- Think about what happens to a piece when it is split into several equal smaller pieces. - Check whether putting all the equal pieces back together would recreate the original length.

1. The \(\frac{1}{3}\) yard is shared equally among \(4\) pieces, so the calculation is \(\frac{1}{3}\div4\). 2. Splitting one third into \(4\) equal parts gives \(\frac{1}{12}\).

Each piece is \(\frac{1}{12}\) yard long.

- The quotient should be smaller than the original unit fraction. - Test a choice by asking whether three equal copies rebuild the starting fraction.

1. Dividing \(\frac{1}{7}\) into \(3\) equal shares makes each share \(\frac{1}{21}\). 2. Check: \(\frac{1}{21}\times3=\frac{3}{21}=\frac{1}{7}\).

The value is \(\frac{1}{21}\). The check is \(\frac{1}{21}\times3=\frac{1}{7}\), so three equal shares rebuild the shaded \(\frac{1}{7}\).

- Treat the opening time as one quantity being shared equally. - Check that five equal clip lengths add back to the full opening time.

1. The total opening time is \(\frac{1}{3}\) minute and it is split into \(5\) equal parts. 2. \(\frac{1}{3}\div5=\frac{1}{15}\).

Each clip lasts \(\frac{1}{15}\) minute.

- Read the starting fraction from panel a). - In panel b), count how many smaller shaded parts make the same amount. - One of those equal shaded parts is one person's share.

1. Panel a) has \(1\) of \(7\) equal parts shaded, so the amount being shared is \(\frac{1}{7}\) of the whole. 2. In panel b), that same one-seventh region contains \(4\) equal cells, one for each share. 3. The whole grid has \(28\) equal cells, so each person receives \(\frac{1}{28}\) of the whole. Therefore \(\frac{1}{7}\div4=\frac{1}{28}\).

a) \(\frac{1}{7}\) b) \(\frac{1}{28}\) c) \(\frac{1}{7}\div4=\frac{1}{28}\). The shaded cells together equal the original seventh, and one shaded cell is one person's share.

- Count how many equal columns make the whole. - Look at how the shaded column is split. - Compare one shaded cell with all the equal cells in the rectangle.

1. The whole has \(11\) equal columns, so the shaded column is \(\frac{1}{11}\) of the whole. 2. Each column is split into \(2\) equal cells, making \(22\) equal cells in the whole. 3. The two shaded cells are the two equal shares. Each cell is \(\frac{1}{22}\) of the whole. 4. Therefore \(\frac{1}{11}\div2=\frac{1}{22}\).

a) \(\frac{1}{11}\) b) \(\frac{1}{22}\) c) \(\frac{1}{11}\div2=\frac{1}{22}\). The shaded column is split into two equal cells, one for each share.

- Read the starting amount from panel a). - In panel b), identify how many fourths make the same amount as the half. - Split those fourths equally between the two cups.

1. Panel a) shows \(\frac{1}{2}\) cup. 2. In panel b), the whole is divided into \(4\) equal parts, and the original half covers \(2\) fourths. Sharing those two fourths equally between two cups gives one fourth per cup. 3. Thus \(\frac{1}{2}\div2=\frac{1}{4}\).

a) \(\frac{1}{2}\) cup b) \(\frac{1}{4}\) cup c) \(\frac{1}{2}\div2=\frac{1}{4}\). Panel b) shows the original half as two fourths, one for each equal share.

- Read the starting fraction from panel a). - In panel b), count how many finer parts cover exactly the same amount. - One of those finer parts is one of the four equal shares.

1. Panel a) shows \(1\) of \(5\) equal parts, so the starting amount is \(\frac{1}{5}\). 2. Panel b) partitions the whole into \(20\) equal parts. The original fifth occupies \(4\) of those parts, so each of its \(4\) equal shares is one twentieth. 3. Thus \(\frac{1}{5}\div4=\frac{1}{20}\).

a) \(\frac{1}{5}\) b) \(\frac{1}{20}\) c) \(\frac{1}{5}\div4=\frac{1}{20}\). Panel b) shows the original fifth as four twentieths, so one of the four equal shares is one twentieth.

- Read the starting fraction from panel a). - In panel b), compare the shaded region with all the equal cells in the whole. - Think about how the shaded cells can be shared equally among the \(5\) plant types.

1. Panel a) has \(1\) of \(6\) equal parts shaded, so \(\frac{1}{6}\) of the garden is being divided. 2. In panel b), the same one-sixth region contains \(5\) equal cells. Sharing those cells among \(5\) plant types gives one cell per share. 3. The whole grid has \(30\) equal cells, so one share is \(\frac{1}{30}\) of the whole. Therefore \(\frac{1}{6}\div5=\frac{1}{30}\).

a) \(\frac{1}{6}\) b) \(\frac{1}{30}\) c) \(\frac{1}{6}\div5=\frac{1}{30}\). The shaded sixth is split into \(5\) equal cells, one for each share.

- Both problems start with the same amount. - Think about whether dividing that amount among more groups makes each share larger or smaller.

1. \(\frac{1}{4}\div2=\frac{1}{8}\). 2. \(\frac{1}{4}\div4=\frac{1}{16}\). 3. Splitting the same unit fraction into fewer equal shares makes each share larger, so \(\frac{1}{8}>\frac{1}{16}\).

\(\frac{1}{4}\div2=\frac{1}{8}\) is greater than \(\frac{1}{4}\div4=\frac{1}{16}\).

- Imagine the whole first divided into ninths. - Then subdivide one of those ninths into six equal parts.

1. One ninth is one of \(9\) equal parts of a whole. 2. Splitting that one ninth into \(6\) equal shares makes \(9\times6=54\) equal shares of the whole. 3. Each share is \(\frac{1}{54}\).

\(\frac{1}{9}\div6=\frac{1}{54}\). One ninth is one of \(9\) equal parts of the whole; splitting that ninth into \(6\) equal shares makes \(54\) equal parts in the whole, so each share is \(\frac{1}{54}\).

- The whole-number divisor tells how many equal shares the fractional area is split into. - Your answer should be smaller than the original unit fraction. - Multiply one share by the number of varieties to check that you recover the original area.

1. Divide the unit fraction by the number of equal shares: \(\frac{1}{8}\div6=\frac{1}{48}\). 2. Each variety receives \(\frac{1}{48}\) acre. 3. Check: \(6\times\frac{1}{48}=\frac{6}{48}=\frac{1}{8}\).

Each variety receives \(\frac{1}{48}\) acre. Check: \(6\times\frac{1}{48}=\frac{1}{8}\) acre.

- Six equal container amounts must combine to the original \(\frac{1}{5}\) cup. - Use the relationship between one share and the total to judge the quotient's size. - A valid multiplication check should rebuild the starting amount exactly.

1. Divide the unit fraction by \(6\): \(\frac{1}{5}\div6=\frac{1}{30}\) cup. 2. The reverse multiplication check is \(6\times\frac{1}{30}=\frac{6}{30}=\frac{1}{5}\). 3. The check confirms that six equal shares rebuild the original amount.

Each container gets \(\frac{1}{30}\) cup, and \(6\times\frac{1}{30}=\frac{1}{5}\).

- Simplify the shaded fraction first. - Think about splitting that shaded amount into \(7\) equal shares. - Ask how many equal small parts the whole liter would have after the split.

1. The model shades \(2\) of \(8\) equal parts, so the sample is \(\frac{2}{8}=\frac{1}{4}\,\text{liter}\). 2. Split that one-fourth-liter amount into \(7\) equal shares. This makes \(4\times7=28\) equal parts in one liter. 3. Each container receives \(\frac{1}{28}\,\text{liter}\), so \(\frac{1}{4}\div7=\frac{1}{28}\).

The shaded sample is \(\frac{1}{4}\,\text{liter}\). Each container receives \(\frac{1}{28}\,\text{liter}\).

- Compare the entire colored region with all equal parts in the bar before focusing on one share. - Count how many equal pieces the colored region has been split into. - One final share is one of the smallest equal parts shown in the whole bar.

1. The colored region covers \(4\) of the \(12\) equal parts, so it represents \(\frac{4}{12}=\frac{1}{3}\) of the whole. 2. That \(\frac{1}{3}\) region is split into \(4\) equal shares. 3. Each share is one of the \(12\) equal parts of the whole, so \(\frac{1}{3}\div4=\frac{1}{12}\).

The original unit fraction is \(\frac{1}{3}\). The model represents \(\frac{1}{3}\div4=\frac{1}{12}\), so one share is \(\frac{1}{12}\) of the whole.

- First identify the fraction shown in panel a). - Compare the number of equal parts in the whole before and after the subdivision. - In panel b), focus on the size of one of the three equal pieces that replace the original shaded part.

1. Panel a) shows \(1\) of \(5\) equal parts, so the starting amount is \(\frac{1}{5}\). 2. In panel b), the whole is divided into \(15\) equal parts, and the original one-fifth region occupies \(3\) of those parts. 3. Splitting those \(3\) small parts into \(3\) equal shares gives one \(\frac{1}{15}\)-sized part per share. 4. Therefore \(\frac{1}{5}\div3=\frac{1}{15}\).

\(\frac{1}{5}\div3=\frac{1}{15}\). Panel b) shows that subdividing the original fifth into \(3\) equal shares makes \(15\) equal parts in the whole, so one share is \(\frac{1}{15}\).

- Rewrite the starting unit fraction using twentieths. - Ask how many copies of the quotient make the original fraction.

1. The quotient \(\frac{1}{20}\) must be repeated \(n\) times to rebuild \(\frac{1}{4}\). 2. Since \(\frac{1}{4}=\frac{5}{20}\), five copies of \(\frac{1}{20}\) make \(\frac{1}{4}\). 3. Therefore \(n=5\).

\(n=5\).

- Make the unit fraction the total amount being shared. - Make the whole number the number of equal shares. - Interpret the quotient as the size of one share.

1. One valid story is: \(\frac{1}{5}\) pound of clay is shared equally among \(6\) students. 2. \(\frac{1}{5}\div6=\frac{1}{30}\). 3. In this story, each student receives \(\frac{1}{30}\) pound of clay.

Answers will vary. A valid story must use \(\frac{1}{5}\) as the total amount, \(6\) as the number of equal shares, and \(\frac{1}{30}\) as the size of each share.

- All three expressions start with the same amount. - Think about how the size of one share changes when the number of shares increases.

1. The same amount \(\frac{1}{3}\) is split into different numbers of equal shares, so fewer shares give a larger share. 2. \(\frac{1}{3}\div2=\frac{1}{6}\), \(\frac{1}{3}\div5=\frac{1}{15}\), and \(\frac{1}{3}\div8=\frac{1}{24}\). 3. Therefore \(\frac{1}{6}>\frac{1}{15}>\frac{1}{24}\).

\(\frac{1}{3}\div2>\frac{1}{3}\div5>\frac{1}{3}\div8\), with quotients \(\frac{1}{6},\frac{1}{15},\frac{1}{24}\).

- Think about subdividing one equal part into three smaller equal parts. - Compare each original denominator with the denominator of its quotient.

1. \(\frac{1}{2}\div3=\frac{1}{6}\). 2. \(\frac{1}{5}\div3=\frac{1}{15}\). 3. \(\frac{1}{8}\div3=\frac{1}{24}\). 4. Each original unit fraction is split into \(3\) equal parts, so the whole is partitioned into three times as many equal parts.

The quotients are \(\frac{1}{6},\frac{1}{15},\frac{1}{24}\). Each denominator is \(3\) times the original denominator because each original equal part is split into \(3\) smaller equal parts.

- Read the multiplication as a statement about equal copies. - Reverse the relationship to describe one share of the total.

1. The multiplication says seven equal copies of \(\frac{1}{56}\) make \(\frac{1}{8}\). 2. Therefore splitting \(\frac{1}{8}\) into \(7\) equal shares gives \(\frac{1}{56}\). 3. The division equation is \(\frac{1}{8}\div7=\frac{1}{56}\).

\(\frac{1}{8}\div7=\frac{1}{56}\).

- Think about how many equal pieces the whole has after both stages of partitioning. - Compare the total number of final equal pieces in the two expressions.

1. Splitting \(\frac{1}{5}\) into \(7\) equal shares gives \(\frac{1}{35}\). 2. Splitting \(\frac{1}{7}\) into \(5\) equal shares also gives \(\frac{1}{35}\). 3. In both cases, the whole ends up partitioned into \(5\times7=35\) equal parts.

The quotients are equal; both are \(\frac{1}{35}\). In either order, the two stages partition the whole into \(5\times7=35\) equal parts, so one final share is \(\frac{1}{35}\).

- Think about what happens to the size of a piece when a unit fraction is shared among several cups. - The quotient should be smaller than the original \(\frac{1}{6}\) liter. - Look for a model whose single shaded part has the correct size.

1. The division is \(\frac{1}{6}\div4\). 2. Splitting one sixth into \(4\) equal pieces creates \(24\) equal pieces in the whole liter, so one share is \(\frac{1}{24}\) liter. 3. Model a) shows one of \(24\) equal parts, so it represents one cup. Model b) shows \(\frac{4}{6}\), not the quotient.

Model a) is correct; each cup contains \(\frac{1}{24}\) liter. One sixth split into \(4\) equal shares makes \(24\) equal parts in the whole liter, so one share matches one of the \(24\) parts in model a).

- Find the size of one equal share in each situation. - Both answers are unit fractions, so compare their denominators carefully. - Remember that a larger denominator means smaller equal parts when the numerator is \(1\).

1. In a), \(\frac{1}{4}\div2=\frac{1}{8}\). 2. In b), \(\frac{1}{6}\div3=\frac{1}{18}\). 3. Since \(\frac{1}{8}>\frac{1}{18}\), a person in a) receives the larger share.

a) gives the larger share: \(\frac{1}{8}\) compared with \(\frac{1}{18}\).

- First find the size of one equal share of the original fraction of an hour. - After you have the fraction of an hour, use the number of minutes in one hour to interpret it.

1. Divide the unit fraction by the number of clips: \(\frac{1}{12}\div5=\frac{1}{60}\) hour. 2. One hour is \(60\) minutes, so \(\frac{1}{60}\) hour is \(1\) minute.

Each clip is \(\frac{1}{60}\) hour, or \(1\) minute, long.

- Focus on what it means to share one fixed amount between two people. - Compare the size of one new share with the shaded amount you started with. - Use the model to think about how many equal pieces the whole would have after the split.

1. The shaded fifth is the amount being split into \(2\) equal shares. 2. Each share must be smaller than the original \(\frac{1}{5}\). 3. Splitting one fifth into \(2\) equal parts gives \(\frac{1}{10}\), so \(\frac{1}{5}\div2=\frac{1}{10}\).

Tara’s claim is false. \(\frac{1}{5}\div2=\frac{1}{10}\), which is smaller than \(\frac{1}{5}\).

- Find the equal share produced in each situation. - Both answers will be unit fractions, which makes comparing them easier. - For unit fractions, think about how the denominator affects the size of one part.

1. In a), \(\frac{1}{8}\div2=\frac{1}{16}\). 2. In b), \(\frac{1}{4}\div8=\frac{1}{32}\). 3. Since \(\frac{1}{16}>\frac{1}{32}\), the share in a) is larger.

a) \(\frac{1}{16}\) b) \(\frac{1}{32}\) The share in a) is larger.

- Keep separate the piece's length in meters and its size relative to the original strip. - One of \(7\) equal pieces is \(\frac{1}{7}\) of the strip, but the strip itself is only \(\frac{1}{9}\) meter. - Use multiplication to test whether seven copies of your length rebuild the strip.

1. Divide the original unit fraction by \(7\): \(\frac{1}{9}\div7=\frac{1}{63}\) meter. 2. Because the strip is divided into \(7\) equal pieces, one piece is \(\frac{1}{7}\) of the original strip. 3. Thus each piece is \(\frac{1}{63}\) meter and \(\frac{1}{7}\) of the starting strip.

Each piece is \(\frac{1}{63}\) meter, which is \(\frac{1}{7}\) of the original strip.

- First divide the original unit fraction into the stated number of equal zones. - Then compare the two unit fractions by considering the size of equal parts with different denominators.

1. One zone has length \(\frac{1}{3}\div4=\frac{1}{12}\) mile. 2. Compare \(\frac{1}{12}\) and \(\frac{1}{10}\). With the same numerator \(1\), the fraction with denominator \(12\) is smaller. 3. Therefore each zone is shorter than \(\frac{1}{10}\) mile.

Each zone is \(\frac{1}{12}\) mile long, which is shorter than \(\frac{1}{10}\) mile.

- Start with one of the equal parts represented by the original unit fraction. - Think about how splitting that single part again changes the total number of equal pieces in the whole. - The quotient should be smaller than \(\frac{1}{11}\).

1. Divide the unit fraction by \(9\): \(\frac{1}{11}\div9=\frac{1}{99}\). 2. One eleventh of the mile is one of \(11\) equal parts. Splitting that part into \(9\) equal pieces makes \(11\times9=99\) equal pieces in the whole mile. 3. Therefore one zone is \(\frac{1}{99}\) mile.

One zone is \(\frac{1}{99}\) mile. The denominator is \(99\) because one of \(11\) equal parts is split into \(9\) equal pieces, making \(11\times9=99\) equal pieces in the whole mile.

- Divide the original unit fraction into the stated number of equal shares. - The result will still be a unit fraction. - Compare unit fractions by thinking about the size of equal parts with different denominators.

1. One share is \(\frac{1}{12}\div4=\frac{1}{48}\) mile. 2. Compare \(\frac{1}{48}\) and \(\frac{1}{50}\). For unit fractions, the smaller denominator gives the larger fraction. 3. Therefore \(\frac{1}{48}>\frac{1}{50}\).

One part is \(\frac{1}{48}\) mile, which is greater than \(\frac{1}{50}\) mile.

- Work backward by combining all of the equal shares. - Multiplication reverses the equal-sharing division. - Simplify the rebuilt fraction before stating the original unit fraction.

1. Four equal shares of \(\frac{1}{44}\) rebuild the original fraction. 2. Multiply: \(4\times\frac{1}{44}=\frac{4}{44}=\frac{1}{11}\). 3. Therefore the original division was \(\frac{1}{11}\div4=\frac{1}{44}\).

The original unit fraction was \(\frac{1}{11}\). Multiplication reverses the sharing: \(4\times\frac{1}{44}=\frac{1}{11}\), so \(\frac{1}{11}\div4=\frac{1}{44}\).

- First find the size of one equal share of the original unit fraction. - Then combine the stated number of equal shares. - Simplify the resulting fraction of the whole container.

1. One cup share is \(\frac{1}{3}\div4=\frac{1}{12}\) of the whole container. 2. Three such shares contain \(3\times\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\). 3. Therefore the combined amount is \(\frac{1}{4}\) of the whole container.

The \(3\) shares together make \(\frac{1}{4}\) of the whole container.

- Think about how the original denominator combines with the number of equal shares. - Compare the products formed by each denominator and whole-number divisor. - Equal final denominators produce equal unit-fraction quotients.

1. a) \(\frac{1}{6}\div4=\frac{1}{24}\). 2. b) \(\frac{1}{8}\div3=\frac{1}{24}\). 3. c) \(\frac{1}{12}\div2=\frac{1}{24}\). 4. In each case, splitting one unit-fraction part into the stated number of equal shares creates \(24\) equal parts in the whole because \(6\times4=8\times3=12\times2=24\).

a) \(\frac{1}{24}\) b) \(\frac{1}{24}\) c) \(\frac{1}{24}\) They are equal because \(6\times4=8\times3=12\times2=24\), so each division creates \(24\) equal parts in the whole.

- Think about what it means to divide one amount into two equal shares. - Compare that idea with selecting one of two equal parts of the same amount. - Check that both descriptions lead to the same unit fraction.

1. Split one fifteenth into \(2\) equal shares: \(\frac{1}{15}\div2=\frac{1}{30}\). 2. Taking one half of \(\frac{1}{15}\) gives \(\frac{1}{2}\times\frac{1}{15}=\frac{1}{30}\). 3. Both operations find one of two equal parts of the original amount.

\(\frac{1}{15}\div2=\frac{1}{30}\). This is the same as \(\frac{1}{2}\times\frac{1}{15}\) because both operations find one of two equal parts of the original \(\frac{1}{15}\).

- Interpret \(\frac{1}{9}\) as one equal part of a whole. - Subdivide that one part into the eight equal shares named by the divisor. - Check that the quotient is smaller than the amount being shared.

1. The original amount is one of \(9\) equal parts of a whole. 2. Splitting that one part into \(8\) equal shares makes \(9\times8=72\) equal shares in the whole, not \(9+8=17\). 3. Therefore \(\frac{1}{9}\div8=\frac{1}{72}\). 4. The quotient is smaller than \(\frac{1}{9}\), as an equal share must be.

Sofia's rule is incorrect. Equal sharing subdivides one ninth into \(8\) smaller equal parts, making \(72\) equal parts in the whole, so \(\frac{1}{9}\div8=\frac{1}{72}\).

- Follow the sharing process one stage at a time. - Count how many final equal packets result from the two stages together. - Compare the two-stage quotient with dividing by that total number of packets.

1. First share: \(\frac{1}{5}\div3=\frac{1}{15}\). 2. Split each share into \(4\) packets: \(\frac{1}{15}\div4=\frac{1}{60}\). 3. There are \(3\times4=12\) equal packets altogether, and \(\frac{1}{5}\div12=\frac{1}{60}\). 4. Both procedures give the same packet size.

One packet is \(\frac{1}{60}\) of the whole strip, and \(\frac{1}{5}\div12=\frac{1}{60}\).

- Find each equal share before comparing the situations. - Both answers are unit fractions, so their denominators can be compared directly. - Consider both the starting fraction and the number of people sharing it.

1. In a), \(\frac{1}{7}\div4=\frac{1}{28}\). 2. In b), \(\frac{1}{8}\div3=\frac{1}{24}\). 3. Since \(\frac{1}{24}>\frac{1}{28}\), b) gives the larger share. 4. Although \(\frac{1}{8}<\frac{1}{7}\), it is split among fewer people, which more than offsets the smaller starting amount.

b) gives the larger share: \(\frac{1}{24}\) compared with \(\frac{1}{28}\). The smaller starting fraction still gives the larger share because it is split among only \(3\) people instead of \(4\).

- Describe the denominator of one share using \(8\) and the unknown number of students. - For unit fractions, a larger denominator means a smaller value. - Find the multiple of \(8\) that lies strictly between the two benchmark denominators.

1. If the ribbon is shared among a whole-number count of students, each share has the form \(\frac{1}{8}\div\text{(student count)}\). 2. The final denominator is \(8\) times the student count. 3. For the share to lie strictly between \(\frac{1}{56}\) and \(\frac{1}{40}\), that denominator must lie strictly between \(40\) and \(56\). 4. The only multiple of \(8\) strictly between \(40\) and \(56\) is \(48\), and \(48=8\times6\). 5. Therefore \(6\) students share the ribbon, and each receives \(\frac{1}{48}\) yard.

\(6\) students; each share is \(\frac{1}{48}\) yard. No other whole-number count works because \(48\) is the only multiple of \(8\) strictly between \(40\) and \(56\).