- Count the curved arrows. - Use the fraction equivalent of the jump label. - The whole-number factor tells how many equal jumps are used.

1. One jump is \(\frac{2}{5}\). 2. The picture shows \(4\) equal jumps. 3. Four copies of \(\frac{2}{5}\) are represented by \(\frac{2}{5}\times4\).

\(\frac{2}{5}\times4\)

- Count the number of equal fractional addends. - The whole-number factor tells how many copies of the fraction are used. - Simplify the fraction after adding.

1. The sum has three copies of \(\frac{1}{6}\), so it matches \(\frac{1}{6}\times3\). 2. \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\).

A) \(\frac{1}{6}\times3=\frac{1}{2}\).

- Read the shaded fraction from the bar. - Ask which expression means that fraction of the full length. - You do not need to calculate the length.

1. The bar has \(4\) equal parts and \(3\) are shaded, so the shaded part is \(\frac{3}{4}\) of the whole. 2. Therefore the shaded length is represented by \(\frac{3}{4}\times12\).

A) \(\frac{3}{4}\times12\).

- The denominator tells how many equal parts to split the whole-number amount into. - After finding one equal part, use the numerator to decide how many of those parts to take. - Keep the gallon unit attached to the quantity being partitioned.

1. Divide the \(6\) gallons into \(4\) equal parts: \(6\div4=\frac{3}{2}\) gallons in each part. 2. Take \(3\) of those equal parts: \(3\times\frac{3}{2}=\frac{9}{2}\) gallons. 3. \(\frac{9}{2}=4\frac{1}{2}\), so \(4\frac{1}{2}\) gallons are used.

\(4\frac{1}{2}\) gallons are used.

- Let the denominator tell you how many equal groups to make. - Then use the numerator to decide how many of those groups to take.

1. Divide \(12\) tiles into \(4\) equal groups: each group has \(3\) tiles. 2. Take \(3\) groups: \(3\times3=9\) tiles.

\(9\) tiles are selected.

- Use the denominator to decide how many equal groups the set needs. - Use the numerator to decide how many groups count toward the product.

1. Partition \(15\) counters into \(5\) equal groups: \(15\div5=3\) counters per group. 2. Select \(2\) groups: \(2\times3=6\) counters.

\(6\) counters are selected.

- Let the denominator tell you how many equal groups to make. - Then use the numerator to decide how many groups are included.

1. The denominator \(7\) partitions \(21\) cards into \(7\) equal groups of \(3\). 2. The numerator \(4\) selects \(4\) groups. 3. \(4\times3=12\) cards are selected.

\(12\) cards are selected.

- Interpret the denominator as a number of equal groups. - Interpret the numerator as the number of groups to use.

1. The denominator \(8\) partitions \(24\) into \(8\) equal parts: \(24\div8=3\). 2. The numerator \(5\) selects \(5\) of those parts: \(5\times3=15\). 3. Therefore \(\frac{5}{8}\times24=15\).

\(15\)

- Use the denominator to decide the number of equal groups. - Use the numerator to decide how many groups are selected.

1. Divide \(36\) objects into \(12\) equal groups: each group has \(3\) objects. 2. Take \(7\) groups: \(7\times3=21\).

\(21\) objects.

- Interpret the denominator as the number of equal parts in the whole-number amount. - Interpret the numerator as the number of those equal parts being taken. - Compare those roles with the order of multiplication and division in the proposed expression.

1. Taking \(\frac{5}{6}\) of \(18\) means partitioning \(18\) into \(6\) equal parts and taking \(5\) parts. 2. Multiplying by \(5\) and dividing by \(6\) gives the same amount: \((5\times18)\div6=15\). 3. Therefore \(\frac{5}{6}\times18=15\).

The expression works because multiplying by \(5\) counts five copies and dividing by \(6\) makes each copy one sixth of the original amount. Thus, \(\frac{5}{6}\times18=15\).

- Use the denominator to determine the number of equal groups. - Use the numerator to determine how many groups to select.

1. Because \(28=7\times4\), one seventh of the \(28\) counters is \(4\) counters. 2. Three sevenths is three groups of \(4\), so \(3\times4=12\) counters are selected. 3. Therefore \(\frac{3}{7}\times28=12\).

\(12\) counters.

- Use the denominator to decide how many equal groups the whole collection should form. - After finding one equal group, think about what the numerator tells you to select.

1. Partition the \(30\) counters into \(6\) equal groups, so each group has \(30\div6=5\) counters. 2. The numerator \(5\) means take \(5\) of those groups. 3. The selected amount is \(5\times5=25\) counters.

\(\frac{5}{6}\times30=25\).

- Let the denominator tell you how many equal groups to make from the total time. - Then use the numerator to decide how many of those groups belong to the selected part.

1. Split \(45\) minutes into \(9\) equal groups: \(45\div9=5\) minutes per group. 2. Take \(4\) of those groups: \(4\times5=20\). 3. Therefore \(\frac{4}{9}\times45=20\).

The skill stations use \(20\) minutes.

- First compare the two fraction multipliers themselves. - Think about whether equivalent fractions select the same portion of one fixed whole collection. - Use one convenient grouping to find the selected amount.

1. For \(\frac{2}{3}\times30\), divide \(30\) into \(3\) equal groups of \(10\), then take \(2\) groups: \(20\). 2. For \(\frac{4}{6}\times30\), divide \(30\) into \(6\) equal groups of \(5\), then take \(4\) groups: \(20\). 3. The fractions \(\frac{2}{3}\) and \(\frac{4}{6}\) are equivalent, so they select the same portion of the collection.

Both products equal \(20\).

- Count the equal rows in the model and connect them to the denominator. - What does the numerator tell you to do with those equal groups? - Use the size of one group to find the selected amount.

1. The denominator \(12\) tells how many equal groups the whole amount is partitioned into, so a) is correct. 2. Each group has \(36\div12=3\) objects. 3. Take \(5\) groups: \(5\times3=15\). 4. Therefore \(\frac{5}{12}\times36=15\).

a) is correct, and the product is \(15\).

- Let the denominator determine the equal partition of the whole amount. - Let the numerator determine how many equal groups are selected. - Translate those two roles into operations in the same order.

1. The denominator \(5\) means divide the \(40\) objects into \(5\) equal groups: \(40\div5=8\). 2. The numerator \(3\) means take \(3\) of those groups. 3. Thus \(\frac{3}{5}\times40=3\times8=24\).

\(\frac{3}{5}\times40=3\times(40\div5)=24\).

- Use the denominator to decide how many equal groups the collection should form. - Find the size of one group before using the numerator. - The numerator tells how many of those equal groups are selected.

1. Divide \(54\) into \(9\) equal groups: \(54\div9=6\) items per group. 2. Select \(7\) groups: \(7\times6=42\). 3. Therefore \(\frac{7}{9}\times54=42\).

\(42\) items are selected.

- Use the denominator to decide how many equal groups to make from the total time. - Find the size of one group before using the numerator. - The numerator tells how many equal time groups are included.

1. The denominator \(8\) means to divide the \(40\) minutes into \(8\) equal groups. 2. Each group has \(40\div8=5\) minutes. 3. The numerator \(7\) means to select \(7\) of those groups: \(7\times5=35\). 4. Therefore \(\frac{7}{8}\times40=35\).

The activities use \(35\) minutes.

- Use the number of equal parts in one whole to identify the denominator of the multiplier. - Count the colored parts across both wholes to identify the numerator. - Determine the number of counters represented by one equal part before combining the colored parts.

1. Each whole is divided into \(4\) equal parts, so each colored part represents one fourth of \(8\), or \(2\) counters. 2. The model colors \(7\) quarter-parts, so the multiplier is \(\frac{7}{4}\). 3. Seven groups of \(2\) counters make \(14\) counters, so \(\frac{7}{4}\times8=14\).

The multiplier is \(\frac{7}{4}\), and \(\frac{7}{4}\times8=14\) counters.

- For the repeated-addition view, add four equal fractions. - For the scaling view, find one fourth of \(4\) first. - Compare the two results rather than treating the explanations as different operations.

1. Maya's repeated addition gives \(\frac{12}{4}=3\). 2. For Owen's scaling view, one fourth of \(4\) is \(1\), so three fourths of \(4\) is \(3\). 3. Both interpretations give \(\frac{3}{4}\times4=3\).

Both interpretations give \(3\): \(\frac{3}{4}+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=3\) and \(\frac{3}{4}\) of \(4\) is \(3\).

- Look at one bar to name the size of one small part. - Count the shaded small parts across both bars. - Use that count as the whole-number factor.

1. Each bar is split into \(5\) equal parts, so one small part is \(\frac{1}{5}\) of a whole. 2. There are \(7\) shaded one-fifth parts. 3. Seven copies of \(\frac{1}{5}\) give \(\frac{1}{5}\times7=\frac{7}{5}\). 4. The factor \(7\) counts the shaded one-fifth parts.

\(\frac{1}{5}\times7=\frac{7}{5}\). The factor \(7\) counts the shaded one-fifth parts.

- Think about the roles of the numerator and denominator in the fraction. - Test whether each expression partitions into fifths and selects three of those parts.

1. \(\frac{3}{5}\times20\) means divide \(20\) into \(5\) equal parts and take \(3\), so A is equivalent. 2. Multiplying \(20\) by \(3\) and then dividing by \(5\) gives the same value, so B is equivalent. 3. C divides by \(3\) and multiplies by \(5\), reversing the roles of numerator and denominator.

A and B are equal to \(\frac{3}{5}\times20\).

- Choose a whole-number collection that can naturally be partitioned into ten equal groups. - Make the numerator describe how many of those equal groups are selected. - Check that your story leads to the same multiplication expression.

1. One valid story is: A display has \(30\) photos, and \(\frac{7}{10}\) of them are landscape photos. 2. Partition \(30\) into \(10\) equal groups: \(30\div10=3\). 3. Take \(7\) groups: \(7\times3=21\).

Answers will vary. A valid story must identify \(30\) as the whole collection, partition it into \(10\) equal groups, select \(7\) groups, and give the product \(21\).

- Identify how many equal groups make the original whole set. - Use the numerator to determine how many of those equal-sized groups are selected altogether. - Compare the number of selected groups with the number of groups in one whole set.

1. The original \(8\) objects are partitioned into \(4\) equal groups, so each fourth-group contains \(8\div4=2\) objects. 2. The model extends to show \(7\) such groups altogether. Taking \(7\) groups gives \(7\times2=14\). 3. Because \(\frac{7}{4}>1\), the model must extend beyond the original \(4\) groups, so a product greater than \(8\) is reasonable.

\(\frac{7}{4}\times8=14\). The product is greater than \(8\) because \(\frac{7}{4}\) selects seven fourth-groups, which is three groups more than one whole set.

- Use the denominator to find the size of one equal part of \(10\). - The numerator may count more parts than make one whole when the fraction is improper.

1. Partition \(10\) into \(5\) equal parts: each part is \(2\). 2. Take \(9\) parts: \(9\times2=18\). 3. Since \(\frac{9}{5}=1\frac{4}{5}\), the multiplier represents one whole amount plus four fifths more.

\(18\)

- The denominator controls how the whole-number quantity is partitioned. - The numerator controls how many equal parts are taken. - Check whether both calculation orders preserve those roles.

1. Partition first: \(40\div8=5\), then \(5\times3=15\). 2. Multiply first: \(40\times3=120\), then \(120\div8=15\). 3. Both orders represent taking \(3\) of \(8\) equal parts of \(40\).

\(15\) seats are reserved.

- The denominator tells how many equal parts to make from \(12\). - The numerator tells how many parts to count, even when that is more than one whole set.

1. Partition \(12\) into \(6\) equal parts: each part is \(2\). 2. Take \(11\) parts: \(11\times2=22\). 3. The improper fraction \(\frac{11}{6}\) represents more than one full set of six parts, so the product is greater than \(12\).

\(22\)

- Count the equal rows in the model to connect the picture with sixths. - Find how many objects are in one row. - Compare \(20\) with one row to decide how many rows are selected.

1. Divide the \(30\) tokens into \(6\) equal groups: \(30\div6=5\) tokens per group. 2. To select \(20\) tokens, the number of groups must satisfy \(a\times5=20\), so \(a=4\). 3. The numerator \(4\) means selecting \(4\) of the \(6\) equal groups.

\(a=4\), so the fraction is \(\frac{4}{6}\). The numerator tells how many of the six equal groups are selected.

- What do the \(3\) and the \(8\) each tell you about the equal groups? - Check whether Leila's \(4\) groups match the fraction she is using. - After deciding how many equal groups the whole should have, decide how many of those groups to take.

1. The denominator \(8\) tells how many equal groups the \(24\) objects should be split into. Leila made \(4\) groups instead of \(8\). 2. Divide \(24\) into \(8\) equal groups: \(24\div8=3\) objects per group. 3. Take \(3\) groups because the numerator is \(3\): \(3\times3=9\). 4. Therefore \(\frac{3}{8}\times24=9\).

Leila should make \(8\) equal groups and take \(3\) of them. The product is \(9\).

- In each model, identify how many equal groups make one original whole amount. - Use the numerator to decide how many equal-sized groups are selected. - Compare each multiplier with \(1\) and compare the selected region with the original amount.

1. In model a), \(20\) is split into \(5\) equal groups of \(4\), and \(4\) groups are selected. Thus \(\frac{4}{5}\times20=16\). 2. In model b), \(20\) is split into \(4\) equal groups of \(5\), and the model extends to a fifth group. Thus \(\frac{5}{4}\times20=25\). 3. The first multiplier is less than \(1\), so it shrinks the amount. The second is greater than \(1\), so it enlarges the amount.

\(\frac{4}{5}\times20=16\), which is less than \(20\). \(\frac{5}{4}\times20=25\), which is greater than \(20\).

- The denominator tells how many equal parts the full length should be divided into. - The equal parts do not need to have whole-number lengths. - After finding one part, use the numerator to determine how many parts are included.

1. Partition \(14\) feet into \(6\) equal lengths: \(14\div6=\frac{14}{6}=\frac{7}{3}\) feet per part. 2. Take \(5\) parts: \(5\times\frac{7}{3}=\frac{35}{3}=11\frac{2}{3}\). 3. Therefore \(\frac{5}{6}\times14=11\frac{2}{3}\) feet.

Workers finish \(11\frac{2}{3}\) feet of trail.

- Count the equal rows in the model and connect them to ninths. - Decide what one row represents. - Then use the numerator to decide how many rows are needed.

1. Dividing \(45\) by \(9\) correctly finds one ninth of the collection: \(5\). 2. The numerator \(2\) means two of those one-ninth groups are needed. 3. Multiply \(2\times5=10\). 4. Therefore \(\frac{2}{9}\times45=10\).

Priya correctly found one ninth but forgot to take two such groups. The product is \(10\).

- The denominator tells how many equal parts the original length is divided into. - One equal part does not have to be a whole-number length. - Use the numerator to combine the needed number of equal parts.

1. One ninth of \(22\) feet is \(22\div9=\frac{22}{9}\) feet. 2. Four such parts give \(4\times\frac{22}{9}=\frac{88}{9}\) feet. 3. Convert \(\frac{88}{9}=9\frac{7}{9}\) feet.

The scaled length is \(9\frac{7}{9}\) feet.

- The number of equal groups created from the whole becomes the denominator. - The number of groups selected becomes the numerator. - Use the group size to find the selected number of objects.

1. Dividing the whole collection into \(8\) equal groups means the denominator is \(8\). 2. Selecting \(5\) groups means the numerator is \(5\), so the expression is \(\frac{5}{8}\times32\). 3. Each group contains \(32\div8=4\) objects. 4. Five groups contain \(5\times4=20\) objects.

The expression is \(\frac{5}{8}\times32\), and \(20\) objects are selected.

- The denominator of the multiplier tells how many equal parts each whole must have. - Check the partition size before counting colored parts. - After choosing the model, find the size of one fractional group of \(12\).

1. A multiplier of \(\frac{5}{3}\) requires thirds, so model a), where each whole is divided into \(3\) equal parts and \(5\) such parts are colored, is correct. 2. Model b) divides each whole into fourths, so its five colored parts represent \(\frac{5}{4}\), not \(\frac{5}{3}\). 3. One third of \(12\) feet is \(4\) feet, and five such groups make \(20\) feet. 4. Therefore \(\frac{5}{3}\times12=20\).

Model a) is correct. Model b) uses fourths instead of thirds. The product is \(20\) feet.

- Work backward from the selected number of objects to find the size of one selected group. - Then determine how many equal groups of that size make the full collection. - The denominator records the total number of equal groups in the whole.

1. Taking \(3\) equal groups gives \(9\) objects, so each group must contain \(9\div3=3\) objects. 2. Dividing \(24\) objects into groups of \(3\) creates \(24\div3=8\) equal groups. 3. Therefore \(b=8\), and \(\frac{3}{8}\times24=9\).

\(b=8\).

- Work backward from the amount represented by the selected groups. - Find the size of one equal group before rebuilding all the groups in the whole. - Use the original fraction multiplication equation to check the result.

1. Five equal groups total \(15\), so one group contains \(15\div5=3\). 2. The whole amount has \(7\) such groups, so \(q=7\times3=21\). 3. Check: \(\frac{5}{7}\times21=15\).

\(q=21\).

- Treat \(18\,\text{yd}^2\) as three equal fourths of the whole area. - Find the area of one fourth before rebuilding all four fourths. - Check the total by finding three fourths of it.

1. Three fourths of the garden area equal \(18\,\text{yd}^2\). 2. One fourth of the area is \(18\div3=6\,\text{yd}^2\). 3. Four fourths make the whole garden, so the total area is \(4\times6=24\,\text{yd}^2\). 4. Check: \(\frac{3}{4}\times24=18\).

The garden has a total area of \(24\,\text{yd}^2\).