Free Math Worksheets

- How many equal intervals are between consecutive whole numbers? - How does that number determine the denominator? - Count the small intervals from the whole number immediately before each point. - A value greater than \(1\) can be written as a mixed number. - On a number line, values increase from left to right.

1. The interval from \(0\) to \(1\) is divided into \(5\) equal parts, so each tick represents \(\frac{1}{5}\). 2. Point \(A\) is at the third tick to the right of \(0\), so \(A = \frac{3}{5}\). 3. Point \(B\) is at the third tick to the right of \(1\), so \(B = 1\frac{3}{5} = \frac{8}{5}\). 4. Since \(A\) is to the left of \(B\) on the number line, \(A < B\).

\(A = \frac{3}{5}\) \(B = 1\frac{3}{5}\), or \(\frac{8}{5}\) \(A < B\)

- Rewrite all fractions in eighths. - Choose a unit length that can be divided into eight equal parts. - Compare the distances from \(\frac{3}{8}\) on the number line.

1. The least common denominator is \(8\). Rewrite the fractions: \(\frac{1}{4}=\frac{2}{8}\), \(\frac{1}{2}=\frac{4}{8}\), \(\frac{5}{8}\), and \(\frac{3}{4}=\frac{6}{8}\). 2. A useful unit length is \(8\,\text{cm}\) or \(8\) grid spaces, because each eighth then has a length of \(1\,\text{cm}\) or one grid space. 3. The distance from \(\frac{3}{8}\) to \(\frac{1}{4}=\frac{2}{8}\) is \(\frac{1}{8}\). The distance to \(\frac{1}{2}=\frac{4}{8}\) is also \(\frac{1}{8}\). 4. Therefore, \(\frac{1}{4}\) and \(\frac{1}{2}\) are equally close to \(\frac{3}{8}\).

A useful unit length is \(8\,\text{cm}\) or \(8\) grid spaces. The fractions \(\frac{1}{4}\) and \(\frac{1}{2}\) are both closest to \(\frac{3}{8}\).

- Find the least common multiple of the denominators. - Choose a unit length that is easy to divide into twelfths. - Rewrite all fractions in twelfths before ordering them.

1. The least common multiple of \(3\), \(6\), \(4\), and \(12\) is \(12\). 2. With a \(12\,\text{cm}\) unit interval, each twelfth is exactly \(1\,\text{cm}\). With a \(10\,\text{cm}\) unit interval, each twelfth is \(\frac{5}{6}\,\text{cm}\), which is harder to mark accurately. 3. Another useful unit length is \(6\,\text{cm}\), because each twelfth is \(\frac{1}{2}\,\text{cm}\). A \(24\,\text{cm}\) interval would also work. 4. Rewrite the fractions in twelfths: \(\frac{1}{6}=\frac{2}{12}\), \(\frac{5}{12}\), \(\frac{2}{3}=\frac{8}{12}\), and \(\frac{3}{4}=\frac{9}{12}\). 5. Therefore, \(\frac{1}{6}<\frac{5}{12}<\frac{2}{3}<\frac{3}{4}\).

a) A \(12\,\text{cm}\) interval makes each twelfth exactly \(1\,\text{cm}\). b) One possible length is \(6\,\text{cm}\). c) \(\frac{1}{6}<\frac{5}{12}<\frac{2}{3}<\frac{3}{4}\)

- Rewrite both fractions using the same larger denominator. - Look for whole-number numerators between the two new numerators. - Check that each result is greater than \(\frac{3}{7}\) and less than \(\frac{4}{7}\).

1. Rewrite both fractions with denominator \(28\): \(\frac{3}{7}=\frac{12}{28}\) and \(\frac{4}{7}=\frac{16}{28}\). 2. The numerators \(13\), \(14\), and \(15\) lie between \(12\) and \(16\). 3. Therefore, three possible fractions are \(\frac{13}{28}\), \(\frac{14}{28}\), and \(\frac{15}{28}\). Each would be located between the two given fractions on a number line.

One possible answer is \(\frac{13}{28}\), \(\frac{14}{28}\), and \(\frac{15}{28}\).

- Rewrite both fractions with a larger common denominator. - Choose a denominator that creates at least two integer numerators between the endpoints. - The same idea can be repeated with an even larger denominator.

1. Rewrite both fractions with denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{4}{4}=\frac{12}{12}\). 2. The fractions \(\frac{10}{12}\) and \(\frac{11}{12}\) lie strictly between \(\frac{9}{12}\) and \(\frac{12}{12}\). 3. In general, first rewrite two fractions with a common denominator. If no integer numerator lies between the new numerators, multiply both fractions by another common factor to create more numerator values between them.

Two possible fractions are \(\frac{10}{12}\) and \(\frac{11}{12}\).

- Count the equal intervals between \(0\) and \(1\) to determine the denominator. - For points to the right of \(1\) or \(2\), use mixed numbers. - In part b), count the small intervals from each point to the next whole number. - For part c), remember that values increase from left to right on a number line.

1. Each interval between consecutive whole numbers is divided into \(4\) equal parts, so each small interval is \(\frac{1}{4}\). 2. The marked values are \(X = \frac{3}{4}\), \(Y = 1\frac{2}{4} = 1\frac{1}{2}\), and \(Z = 2\frac{1}{4}\). 3. From \(X\) to \(1\), the missing fraction is \(\frac{1}{4}\). 4. From \(Y\) to \(2\), the missing fraction is \(\frac{1}{2}\), or \(\frac{2}{4}\). 5. From \(Z\) to \(3\), the missing fraction is \(\frac{3}{4}\). 6. Reading the number line from left to right gives \(X < Y < Z\).

a) \(X = \frac{3}{4}\); \(Y = 1\frac{1}{2}\); \(Z = 2\frac{1}{4}\) b) From \(X\), \(\frac{1}{4}\) is needed; from \(Y\), \(\frac{1}{2}\) is needed; from \(Z\), \(\frac{3}{4}\) is needed. c) \(X < Y < Z\)