Free math worksheets

- Determine the fractional size of one small interval. - Express \(\frac{3}{4}\) using that denominator. - Adjacent ticks change the numerator by one when the denominator stays fixed.

1. The number line is marked in eighths, and \(\frac{3}{4}=\frac{6}{8}\). 2. The tick immediately before is \(\frac{5}{8}\). 3. The tick immediately after is \(\frac{7}{8}\).

\(\frac{5}{8}\) and \(\frac{7}{8}\)

- Count the equal intervals from \(0\) to \(1\). - Determine the unit fraction represented by one interval. - Count how many intervals from \(0\) reach Q.

1. The interval from \(0\) to \(1\) is divided into \(4\) equal parts, so each interval is \(\frac{1}{4}\). 2. Q is \(3\) intervals to the right of \(0\). 3. Therefore, Q represents \(\frac{3}{4}\).

\(\frac{3}{4}\)

- Determine the size of one tick interval. - Count the number of half-sized intervals from \(0\) to R. - Regroup two halves as one whole for the mixed-number form.

1. Each tick interval is \(\frac{1}{2}\). 2. R is three half-sized intervals from \(0\), so it represents \(\frac{3}{2}\). 3. Two halves make \(1\) whole, with one half left, so \(\frac{3}{2}=1\frac{1}{2}\).

\(1\frac{1}{2}=\frac{3}{2}\)

- Count the equal intervals from \(0\) to \(1\) to identify the unit fraction for each step. - Rewrite the listed fractions in the same fractional unit as the number line. - For part b), compare how many equal intervals separate \(\frac{3}{8}\) from each nearby marked point.

1. The number line is divided into eighths. Rewrite the fractions in eighths: \(\frac{1}{4}=\frac{2}{8}\), \(\frac{1}{2}=\frac{4}{8}\), \(\frac{5}{8}\), and \(\frac{3}{4}=\frac{6}{8}\). 2. Reading the marked positions gives \(A=\frac{1}{4}\), \(B=\frac{1}{2}\), \(C=\frac{5}{8}\), and \(D=\frac{3}{4}\). 3. The point \(\frac{3}{8}\) is one eighth from both \(\frac{2}{8}=\frac{1}{4}\) and \(\frac{4}{8}=\frac{1}{2}\). 4. Therefore, \(\frac{1}{4}\) and \(\frac{1}{2}\) are equally close to \(\frac{3}{8}\).

a) \(A=\frac{1}{4}\), \(B=\frac{1}{2}\), \(C=\frac{5}{8}\), \(D=\frac{3}{4}\) b) \(\frac{1}{4}\) and \(\frac{1}{2}\)

- 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 \(\frac{8}{5}\) \(A < B\)

- Count the equal intervals from \(0\) to \(1\). - Decide what fraction the red marker actually represents. - Rename \(\frac{3}{4}\) using the number line's fractional unit.

1. The interval from \(0\) to \(1\) is divided into \(8\) equal parts, so each interval represents \(\frac{1}{8}\). 2. Marker S is at the third tick, so it represents \(\frac{3}{8}\), not \(\frac{3}{4}\). 3. Rewrite \(\frac{3}{4}\) in eighths: \(\frac{3}{4}=\frac{6}{8}\). 4. Therefore, \(\frac{3}{4}\) belongs at the sixth tick after \(0\).

Sam treated the numerator as the tick number without matching the denominator. S is at \(\frac{3}{8}\); \(\frac{3}{4}\) belongs at the sixth tick after \(0\).

- Determine the value of one interval from the number line. - Read A's location in tenths. - Moving one tick to the right adds one interval.

1. The number line is marked in tenths, and point A is at \(\frac{4}{10}=\frac{2}{5}\). 2. One tick to the right adds \(\frac{1}{10}\). 3. \(\frac{4}{10}+\frac{1}{10}=\frac{5}{10}\).

\(\frac{5}{10}\)

- Determine the fractional size of one small interval. - Read both marked locations using that same unit fraction. - Distance can be found by counting the equal intervals between the points.

1. Each small interval is \(\frac{1}{6}\) unit. Point A is at \(\frac{5}{6}\), and point B is at \(1\frac{1}{3}=\frac{8}{6}\). 2. Count the sixth-intervals from A to B: there are \(3\). 3. Therefore, the distance is \(\frac{3}{6}\) unit, which is equivalent to \(\frac{1}{2}\) unit.

\(\frac{3}{6}\) unit

- Determine the value of one small interval. - Express both labeled positions using that same fractional unit. - Count spaces between the positions, not both endpoint ticks.

1. The number line is marked in tenths. The labeled fractions are \(\frac{3}{10}\) and \(\frac{4}{5}=\frac{8}{10}\). 2. Their positions are the third and eighth tenths ticks. 3. The number of intervals between them is \(8-3=5\).

\(5\) tick intervals

- Read the labeled values from left to right. - Identify the third position only after determining the full order. - For part b), use the labels directly before and after the third card.

1. The four labeled values are \(\frac{1}{6}\), \(\frac{1}{4}\), \(\frac{1}{3}\), and \(\frac{1}{2}\). 2. Their left-to-right order is \(\frac{1}{6},\frac{1}{4},\frac{1}{3},\frac{1}{2}\). 3. The third card is \(\frac{1}{3}\). Its immediate neighbors are \(\frac{1}{4}\) and \(\frac{1}{2}\).

a) \(\frac{1}{3}\) b) \(\frac{1}{4}\) and \(\frac{1}{2}\)

- A forward move adds sixths; a backward move subtracts sixths. - All fractions already have the same denominator. - Keep the final position in sixths, as requested.

1. Represent the forward move by addition and the backward move by subtraction: \(\frac{5}{6}+\frac{2}{6}-\frac{3}{6}\). 2. Add first: \(\frac{5}{6}+\frac{2}{6}=\frac{7}{6}\). 3. Then subtract: \(\frac{7}{6}-\frac{3}{6}=\frac{4}{6}\). 4. Therefore, the marker finishes at \(\frac{4}{6}\).

\(\frac{5}{6}+\frac{2}{6}-\frac{3}{6}=\frac{4}{6}\)

- Read the two labeled endpoint fractions from the image. - Express both endpoints in twelfths. - Strictly between means neither endpoint is included.

1. Read the labeled endpoints as \(\frac{1}{3}\) and \(\frac{3}{4}\). The displayed number line is divided into twelfths. 2. Rewrite the endpoints in twelfths: \(\frac{1}{3}=\frac{4}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). 3. The marks strictly between them are \(\frac{5}{12}\), \(\frac{6}{12}\), \(\frac{7}{12}\), and \(\frac{8}{12}\), so there are \(4\) such marks.

a) \(4\) tick marks b) \(\frac{5}{12},\frac{6}{12},\frac{7}{12},\frac{8}{12}\)

- Compare the size of each change between neighboring displayed labels. - Recall which direction values increase on a standard number line. - Keep the equal spacing while putting the values into increasing order.

1. Each neighboring pair differs by \(\frac{1}{4}\), so the numerical spacing is equal. 2. On a standard number line, values increase as you move to the right. Rina's labels decrease to the right. 3. Reversing the label order gives \(0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\).

a) Yes; neighboring labels differ by \(\frac{1}{4}\). b) The values decrease instead of increase from left to right. c) \(0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\)

- Count the equal intervals from \(0\) to the labeled \(\frac{1}{2}\) tick. - Find the unit fraction that repeats that many times to make one half. - Use that interval size to count from \(0\) to P.

1. There are two equal tick intervals from \(0\) to \(\frac{1}{2}\). 2. Therefore, one interval is \(\frac{1}{4}\), because two copies of \(\frac{1}{4}\) make \(\frac{1}{2}\). 3. P is three intervals to the right of \(0\), so P represents \(\frac{3}{4}\).

P represents \(\frac{3}{4}\). The \(\frac{1}{2}\) label is two equal intervals from \(0\), so each interval is \(\frac{1}{4}\).

- 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) \(\frac{1}{6}<\frac{5}{12}<\frac{2}{3}<\frac{3}{4}\)

- Rename both endpoint fractions as twelfths. - Look for whole-number numerators strictly between the two new numerators. - Check that each chosen fraction is greater than the lower endpoint and less than the upper endpoint.

1. Rewrite both endpoints with denominator \(12\): \(\frac{1}{3}=\frac{4}{12}\) and \(\frac{2}{3}=\frac{8}{12}\). 2. The numerators \(5\), \(6\), and \(7\) lie strictly between \(4\) and \(8\). 3. Therefore, \(\frac{5}{12}\), \(\frac{6}{12}\), and \(\frac{7}{12}\) are all strictly between the two given fractions.

\(\frac{1}{3}=\frac{4}{12}<\frac{5}{12}<\frac{6}{12}<\frac{7}{12}<\frac{8}{12}=\frac{2}{3}\)

- Rename both endpoints using denominator \(12\). - Look at the integer numerators strictly between the new endpoint numerators. - Explain why using equivalent fractions can reveal values between two fractions that looked consecutive in fourths.

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. Equivalent fractions can name the same endpoints using smaller equal parts, which reveals additional fraction values between the original fourths.

Two fractions are \(\frac{10}{12}\) and \(\frac{11}{12}\). Since \(\frac{3}{4}=\frac{9}{12}\) and \(1=\frac{12}{12}\), both values lie strictly between the endpoints.

- 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) \(X < Y < Z\)

- Use the labeled \(\frac{1}{4}\) to determine the value of one small interval. - Express the marked values, \(\frac{2}{5}\), and \(\frac{1}{2}\) with a common denominator. - Your final ordered list must contain fraction values rather than the letters \(D\), \(E\), and \(F\).

1. There are \(5\) equal intervals from \(0\) to \(\frac{1}{4}\). Rewrite \(\frac{1}{4}\) as \(\frac{5}{20}\), so each small interval represents \(\frac{1}{20}\). 2. Therefore, \(D=\frac{3}{20}\), \(E=\frac{7}{20}\), and \(F=\frac{12}{20}=\frac{3}{5}\). 3. Rewrite the two additional fractions in twentieths: \(\frac{2}{5}=\frac{8}{20}\) and \(\frac{1}{2}=\frac{10}{20}\). 4. Comparing the numerators gives \(\frac{3}{20}<\frac{7}{20}<\frac{8}{20}<\frac{10}{20}<\frac{12}{20}\).

\(D=\frac{3}{20}\), \(E=\frac{7}{20}\), \(F=\frac{3}{5}\) Order: \(\frac{3}{20}<\frac{7}{20}<\frac{2}{5}<\frac{1}{2}<\frac{3}{5}\)

- Determine the value of one small interval from the number line. - Count the equal intervals between A and B. - Move half that number of intervals from either endpoint.

1. The unit interval is divided into twelfths. Point A is at \(\frac{4}{12}=\frac{1}{3}\), and point B is at \(\frac{10}{12}=\frac{5}{6}\). 2. There are \(6\) twelfth-steps between A and B, so halfway is \(3\) twelfth-steps from either endpoint. 3. \(\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\).

\(\frac{7}{12}\)

- Determine the fractional size of one interval between \(1\) and \(2\). - Count from \(1\) to locate \(K\). - For part b), rename the two values with a useful common denominator before comparing them.

1. The interval from \(1\) to \(2\) is divided into sixths. Point \(K\) is four sixths to the right of \(1\), so \(K=1\frac{4}{6}=1\frac{2}{3}=\frac{5}{3}\). 2. Compare \(\frac{5}{3}\) and \(\frac{7}{4}\) using twelfths: \(\frac{5}{3}=\frac{20}{12}\) and \(\frac{7}{4}=\frac{21}{12}\). 3. Since \(20<21\), point \(K\) is to the left of \(\frac{7}{4}\).

a) \(K=1\frac{2}{3}\) b) \(K\) is to the left of \(\frac{7}{4}\).

- Fractions equally far from \(\frac{1}{2}\) appear on opposite sides of it. - Count or compare equal intervals from \(\frac{1}{2}\) to each labeled value. - Match values whose distances from the benchmark are equal.

1. The distance from \(\frac{1}{4}\) to \(\frac{1}{2}\) is \(\frac{1}{4}\), and the distance from \(\frac{3}{4}\) to \(\frac{1}{2}\) is also \(\frac{1}{4}\). 2. The distance from \(\frac{1}{3}\) to \(\frac{1}{2}\) is \(\frac{1}{6}\), and the distance from \(\frac{2}{3}\) to \(\frac{1}{2}\) is also \(\frac{1}{6}\). 3. Therefore, there are two pairs equally far from \(\frac{1}{2}\).

\(\frac{1}{4}\) and \(\frac{3}{4}\); \(\frac{1}{3}\) and \(\frac{2}{3}\)

- Determine the size of one small interval on the number line. - Compare the number of intervals separating each pair of marked points. - The pair with the fewest intervals between them is closest.

1. The number line is marked in thirds. Reading the positions gives \(A=\frac{2}{3}\), \(B=\frac{4}{3}\), and \(C=\frac{5}{3}\). 2. From A to B is \(2\) one-third intervals, so the distance is \(\frac{2}{3}\). 3. From B to C is \(1\) one-third interval, so the distance is \(\frac{1}{3}\). 4. From A to C is \(3\) one-third intervals, so the distance is \(1\). 5. Therefore, B and C are closest together, \(\frac{1}{3}\) unit apart.

Points \(B\) and \(C\), with distance \(\frac{1}{3}\) unit

- Read the two labeled fraction values from the aligned number lines. - Compare how far each value is below \(1\). - Explain why comparing a numerator only with its own denominator does not order two different fractions.

1. Line A shows \(\frac{3}{4}\), which is \(\frac{1}{4}\) below \(1\). Line B shows \(\frac{4}{5}\), which is \(\frac{1}{5}\) below \(1\). 2. Since \(\frac{1}{5}<\frac{1}{4}\), \(\frac{4}{5}\) is closer to \(1\). Therefore, \(\frac{3}{4}\) is to the left of \(\frac{4}{5}\). 3. Nora's reasoning does not compare the two fraction values; a numerator being less than its own denominator only shows that the fraction is less than \(1\).

No. The correct order is \(\frac{3}{4}<\frac{4}{5}\).

- Rewrite each improper fraction as a mixed number. - Notice that all four whole-number parts are equal. - Compare the remaining positive unit fractions by the sizes of their parts.

1. Rewrite the improper fractions as mixed numbers: \(\frac{9}{8}=1\frac{1}{8}\) and \(\frac{4}{3}=1\frac{1}{3}\). 2. All four values have whole-number part \(1\), so compare \(\frac{1}{8}\), \(\frac{1}{4}\), \(\frac{1}{3}\), and \(\frac{1}{2}\). 3. For positive unit fractions, a larger denominator gives a smaller fraction: \(\frac{1}{8}<\frac{1}{4}<\frac{1}{3}<\frac{1}{2}\). 4. Therefore, \(\frac{9}{8}<1\frac{1}{4}<\frac{4}{3}<1\frac{1}{2}\).

\(\frac{9}{8}<1\frac{1}{4}<\frac{4}{3}<1\frac{1}{2}\)

- Determine the fractional size of one tick interval. - Express the requested distance using that same fractional unit. - Move the same number of intervals left and right from C.

1. The number line is marked in twelfths. Point \(C\) is at \(\frac{8}{12}=\frac{2}{3}\). 2. A distance of \(\frac{1}{4}\) equals \(\frac{3}{12}\), so move three tick intervals in each direction from C. 3. Three ticks left lands at \(\frac{5}{12}\); three ticks right lands at \(\frac{11}{12}\).

\(\frac{5}{12}\) and \(\frac{11}{12}\)

- Use the tick spacing to express each labeled value in twelfths. - Count intervals in the first gap and then in the second gap. - Equal spacing requires the two interval counts to match.

1. The labeled points are \(\frac{1}{4}=\frac{3}{12}\), \(\frac{1}{2}=\frac{6}{12}\), and \(\frac{5}{6}=\frac{10}{12}\). 2. From \(\frac{3}{12}\) to \(\frac{6}{12}\) there are \(3\) twelfth-sized intervals. 3. From \(\frac{6}{12}\) to \(\frac{10}{12}\) there are \(4\) twelfth-sized intervals. 4. Because \(3\ne4\), the points are not equally spaced.

No. The first gap is \(3\) twelfth-sized intervals and the second gap is \(4\) twelfth-sized intervals.

- Find the total distance from A to B using the two stated fractions. - Count the equal intervals between the two markers in the image. - For part b), move two of those interval lengths to the right from A.

1. The distance from A to B is \(\frac{7}{8}-\frac{1}{4}=\frac{7}{8}-\frac{2}{8}=\frac{5}{8}\). 2. The image shows \(5\) equal intervals from A to B, so one interval is \(\frac{1}{8}\). 3. Two intervals to the right of A is \(\frac{1}{4}+\frac{2}{8}=\frac{4}{8}=\frac{1}{2}\).

a) \(\frac{1}{8}\) unit b) \(\frac{1}{2}\)

- Count the spaces between neighboring tick marks rather than the marks themselves. - The denominator tells how many equal intervals make one whole. - Count three interval lengths from \(0\) for the requested tick.

1. The displayed line has \(9\) tick marks but only \(8\) spaces between neighboring marks. 2. The unit from \(0\) to \(1\) is therefore divided into eighths, so each interval is \(\frac{1}{8}\). 3. The third interior tick after \(0\) is \(\frac{3}{8}\).

Jonah counted marks instead of intervals. Each interval is \(\frac{1}{8}\), and the third interior tick is \(\frac{3}{8}\).

- For part a), the interval count must be divisible by all three denominators. - Find the least common multiple of \(3\), \(4\), and \(6\). - For part b), rewrite each fraction with the common denominator.

1. The number of intervals must be divisible by \(3\), \(4\), and \(6\). 2. The least such number is \(12\), so the line should be divided into twelfths. 3. \(\frac{1}{4}=\frac{3}{12}\), so it is at tick \(3\). \(\frac{1}{3}=\frac{4}{12}\), so it is at tick \(4\). \(\frac{5}{6}=\frac{10}{12}\), so it is at tick \(10\).

a) \(12\) equal intervals b) \(\frac{1}{4}\): tick \(3\); \(\frac{1}{3}\): tick \(4\); \(\frac{5}{6}\): tick \(10\)

- Read the two displayed endpoint values first. - Count how many equal spaces divide the displayed interval. - Express the left endpoint in eighths and move one interval at a time.

1. The displayed endpoints are \(\frac{1}{4}\) and \(\frac{3}{4}\), so the shown interval has length \(\frac{1}{2}\). 2. The image divides that interval into \(4\) equal parts, so each small interval is \(\frac{1}{8}\) unit. 3. Starting at \(\frac{1}{4}=\frac{2}{8}\), the interior ticks are \(\frac{3}{8}\), \(\frac{4}{8}=\frac{1}{2}\), and \(\frac{5}{8}\).

a) \(\frac{1}{8}\) unit b) \(\frac{3}{8},\frac{1}{2},\frac{5}{8}\)

- First determine the size of one interval from the displayed number line. - Rename the two nearby tick fractions in twelfths. - Compare \(\frac{5}{12}\) with those neighboring values.

1. The displayed unit is divided into sixths. The neighboring ticks around \(\frac{5}{12}\) are \(\frac{2}{6}\) and \(\frac{3}{6}\). 2. Rename them as twelfths: \(\frac{2}{6}=\frac{4}{12}=\frac{1}{3}\) and \(\frac{3}{6}=\frac{6}{12}=\frac{1}{2}\). 3. The fraction \(\frac{5}{12}\) is not on either tick. It is \(\frac{1}{12}\) above \(\frac{4}{12}\) and \(\frac{1}{12}\) below \(\frac{6}{12}\), so it lies halfway between the two ticks.

No. \(\frac{5}{12}\) lies exactly halfway between \(\frac{1}{3}\) and \(\frac{1}{2}\).

- Read each point's starting position from the number line before shifting it. - Apply the same rightward shift to all three points. - Compare the three new positions from left to right.

1. Read the starting positions from the image: A is at \(\frac{1}{4}\), B is at \(\frac{1}{2}\), and C is at \(\frac{3}{4}\). 2. Shift each point \(\frac{1}{4}\) unit right: A moves to \(\frac{1}{2}\), B moves to \(\frac{3}{4}\), and C moves to \(1\). 3. The new positions satisfy \(\frac{1}{2}<\frac{3}{4}<1\). Moving every point the same distance preserves their order.

a) A: \(\frac{1}{2}\), B: \(\frac{3}{4}\), C: \(1\) b) No. Every point moves the same distance to the right, so their order is unchanged.

- Check the numerical change from each label to the next. - Equal spacing requires the same change across every interval. - Also check whether two different marks have been given the same value.

1. Proposal A increases by \(\frac{1}{4}\) at every step, so it represents four equal intervals. 2. In proposal B, the fourth mark is labeled \(\frac{4}{4}=1\), so the last two marks have the same value instead of being one interval apart. The fourth mark should be \(\frac{3}{4}\). 3. In proposal C, the first three increases are \(\frac{1}{5}\), but the final increase from \(\frac{3}{5}\) to \(1\) is \(\frac{2}{5}\). The intervals are not equal.

Proposal A is correct. In B, the fourth mark repeats the endpoint value \(1\); it should be \(\frac{3}{4}\). In C, the final interval is \(\frac{2}{5}\), while the earlier intervals are \(\frac{1}{5}\).

- Read point \(P\)'s starting position from the number line. - Count the stated number of eighth-sized intervals for each rightward move. - For part b), compare \(R\)'s final position with one whole.

1. Read \(P=\frac{1}{8}\) from the displayed number line. 2. Move three eighth-sized intervals right: \(Q=\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\). 3. Move two more eighth-sized intervals right: \(R=\frac{4}{8}+\frac{2}{8}=\frac{6}{8}=\frac{3}{4}\). 4. The distance from \(R\) to \(1\) is \(1-\frac{3}{4}=\frac{1}{4}\).

a) \(Q=\frac{1}{2}\), \(R=\frac{3}{4}\) b) \(\frac{1}{4}\) unit

- Read the labeled major marks from the displayed line. - Notice how many equal small spaces lie inside each major interval. - Continue the same small interval size from \(0\) to name the unlabeled ticks.

1. The labeled major marks are \(0\), \(\frac{1}{3}\), \(\frac{2}{3}\), and \(1\), so each major interval is \(\frac{1}{3}\) unit. 2. The image places one unlabeled tick halfway through each major interval. Each third is therefore split into two equal pieces, giving \(6\) equal intervals across the whole line. 3. Each small interval is \(\frac{1}{6}\) unit, and the unlabeled interior ticks are \(\frac{1}{6}\), \(\frac{1}{2}\), and \(\frac{5}{6}\).

a) \(\frac{1}{6}\) unit b) \(\frac{1}{6},\frac{1}{2},\frac{5}{6}\)

- Count the intervals and locate \(P\) on line a). - Rename that fraction using the number of intervals on line b), then on line c). - Equivalent fractions should land at the same distance from \(0\).

1. Count the intervals on line a). It is divided into thirds, and \(P\) is at tick \(2\), so \(P=\frac{2}{3}\). 2. Line b) is divided into sixths. Since \(\frac{2}{3}=\frac{4}{6}\), \(P\) belongs at tick \(4\) on line b). 3. Line c) is divided into twelfths. Since \(\frac{2}{3}=\frac{8}{12}\), \(P\) belongs at tick \(8\) on line c). 4. Equivalent fractions name the same number-line position even when the interval sizes differ.

a) Tick \(4\) on line b) and tick \(8\) on line c) b) \(\frac{2}{3}=\frac{4}{6}=\frac{8}{12}\), so all three fractions represent the same point.

- Determine the size of one interval from the displayed line. - Work backward from the farthest allowed endpoint, \(1\). - Include every displayed tick at or before the greatest possible start.

1. The displayed line is divided into tenths, so moving three intervals means moving \(\frac{3}{10}\) unit to the right. 2. The starting position must be at or left of \(1-\frac{3}{10}=\frac{7}{10}\). 3. The possible starting ticks are \(0,\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10},\frac{5}{10},\frac{6}{10},\frac{7}{10}\). 4. There are \(8\) possible starting positions. Starting at the next tick would move the marker past \(1\).

a) \(0,\frac{1}{10},\frac{2}{10},\frac{3}{10},\frac{4}{10},\frac{5}{10},\frac{6}{10},\frac{7}{10}\) b) \(8\) positions

- Compare the two labeled values and count the equal spaces between them. - Express both labeled values in sixths before matching their difference to the interval count. - After finding one interval, work left from the nearer labeled tick and then check by moving across the whole line.

1. From \(\frac{5}{6}\) to \(\frac{4}{3}=\frac{8}{6}\), the number line shows \(3\) equal intervals. 2. The labeled values differ by \(\frac{8}{6}-\frac{5}{6}=\frac{3}{6}\). Three equal intervals make \(\frac{3}{6}\), so one interval is \(\frac{1}{6}\). 3. The leftmost tick is two intervals left of \(\frac{5}{6}\): \(\frac{5}{6}-\frac{2}{6}=\frac{3}{6}=\frac{1}{2}\). 4. Moving right by \(\frac{1}{6}\) gives \(\frac{1}{2},\frac{2}{3},\frac{5}{6},1,\frac{7}{6},\frac{4}{3}\).

a) \(\frac{1}{6}\) b) \(\frac{1}{2}\) c) \(\frac{1}{2},\frac{2}{3},\frac{5}{6},1,\frac{7}{6},\frac{4}{3}\)

- Read the marked fractions from the number line before comparing them with the two boundaries. - Decide what “at least” means for equality at the lower endpoint. - Decide what “less than” means for equality at the upper endpoint.

1. The marked fractions are \(\frac{11}{12}\), \(\frac{7}{6}\), \(\frac{5}{4}\), \(\frac{3}{2}\), and \(\frac{5}{3}\). 2. The fractions \(\frac{7}{6}=1\frac{1}{6}\) and \(\frac{5}{4}=1\frac{1}{4}\) are at least \(1\) and less than \(1\frac{1}{2}\). 3. The fraction \(\frac{11}{12}<1\), so it is excluded. The fraction \(\frac{3}{2}=1\frac{1}{2}\) equals the excluded upper endpoint, and \(\frac{5}{3}=1\frac{2}{3}\) is above it. 4. The lower endpoint \(1\) would be included because “at least” allows equality, while the upper endpoint is excluded because “less than” does not.

\(\frac{7}{6}\) and \(\frac{5}{4}\). The lower endpoint \(1\) is included, but the upper endpoint \(1\frac{1}{2}\) is excluded.

- Read the candidate fractions from the displayed line. - For a candidate between \(0\) and \(1\), compare its distance from each endpoint. - Look for a candidate whose first distance is two copies of its second distance.

1. Test the marked point \(\frac{2}{3}\). Its distance from \(0\) is \(\frac{2}{3}\). 2. Its distance from \(1\) is \(1-\frac{2}{3}=\frac{1}{3}\). 3. Since \(\frac{2}{3}=2\times\frac{1}{3}\), the distance from \(0\) is twice the distance from \(1\). 4. The other marked fractions do not satisfy this exact relationship.

\(\frac{2}{3}\). Its distance from \(0\) is \(\frac{2}{3}\), and its distance from \(1\) is \(\frac{1}{3}\), so the first distance is twice the second.

- Compare the endpoint values of the two complete number lines. - Count how many equal intervals make each whole and how many intervals reach the marked point. - The same fraction of two different wholes need not have the same numerical value.

1. On line a), the whole from \(0\) to \(1\) is divided into eighths, and \(P\) is three intervals from \(0\), so \(P=\frac{3}{8}\). 2. On line b), the whole from \(0\) to \(2\) is divided into eight equal intervals. Each interval has value \(\frac{2}{8}=\frac{1}{4}\). 3. Point \(Q\) is three such intervals from \(0\), so \(Q=3\times\frac{1}{4}=\frac{3}{4}\). 4. The points occupy the same fraction of their respective wholes, but line b)'s whole has twice the value of line a)'s whole.

a) \(Q=\frac{3}{4}\) b) They have the same relative position, but the whole on line b) has value \(2\) instead of \(1\), so the same fraction of that whole represents twice as much.

- Read the starting position from the displayed eighths scale. - Rewrite the jump size using eighths and add the same amount repeatedly. - Compare each position with \(1=\frac{8}{8}\).

1. Read the starting point as \(\frac{1}{8}\). Rewrite the jump as eighths: \(\frac{1}{4}=\frac{2}{8}\). 2. Add \(\frac{2}{8}\) repeatedly: \(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{9}{8},\frac{11}{8}\). 3. Since \(1=\frac{8}{8}\), the first position greater than \(1\) is \(\frac{9}{8}\).

a) \(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{9}{8},\frac{11}{8}\) b) \(\frac{9}{8}\)

- Count the equal intervals from \(0\) to the labeled \(\frac{1}{3}\) tick. - Ask what unit fraction must repeat that many times to make \(\frac{1}{3}\). - Once you know the interval size, count the intervals from \(0\) to P.

1. The labeled \(\frac{1}{3}\) tick is four equal intervals to the right of \(0\). 2. Since \(\frac{1}{3}=\frac{4}{12}\), one interval is \(\frac{1}{12}\). 3. P is nine intervals to the right of \(0\), so P is \(\frac{9}{12}=\frac{3}{4}\).

a) \(\frac{1}{12}\) b) \(\frac{3}{4}\)

- Read the left endpoint of the displayed interval before interpreting the first interior tick. - The equal interval size tells how far each tick is from the previous one, not its coordinate by itself. - Check that both corrected values lie between the displayed endpoints.

1. The displayed interval runs from \(1\) to \(2\) and is divided into three equal parts, so each interval is \(\frac{1}{3}\) unit long. 2. The first interior tick is \(1+\frac{1}{3}=1\frac{1}{3}=\frac{4}{3}\). 3. The second interior tick is \(1+\frac{2}{3}=1\frac{2}{3}=\frac{5}{3}\). 4. Tessa named the distance traveled from \(1\), not the actual number-line coordinate.

Tessa left out the starting value \(1\). The interior ticks are \(\frac{4}{3}\) and \(\frac{5}{3}\).

- Read A through D from the displayed eighths scale to confirm the alternating gaps. - Continue with the other gap size after point D. - Keep the increments in eighths while extending the pattern.

1. Read the marked positions from the eighths scale: \(A=\frac{1}{8}\), \(B=\frac{3}{8}\), \(C=\frac{1}{2}\), and \(D=\frac{3}{4}\). 2. After the \(\frac{1}{4}\) gap from C to D, the next gap is \(\frac{1}{8}\). Thus \(E=\frac{3}{4}+\frac{1}{8}=\frac{7}{8}\). 3. The following gap is \(\frac{1}{4}=\frac{2}{8}\), so \(F=\frac{7}{8}+\frac{2}{8}=\frac{9}{8}=1\frac{1}{8}\).

\(E=\frac{7}{8}\) and \(F=\frac{9}{8}=1\frac{1}{8}\)

- Count the equal spaces covered by the highlighted segment on each line. - Compare the size of one interval on line a) with one interval on line b). - A fixed distance requires more intervals when each interval is smaller.

1. On line a), the highlighted segment runs from the first fourth-sized tick to the third fourth-sized tick, so it covers \(2\) intervals and has length \(\frac{2}{4}=\frac{1}{2}\). 2. On line b), the same endpoints occur at the third and ninth twelfth-sized ticks, so the highlighted segment covers \(6\) intervals and has length \(\frac{6}{12}=\frac{1}{2}\). 3. Twelfth-sized intervals are smaller than fourth-sized intervals, so more of them are needed to cover the same distance.

The distance covers \(2\) intervals on line a) and \(6\) intervals on line b). The smaller twelfth-sized intervals require a larger interval count for the same distance.

- Determine the unit-fraction size of one displayed interval. - Adjacent fractions with consecutive numerators differ by one copy of that unit fraction. - After finding the denominator, move three equal intervals from the second labeled tick.

1. Count the equal intervals from \(0\) to \(1\). There are \(8\), so adjacent ticks are \(\frac{1}{8}\) unit apart. 2. The adjacent labels have numerators \(5\) and \(6\), so they must be \(\frac{5}{8}\) and \(\frac{6}{8}\). The missing denominator is \(8\). 3. Three ticks to the right of \(\frac{6}{8}\) is \(\frac{9}{8}=1\frac{1}{8}\).

a) \(8\) b) \(\frac{9}{8}=1\frac{1}{8}\)

- Read point S from the displayed eighths scale. - Combine Marker A's right and left moves to find its endpoint. - From Marker B's intermediate point, count equal intervals to that same endpoint.

1. Read \(S=\frac{3}{8}\) from the number line. Marker A has a net movement of three eighth-sized intervals right, so it finishes at \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\). 2. Marker B first moves one interval left from \(\frac{3}{8}\) to \(\frac{2}{8}\). 3. To reach \(\frac{6}{8}\), Marker B must move four eighth-sized intervals to the right.

a) \(4\) intervals b) \(\frac{3}{4}\)

- Count the equal intervals on each displayed line to identify the unit fractions. - Look for positions that can be named with both denominators. - The numerator tells the tick number when counting equal intervals from \(0\).

1. Count the equal intervals. Line a) has \(8\) intervals, so its interior ticks are eighths. Line b) has \(12\) intervals, so its interior ticks are twelfths. 2. Compare equivalent positions. The shared interior values are \(\frac{2}{8}=\frac{3}{12}=\frac{1}{4}\), \(\frac{4}{8}=\frac{6}{12}=\frac{1}{2}\), and \(\frac{6}{8}=\frac{9}{12}=\frac{3}{4}\). 3. Therefore, the tick-number pairs are \(2\) and \(3\), \(4\) and \(6\), and \(6\) and \(9\). Checking all interior eighths shows there are no other shared interior ticks.

\(\frac{1}{4}\): tick \(2\) on line a) and tick \(3\) on line b) \(\frac{1}{2}\): tick \(4\) on line a) and tick \(6\) on line b) \(\frac{3}{4}\): tick \(6\) on line a) and tick \(9\) on line b)

- Read all three positions from the displayed number line. - For each possible meeting point, find the distances the other two people would walk. - Compare the three total distances using a common denominator if helpful.

1. Read the positions from the number line: Ava is at \(\frac{1}{4}=\frac{3}{12}\), Ben is at \(\frac{1}{2}=\frac{6}{12}\), and Carlos is at \(\frac{5}{6}=\frac{10}{12}\). 2. If they meet at Ava's position, Ben and Carlos walk \(\frac{3}{12}\) and \(\frac{7}{12}\), for a total of \(\frac{10}{12}=\frac{5}{6}\). 3. If they meet at Ben's position, Ava and Carlos walk \(\frac{3}{12}\) and \(\frac{4}{12}\), for a total of \(\frac{7}{12}\). 4. If they meet at Carlos's position, Ava and Ben walk \(\frac{7}{12}\) and \(\frac{4}{12}\), for a total of \(\frac{11}{12}\). The least total is \(\frac{7}{12}\), at Ben's position.

They should meet at Ben's position, \(\frac{1}{2}\). The other two students walk a total of \(\frac{7}{12}\) of the path.