Grade 5 · 4.NF.B.3C · 5.NF.A.1

Subtracting mixed numbers

Khanh’s garden wagon holds five and one quarter cubic feet of mulch, and he shovels out two and two thirds cubic feet. How can he take two thirds away when there is only one quarter beside the wholes?

You need

Adding and subtracting fractions with unlike denominatorsGrade 5

Where you are 4 in Mixed numbers and improper fractions

Fractions, ratio and percent3

Fractions15

Mixed numbers and improper fractions4

Mixed numbers and improper fractions: rewriting and comparing Adding and subtracting mixed numbers with like denominators Adding mixed numbers

Subtracting mixed numbers

When you reach for it

Khanh, Nanette and Ojas keep a community garden. A big pile of mulch sits by the gate, and they haul it to the beds in a garden wagon. The wagon is measured in cubic feet, and the loads almost never come out even. A load might be 6\tfrac{7}{8} cubic feet, and a bed might take 3\tfrac{1}{4} of it.

A mixed number such as 6\tfrac{7}{8} is a whole number and a fraction together: 6 wholes and \tfrac{7}{8} of another. You reach for this lesson whenever one mixed number is taken from another, or a mixed number is taken from a whole number. It has three parts, one for each day:

Part 1: wholes from wholes and fractions from fractions, when the fractions have different denominators.
Part 2: when the fraction being taken is bigger than the fraction you have, trading one whole for fraction pieces, and subtracting from a whole number.
Part 3: the improper-fraction route, counting up as a check, and stories that ask how much is left or how much more.

The procedure

Part 1: wholes from wholes, fractions from fractions

Khanh’s wagon holds 6\tfrac{7}{8} cubic feet of mulch. He spreads 3\tfrac{1}{4} cubic feet on the bean bed. How much is still in the wagon?

A mixed number means a sum. 6\tfrac{7}{8} means 6 + \tfrac{7}{8}, and 3\tfrac{1}{4} means 3 + \tfrac{1}{4}. Taking away 3\tfrac{1}{4} means taking away 3 wholes and also taking away \tfrac{1}{4}. Khanh can do those two jobs one after the other, in either order, and the wagon ends up the same. So subtract the wholes from the wholes and the fraction from the fraction.

The fractions have different denominators, so the pieces do not match yet. Rename them in one size first, exactly as in Adding and subtracting fractions with unlike denominators. Here 8 is a multiple of 4, so eighths will do, and \tfrac{1}{4} = \tfrac{2}{8}. Renaming the fraction part does not change the whole part. 3\tfrac{1}{4} and 3\tfrac{2}{8} are the same amount.

6\tfrac{7}{8} - 3\tfrac{1}{4} = 6\tfrac{7}{8} - 3\tfrac{2}{8} = 3\tfrac{5}{8}

The wholes give 6 - 3 = 3. The fractions give \tfrac{7}{8} - \tfrac{2}{8} = \tfrac{5}{8}. Then put the two parts back together: 3\tfrac{5}{8} cubic feet are left.

Figure 1Start at 6\tfrac{7}{8}. Step back 3 wholes, then step back \tfrac{2}{8}, which is \tfrac{1}{4}. You land on 3\tfrac{5}{8}.

The number line shows the same thing. Start at 6\tfrac{7}{8}. Step back 3 wholes to 3\tfrac{7}{8}. Then step back \tfrac{2}{8} to 3\tfrac{5}{8}.

Writing the answer. An answer is finished when its fraction part is less than 1 and in simplest form. If the fractions give \tfrac{3}{6}, write \tfrac{1}{2} instead.

Checking. Subtraction undoes addition. If 3\tfrac{5}{8} is left and 3\tfrac{2}{8} was spread, the two together must make the load Khanh started with: 3\tfrac{5}{8} + 3\tfrac{2}{8} = 6\tfrac{7}{8}. Adding mixed numbers has its own lesson, Adding mixed numbers.

Part 2: trading one whole for pieces

Now Nanette’s wagon holds 5\tfrac{1}{4} cubic feet, and she spreads 2\tfrac{2}{3} cubic feet around the squash. Rename the fractions first. Quarters and thirds share no factor except 1, so use 4 \times 3 = 12: \tfrac{1}{4} = \tfrac{3}{12} and \tfrac{2}{3} = \tfrac{8}{12}.

5\tfrac{1}{4} - 2\tfrac{2}{3} = 5\tfrac{3}{12} - 2\tfrac{8}{12}

The fraction part she has is 3 twelfths. The fraction part she must take is 8 twelfths. You cannot take 8 pieces from 3 pieces. But Nanette has more than 3 twelfths of mulch. She has 5 whole cubic feet too.

So she trades. She takes one of the 5 wholes and cuts it into twelfths. One whole is \tfrac{12}{12}. That leaves 4 wholes, and the fraction part grows from 3 twelfths to 12 + 3 = 15 twelfths.

5\tfrac{3}{12} = 4 + \tfrac{12}{12} + \tfrac{3}{12} = 4\tfrac{15}{12}

Figure 2Cutting one whole into 12 twelfths does not change the amount. 5\tfrac{3}{12} and 4\tfrac{15}{12} are the same mulch, named two ways.

Nothing was added or taken away. The same mulch is now counted with fewer wholes and more pieces. It is the move you already know from whole-number subtraction, where one ten is traded for ten ones. Here one whole is traded for as many pieces as the denominator says: 12 twelfths, or 8 eighths, or 5 fifths.

Now there are enough twelfths, and the subtraction goes as in Part 1.

\begin{aligned} 5\tfrac{1}{4} - 2\tfrac{2}{3} &= 5\tfrac{3}{12} - 2\tfrac{8}{12} \\ &= 4\tfrac{15}{12} - 2\tfrac{8}{12} \\ &= 2\tfrac{7}{12} \end{aligned}

The fractions give \tfrac{15}{12} - \tfrac{8}{12} = \tfrac{7}{12}, and the wholes give 4 - 2 = 2. Nanette has 2\tfrac{7}{12} cubic feet left. The form 4\tfrac{15}{12} is only a step on the way. It is never the final answer, because its fraction part is more than 1.

When to trade. Rename the fractions, then compare the two fraction parts. If the first is at least as big as the second, no trade is needed. If it is smaller, trade one whole. One trade is always enough. After the trade, the first fraction part is at least 1, and the fraction being taken is less than 1.

Subtracting from a whole number. A full wagon might hold exactly 7 cubic feet, with no fraction part at all. Then there is nothing to take the fraction from, so trade right away. Write one of the wholes as pieces of the size being taken. If fifths are being taken, 7 = 6\tfrac{5}{5}. Example 4 does this.

Part 3: another route, a check, and stories

The improper-fraction route. A mixed number can be written as one fraction whose top is bigger than its bottom. Multiply the whole part by the denominator and add the numerator (Mixed numbers and improper fractions: rewriting and comparing). So 5\tfrac{1}{4} = \tfrac{21}{4}, because 5 \times 4 + 1 = 21, and 2\tfrac{2}{3} = \tfrac{8}{3}, because 2 \times 3 + 2 = 8. Now there are only two fractions, and they subtract like any two fractions.

\frac{21}{4} - \frac{8}{3} = \frac{63}{12} - \frac{32}{12} = \frac{31}{12}

Last, turn the answer back into a mixed number: 31 \div 12 = 2 remainder 7, so \tfrac{31}{12} = 2\tfrac{7}{12}. This matches Part 2.

This route never needs a trade. Every whole is already cut into pieces, so there are always enough pieces to take. The cost is bigger numbers. With large whole parts, such as 48\tfrac{1}{3}, the numerators get large, and the Part 2 route is usually quicker.

Counting up. A difference is the distance between two numbers on the number line. So you can also find it by starting at the smaller number and counting up to the larger one. Count up in easy jumps: first to the next whole number, then by wholes, then the last bit.

Figure 3Counting up from 2\tfrac{2}{3} to 5\tfrac{1}{4} takes \tfrac{1}{3}, then 2, then \tfrac{1}{4}. Together that is 2\tfrac{7}{12}.

From 2\tfrac{2}{3}, a jump of \tfrac{1}{3} reaches 3. A jump of 2 reaches 5. A jump of \tfrac{1}{4} reaches 5\tfrac{1}{4}. The jumps add to

2 + \tfrac{1}{3} + \tfrac{1}{4} = 2 + \tfrac{4}{12} + \tfrac{3}{12} = 2\tfrac{7}{12}

Counting up is a good check because it uses different steps from the trading route. If both routes give the same answer, a slip in one of them would have to be matched by the same slip in the other.

A size check. Round each mixed number to the nearest half before you start (Estimating with fractions). 5\tfrac{1}{4} is between 5 and 5\tfrac{1}{2}, and 2\tfrac{2}{3} is close to 2\tfrac{1}{2}, so the answer should be close to 2\tfrac{1}{2}, somewhere between 5 - 2\tfrac{1}{2} and 5\tfrac{1}{2} - 2\tfrac{1}{2}. 2\tfrac{7}{12} is.

Stories. Two kinds of story in this lesson are subtraction. “How much is left?” takes one amount away from another: mulch spread from a load. “How much more?” compares two amounts: how much bigger one load is than another. In both, the answer is the larger amount minus the smaller one. Stories where you must first decide whether amounts are joined or taken away belong to Word problems that add or subtract fractions.

Worked examples

Examples 1 and 2 belong to Part 1, Examples 3 to 5 to Part 2, and Examples 6 and 7 to Part 3.

Example 1

Thirds inside sixths

Nanette loads 8\tfrac{5}{6} cubic feet of mulch into the wagon and spreads 2\tfrac{1}{3} cubic feet under the roses. How much is left?

Solution

Taking mulch out of the wagon is subtraction: 8\tfrac{5}{6} - 2\tfrac{1}{3}. Rename the fractions. 6 \div 3 = 2, a whole number, so sixths will do, and \tfrac{1}{3} = \tfrac{2}{6}.

8\tfrac{5}{6} - 2\tfrac{1}{3} = 8\tfrac{5}{6} - 2\tfrac{2}{6} = 6\tfrac{3}{6}

The fractions give \tfrac{5}{6} - \tfrac{2}{6} = \tfrac{3}{6}. Five sixths was more than two sixths, so no trade was needed. The wholes give 8 - 2 = 6.

The fraction part is not in simplest form: 3 and 6 can both be divided by 3, and \tfrac{3}{6} = \tfrac{1}{2}. So 6\tfrac{1}{2} cubic feet are left. Check by adding back: 6\tfrac{3}{6} + 2\tfrac{2}{6} = 8\tfrac{5}{6}.

Example 2

Both fractions change

Find 9\tfrac{7}{10} - 4\tfrac{1}{4}.

Solution

Neither 10 nor 4 is a multiple of the other, and they share the factor 2. List multiples of 10: 10, 20. 4 divides 20, so twentieths will do. Cut each tenth into 2 and each quarter into 5:

\tfrac{7}{10} = \tfrac{14}{20} \qquad \tfrac{1}{4} = \tfrac{5}{20}

14 twentieths is more than 5 twentieths, so no trade is needed.

9\tfrac{7}{10} - 4\tfrac{1}{4} = 9\tfrac{14}{20} - 4\tfrac{5}{20} = 5\tfrac{9}{20}

9 and 20 share no factor except 1, so 5\tfrac{9}{20} is in simplest form.

Example 3

A trade with tenths

Khanh loads 6\tfrac{1}{5} cubic feet and dumps 3\tfrac{1}{2} cubic feet along the fence. How much is still in the wagon?

Solution

Rename in tenths, since 5 and 2 share no factor except 1 and 5 \times 2 = 10: \tfrac{1}{5} = \tfrac{2}{10} and \tfrac{1}{2} = \tfrac{5}{10}. Now compare. Khanh has 2 tenths beside his wholes and must take 5 tenths, so he trades one whole for \tfrac{10}{10}:

6\tfrac{2}{10} = 5 + \tfrac{10}{10} + \tfrac{2}{10} = 5\tfrac{12}{10}

Then subtract as before.

\begin{aligned} 6\tfrac{1}{5} - 3\tfrac{1}{2} &= 6\tfrac{2}{10} - 3\tfrac{5}{10} \\ &= 5\tfrac{12}{10} - 3\tfrac{5}{10} \\ &= 2\tfrac{7}{10} \end{aligned}

2\tfrac{7}{10} cubic feet are left. As a size check, 6\tfrac{1}{5} is a little more than 6 and 3\tfrac{1}{2} is exactly 3\tfrac{1}{2}, so the answer should be a little more than 2\tfrac{1}{2}, and it is.

Example 4

Taking from a whole number

Ojas orders 7 cubic feet of mulch. His first wagon load takes 2\tfrac{3}{5} cubic feet. How much of the order is still by the gate?

Solution

The story is 7 - 2\tfrac{3}{5}. The number 7 has no fraction part, so there are no fifths to take from. Trade one whole for \tfrac{5}{5}, since fifths are the pieces being taken:

7 = 6 + \tfrac{5}{5} = 6\tfrac{5}{5}

Now the fractions give \tfrac{5}{5} - \tfrac{3}{5} = \tfrac{2}{5} and the wholes give 6 - 2 = 4.

7 - 2\tfrac{3}{5} = 6\tfrac{5}{5} - 2\tfrac{3}{5} = 4\tfrac{2}{5}

4\tfrac{2}{5} cubic feet are still by the gate. Check by counting up from 2\tfrac{3}{5}: a jump of \tfrac{2}{5} reaches 3, and a jump of 4 reaches 7.

Example 5

A trade, then simplifying

Find 4\tfrac{1}{6} - 1\tfrac{2}{3}.

Solution

Sixths will do, since 6 \div 3 = 2, so \tfrac{2}{3} = \tfrac{4}{6}. One sixth is less than four sixths, so trade one of the 4 wholes for \tfrac{6}{6}. The fraction part becomes 6 + 1 = 7 sixths.

\begin{aligned} 4\tfrac{1}{6} - 1\tfrac{2}{3} &= 4\tfrac{1}{6} - 1\tfrac{4}{6} \\ &= 3\tfrac{7}{6} - 1\tfrac{4}{6} \\ &= 2\tfrac{3}{6} \end{aligned}

The fraction part simplifies, since 3 and 6 can both be divided by 3. The answer is 2\tfrac{1}{2}.

Example 6

How much more, by improper fractions

Ojas’s wagon carries 5\tfrac{3}{8} cubic feet. Khanh’s carries 2\tfrac{5}{6} cubic feet. How much more does Ojas’s wagon carry? Use the improper-fraction route, then check by counting up.

Solution

“How much more” compares two loads, so subtract the smaller from the larger: 5\tfrac{3}{8} - 2\tfrac{5}{6}. Write each as one fraction. 5 \times 8 + 3 = 43 and 2 \times 6 + 5 = 17, so the problem is \tfrac{43}{8} - \tfrac{17}{6}.

Multiples of 8 go 8, 16, 24, and 6 divides 24. Cut each eighth into 3 and each sixth into 4:

\frac{43}{8} - \frac{17}{6} = \frac{129}{24} - \frac{68}{24} = \frac{61}{24}

61 \div 24 = 2 remainder 13, so \tfrac{61}{24} = 2\tfrac{13}{24}. Ojas’s wagon carries 2\tfrac{13}{24} cubic feet more.

Check by counting up from 2\tfrac{5}{6}. A jump of \tfrac{1}{6}, which is \tfrac{4}{24}, reaches 3. A jump of 2 reaches 5. A jump of \tfrac{3}{8}, which is \tfrac{9}{24}, reaches 5\tfrac{3}{8}. The jumps add to 2 + \tfrac{4}{24} + \tfrac{9}{24} = 2\tfrac{13}{24}, the same answer.

Example 7

Two beds from one order

Nanette starts with 9 cubic feet of mulch. She spreads 2\tfrac{3}{4} cubic feet on the herb bed and then 3\tfrac{5}{6} cubic feet on the pepper bed. How much is left?

Solution

Two amounts are taken away, one after the other, so subtract twice. First the herb bed. Trade one of the 9 wholes for \tfrac{4}{4}:

9 - 2\tfrac{3}{4} = 8\tfrac{4}{4} - 2\tfrac{3}{4} = 6\tfrac{1}{4}

Then the pepper bed. Rename in twelfths: \tfrac{1}{4} = \tfrac{3}{12} and \tfrac{5}{6} = \tfrac{10}{12}. Three twelfths is less than ten twelfths, so trade one of the 6 wholes for \tfrac{12}{12}:

\begin{aligned} 6\tfrac{1}{4} - 3\tfrac{5}{6} &= 6\tfrac{3}{12} - 3\tfrac{10}{12} \\ &= 5\tfrac{15}{12} - 3\tfrac{10}{12} \\ &= 2\tfrac{5}{12} \end{aligned}

2\tfrac{5}{12} cubic feet are left. As a size check, the two beds took about 3 and about 4 cubic feet, so a bit more than 2 should be left, and 2\tfrac{5}{12} is a bit more than 2.

Boundaries

Larger minus smaller. Every difference here takes a smaller mixed number from a larger one. If the whole parts are equal and the first fraction part is the smaller one, the first number is the smaller number, and there is no whole to trade. The answer would be below zero, which needs negative numbers (Operations with rational numbers).
One trade, from proper fraction parts. The rule “one trade is always enough” assumes each fraction part is less than 1, as it is in a mixed number in simplest form. A step such as 4\tfrac{15}{12} is allowed along the way, but it is not a finished answer.
The same unit. Both amounts must measure the same thing in the same unit. Cubic feet of mulch can be taken from cubic feet of mulch. Pounds cannot be taken from cubic feet.
No zero denominators. A denominator says how many equal parts a whole is cut into, so it is never 0.

Where this goes next

Stories with several steps, where you read the situation to decide whether to add or subtract, come next in Word problems that add or subtract fractions. Later, the same trade of one whole for pieces shows up when you subtract measurements such as 5 feet 2 inches minus 3 feet 7 inches, where one foot is traded for 12 inches. And when negative numbers arrive, a smaller mixed number minus a larger one gets an answer too (Operations with rational numbers).

Exercises

Part 1: wholes from wholes, fractions from fractions

Subtract. Write each answer in simplest form.

7\tfrac{5}{6} - 4\tfrac{1}{2}
9\tfrac{3}{4} - 5\tfrac{1}{8}
8\tfrac{2}{3} - 3\tfrac{1}{5}
6\tfrac{9}{10} - 2\tfrac{2}{5}
5\tfrac{7}{12} - 1\tfrac{3}{8}
4\tfrac{5}{6} - 4\tfrac{1}{4}
Khanh’s wagon holds 10\tfrac{3}{4} cubic feet of mulch. He spreads 6\tfrac{1}{6} cubic feet on the corn bed. How much is left in the wagon?
Draw a number line and step back to find 5\tfrac{2}{3} - 2\tfrac{1}{6}.

Part 2: trading one whole

Subtract. Write each answer in simplest form.

6\tfrac{1}{4} - 2\tfrac{1}{2}
8\tfrac{1}{3} - 3\tfrac{5}{6}
7\tfrac{2}{5} - 4\tfrac{2}{3}
9\tfrac{1}{6} - 5\tfrac{3}{8}
5\tfrac{3}{10} - 1\tfrac{3}{4}
6 - 3\tfrac{4}{9}
11 - 7\tfrac{5}{12}
Nanette’s wagon holds 4\tfrac{1}{8} cubic feet of mulch. She unloads 1\tfrac{3}{4} cubic feet at the tomato bed. How much is still in the wagon?
Fill in the blanks: 5\tfrac{1}{3} = 4\tfrac{?}{3} and 5\tfrac{1}{3} = 4\tfrac{?}{6}.

Part 3: improper fractions, counting up, stories

Write each answer as a mixed number in simplest form.

Use the improper-fraction route to find 4\tfrac{2}{5} - 1\tfrac{3}{4}.
Use the improper-fraction route to find 3\tfrac{1}{6} - 1\tfrac{4}{9}.
Count up to find 8\tfrac{1}{6} - 4\tfrac{2}{3}.
Ojas’s bag of cedar mulch weighs 9\tfrac{1}{4} pounds. Khanh’s bag weighs 6\tfrac{7}{10} pounds. How much heavier is Ojas’s bag?
The wagon holds 8\tfrac{1}{2} cubic feet. Nanette spreads 2\tfrac{2}{3} cubic feet on one bed and then 3\tfrac{3}{4} cubic feet on another. How much is left?

Reasoning

Explain why trading one whole is always enough. Why is a second trade never needed when both fraction parts are less than 1?
Using the meaning of a mixed number as a whole number plus a fraction, explain why 7\tfrac{3}{4} - 2\tfrac{1}{4} can be found by working out 7 - 2 and \tfrac{3}{4} - \tfrac{1}{4} separately.
Explain why the improper-fraction route never needs a trade, and say when the trading route is the quicker one.

Answers

3\tfrac{1}{3}. \tfrac{1}{2} = \tfrac{3}{6}, so 7\tfrac{5}{6} - 4\tfrac{3}{6} = 3\tfrac{2}{6}, and \tfrac{2}{6} = \tfrac{1}{3}.
4\tfrac{5}{8}. \tfrac{3}{4} = \tfrac{6}{8}, so 9\tfrac{6}{8} - 5\tfrac{1}{8} = 4\tfrac{5}{8}.
5\tfrac{7}{15}. In fifteenths, 8\tfrac{10}{15} - 3\tfrac{3}{15} = 5\tfrac{7}{15}.
4\tfrac{1}{2}. \tfrac{2}{5} = \tfrac{4}{10}, so 6\tfrac{9}{10} - 2\tfrac{4}{10} = 4\tfrac{5}{10}, and \tfrac{5}{10} = \tfrac{1}{2}.
4\tfrac{5}{24}. In twenty-fourths, 5\tfrac{14}{24} - 1\tfrac{9}{24} = 4\tfrac{5}{24}.
\tfrac{7}{12}. The wholes give 4 - 4 = 0. In twelfths, \tfrac{10}{12} - \tfrac{3}{12} = \tfrac{7}{12}.
4\tfrac{7}{12} cubic feet. In twelfths, 10\tfrac{9}{12} - 6\tfrac{2}{12} = 4\tfrac{7}{12}.
3\tfrac{1}{2}. Mark the line in sixths, so 5\tfrac{2}{3} is 5\tfrac{4}{6}. Step back 2 to 3\tfrac{4}{6}, then \tfrac{1}{6} to 3\tfrac{3}{6}, which is 3\tfrac{1}{2}.
3\tfrac{3}{4}. 6\tfrac{1}{4} - 2\tfrac{2}{4} needs a trade: 5\tfrac{5}{4} - 2\tfrac{2}{4} = 3\tfrac{3}{4}.
4\tfrac{1}{2}. 8\tfrac{2}{6} - 3\tfrac{5}{6} needs a trade: 7\tfrac{8}{6} - 3\tfrac{5}{6} = 4\tfrac{3}{6}, and \tfrac{3}{6} = \tfrac{1}{2}.
2\tfrac{11}{15}. 7\tfrac{6}{15} - 4\tfrac{10}{15} needs a trade: 6\tfrac{21}{15} - 4\tfrac{10}{15} = 2\tfrac{11}{15}.
3\tfrac{19}{24}. 9\tfrac{4}{24} - 5\tfrac{9}{24} needs a trade: 8\tfrac{28}{24} - 5\tfrac{9}{24} = 3\tfrac{19}{24}.
3\tfrac{11}{20}. 5\tfrac{6}{20} - 1\tfrac{15}{20} needs a trade: 4\tfrac{26}{20} - 1\tfrac{15}{20} = 3\tfrac{11}{20}.
2\tfrac{5}{9}. 6 = 5\tfrac{9}{9}, and 5\tfrac{9}{9} - 3\tfrac{4}{9} = 2\tfrac{5}{9}.
3\tfrac{7}{12}. 11 = 10\tfrac{12}{12}, and 10\tfrac{12}{12} - 7\tfrac{5}{12} = 3\tfrac{7}{12}.
2\tfrac{3}{8} cubic feet. 4\tfrac{1}{8} - 1\tfrac{6}{8} needs a trade: 3\tfrac{9}{8} - 1\tfrac{6}{8} = 2\tfrac{3}{8}.
4\tfrac{4}{3} and 4\tfrac{8}{6}. One whole is \tfrac{3}{3}, and 3 + 1 = 4 thirds. In sixths, one whole is \tfrac{6}{6} and \tfrac{1}{3} = \tfrac{2}{6}, so 6 + 2 = 8 sixths.
2\tfrac{13}{20}. \tfrac{22}{5} - \tfrac{7}{4} = \tfrac{88}{20} - \tfrac{35}{20} = \tfrac{53}{20}, and 53 \div 20 = 2 remainder 13.
1\tfrac{13}{18}. \tfrac{19}{6} - \tfrac{13}{9} = \tfrac{57}{18} - \tfrac{26}{18} = \tfrac{31}{18}, and 31 \div 18 = 1 remainder 13.
3\tfrac{1}{2}. From 4\tfrac{2}{3}, jump \tfrac{1}{3} to 5, jump 3 to 8, jump \tfrac{1}{6} to 8\tfrac{1}{6}. The jumps add to 3 + \tfrac{2}{6} + \tfrac{1}{6} = 3\tfrac{3}{6}, which is 3\tfrac{1}{2}.
2\tfrac{11}{20} pounds. “How much heavier” compares, so subtract. In twentieths, 9\tfrac{5}{20} - 6\tfrac{14}{20} needs a trade: 8\tfrac{25}{20} - 6\tfrac{14}{20} = 2\tfrac{11}{20}.
2\tfrac{1}{12} cubic feet. First 8\tfrac{3}{6} - 2\tfrac{4}{6} = 7\tfrac{9}{6} - 2\tfrac{4}{6} = 5\tfrac{5}{6}. Then 5\tfrac{10}{12} - 3\tfrac{9}{12} = 2\tfrac{1}{12}.
A trade is needed only when the first fraction part is smaller than the second. The trade adds one whole, which is a full 1, to the first fraction part, so afterward that part is at least 1. The fraction being taken is less than 1. So after one trade there are always enough pieces, and a second trade is never needed.
5\tfrac{1}{2}. 7\tfrac{3}{4} means 7 + \tfrac{3}{4}, and 2\tfrac{1}{4} means 2 + \tfrac{1}{4}. Taking away 2\tfrac{1}{4} means taking away 2 and taking away \tfrac{1}{4}, and the order of the two take-aways does not change what is left. Taking 2 from the wholes leaves 5, and taking \tfrac{1}{4} from \tfrac{3}{4} leaves \tfrac{2}{4}, so the answer is 5\tfrac{2}{4} = 5\tfrac{1}{2}.
Writing a mixed number as one fraction cuts every whole into pieces, so all the amount is already counted in pieces. The first number is the larger, so its fraction has more pieces than the second once the denominators match, and no trade is needed. The trading route is quicker when the whole parts are large, because the improper fractions then have large numerators.