- The numerators are the same, so compare the size of one part. - Which is a smaller piece of the same whole: an eighth or a fourth?

1. Both fractions count \(3\) equal parts of the same-size whole. 2. An eighth is smaller than a fourth, so \(3\) eighths is less than \(3\) fourths. 3. Therefore, \(\frac{3}{8}<\frac{3}{4}\).

\(\frac{3}{8}<\frac{3}{4}\)

- Read the number of equal parts and shaded parts in each same-size bar. - The numerators match, so focus on the size of one part in each bar. - Dividing the same whole into fewer equal parts makes each part larger.

1. Bar A shows \(\frac{3}{4}\), and bar B shows \(\frac{3}{8}\). 2. Both fractions contain \(3\) shaded parts, but a fourth is larger than an eighth because the same whole is divided into fewer equal parts. 3. Therefore, \(\frac{3}{4}>\frac{3}{8}\).

A: \(\frac{3}{4}\); B: \(\frac{3}{8}\); \(\frac{3}{4}>\frac{3}{8}\).

- Each bar is a whole of the same size. - Compare how large a single shaded piece is in each bar.

1. Bar a) shows \(\frac{1}{3}\). Bar b) shows \(\frac{1}{2}\). 2. One half is a larger piece of the same whole than one third. 3. Thus \(\frac{1}{2}>\frac{1}{3}\).

Bar b): \(\frac{1}{2}\)

- Compare shaded lengths, not just the numbers of pieces. - Each bar represents one whole of the same size. - Think about whether the different cuts change the amount shaded.

1. Bar a) has \(\frac{2}{4}\) shaded. 2. Bar b) has \(\frac{3}{6}\) shaded. 3. Both show half of the same-size whole, so they are equal.

\(\frac{2}{4}=\frac{3}{6}\)

- Picture the halfway point of each same-size bar. - Check whether the shading ends before or after that point.

1. Bar a) shows \(\frac{2}{5}\), less than half. 2. Bar b) shows \(\frac{3}{4}\), greater than half. 3. So bar b) shows more than half.

Bar b), because \(\frac{3}{4}>\frac{1}{2}\).

- The numerators match. - Compare the sizes of one fifth and one eighth of the same whole. - Taking three of the larger pieces gives the larger amount.

1. Both fractions have three pieces. 2. Fifths are larger than eighths when the wholes are the same size. 3. Three fifths is greater than three eighths: \(\frac{3}{5}>\frac{3}{8}\).

\(\frac{3}{5}>\frac{3}{8}\). Both have three parts, but fifths are larger than eighths of the same whole.

- The second fraction already uses sixths. - Rename the half using the same-size sixth pieces. - Compare the numbers of equal pieces.

1. \(\frac{1}{2}=\frac{3}{6}\). 2. Three sixths is greater than two sixths, so \(\frac{1}{2}>\frac{2}{6}\).

\(\frac{1}{2}>\frac{2}{6}\)

- Try simplifying the fraction with sixths. - Equal fractions can have different numerators and denominators.

1. Divide the numerator and denominator of \(\frac{4}{6}\) by \(2\). 2. This gives \(\frac{2}{3}\), so both fractions are equal.

\(\frac{4}{6}=\frac{2}{3}\), because dividing the numerator and denominator of \(\frac{4}{6}\) by \(2\) gives \(\frac{2}{3}\).

- Write each part-to-whole relationship as a fraction. - Simplify both fractions before comparing them. - When fractions have the same numerator, think about what a smaller denominator means for the size of each part.

1. For Group A, the fraction is \(\frac{6}{8}=\frac{3}{4}\). 2. For Group B, the fraction is \(\frac{6}{10}=\frac{3}{5}\). 3. The simplified fractions have the same numerator. With the same numerator, the fraction with the smaller denominator is greater, so \(\frac{3}{4}>\frac{3}{5}\). 4. Therefore, Group A has the greater fraction of students with a pet.

Group A, because \(\frac{6}{8}=\frac{3}{4}\), \(\frac{6}{10}=\frac{3}{5}\), and \(\frac{3}{4}>\frac{3}{5}\).

- Find the least common multiple of the two denominators. - Rename only the fraction that does not already have the common denominator. - Once the denominators match, compare the numerators.

1. The least common multiple of \(6\) and \(12\) is \(12\). 2. Rewrite \(\frac{5}{6}\) as \(\frac{10}{12}\). The other fraction is already \(\frac{7}{12}\). 3. Since \(10>7\), \(\frac{10}{12}>\frac{7}{12}\). Therefore, \(\frac{5}{6}>\frac{7}{12}\).

\(\frac{5}{6}=\frac{10}{12}\), and \(\frac{5}{6}>\frac{7}{12}\).

- Write each number of made shots over the total attempts. - Simplify or rename the fractions so they can be compared using twelfths. - Compare the numerators once the denominators match.

1. Lucas's shooting rate is \(\frac{6}{8}=\frac{3}{4}\). Sarah's shooting rate is \(\frac{8}{12}=\frac{2}{3}\). 2. Rewrite \(\frac{3}{4}\) in twelfths: \(\frac{3}{4}=\frac{9}{12}\). 3. Since \(\frac{9}{12}>\frac{8}{12}\), Lucas has the higher shooting rate.

Lucas has the higher shooting rate because \(\frac{6}{8}=\frac{9}{12}>\frac{8}{12}\).

- Write the shaded portion of each figure as a fraction. - Simplify before comparing. - Rename the three simplified fractions as twelfths.

1. Figure a) shows \(\frac{3}{6}=\frac{1}{2}\). 2. Figure b) shows \(\frac{1}{4}\). 3. Figure c) shows \(\frac{2}{6}=\frac{1}{3}\). 4. Rename the fractions as twelfths: \(\frac{1}{2}=\frac{6}{12}\), \(\frac{1}{4}=\frac{3}{12}\), and \(\frac{1}{3}=\frac{4}{12}\). 5. Therefore, \(\frac{1}{4}<\frac{1}{3}<\frac{1}{2}\).

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

- Both fractions are close to \(1\). - Find the unit-fraction gap between each fraction and \(1\). - The fraction with the smaller gap to \(1\) is greater.

1. \(\frac{5}{6}\) is \(\frac{1}{6}\) less than \(1\). 2. \(\frac{7}{8}\) is \(\frac{1}{8}\) less than \(1\). 3. Since \(\frac{1}{8}<\frac{1}{6}\), \(\frac{7}{8}\) is closer to \(1\). 4. Therefore, \(\frac{7}{8}>\frac{5}{6}\).

\(\frac{7}{8}>\frac{5}{6}\). The gaps to \(1\) are \(\frac{1}{8}\) and \(\frac{1}{6}\), and \(\frac{1}{8}<\frac{1}{6}\).

- The changed fraction is larger than it was before. - Rewrite the fifths as tenths. - Recheck the comparison rather than keeping the original sign.

1. Rewrite \(\frac{4}{5}\) in tenths: \(\frac{4}{5}=\frac{8}{10}\). 2. Compare \(\frac{8}{10}\) with \(\frac{7}{10}\). 3. Since \(8>7\), the new comparison is \(\frac{4}{5}>\frac{7}{10}\).

\(\frac{4}{5}>\frac{7}{10}\)

- The denominator \(12\) is a multiple of \(6\). - Rewrite the fraction in sixths as an equivalent fraction in twelfths. - Compare the numerators once the denominators match.

1. Rewrite \(\frac{5}{6}\) in twelfths: \(\frac{5}{6}=\frac{10}{12}\). 2. Compare \(\frac{10}{12}\) and \(\frac{11}{12}\). 3. Since \(11>10\), \(\frac{11}{12}>\frac{5}{6}\).

\(\frac{5}{6}<\frac{11}{12}\)

- Both denominators can be scaled to \(100\). - Keep each fraction’s value unchanged while renaming it. - Compare the hundredths after converting.

1. Rewrite \(\frac{7}{10}\) as hundredths: \(\frac{7}{10}=\frac{70}{100}=0.70\). 2. Rewrite \(\frac{3}{4}\) as hundredths: \(\frac{3}{4}=\frac{75}{100}=0.75\). 3. Since \(70<75\), \(0.70<0.75\). 4. Therefore, \(\frac{7}{10}<\frac{3}{4}\).

\(\frac{7}{10}=0.70<0.75=\frac{3}{4}\)

- Read each shaded fraction from its same-size bar. - Rename the benchmark using tenths so all three amounts use equal-sized parts. - Compare the counts of tenths on either side of the benchmark.

1. Bar H shows \(\frac{1}{2}\). Bar B shows \(\frac{4}{10}\), and bar C shows \(\frac{6}{10}\). 2. Rename the benchmark as \(\frac{5}{10}\). 3. Since \(\frac{4}{10}<\frac{5}{10}<\frac{6}{10}\), B is below the benchmark and C is above it.

a) H: \(\frac{1}{2}\); B: \(\frac{4}{10}\); C: \(\frac{6}{10}\) b) B is less than \(\frac{1}{2}\); C is greater than \(\frac{1}{2}\).

- The two bars have different numbers of equal pieces. - How could you rewrite the fractions using pieces of the same size? - Explain why comparing numerators alone fails here.

1. Bar a) shows \(\frac{5}{8}\); bar b) shows \(\frac{3}{4}\). 2. Rename \(\frac{3}{4}\) as \(\frac{6}{8}\). 3. Since \(\frac{5}{8}<\frac{6}{8}\), bar b) has more shading. Maya compared numerators without checking piece size.

No. \(\frac{5}{8}<\frac{3}{4}=\frac{6}{8}\). Bar b) shows more.

- Look for a common denominator for all three fractions. - Rewrite each fraction with that denominator. - Compare the numerators once the denominators match.

1. A common denominator for \(3\), \(4\), and \(6\) is \(12\). 2. Rewrite each fraction: \(\frac{2}{3}=\frac{8}{12}\), \(\frac{3}{4}=\frac{9}{12}\), and \(\frac{5}{6}=\frac{10}{12}\). 3. Since \(8<9<10\), the order is Class A, Class B, Class C. Class C filled the greatest fraction.

Class C filled the greatest fraction. Least to greatest: Class A \(\left(\frac{2}{3}\right)\), Class B \(\left(\frac{3}{4}\right)\), Class C \(\left(\frac{5}{6}\right)\).

- Find the numerator that would make each fraction exactly \(\frac{1}{2}\). - Decide whether each given numerator is above or below its halfway numerator. - Use those benchmark positions to compare the two fractions.

1. Half of \(8\) is \(4\). Since \(3<4\), \(\frac{3}{8}<\frac{1}{2}\). 2. Half of \(12\) is \(6\). Since \(7>6\), \(\frac{7}{12}>\frac{1}{2}\). 3. One fraction is below \(\frac{1}{2}\) and the other is above it, so \(\frac{7}{12}>\frac{3}{8}\).

\(\frac{7}{12}>\frac{3}{8}\)

- Look for a denominator that \(2\), \(3\), and \(4\) all divide evenly. - Rename each fraction with that denominator. - Compare the numerators after the denominators match.

1. A common denominator for \(2\), \(3\), and \(4\) is \(12\). 2. Rewrite the fractions: \(\frac{1}{2}=\frac{6}{12}\), \(\frac{2}{3}=\frac{8}{12}\), and \(\frac{3}{4}=\frac{9}{12}\). 3. Since \(6<8<9\), the order is \(\frac{1}{2}<\frac{2}{3}<\frac{3}{4}\).

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

- Decide whether \(\frac{1}{2}\) or \(1\) separates either pair. - A fraction with a larger numerator than denominator is greater than \(1\). - For unit fractions, compare the sizes of the equal pieces.

1. For a), \(\frac{2}{5}<\frac{1}{2}\), while \(\frac{5}{8}>\frac{1}{2}\). Therefore, \(\frac{5}{8}\) is greater. 2. For b), \(\frac{11}{10}>1\), while \(\frac{5}{6}<1\). Therefore, \(\frac{11}{10}\) is greater. 3. For c), both fractions have numerator \(1\). A smaller denominator means a larger piece of the same whole, so \(\frac{1}{6}>\frac{1}{8}\).

a) \(\frac{5}{8}\) b) \(\frac{11}{10}\) c) \(\frac{1}{6}\)

- 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.

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

- With equal numerators, compare the sizes of the unit fractions. - For the greatest amount, compare the strongest candidates using tenths. - For the least amount, first compare the two amounts already written in eighths, then use \(\frac{1}{2}\) as a benchmark for the others.

1. Anna and Ben have fractions with equal numerators. Since fifths are larger parts than eighths, \(\frac{3}{5}>\frac{3}{8}\), so Ben has more than Anna. 2. David has \(\frac{7}{10}\,\text{L}\). Rewrite Ben's amount in tenths: \(\frac{3}{5}=\frac{6}{10}\). Since \(\frac{7}{10}>\frac{6}{10}\), David has more than Ben and therefore has the most. 3. Clara has \(\frac{2}{8}\,\text{L}\), which is less than Anna's \(\frac{3}{8}\,\text{L}\). Ben and David each have more than \(\frac{1}{2}\,\text{L}\), so Clara has the least.

a) Ben b) David, with \(\frac{7}{10}\,\text{L}\) c) Clara, with \(\frac{2}{8}\,\text{L}\)

- Subtract each amount eaten from \(1\) to find the amount left. - Compare the three unit fractions. - For unit fractions, think about how the denominator affects the size of one piece.

1. Anthony leaves \(1-\frac{5}{6}=\frac{1}{6}\). 2. Brianna leaves \(1-\frac{7}{8}=\frac{1}{8}\). 3. Carlos leaves \(1-\frac{11}{12}=\frac{1}{12}\). 4. For unit fractions, a larger denominator means a smaller fraction. Since \(\frac{1}{12}<\frac{1}{8}<\frac{1}{6}\), Carlos leaves the smallest piece.

Carlos leaves the smallest piece: \(\frac{1}{12}\) of a pizza.

- 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 shaded parts and total equal parts in each figure. - Simplify all three fractions. - Rewrite the two different simplified fractions with a common denominator.

1. Figure a) shows \(\frac{4}{6}=\frac{2}{3}\). 2. Figure b) shows \(\frac{6}{8}=\frac{3}{4}\). 3. Figure c) shows \(\frac{8}{12}=\frac{2}{3}\). 4. Figures a) and c) are equivalent and represent \(\frac{2}{3}\). 5. Compare \(\frac{2}{3}=\frac{8}{12}\) with \(\frac{3}{4}=\frac{9}{12}\). Since \(\frac{9}{12}>\frac{8}{12}\), figure b) shows the greatest fraction.

1) Figures a) and c) 2) \(\frac{2}{3}\) 3) Figure b), which shows \(\frac{3}{4}\)

- Read each remaining fraction from the same-size bars. - Find a common denominator for the two fractions. - Compare the numerators after the denominators match.

1. Bar 1 shows \(2\) of \(3\) equal pieces remaining, so its fraction is \(\frac{2}{3}\). 2. Bar 2 shows \(3\) of \(4\) equal pieces remaining, so its fraction is \(\frac{3}{4}\). 3. Rewrite both fractions with denominator \(12\): \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). 4. Since \(\frac{9}{12}>\frac{8}{12}\), Bar 2 has the greater fraction remaining.

a) Bar 1: \(\frac{2}{3}\); Bar 2: \(\frac{3}{4}\) b) Bar 2, because \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{3}{4}=\frac{9}{12}\).

- Read each fraction from the circle models. - Look for the least denominator that both \(6\) and \(4\) divide evenly. - Compare the numerators after renaming both fractions.

1. Circle 1 has \(5\) of \(6\) equal parts shaded, so its fraction is \(\frac{5}{6}\). 2. Circle 2 has \(3\) of \(4\) equal parts shaded, so its fraction is \(\frac{3}{4}\). 3. The least common denominator is \(12\). Rewrite the fractions: \(\frac{5}{6}=\frac{10}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). 4. Since \(\frac{10}{12}>\frac{9}{12}\), Circle 1 has the greater shaded fraction.

a) Circle 1: \(\frac{5}{6}\); Circle 2: \(\frac{3}{4}\) b) Circle 1, because \(\frac{5}{6}=\frac{10}{12}>\frac{9}{12}=\frac{3}{4}\).

- Count the equal parts of each separate whole. - Find a way to name fifths in tenths. - Compare how many tenths are shaded.

1. Figure 1 has \(3\) of \(5\) parts shaded: \(\frac{3}{5}\). 2. Figure 2 has \(7\) of \(10\) parts shaded: \(\frac{7}{10}\). 3. Rename fifths as tenths: \(\frac{3}{5}=\frac{6}{10}\). Since \(\frac{6}{10}<\frac{7}{10}\), Figure 2 has the greater shaded fraction.

a) Figure 1: \(\frac{3}{5}\); Figure 2: \(\frac{7}{10}\) b) Figure 2, because \(\frac{3}{5}=\frac{6}{10}<\frac{7}{10}\).

- Count the shaded parts and total equal parts in each circle. - Simplify both fractions. - Rewrite the fractions with a common denominator before comparing.

1. Circle 1 has \(4\) of \(6\) equal parts shaded, so \(\frac{4}{6}=\frac{2}{3}\). 2. Circle 2 has \(6\) of \(8\) equal parts shaded, so \(\frac{6}{8}=\frac{3}{4}\). 3. Rewrite the fractions with denominator \(12\): \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). 4. Since \(\frac{8}{12}<\frac{9}{12}\), \(\frac{2}{3}<\frac{3}{4}\).

a) Circle 1: \(\frac{2}{3}\); Circle 2: \(\frac{3}{4}\) b) \(\frac{2}{3}<\frac{3}{4}\)

- Read and simplify each shaded fraction first. - Choose a common denominator for the four fractions. - Compare the numerators after renaming the fractions.

1. Figure a) shows \(\frac{4}{6}=\frac{2}{3}\). 2. Figure b) shows \(\frac{3}{4}\). 3. Figure c) shows \(\frac{1}{3}\). 4. Figure d) shows \(\frac{2}{4}=\frac{1}{2}\). 5. Rename the fractions as twelfths: \(\frac{2}{3}=\frac{8}{12}\), \(\frac{3}{4}=\frac{9}{12}\), \(\frac{1}{3}=\frac{4}{12}\), and \(\frac{1}{2}=\frac{6}{12}\). 6. Therefore, \(\frac{1}{3}<\frac{1}{2}<\frac{2}{3}<\frac{3}{4}\).

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

- Read the shaded portion of each figure and simplify if needed. - Look for a common denominator. - Compare the numerators after renaming both fractions.

1. Figure 1 has \(3\) of \(4\) equal parts shaded, so its fraction is \(\frac{3}{4}\). 2. Figure 2 has \(4\) of \(6\) squares shaded, so \(\frac{4}{6}=\frac{2}{3}\). 3. Rewrite both fractions with denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{2}{3}=\frac{8}{12}\). 4. Since \(\frac{9}{12}>\frac{8}{12}\), Figure 1 has the greater shaded fraction.

a) Figure 1: \(\frac{3}{4}\); Figure 2: \(\frac{2}{3}\) b) Figure 1, because \(\frac{3}{4}=\frac{9}{12}>\frac{8}{12}=\frac{2}{3}\).

- Read each remaining fraction from the three same-size cake models. - Simplify each fraction first. - Rename the simplified fractions as twelfths before ordering them.

1. Ava has \(9\) of \(12\) parts left, so \(\frac{9}{12}=\frac{3}{4}\). 2. Ben has \(2\) of \(4\) parts left, so \(\frac{2}{4}=\frac{1}{2}\). 3. Charlotte has \(4\) of \(6\) parts left, so \(\frac{4}{6}=\frac{2}{3}\). 4. Rename the fractions as twelfths: \(\frac{3}{4}=\frac{9}{12}\), \(\frac{1}{2}=\frac{6}{12}\), and \(\frac{2}{3}=\frac{8}{12}\). 5. Therefore, \(\frac{1}{2}<\frac{2}{3}<\frac{3}{4}\), and Ava has the most cake left.

a) Ava: \(\frac{3}{4}\); Ben: \(\frac{1}{2}\); Charlotte: \(\frac{2}{3}\) b) \(\frac{1}{2}<\frac{2}{3}<\frac{3}{4}\). Ava has the most left.

- Read each planted fraction from the diagram. - Find one denominator that \(4\), \(3\), and \(6\) all divide evenly. - Compare the numerators after renaming the fractions.

1. Bed a) shows \(\frac{3}{4}\), bed b) shows \(\frac{2}{3}\), and bed c) shows \(\frac{5}{6}\). 2. Rewrite all three fractions with denominator \(12\): \(\frac{3}{4}=\frac{9}{12}\), \(\frac{2}{3}=\frac{8}{12}\), and \(\frac{5}{6}=\frac{10}{12}\). 3. Since \(10>9>8\), \(\frac{5}{6}>\frac{3}{4}>\frac{2}{3}\). 4. Bed c) has the greatest planted area.

Bed a): \(\frac{3}{4}\); bed b): \(\frac{2}{3}\); bed c): \(\frac{5}{6}\). Bed c) has the greatest planted area.

- Write the shaded fraction for each bar. - Simplify before comparing. - Rewrite the fractions with a common denominator.

1. Figure a) shows \(\frac{3}{5}\) remaining. 2. Figure b) shows \(\frac{2}{4}=\frac{1}{2}\) remaining. 3. Rewrite the fractions with denominator \(10\): \(\frac{3}{5}=\frac{6}{10}\) and \(\frac{1}{2}=\frac{5}{10}\). 4. Since \(\frac{6}{10}>\frac{5}{10}\), figure a) has the greater fraction remaining.

Figure a), because \(\frac{3}{5}>\frac{1}{2}\).

- Read the shaded fraction in each pictured figure before comparing the figures. - Simplify any fraction that is not already in simplest form. - A common denominator can help you compare all four fractions.

1. Figure K shows \(\frac{2}{4}=\frac{1}{2}\). 2. Figure M shows \(\frac{1}{6}\). 3. Figure R shows \(\frac{2}{3}\). 4. Figure T shows \(\frac{1}{3}\). 5. Using denominator \(6\), the fractions are \(\frac{1}{6}\), \(\frac{2}{6}\), \(\frac{3}{6}\), and \(\frac{4}{6}\). 6. Therefore, the labels from least to greatest are M, T, K, R.

M, T, K, R \(\frac{1}{6}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\)

- For each field, count the shaded dots and the total dots shown. - Compare the two fractions separately from comparing the two shaded counts. - Ask whether the two fractions refer to wholes of the same size.

1. Field A has \(30\) shaded dots out of \(40\), so its shaded fraction is \(\frac{30}{40}=\frac{3}{4}\). 2. Field B has \(40\) shaded dots out of \(60\), so its shaded fraction is \(\frac{40}{60}=\frac{2}{3}\). 3. Compare the fractions: \(\frac{3}{4}=\frac{9}{12}\) and \(\frac{2}{3}=\frac{8}{12}\), so \(\frac{3}{4}>\frac{2}{3}\). 4. Compare the actual shaded counts: Field B has \(40\) shaded dots and Field A has \(30\), so Field B has \(10\) more shaded dots. 5. The fractions describe portions of different-sized wholes. A greater fraction of a smaller whole can still represent fewer objects than a smaller fraction of a larger whole.

a) A: \(\frac{3}{4}\); B: \(\frac{2}{3}\) b) Field A has the greater shaded fraction. c) Field B has \(10\) more shaded dots. d) The fields have different total numbers of dots, so comparing the fractions alone does not compare the actual shaded counts.

- Fractions with different denominators do not have equal-sized parts. - Compare each fraction with the benchmark \(\frac{1}{2}\). - Then use the benchmark positions to compare the two fractions.

1. Rewrite \(\frac{4}{8}\) as \(\frac{1}{2}\). 2. Since \(\frac{5}{12}<\frac{6}{12}=\frac{1}{2}\), \(\frac{5}{12}<\frac{4}{8}\). 3. Comparing only the numerators is not valid when the denominators are different.

Maya is not correct. \(\frac{5}{12}<\frac{4}{8}\).

- Compare each fraction with the benchmark \(\frac{1}{2}\). - Express the benchmark using each fraction’s denominator. - The smaller missing part identifies the closer fraction.

1. Rewrite \(\frac{1}{2}\) in twelfths: \(\frac{1}{2}=\frac{6}{12}\). The distance from \(\frac{5}{12}\) to \(\frac{1}{2}\) is \(\frac{1}{12}\). 2. Rewrite \(\frac{1}{2}\) in eighths: \(\frac{1}{2}=\frac{4}{8}\). The distance from \(\frac{3}{8}\) to \(\frac{1}{2}\) is \(\frac{1}{8}\). 3. Since \(\frac{1}{12}<\frac{1}{8}\), \(\frac{5}{12}\) is closer to \(\frac{1}{2}\).

\(\frac{5}{12}\) is closer to \(\frac{1}{2}\).

- The ribbons began with equal whole lengths, so their fractions can be compared directly. - Rewrite the fraction in fifths as an equivalent fraction in tenths. - Subtract the smaller number of tenths from the larger number of tenths.

1. Rewrite \(\frac{3}{5}\) in tenths: \(\frac{3}{5}=\frac{6}{10}\). 2. Since \(7>6\), the blue ribbon has the greater fraction left. 3. The difference is \(\frac{7}{10}-\frac{6}{10}=\frac{1}{10}\).

The blue ribbon has the greater fraction left by \(\frac{1}{10}\) of the original length.

- Rewrite both boundary fractions with denominator \(12\). - The inequality is strict, so the unknown fraction cannot equal either boundary. - Look for a whole number between the two resulting numerators.

1. Rewrite the bounds in twelfths: \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{5}{6}=\frac{10}{12}\). 2. The only whole-number numerator strictly between \(8\) and \(10\) is \(9\). 3. Therefore, \(\frac{n}{12}=\frac{9}{12}\), so \(n=9\).

\(n=9\)

- Separate each improper fraction into a whole and a fractional part. - The whole-number parts match, so only the leftover parts need comparison. - Compare both leftover parts with the benchmark \(\frac{1}{5}\).

1. \(\frac{13}{10}=1\frac{3}{10}\). 2. \(\frac{9}{8}=1\frac{1}{8}\). 3. The whole-number parts are equal. Compare the leftover parts with \(\frac{1}{5}\): \(\frac{3}{10}>\frac{2}{10}=\frac{1}{5}\), and \(\frac{1}{5}>\frac{1}{8}\). 4. Since \(\frac{3}{10}>\frac{1}{8}\), \(\frac{13}{10}>\frac{9}{8}\).

\(\frac{13}{10}>\frac{9}{8}\)

- Test each candidate against both boundaries. - Rewrite thirds and halves in twelfths when checking choice B. - Eliminate any fraction that fails either comparison.

1. Choice A is below the lower bound because \(\frac{1}{4}<\frac{1}{3}\). 2. Rewrite the bounds in twelfths: \(\frac{1}{3}=\frac{4}{12}<\frac{5}{12}<\frac{6}{12}=\frac{1}{2}\). 3. Choices C and D are both greater than \(\frac{1}{2}\). 4. Therefore, only \(\frac{5}{12}\) lies strictly between the bounds.

B. \(\frac{5}{12}\)

- All three fractions have the same numerator. - Think about what happens to a fraction with numerator \(5\) when its denominator gets larger. - Test the listed candidates against both comparison signs rather than narrowing the interval first.

1. The three fractions have the same numerator. With equal numerators, a larger denominator makes a smaller fraction. 2. For \(\frac{5}{d}\) to be less than \(\frac{5}{8}\), \(d\) must be greater than \(8\). 3. For \(\frac{5}{d}\) to be greater than \(\frac{5}{12}\), \(d\) must be less than \(12\). 4. Of \(6\), \(8\), \(10\), and \(12\), only \(10\) satisfies both comparisons. Therefore, \(d=10\).

\(d=10\)

- A denominator describes part size, but the numerator describes how many parts are used. - Rewrite \(\frac{6}{8}\) in fourths and compare how far each fraction is below \(1\). - The fraction with the smaller missing part is greater.

1. Smaller eighth-sized parts do not automatically make the fraction greater; the fractions contain different numbers of parts. 2. Rewrite \(\frac{6}{8}\) as \(\frac{3}{4}\). The fraction \(\frac{5}{6}\) is \(\frac{1}{6}\) below \(1\), while \(\frac{3}{4}\) is \(\frac{1}{4}\) below \(1\). 3. Since \(\frac{1}{6}<\frac{1}{4}\), \(\frac{5}{6}\) is closer to \(1\), so \(\frac{5}{6}>\frac{6}{8}\).

Jules compared part sizes without accounting for the number of parts. The correct comparison is \(\frac{5}{6}>\frac{6}{8}\).

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

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

- Separate what the benchmark tells you from what it does not tell you. - Two numbers can both be greater than the same number without being equal. - Compare how far each fraction is above \(\frac{1}{2}\).

1. It is valid that both fractions are greater than \(\frac{1}{2}\). 2. Being on the same side of a benchmark does not prove that two fractions are equal. 3. Write each fraction as an amount above \(\frac{1}{2}\): \(\frac{5}{8}=\frac{1}{2}+\frac{1}{8}\) and \(\frac{7}{12}=\frac{1}{2}+\frac{1}{12}\). 4. Since \(\frac{1}{8}>\frac{1}{12}\), \(\frac{5}{8}>\frac{7}{12}\).

Mia is correct that both fractions are greater than \(\frac{1}{2}\), but that does not make them equal. \(\frac{5}{8}>\frac{7}{12}\).

- A counterexample must satisfy the stated condition but make the conclusion false. - Check the numerator and denominator condition in every pair. - Compare each pair’s actual values, using \(\frac{1}{2}\) as a benchmark when helpful.

1. In every choice, the first fraction has both the greater numerator and the greater denominator. 2. Choices a), b), and d) also have the greater first fraction, so they do not disprove the claim. 3. For choice c), write each fraction as an amount above \(\frac{1}{2}\): \(\frac{7}{12}=\frac{1}{2}+\frac{1}{12}\) and \(\frac{5}{8}=\frac{1}{2}+\frac{1}{8}\). 4. Since \(\frac{1}{12}<\frac{1}{8}\), \(\frac{7}{12}<\frac{5}{8}\). Choice c) is a counterexample.

c) \(\frac{7}{12}<\frac{5}{8}\), even though \(7>5\) and \(12>8\).

- Read the two comparison statements as one ordered chain. - Use the middle fraction to connect the two outside fractions. - Verify the conclusion by rewriting thirds as sixths.

1. The chain places \(\frac{3}{4}\) between the other two fractions. 2. Therefore, \(\frac{2}{3}<\frac{5}{6}\) by the order shown in the chain. 3. Verify using sixths: \(\frac{2}{3}=\frac{4}{6}\). 4. Since \(\frac{4}{6}<\frac{5}{6}\), the comparison is confirmed.

\(\frac{2}{3}<\frac{5}{6}\) because \(\frac{2}{3}=\frac{4}{6}<\frac{5}{6}\).

- List the fractions with denominator \(8\) in increasing order. - Use a familiar benchmark such as \(\frac{1}{2}\) to compare an eighth near \(\frac{2}{3}\). - Once you locate \(\frac{2}{3}\) between two consecutive eighths, the remaining numerator values can be sorted by order.

1. Compare nearby eighths with \(\frac{2}{3}\). Since \(\frac{5}{8}=\frac{1}{2}+\frac{1}{8}\) and \(\frac{2}{3}=\frac{1}{2}+\frac{1}{6}\), \(\frac{5}{8}<\frac{2}{3}\). 2. Also, \(\frac{6}{8}=\frac{3}{4}>\frac{2}{3}\). 3. Therefore, \(\frac{2}{3}\) lies strictly between \(\frac{5}{8}\) and \(\frac{6}{8}\), so no value of \(n\) makes the fractions equal. 4. Thus, \(n=0,1,2,3,4,5\) give a smaller fraction, and \(n=6,7,8\) give a greater fraction.

Less than: \(n=0,1,2,3,4,5\) Equal: none Greater than: \(n=6,7,8\)

- Look for the shared benchmark \(\frac{1}{2}\). - Rewrite each fraction as that benchmark plus one leftover unit fraction. - Compare only the two leftover unit fractions.

1. Decompose the fractions: \(\frac{2}{3}=\frac{1}{2}+\frac{1}{6}\) and \(\frac{7}{12}=\frac{1}{2}+\frac{1}{12}\). 2. Both fractions contain \(\frac{1}{2}\). Since \(\frac{1}{6}>\frac{1}{12}\), \(\frac{2}{3}>\frac{7}{12}\).

\(\frac{2}{3}>\frac{7}{12}\) because \(\frac{1}{6}>\frac{1}{12}\) after the shared \(\frac{1}{2}\) is removed.

- There are four allowed changes: raise or lower either numerator by \(1\). - Keep both denominators fixed while testing. - Use \(\frac{1}{2}\) as a benchmark to check the promising repairs. - Explain why the other two changes move the fractions in the wrong directions.

1. Decreasing the left numerator gives \(\frac{4}{8}<\frac{7}{12}\). Since \(\frac{4}{8}=\frac{1}{2}=\frac{6}{12}<\frac{7}{12}\), this repair works. 2. Increasing the right numerator gives \(\frac{5}{8}<\frac{8}{12}\). The fractions are \(\frac{1}{2}+\frac{1}{8}\) and \(\frac{1}{2}+\frac{1}{6}\), and \(\frac{1}{8}<\frac{1}{6}\), so this repair works. 3. Increasing the left numerator makes the left fraction larger, and decreasing the right numerator makes the right fraction smaller, so those two changes cannot repair the comparison. 4. All four allowed numerator changes have been considered.

\(\frac{4}{8}<\frac{7}{12}\) \(\frac{5}{8}<\frac{8}{12}\)

- Express both benchmarks using twelfths. - Notice that the benchmark numerators are consecutive whole numbers. - Determine on which side of the boundary each possible numerator lies.

1. Rewrite the benchmarks in twelfths: \(\frac{2}{3}=\frac{8}{12}\) and \(\frac{3}{4}=\frac{9}{12}\). 2. The benchmark points are neighboring twelfths. Any fraction \(\frac{n}{12}\) with \(n\le8\) is closer to \(\frac{8}{12}\) than to \(\frac{9}{12}\). 3. Any fraction \(\frac{n}{12}\) with \(n\ge9\) is closer to \(\frac{9}{12}\) than to \(\frac{8}{12}\). 4. Therefore, the required values are \(n=9,10,11,12\).

\(n=9,10,11,12\)

- Check each line in order and stop at the first line that changes a fraction’s value. - Verify each fraction separately before comparing them. - Once both fractions are in eighths, compare the numerators.

1. The first conversion is correct because \(\frac{3}{4}=\frac{6}{8}\). 2. The second line is the first incorrect line. The fraction \(\frac{5}{8}\) is already written in eighths, so it remains \(\frac{5}{8}\), not \(\frac{7}{8}\). 3. Since \(6>5\), \(\frac{3}{4}>\frac{5}{8}\).

The first incorrect line is \(\frac{5}{8}=\frac{7}{8}\). It should remain \(\frac{5}{8}\), so \(\frac{3}{4}>\frac{5}{8}\).

- What does comparing each fraction with one half actually tell you? - If both are on the same side of that benchmark, what else could you compare? - Try equal-sized pieces, or think about the part missing from one whole.

1. Both fractions are greater than \(\frac{1}{2}\). That fact alone does not tell which fraction is greater. 2. Rename them in twenty-fourths: \(\frac{5}{6}=\frac{20}{24}\) and \(\frac{7}{8}=\frac{21}{24}\). 3. Since \(21>20\), \(\frac{7}{8}>\frac{5}{6}\). Another way is to notice that they are \(\frac{1}{8}\) and \(\frac{1}{6}\) short of one whole, respectively.

No. Both fractions are greater than \(\frac{1}{2}\), but that does not settle their order. \(\frac{7}{8}>\frac{5}{6}\), since \(\frac{21}{24}>\frac{20}{24}\).

- Find a common denominator for the two bounds and for tenths. - Think about which numerators in that common denominator come from tenths. - Check that both inequalities are strict.

1. Rewrite the bounds in twentieths: \(\frac{1}{4}=\frac{5}{20}\) and \(\frac{2}{5}=\frac{8}{20}\). 2. A fraction in tenths can be renamed in twentieths by doubling its numerator and denominator. 3. Only the even numerator \(6\) lies strictly between \(5\) and \(8\). So \(\frac{6}{20}=\frac{3}{10}\) is the only answer.

\(\frac{3}{10}\). It is the only fraction in tenths strictly between \(\frac{5}{20}\) and \(\frac{8}{20}\).

- Check whether the two stated ranges overlap. - Try choosing one pair that makes \(M\) greater. - Then choose another valid pair that reverses the order.

1. The bounds overlap because some fractions are greater than \(\frac{3}{4}\) but still less than \(\frac{4}{5}\). 2. Example with \(M>N\): let \(M=\frac{7}{8}\) and \(N=\frac{3}{4}\). Then \(M>\frac{3}{4}\), \(N<\frac{4}{5}\), and \(M>N\). 3. Example with \(M<N\): let \(M=\frac{77}{100}\) and \(N=\frac{79}{100}\). Then \(M>\frac{75}{100}=\frac{3}{4}\), \(N<\frac{80}{100}=\frac{4}{5}\), and \(M<N\). 4. Since both orders are possible, the information is not enough.

No. For example, \(M=\frac{7}{8}\) and \(N=\frac{3}{4}\) give \(M>N\), while \(M=\frac{77}{100}\) and \(N=\frac{79}{100}\) give \(M<N\). Both pairs satisfy the stated bounds.

- List every possible ordered numerator pair with \(a<c\). - Eliminate pairs that cannot leave denominators larger than both numerators. - For each remaining pair, test every proper assignment of the two denominator cards. - Keep a checklist so you can justify that every possible arrangement was considered.

1. The possible numerator pairs with \(a<c\) are \((4,5)\), \((4,6)\), \((4,8)\), \((5,6)\), \((5,8)\), and \((6,8)\). 2. The proper-fraction condition eliminates every pair with numerator \(8\), and it eliminates \((5,6)\) because the remaining denominators \(4\) and \(8\) cannot make both fractions proper. 3. For numerators \((4,5)\), the two proper arrangements are \(\frac{4}{6}\) and \(\frac{5}{8}\), or \(\frac{4}{8}\) and \(\frac{5}{6}\). The first works because \(\frac{4}{6}=\frac{2}{3}=\frac{1}{2}+\frac{1}{6}\), while \(\frac{5}{8}=\frac{1}{2}+\frac{1}{8}\), and \(\frac{1}{6}>\frac{1}{8}\). The second does not work because \(\frac{4}{8}=\frac{1}{2}<\frac{5}{6}\). 4. For numerators \((4,6)\), the only proper arrangement is \(\frac{4}{5}\) and \(\frac{6}{8}\). Since \(\frac{4}{5}\) is \(\frac{1}{5}\) below \(1\) and \(\frac{6}{8}=\frac{3}{4}\) is \(\frac{1}{4}\) below \(1\), \(\frac{4}{5}>\frac{6}{8}\). 5. Every possible numerator pair and proper denominator assignment has been checked, so these are the only two comparisons.

\(\frac{4}{6}>\frac{5}{8}\) \(\frac{4}{5}>\frac{6}{8}\)