Free math worksheets

- Use the selected column to identify one third of the whole. - Look at how much of that column lies in the selected row. - Compare the single overlap cell with all the equal cells.

1. The overlap represents \(\frac{1}{2}\) of \(\frac{1}{3}\). 2. The grid has \(2\times3=6\) equal cells, and exactly \(1\) cell is in both selected regions. 3. Therefore \(\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\).

\(\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}\), so the overlap covers \(\frac{1}{6}\) of the whole rectangle.

- Read one factor from the vertical partition and the other from the horizontal partition. - The overlap is the part that satisfies both fractional conditions. - Compare the number of overlap cells with the total number of equal cells in the whole grid.

1. The vertical region covers \(5\) of \(8\) equal columns, so \(\frac{5}{8}\) of the animals are dogs. 2. The horizontal region covers \(2\) of \(5\) equal rows, so the model takes \(\frac{2}{5}\) of the dog region. 3. The overlap covers \(10\) of the \(40\) equal cells, so \(\frac{2}{5}\times\frac{5}{8}=\frac{10}{40}=\frac{1}{4}\).

\(\frac{2}{5}\times\frac{5}{8}=\frac{1}{4}\). Puppies make up \(\frac{1}{4}\) of all the animals.

- Read the width fraction by comparing green columns with all columns. - Read the height fraction by comparing green rows with all rows. - The green rectangle's area is the product of those two side fractions.

1. The pasture is \(5\) equal columns wide, and the green rectangle spans \(4\) columns, so its width is \(\frac{4}{5}\) of the pasture width. 2. The pasture is \(3\) equal rows tall, and the green rectangle spans \(2\) rows, so its height is \(\frac{2}{3}\) of the pasture height. 3. The area fraction is \(\frac{4}{5}\times\frac{2}{3}=\frac{8}{15}\).

a) \(\frac{4}{5}\) b) \(\frac{2}{3}\) c) \(\frac{4}{5}\times\frac{2}{3}=\frac{8}{15}\)

- One fraction controls columns and the other controls rows. - The cells that satisfy both conditions form the product area.

1. The overlap represents \(\frac{7}{8}\times\frac{3}{5}\). 2. Multiply: \(\frac{7}{8}\times\frac{3}{5}=\frac{21}{40}\). 3. The grid shows \(21\) overlap cells out of \(40\) total cells.

\(\frac{21}{40}\) of the rectangle lies in both regions.

- Read one factor from the fraction of columns in the vertical region. - Read the other factor from the fraction of rows in the horizontal region. - Compare the overlap cell count with all equal cells in the grid.

1. The vertical region covers \(5\) of \(9\) columns, so it represents \(\frac{5}{9}\). 2. The horizontal region covers \(3\) of \(5\) rows, so it represents \(\frac{3}{5}\). 3. Their overlap contains \(15\) of the \(45\) equal cells, so \(\frac{3}{5}\times\frac{5}{9}=\frac{15}{45}=\frac{1}{3}\).

The overlap is \(15\) of \(45\) cells, so \(\frac{3}{5}\times\frac{5}{9}=\frac{1}{3}\).

- In panel a), compare selected columns with all columns. - In panel b), compare selected rows with all rows. - The overlap panel must represent cells belonging to both factor regions.

1. Panel a) has \(5\) selected columns out of \(7\), so it represents \(\frac{5}{7}\). 2. Panel b) has \(3\) selected rows out of \(4\), so it represents \(\frac{3}{4}\). 3. Panel c) contains \(15\) shaded overlap cells out of \(28\) total cells. 4. Therefore \(\frac{5}{7}\times\frac{3}{4}=\frac{15}{28}\).

a) \(\frac{5}{7}\) and \(\frac{3}{4}\) b) \(15\) overlap cells out of \(28\) c) \(\frac{5}{7}\times\frac{3}{4}=\frac{15}{28}\)

- Treat orange as belonging to both shaded directions. - Count columns for the width fraction and rows for the height fraction. - Compare orange overlap cells with all cells to verify the product.

1. The grid has \(5\) columns, and the width region spans \(4\), so the width is \(\frac{4}{5}\) unit. 2. The grid has \(3\) rows, and the height region spans \(2\), so the height is \(\frac{2}{3}\) unit. 3. The orange overlap has \(8\) of the \(15\) cells, so \(\frac{4}{5}\times\frac{2}{3}=\frac{8}{15}\) square unit.

a) Width \(\frac{4}{5}\) unit; height \(\frac{2}{3}\) unit b) \(\frac{4}{5}\times\frac{2}{3}=\frac{8}{15}\) square unit

- Use one factor to select columns and the other to select rows. - Write the overlap cell count over the total number of grid cells. - Simplify the resulting fraction.

1. The grid has \(36\) equal cells. 2. Eight of the \(9\) columns and \(3\) of the \(4\) rows overlap in \(8\times3=24\) cells. 3. The overlap fraction is \(\frac{24}{36}\). 4. Simplify \(\frac{24}{36}=\frac{2}{3}\).

\(\frac{8}{9}\times\frac{3}{4}=\frac{2}{3}\).

- Read the vertical factor from the number of columns in its band compared with all columns. - Read the horizontal factor from the number of rows in its band compared with all rows. - Count the overlap cells as a fraction of every cell in the grid, then simplify.

1. The vertical band covers \(3\) of \(5\) columns, so one factor is \(\frac{3}{5}\). 2. The horizontal band covers \(2\) of \(4\) rows, so the other factor is \(\frac{2}{4}=\frac{1}{2}\). 3. The overlap covers \(6\) of the \(20\) cells, so the product is \(\frac{6}{20}=\frac{3}{10}\). 4. Therefore \(\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}\).

The factors are \(\frac{3}{5}\) and \(\frac{1}{2}\), and the overlap is \(\frac{3}{10}\) of the whole grid.

- Find the complement of the first fraction used. - Find the complement of the fraction used from the remainder. - The final amount is a fraction of a fraction of the original whole.

1. After the first use, \(1-\frac{4}{9}=\frac{5}{9}\) of the original sugar remains. 2. After the second use, \(1-\frac{7}{10}=\frac{3}{10}\) of that remainder is left. 3. Therefore the final fraction of the original amount is \(\frac{3}{10}\times\frac{5}{9}=\frac{15}{90}=\frac{1}{6}\).

After the first use, \(\frac{5}{9}\) remains. Keeping \(\frac{3}{10}\) of that gives \(\frac{3}{10}\times\frac{5}{9}=\frac{1}{6}\) of the original amount.

- The long jumpers are a fraction of a group that is itself a fraction of all students. - For one method, combine the two fractions before using the total number of students. - For the second method, find the track-and-field group first and then take half of it.

1. The long jumpers are \(\frac{1}{2}\) of the \(\frac{2}{5}\) who choose track and field, so \(\frac{1}{2}\times\frac{2}{5}=\frac{1}{5}\) of all students are long jumpers. 2. One method is to find \(\frac{1}{5}\) of \(60\), which is \(12\). 3. Another method is to find \(\frac{2}{5}\) of \(60\), which is \(24\), and then take half of \(24\), which is \(12\).

a) \(\frac{1}{2}\times\frac{2}{5}=\frac{1}{5}\) of all students b) \(12\) students. Either find \(\frac{1}{5}\) of \(60\), or find \(\frac{2}{5}\) of \(60\) and then take one half.

- Read one factor from the vertical partition and the other from the horizontal partition in each model. - The overlap represents the fraction that satisfies both conditions. - Simplify both overlap fractions before comparing them.

1. In Model A, the vertical region is \(\frac{4}{5}\) of the pitcher and the horizontal region is \(\frac{3}{8}\) of that amount. The overlap is \(12\) of \(40\) cells, so \(\frac{3}{8}\times\frac{4}{5}=\frac{12}{40}=\frac{3}{10}\). 2. In Model B, the vertical region is \(\frac{3}{5}\) of the pitcher and the horizontal region is \(\frac{1}{2}\) of that amount. The overlap is \(3\) of \(10\) cells, so \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\). 3. Both overlaps represent \(\frac{3}{10}\) of a full pitcher, so the amounts are equal.

a) Model A: \(\frac{3}{8}\times\frac{4}{5}=\frac{3}{10}\) Model B: \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\) b) The amounts are equal.

- Use the colored regions in the grid; the row and column counts are not stated in the text. - First compare the non-pond columns with all columns. - Then compare the grass rows within the non-pond region before writing the fraction-of-a-fraction product.

1. The pond occupies one of three equal columns, so \(\frac{2}{3}\) of the garden remains. 2. Within the two remaining columns, grass occupies two of five equal rows, so grass is \(\frac{2}{5}\) of the remaining region. 3. Therefore \(\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}\). The four green cells out of fifteen total cells confirm the product.

a) \(\frac{2}{3}\) b) \(\frac{2}{5}\) c) \(\frac{2}{5}\times\frac{2}{3}=\frac{4}{15}\)

- Compare the overlap cells with the cells in the first selected region. - Then compare the overlap cells with all cells in the whole grid. - Connect those two fractions with a multiplication statement.

1. The \(\frac{3}{5}\) region contains \(12\) cells, and \(6\) are in the overlap, so \(\frac{6}{12}=\frac{1}{2}\) of that region was taken. 2. The overlap is \(6\) of \(20\) total cells: \(\frac{6}{20}=\frac{3}{10}\). 3. The model shows \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\).

\(\frac{1}{2}\) of the region was taken, and \(\frac{1}{2}\times\frac{3}{5}=\frac{3}{10}\).

- In each panel, read one factor from columns and the other from rows. - Count the orange overlap cells and all cells in each panel. - Compare what changes and what stays the same when the two factors switch directions.

1. In model a), the vertical region spans \(3\) of \(4\) columns and the horizontal region spans \(2\) of \(3\) rows, so it represents \(\frac{3}{4}\times\frac{2}{3}\). The overlap is \(6\) of \(12\) cells, or \(\frac{1}{2}\). 2. In model b), the vertical region spans \(2\) of \(3\) columns and the horizontal region spans \(3\) of \(4\) rows, so it represents \(\frac{2}{3}\times\frac{3}{4}\). Its overlap is also \(6\) of \(12\) cells, or \(\frac{1}{2}\). 3. Reversing the factor directions does not change the size of the overlap.

a) \(\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}\) b) \(\frac{2}{3}\times\frac{3}{4}=\frac{1}{2}\) Both models have \(6\) overlap cells out of \(12\), so the products are equal.

- In each panel, treat orange as part of both shaded factor regions. - Read one fraction from rows and the other from columns before counting the overlap. - Once both products are known, use a common denominator for their exact difference.

1. Model a) shows \(\frac{1}{2}\) of the rows and \(\frac{5}{6}\) of the columns. Its overlap is \(15\) of \(36\) cells, so the product is \(\frac{5}{12}\). 2. Model b) shows \(\frac{1}{3}\) of the rows and \(\frac{5}{6}\) of the columns. Its overlap is \(10\) of \(36\) cells, so the product is \(\frac{5}{18}\). 3. Using denominator \(36\), \(\frac{5}{12}-\frac{5}{18}=\frac{15}{36}-\frac{10}{36}=\frac{5}{36}\).

a) \(\frac{1}{2}\times\frac{5}{6}=\frac{5}{12}\) b) \(\frac{1}{3}\times\frac{5}{6}=\frac{5}{18}\) The product in a) is greater by \(\frac{5}{36}\).

- Read one factor from the number of selected columns and the other from the selected rows. - The orange cells belong to both regions. - Compare overlap cells with all cells instead of adding the factor numerators and denominators.

1. The vertical region spans \(3\) of \(4\) columns, so one factor is \(\frac{3}{4}\). The horizontal region spans \(2\) of \(5\) rows, so the other factor is \(\frac{2}{5}\). 2. Adding numerators and denominators does not describe the intersection of row and column regions. 3. The overlap has \(6\) cells out of \(20\), so \(\frac{2}{5}\times\frac{3}{4}=\frac{6}{20}=\frac{3}{10}\).

a) The factors are \(\frac{2}{5}\) and \(\frac{3}{4}\). b) The product is represented by the overlap, not by adding fraction parts. The overlap is \(\frac{6}{20}=\frac{3}{10}\).

- In panel a), compare selected columns with all columns. - In panel b), count how many rows the overlap occupies and compare that with the total number of rows. - Use the overlap dimensions to check the product after identifying both factors.

1. Panel a) has \(6\) selected columns out of \(8\), so the known factor is \(\frac{6}{8}=\frac{3}{4}\). 2. In panel b), the overlap is \(3\) rows high out of \(6\) total rows, so the missing factor is \(\frac{3}{6}=\frac{1}{2}\). 3. The overlap has \(3\times6=18\) cells out of \(48\), giving \(\frac{18}{48}=\frac{3}{8}\). 4. Thus \(\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}\).

a) \(\frac{3}{4}\) b) \(\frac{3}{6}=\frac{1}{2}\) c) \(\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}\)

- Compare the row and column partitions in each model with the denominators in the factors. - After choosing a model, use the numerators to decide how many rows and columns belong to the overlap. - Write the overlap cell count over the total cell count.

1. Model a) partitions one dimension into \(7\) equal parts and the other into \(9\) equal parts, matching the denominators. 2. Model b) uses \(4\) and \(5\) as partition counts, but those are the numerators, not the denominators. 3. Model a) has \(7\times9=63\) equal cells. Selecting \(4\) of the \(7\) rows and \(5\) of the \(9\) columns gives \(4\times5=20\) overlap cells. 4. Therefore \(\frac{4}{7}\times\frac{5}{9}=\frac{20}{63}\).

Model a) is appropriate because its \(7\) row parts and \(9\) column parts match the denominators. The product is \(\frac{20}{63}\).

- Read the vertical fraction from columns and the horizontal fraction from rows. - Count orange overlap cells relative to all cells. - Compare the overlap count with half the total cell count.

1. The vertical factor spans \(7\) of \(9\) columns, and the horizontal factor spans \(2\) of \(3\) rows, so the model represents \(\frac{7}{9}\times\frac{2}{3}\). 2. The orange overlap contains \(14\) of \(27\) cells, so the product is \(\frac{14}{27}\). 3. Half of \(27\) cells is \(13\frac{1}{2}\), so \(14\) cells is more than half. Thus \(\frac{14}{27}>\frac{1}{2}\).

The model represents \(\frac{7}{9}\times\frac{2}{3}=\frac{14}{27}\), and the overlap is greater than \(\frac{1}{2}\).

- Count only the orange cells for part a). - A fraction of a whole requires a numerator count and a total equal-part count. - Read the factor fractions from selected columns and rows, including orange in both regions.

1. There are \(20\) orange overlap cells. 2. A product represented as a fraction of the whole needs the overlap count compared with all equal cells, not just the count alone. 3. The model shows \(\frac{4}{9}\) of the columns and \(\frac{5}{6}\) of the rows. The whole grid has \(54\) cells, so the product is \(\frac{20}{54}=\frac{10}{27}\).

a) \(20\) cells b) \(20\) is only a count; it must be compared with all \(54\) equal cells. c) \(\frac{4}{9}\times\frac{5}{6}=\frac{20}{54}=\frac{10}{27}\).

- Include orange cells when reading both the selected row region and selected column region. - Compare orange cells with all cells for the product. - Convert the benchmark fraction into a cell count before finding the gap.

1. The model shows \(7\) of \(8\) columns and \(5\) of \(6\) rows, so the expression is \(\frac{7}{8}\times\frac{5}{6}\). 2. The orange overlap has \(35\) of \(48\) cells, so the product is \(\frac{35}{48}\). 3. Three fourths of \(48\) cells is \(36\) cells, so one additional overlap cell is needed.

a) \(\frac{7}{8}\times\frac{5}{6}\) b) \(\frac{35}{48}\) c) \(1\) additional cell

- Read one factor from columns and one from rows, counting orange as part of both shaded regions. - Express orange cells over all equal cells and simplify. - Translate \(\frac{1}{4}\) of the whole grid into a target number of cells.

1. The model shows \(5\) of \(8\) columns and \(2\) of \(7\) rows, so the expression is \(\frac{5}{8}\times\frac{2}{7}\). 2. The orange overlap has \(10\) cells out of \(56\), so the product is \(\frac{10}{56}=\frac{5}{28}\). 3. One fourth of \(56\) cells is \(14\), so the overlap needs \(14-10=4\) more cells.

a) \(\frac{5}{8}\times\frac{2}{7}\) b) \(10\) cells, representing \(\frac{5}{28}\) of the grid c) \(4\) more cells

- Use the horizontal band's height to identify its fraction of the total rows. - The product is represented only by cells that belong to both bands. - Compare the number of overlap cells with the total number of cells in the grid.

1. The horizontal band covers \(2\) of \(5\) rows, so the second factor is \(\frac{2}{5}\). 2. The overlap covers \(2\times4=8\) cells out of \(5\times7=35\) cells. 3. Therefore the overlap is \(\frac{8}{35}\), matching \(\frac{4}{7}\times\frac{2}{5}=\frac{8}{35}\).

The second factor is \(\frac{2}{5}\), and the product is \(\frac{8}{35}\). The grid verifies this because \(8\) of its \(35\) cells lie in both factor regions.