product. It gives a range to land in, not the answer.

Where this goes next

Every product here could be judged before it was worked out. A factor greater than 1 made the product bigger than the other factor, and a factor less than 1, like the \displaystyle \frac{2}{3} of Example 2, made it smaller. That is the scaling idea of Multiplying as scaling, and it carries on to every product of fractions.

Dividing comes later. Sharing \displaystyle 6\frac{3}{4} cups of seed into portions of \displaystyle 2\frac{1}{4} cups undoes the product that opened this lesson, and it uses the same first move: rewrite the mixed numbers as fractions (Dividing mixed numbers). The areas of Day 3 also grow into boxes whose edges are mixed numbers (Volume of a box with fraction edge lengths).

Exercises

Exercises 1 to 8 go with Day 1, 9 to 16 with Day 2, 17 to 22 with Day 3, and the reasoning group draws on all three days. Write answers greater than 1 as mixed numbers or whole numbers.

Day 1: split the mixed number

\displaystyle 2 \times 3\frac{1}{4}
\displaystyle 5 \times 1\frac{2}{3}
\displaystyle 4 \times 2\frac{5}{6}
\displaystyle \frac{1}{2} \times 6\frac{4}{5}
\displaystyle \frac{3}{4} \times 8\frac{1}{3}
Duc pours \displaystyle 2\frac{1}{3} cups of seed into his feeder each morning. How much does he pour in 6 mornings?
Describe an area model for \displaystyle 5 \times 3\frac{2}{5}: what are the two parts and their areas? Use it to find the product.
Funmi’s big feeder holds \displaystyle 5\frac{1}{2} cups. A tube feeder holds \displaystyle \frac{3}{5} as much. How much does the tube feeder hold?

Day 2: two mixed numbers

\displaystyle 1\frac{1}{2} \times 1\frac{1}{3}
\displaystyle 1\frac{2}{3} \times 2\frac{1}{2}
\displaystyle 2\frac{2}{3} \times 1\frac{1}{8}
\displaystyle 3\frac{1}{2} \times 1\frac{1}{7}
\displaystyle 1\frac{4}{5} \times 2\frac{1}{6}
\displaystyle 5\frac{1}{3} \times 2\frac{1}{4}
Draw a four-part area model for \displaystyle 2\frac{1}{2} \times 1\frac{1}{4}. Give the area of each part and the total, then check with improper fractions.
Anvi’s feeder platform is \displaystyle 3\frac{1}{4} feet long and \displaystyle 1\frac{3}{5} feet wide. What is its area?

Day 3: estimate, then use it

Give a range for \displaystyle 4\frac{1}{5} \times 2\frac{1}{2} from whole numbers, then find the product.
A seed-cake recipe uses \displaystyle 2\frac{2}{3} cups of seed. Funmi makes \displaystyle 2\frac{1}{4} batches. How much seed does she use?
A bag of seed weighs \displaystyle 7\frac{1}{2} pounds. Duc’s feeders went through \displaystyle 2\frac{2}{5} bags in a month. How many pounds was that?
Anvi’s feeders use \displaystyle 1\frac{1}{6} cups of seed a day. How much do they use in \displaystyle 3\frac{1}{2} days? Estimate first.
A patch of yard is \displaystyle 5\frac{1}{4} yards long and \displaystyle 2\frac{2}{3} yards wide. Funmi scatters \displaystyle \frac{1}{2} pound of seed on each square yard. How much seed does the patch need?
Duc works out \displaystyle 2\frac{4}{5} \times 3\frac{1}{8} and gets \displaystyle 5\frac{7}{8}. Use a range from whole numbers to decide whether that can be right, then find the product.

Reasoning

Name the four parts of the area model for \displaystyle 1\frac{1}{2} \times 1\frac{1}{2} and find the product. How much do the two parts where a whole meets a fraction add together, and why do they belong in the product?
Explain why the product of two mixed numbers is always bigger than each of the two mixed numbers.
Find \displaystyle 1\frac{1}{3} \times 1\frac{1}{2} both ways, with four parts and with improper fractions. Explain why the two roads must give the same answer.

Answers

\displaystyle 6\frac{1}{2}. \displaystyle 2 \times 3 + 2 \times \frac{1}{4} = 6 + \frac{2}{4} = 6\frac{1}{2}.
\displaystyle 8\frac{1}{3}. \displaystyle 5 \times 1 + 5 \times \frac{2}{3} = 5 + \frac{10}{3} = 5 + 3\frac{1}{3}.
\displaystyle 11\frac{1}{3}. \displaystyle 4 \times 2 + 4 \times \frac{5}{6} = 8 + \frac{20}{6} = 8 + 3\frac{1}{3}, since \displaystyle \frac{20}{6} = \frac{10}{3}.
\displaystyle 3\frac{2}{5}. Half of 6 is 3, and half of \displaystyle \frac{4}{5} is \displaystyle \frac{2}{5}.
\displaystyle 6\frac{1}{4}. \displaystyle \frac{3}{4} \times 8 = 6, and \displaystyle \frac{3}{4} \times \frac{1}{3} = \frac{3}{12} = \frac{1}{4}.
14 cups. \displaystyle 6 \times 2 + 6 \times \frac{1}{3} = 12 + 2.
17. The rectangle is 5 tall and \displaystyle 3\frac{2}{5} wide. One part is 5 by 3, area 15; the strip is 5 by \displaystyle \frac{2}{5}, area \displaystyle \frac{10}{5} = 2. Together 15 + 2 = 17.
\displaystyle 3\frac{3}{10} cups. \displaystyle \frac{3}{5} \times 5 = 3, and \displaystyle \frac{3}{5} \times \frac{1}{2} = \frac{3}{10}.
2. \displaystyle \frac{3}{2} \times \frac{4}{3}: the 3s become 1 and 1, the 4 and 2 become 2 and 1, so \displaystyle \frac{1}{1} \times \frac{2}{1} = 2.
\displaystyle 4\frac{1}{6}. \displaystyle \frac{5}{3} \times \frac{5}{2} = \frac{25}{6}, and 25 = 24 + 1.
3. \displaystyle \frac{8}{3} \times \frac{9}{8}: the 8s become 1 and 1, the 9 and 3 become 3 and 1.
4. \displaystyle \frac{7}{2} \times \frac{8}{7}: the 7s become 1 and 1, the 8 and 2 become 4 and 1.
\displaystyle 3\frac{9}{10}. \displaystyle \frac{9}{5} \times \frac{13}{6}: the 9 and 6 become 3 and 2, so \displaystyle \frac{3}{5} \times \frac{13}{2} = \frac{39}{10}, and 39 = 30 + 9.
12. \displaystyle \frac{16}{3} \times \frac{9}{4}: the 16 and 4 become 4 and 1, the 9 and 3 become 3 and 1, so 4 \times 3 = 12.
\displaystyle 3\frac{1}{8}. The parts are 2 \times 1 = 2, \displaystyle 2 \times \frac{1}{4} = \frac{1}{2}, \displaystyle \frac{1}{2} \times 1 = \frac{1}{2} and \displaystyle \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}, which add to \displaystyle 3\frac{1}{8}. Check: \displaystyle \frac{5}{2} \times \frac{5}{4} = \frac{25}{8} = 3\frac{1}{8}.
\displaystyle 5\frac{1}{5} square feet. \displaystyle \frac{13}{4} \times \frac{8}{5}: the 8 and 4 become 2 and 1, so \displaystyle \frac{26}{5}, and 26 = 25 + 1.
Between 8 and 15; the product is \displaystyle 10\frac{1}{2}. \displaystyle \frac{21}{5} \times \frac{5}{2}: the 5s become 1 and 1, leaving \displaystyle \frac{21}{2}.
6 cups. \displaystyle \frac{9}{4} \times \frac{8}{3}: the 9 and 3 become 3 and 1, the 8 and 4 become 2 and 1, so 3 \times 2 = 6.
18 pounds. \displaystyle \frac{15}{2} \times \frac{12}{5}: the 15 and 5 become 3 and 1, the 12 and 2 become 6 and 1, so 3 \times 6 = 18.
\displaystyle 4\frac{1}{12} cups. Range: between 1 \times 3 = 3 and 2 \times 4 = 8, near the low end. \displaystyle \frac{7}{6} \times \frac{7}{2} = \frac{49}{12}, and 49 = 48 + 1.
7 pounds. The area is \displaystyle \frac{21}{4} \times \frac{8}{3}: the 21 and 3 become 7 and 1, the 8 and 4 become 2 and 1, so 14 square yards. Then \displaystyle \frac{1}{2} \times 14 = 7 pounds.
It cannot be right; the product is \displaystyle 8\frac{3}{4}. The range is from 2 \times 3 = 6 to 3 \times 4 = 12, and \displaystyle 5\frac{7}{8} is below 6. Exactly: \displaystyle \frac{14}{5} \times \frac{25}{8}, where the 14 and 8 become 7 and 4 and the 25 and 5 become 5 and 1, so \displaystyle \frac{35}{4} = 8\frac{3}{4}.
\displaystyle 2\frac{1}{4}. The parts are 1 \times 1 = 1, \displaystyle 1 \times \frac{1}{2} = \frac{1}{2}, \displaystyle \frac{1}{2} \times 1 = \frac{1}{2} and \displaystyle \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}, total \displaystyle 2\frac{1}{4}. Check: \displaystyle \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}. The two parts \displaystyle 1 \times \frac{1}{2} and \displaystyle \frac{1}{2} \times 1 add to \displaystyle \frac{1}{2} + \frac{1}{2} = 1. They belong because they are strips 1 by \displaystyle \frac{1}{2} inside the square, so they are part of its area; the product is the area of the whole square, every part of it.
A mixed number has at least 1 whole, so it is more than 1. Take two mixed numbers A and B. Since B is 1 plus some more, the distributive property gives A \times B = A \times 1 + A \times (\text{the rest of } B), which is A plus a positive amount, so it is more than A. In the same way it is more than B. In area terms, a rectangle A long and more than 1 wide has more area than A square units.
Both give 2. Four parts: 1 \times 1 = 1, \displaystyle 1 \times \frac{1}{2} = \frac{1}{2}, \displaystyle \frac{1}{3} \times 1 = \frac{1}{3}, \displaystyle \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}, and \displaystyle 1 + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = 2. Improper: \displaystyle \frac{4}{3} \times \frac{3}{2} = \frac{2}{1} \times \frac{1}{1} = 2. They must agree because \displaystyle 1\frac{1}{3} and \displaystyle \frac{4}{3} are the same number, and so are \displaystyle 1\frac{1}{2} and \displaystyle \frac{3}{2}. Both roads find the area of the same rectangle; one cuts it into four parts, the other counts it in sixths.