Free Math Worksheets

- Write each part-to-whole relationship as a fraction. - Simplify both fractions. - When two fractions have the same numerator, compare their denominators.

1. For Class A, the fraction is \(\frac{12}{16}=\frac{3}{4}\). 2. For Class B, the fraction is \(\frac{15}{25}=\frac{3}{5}\). 3. The 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, Class A has the greater fraction of students with a pet.

Class A, because \(\frac{12}{16}=\frac{3}{4}\), \(\frac{15}{25}=\frac{3}{5}\), and \(\frac{3}{4}>\frac{3}{5}\).

- Find the least common multiple of the denominators. - Multiply each numerator and denominator by the factor needed to reach the common denominator. - Once the denominators match, compare the numerators.

1. The least common multiple of \(8\) and \(12\) is \(24\). 2. Rewrite each fraction: \(\frac{5}{8}=\frac{15}{24}\) and \(\frac{7}{12}=\frac{14}{24}\). 3. Since \(15>14\), \(\frac{15}{24}>\frac{14}{24}\). Therefore, \(\frac{5}{8}>\frac{7}{12}\).

\(\frac{5}{8}=\frac{15}{24}\), \(\frac{7}{12}=\frac{14}{24}\), and \(\frac{5}{8}>\frac{7}{12}\).

- Write each number of made shots over the total attempts. - Rewrite the fractions with a common denominator. - Find a common multiple of \(20\) and \(25\).

1. Lucas’s shooting rate is \(\frac{12}{20}=\frac{3}{5}\). Sarah’s shooting rate is \(\frac{18}{25}\). 2. Rewrite \(\frac{3}{5}\) with denominator \(25\): \(\frac{3}{5}=\frac{15}{25}\). 3. Since \(\frac{18}{25}>\frac{15}{25}\), Sarah has the higher shooting rate.

Sarah has the higher shooting rate because \(\frac{18}{25}>\frac{12}{20}=\frac{15}{25}\).

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

1. In figure a), \(9\) of \(15\) squares are shaded, so \(\frac{9}{15}=\frac{3}{5}\). 2. In figure b), \(7\) of \(10\) sectors are shaded, so the fraction is \(\frac{7}{10}\). 3. Rewrite \(\frac{3}{5}\) as \(\frac{6}{10}\). Since \(\frac{6}{10}<\frac{7}{10}\), figure b) has the greater shaded fraction.

a) Figure a): \(\frac{3}{5}\); Figure b): \(\frac{7}{10}\) b) Figure b)

- 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 a) has \(4\) of \(6\) equal parts shaded, so \(\frac{4}{6}=\frac{2}{3}\). 2. Circle b) 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 a): \(\frac{2}{3}\); Circle b): \(\frac{3}{4}\) b) \(<\)

- Write the shaded portion of each figure as a fraction. - Simplify before comparing. - Rewrite all three fractions with denominator \(24\).

1. Figure a) shows \(\frac{3}{8}\). 2. Figure b) shows \(\frac{5}{12}\). 3. Figure c) shows \(\frac{2}{6}=\frac{1}{3}\). 4. Rewrite the fractions with denominator \(24\): \(\frac{3}{8}=\frac{9}{24}\), \(\frac{5}{12}=\frac{10}{24}\), and \(\frac{2}{6}=\frac{8}{24}\). 5. Since \(8<9<10\), \(\frac{1}{3}<\frac{3}{8}<\frac{5}{12}\).

\(\frac{1}{3}<\frac{3}{8}<\frac{5}{12}\)

- Find a number divisible by all three denominators. - Rewrite the fractions with the same denominator. - Compare the numerators after the denominators match.

1. A common denominator for \(3\), \(5\), and \(10\) is \(30\). 2. Rewrite each fraction: \(\frac{2}{3}=\frac{20}{30}\), \(\frac{3}{5}=\frac{18}{30}\), and \(\frac{7}{10}=\frac{21}{30}\). 3. Since \(18<20<21\), the order is Class B, Class A, Class C. Class C filled the greatest fraction.

Class C filled the greatest fraction. Least to greatest: Class B \(\left(\frac{3}{5}\right)\), Class A \(\left(\frac{2}{3}\right)\), Class C \(\left(\frac{7}{10}\right)\).

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

1. Half of \(32\) is \(16\). Since \(15<16\), \(\frac{15}{32}<\frac{1}{2}\). 2. Half of \(30\) is \(15\). Since \(17>15\), \(\frac{17}{30}>\frac{1}{2}\). 3. One fraction is below \(\frac{1}{2}\) and the other is above it, so \(\frac{17}{30}>\frac{15}{32}\).

\(\frac{17}{30}>\frac{15}{32}\)

- Find a common multiple of \(3\), \(5\), and \(10\). - Rewrite each fraction with that denominator. - Compare the numerators after the denominators match.

1. The least common multiple of \(3\), \(5\), and \(10\) is \(30\). 2. Rewrite the fractions: \(\frac{2}{3}=\frac{20}{30}\), \(\frac{3}{5}=\frac{18}{30}\), and \(\frac{7}{10}=\frac{21}{30}\). 3. Since \(18<20<21\), the order is \(\frac{3}{5}<\frac{2}{3}<\frac{7}{10}\).

\(\frac{3}{5}<\frac{2}{3}<\frac{7}{10}\)

- Compare each fraction with \(\frac{1}{2}\) or \(1\) when helpful. - A fraction with numerator greater than its denominator is greater than \(1\). - For unit fractions, the smaller denominator represents the larger piece.

1. For a), \(\frac{4}{9}<\frac{1}{2}\), while \(\frac{5}{8}>\frac{1}{2}\). Therefore, \(\frac{5}{8}\) is greater. 2. For b), \(\frac{11}{10}>1\), while \(\frac{12}{13}<1\). Therefore, \(\frac{11}{10}\) is greater. 3. For c), both fractions have numerator \(1\). A smaller denominator gives a larger unit fraction, so \(\frac{1}{7}>\frac{1}{8}\).

a) \(\frac{5}{8}\) b) \(\frac{11}{10}\) c) \(\frac{1}{7}\)

- Write each building area over the total property area. - Simplify both fractions. - Rewrite the fractions with a common denominator before comparing.

1. Center A uses \(\frac{1200}{5400}=\frac{2}{9}\) of its property for the building. 2. Center B uses \(\frac{900}{4500}=\frac{1}{5}\) of its property for the building. 3. Rewrite with denominator \(45\): \(\frac{2}{9}=\frac{10}{45}\) and \(\frac{1}{5}=\frac{9}{45}\). Since \(\frac{10}{45}>\frac{9}{45}\), Center A uses the greater fraction.

Center A uses a greater fraction: \(\frac{2}{9}>\frac{1}{5}\).

- Find each pond area as a fraction of the total area. - In part b), only the pond area changes. - Simplify the new fraction for Garden Beta. - Use equivalent fractions to compare exactly.

1. Garden Alpha’s pond fraction is \(\frac{250}{1250}=\frac{1}{5}\). Garden Beta’s original pond fraction is \(\frac{320}{1600}=\frac{1}{5}\). 2. After the expansion, Garden Beta’s pond area is \(320+20=340\,\text{ft}^2\). Its new pond fraction is \(\frac{340}{1600}=\frac{17}{80}\). 3. Garden Alpha’s fraction is \(\frac{1}{5}=\frac{16}{80}\). Since \(\frac{17}{80}>\frac{16}{80}\), Garden Beta then has the greater pond fraction.

a) Both gardens initially have a pond fraction of \(\frac{1}{5}\). b) Garden Beta has the greater fraction after the expansion: \(\frac{17}{80}>\frac{1}{5}=\frac{16}{80}\).

- With equal numerators, the fraction with the smaller denominator is greater. - First compare pairs with equal numerators or equal denominators. - Rewrite the remaining pair with a common denominator.

1. Anna and Ben have fractions with equal numerators. Since \(5<7\), \(\frac{3}{5}>\frac{3}{7}\), so Ben has more juice than Anna. 2. For the greatest amount, compare \(\frac{3}{5}\) and \(\frac{4}{7}\): \(\frac{3}{5}=\frac{21}{35}\) and \(\frac{4}{7}=\frac{20}{35}\). Ben has the most. 3. For the least amount, compare \(\frac{2}{5}\) and \(\frac{3}{7}\): \(\frac{2}{5}=\frac{14}{35}\) and \(\frac{3}{7}=\frac{15}{35}\). Clara has the least.

a) Ben b) Ben, with \(\frac{3}{5}\,\text{L}\) c) Clara, with \(\frac{2}{5}\,\text{L}\)

- 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.

- How many pears are one-fourth of \(40\)? - After finding the number of pears, how can you find the number of apples? - Is one-eighth of a whole greater or less than one-fourth? - What happens to the number of apples if there are fewer pears?

1. Find the number of pears: \(\frac{1}{4} \times 40 = 10\). 2. Subtract to find the number of apples: \(40 - 10 = 30\). 3. One-eighth is less than one-fourth. With the same total number of fruit, a smaller fraction of pears means a greater number of apples.

a) There are \(30\) apples. b) The number of apples would be greater because one-eighth is less than one-fourth, so there would be fewer pears.

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

- Write the remaining pieces over the total pieces for each bar. - Find a common denominator for \(3\) and \(5\). - Compare the numerators after rewriting the fractions.

1. Bar a) has \(2\) of \(3\) pieces remaining, so its fraction is \(\frac{2}{3}\). 2. Bar b) has \(3\) of \(5\) pieces remaining, so its fraction is \(\frac{3}{5}\). 3. Rewrite both fractions with denominator \(15\): \(\frac{2}{3}=\frac{10}{15}\) and \(\frac{3}{5}=\frac{9}{15}\). 4. Since \(\frac{10}{15}>\frac{9}{15}\), bar a) has the greater fraction remaining.

a) Bar a): \(\frac{2}{3}\); Bar b): \(\frac{3}{5}\) b) Bar a)

- Count the shaded parts and total equal parts in each circle. - Simplify before comparing. - Find the least common denominator and compare the rewritten numerators.

1. Circle a) has \(5\) of \(8\) equal parts shaded, so its fraction is \(\frac{5}{8}\). 2. Circle b) has \(4\) of \(6\) equal parts shaded, so its fraction is \(\frac{4}{6}=\frac{2}{3}\). 3. The least common denominator of \(8\) and \(3\) is \(24\). Rewrite the fractions: \(\frac{5}{8}=\frac{15}{24}\) and \(\frac{2}{3}=\frac{16}{24}\). 4. Since \(\frac{16}{24}>\frac{15}{24}\), circle b) has the greater shaded fraction.

a) Circle a): \(\frac{5}{8}\); Circle b): \(\frac{2}{3}\) b) Circle b)

- Count the shaded parts and total equal parts in each figure. - Simplify the fractions when possible. - Rewrite all four fractions with a common denominator.

1. Figure a) shows \(\frac{2}{5}\). 2. Figure b) shows \(\frac{1}{4}\). 3. Figure c) shows \(\frac{6}{10}=\frac{3}{5}\). 4. Figure d) shows \(\frac{6}{8}=\frac{3}{4}\). 5. Rewrite the fractions with denominator \(20\): \(\frac{1}{4}=\frac{5}{20}\), \(\frac{2}{5}=\frac{8}{20}\), \(\frac{3}{5}=\frac{12}{20}\), and \(\frac{3}{4}=\frac{15}{20}\). 6. Therefore, \(\frac{1}{4}<\frac{2}{5}<\frac{3}{5}<\frac{3}{4}\).

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

- Count the shaded parts and total equal parts in each figure. - Simplify each fraction. - Rewrite the fractions with a common denominator and compare the numerators.

1. Figure a) has \(5\) of \(8\) equal parts shaded, so its fraction is \(\frac{5}{8}\). 2. Figure b) has \(6\) of \(10\) squares shaded, so \(\frac{6}{10}=\frac{3}{5}\). 3. Rewrite both fractions with denominator \(40\): \(\frac{5}{8}=\frac{25}{40}\) and \(\frac{3}{5}=\frac{24}{40}\). 4. Since \(\frac{25}{40}>\frac{24}{40}\), figure a) has the greater shaded fraction.

a) Figure a): \(\frac{5}{8}\); Figure b): \(\frac{3}{5}\) b) Figure a)

- Count the shaded parts and total equal parts in each figure. - Simplify each fraction. - Rewrite the fractions with a common denominator before ordering.

1. Ava has \(4\) of \(6\) parts left, so \(\frac{4}{6}=\frac{2}{3}\). 2. Ben has \(5\) of \(8\) parts left, so his fraction is \(\frac{5}{8}\). 3. Charlotte has \(9\) of \(12\) parts left, so \(\frac{9}{12}=\frac{3}{4}\). 4. Rewrite the fractions with denominator \(24\): \(\frac{5}{8}=\frac{15}{24}\), \(\frac{2}{3}=\frac{16}{24}\), and \(\frac{3}{4}=\frac{18}{24}\). 5. Therefore, \(\frac{5}{8}<\frac{2}{3}<\frac{3}{4}\), and Charlotte has the most cake left.

a) Ava: \(\frac{2}{3}\); Ben: \(\frac{5}{8}\); Charlotte: \(\frac{3}{4}\) b) \(\frac{5}{8}<\frac{2}{3}<\frac{3}{4}\). Charlotte has the most left.

- Find a denominator that is a multiple of \(4\), \(3\), and \(6\). - Rewrite each fraction with that denominator. - Compare the numerators.

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

Bed c), because \(\frac{5}{6}\) is the greatest fraction.

- 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 \(\frac{3}{5}>\frac{1}{2}\).

- Count the shaded parts and total equal parts in each figure. - Simplify each fraction. - Rewrite the fractions with a common denominator, then read the labels in order.

1. Figure S shows \(\frac{1}{6}\). 2. Figure T shows \(\frac{1}{3}\). 3. Figure A shows \(\frac{2}{4}=\frac{1}{2}\). 4. Figure R shows \(\frac{2}{3}\). 5. Using denominator \(6\), the fractions are \(\frac{1}{6}\), \(\frac{2}{6}\), \(\frac{3}{6}\), and \(\frac{4}{6}\). Thus, \(\frac{1}{6}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\). 6. The labels in order are S, T, A, R, which spell “STAR.”

The word is “STAR.” \(\frac{1}{6}<\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\)