Free math worksheets

- Count the shaded parts and all the equal parts. - Count the unshaded parts. - Compare the numerators because the denominators are the same.

1. Three of the \(6\) equal parts are shaded, so the shaded fraction is \(\frac{3}{6}\). 2. The other \(3\) parts are unshaded, so the unshaded fraction is \(\frac{3}{6}\). 3. The fractions are equal: \(\frac{3}{6}=\frac{3}{6}\).

Shaded: \(\frac{3}{6}\) Unshaded: \(\frac{3}{6}\) \(\frac{3}{6}=\frac{3}{6}\)

- The denominator is \(8\) in every fraction. - Compare the numerators \(1\), \(5\), and \(7\). - Put the smallest numerator first.

1. All three fractions have denominator \(8\), so compare their numerators. 2. Since \(1<5<7\), the fractions are ordered \(\frac{1}{8}<\frac{5}{8}<\frac{7}{8}\).

\(\frac{1}{8}<\frac{5}{8}<\frac{7}{8}\)

- First name one whole using sixths. - Compare each numerator with the denominator. - Use the comparisons with \(1\) to determine the order.

1. One whole is \(\frac{6}{6}\). 2. Since \(4<6\), \(\frac{4}{6}\) is less than \(1\). 3. \(\frac{6}{6}\) is equal to \(1\), and since \(7>6\), \(\frac{7}{6}\) is greater than \(1\). 4. Their order is \(\frac{4}{6}<\frac{6}{6}<\frac{7}{6}\).

Less than \(1\): \(\frac{4}{6}\) Equal to \(1\): \(\frac{6}{6}\) Greater than \(1\): \(\frac{7}{6}\) \(\frac{4}{6}<\frac{6}{6}<\frac{7}{6}\)

- Keep the denominator \(8\) in every fraction. - Find whole numbers greater than \(3\) and less than \(6\). - Use those numbers as the numerators.

1. With denominator \(8\), compare the numerators because every part has the same size. 2. The numerator must be greater than \(3\) and less than \(6\). 3. The whole-number numerators that fit are \(4\) and \(5\). 4. The fractions are \(\frac{4}{8}\) and \(\frac{5}{8}\).

\(\frac{4}{8}\) and \(\frac{5}{8}\)

- Count all equal parts and shaded parts in each badge. - Use the parts that are not shaded to write each unshaded fraction. - With the same denominator, compare the numerators.

1. Badge a) has \(2\) of \(6\) parts shaded, so \(4\) of \(6\) parts are unshaded: \(\frac{4}{6}\). 2. Badge b) has \(5\) of \(6\) parts shaded, so \(1\) of \(6\) parts is unshaded: \(\frac{1}{6}\). 3. The denominators are the same, so compare the numerators. 4. Since \(4>1\), \(\frac{4}{6}>\frac{1}{6}\), and badge a) has the greater unshaded fraction.

Badge a): \(\frac{4}{6}>\frac{1}{6}\).

- Both trays are divided into \(8\) equal spaces. - Write the filled part of each tray as a fraction. - With equal denominators, the greater numerator names the greater fraction.

1. The filled fractions are \(\frac{5}{8}\) and \(\frac{2}{8}\). 2. The denominators are the same, so compare the numerators. 3. Since \(5>2\), \(\frac{5}{8}>\frac{2}{8}\).

\(\frac{5}{8}>\frac{2}{8}\)

- Count the equal parts and shaded parts in each bar. - The shaded pieces can be in different places and still have the same total size. - Use \(=\) when the shaded amounts are equal.

1. Each bar has the same denominator, \(8\), so its parts are the same size. 2. Each bar also has the same numerator, \(4\). 3. The placement of the shaded parts does not change their total amount. 4. Therefore, \(\frac{4}{8}=\frac{4}{8}\).

\(\frac{4}{8}=\frac{4}{8}\)

- Both fractions have denominator \(4\). - Compare the number of fourths in each distance. - The greater numerator names the greater distance.

1. Both distances are measured in fourths of a mile. 2. Compare the numerators \(6\) and \(3\). 3. Since \(6>3\), \(\frac{6}{4}>\frac{3}{4}\). 4. Ava walks farther.

Ava walks farther. \(\frac{6}{4}>\frac{3}{4}\).

- Identify what stays the same in all three fractions. - Think of each numerator as a count of same-size pieces. - Compare each fraction with one whole, \(\frac{8}{8}\).

1. All three fractions count eighth-size parts. 2. Their numerators are \(10\), \(7\), and \(9\). 3. Ordering the numerators gives \(7<9<10\). 4. Therefore, \(\frac{7}{8}<\frac{9}{8}<\frac{10}{8}\).

The student did not order the numerators. The correct order is \(\frac{7}{8}<\frac{9}{8}<\frac{10}{8}\).

- Both fractions have denominator \(8\). - Compare the numerators. - Subtract the smaller numerator from the larger numerator to find the difference in eighths.

1. Both amounts are measured in eighths of a cup. 2. Since \(7>5\), \(\frac{7}{8}>\frac{5}{8}\). 3. The difference is \(7-5=2\) eighths.

The fruit amount is greater: \(\frac{7}{8}>\frac{5}{8}\). It is greater by \(2\) eighths.

- Both containers are divided into eighths. - Compare the numerators. - Subtract the smaller numerator from the larger numerator.

1. Both fractions have denominator \(8\), so compare the numerators. 2. Since \(6>2\), \(\frac{6}{8}>\frac{2}{8}\). 3. The difference is \(6-2=4\) eighths.

The second container is fuller: \(\frac{6}{8}>\frac{2}{8}\). It is fuller by \(4\) eighths.

- The denominator is \(4\) in every fraction. - Compare the numerators \(5\), \(6\), and \(7\). - Put the smallest numerator first.

1. All three fractions have denominator \(4\), so compare their numerators. 2. Since \(5<6<7\), the order is \(\frac{5}{4}<\frac{6}{4}<\frac{7}{4}\).

\(\frac{5}{4}<\frac{6}{4}<\frac{7}{4}\)

- Write \(1\) as sixths. - Move left by one sixth-size space. - Compare the numerators because both fractions have denominator \(6\).

1. One whole is \(\frac{6}{6}\). 2. One sixth-size space left of \(\frac{6}{6}\) is \(\frac{5}{6}\). 3. Since \(5>2\), \(\frac{5}{6}>\frac{2}{6}\).

\(F=\frac{5}{6}\), and \(\frac{5}{6}>\frac{2}{6}\).

- Add one eighth while keeping the denominator \(8\). - Compare the new numerator with the old numerator. - A fraction equals one whole when its numerator equals its denominator.

1. One eighth greater than \(\frac{7}{8}\) is \(\frac{8}{8}\). 2. Fraction B is \(\frac{8}{8}\). 3. Since \(8>7\), \(\frac{8}{8}>\frac{7}{8}\). 4. Also, \(\frac{8}{8}=1\).

Fraction B is \(\frac{8}{8}\). Thus, \(\frac{7}{8}<\frac{8}{8}=1\).

- Find both the filled and empty counts for Tray A. - Use the stated relationship to determine Tray B's empty count. - Convert Tray B's empty count into a filled count before comparing.

1. Tray A has \(3\) filled spaces and \(6-3=3\) empty spaces. 2. Tray B has \(3\) empty spaces, so it also has \(6-3=3\) filled spaces. 3. Both filled fractions are \(\frac{3}{6}\). 4. Therefore, the filled fractions are equal.

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

- One whole is \(\frac{8}{8}\). - Count how many eighth-size spaces each fraction is from \(\frac{8}{8}\). - Then compare the numerators to order the fractions.

1. One whole is \(\frac{8}{8}\). 2. Seven eighths is one eighth below one, and nine eighths is one eighth above one. 3. The fractions are equally close to one. 4. Since \(7<9\), \(\frac{7}{8}<\frac{9}{8}\).

They are equally close to \(1\), and \(\frac{7}{8}<\frac{9}{8}\).

- One whole is \(\frac{6}{6}\). - Count how many sixth-size spaces each fraction is from \(\frac{6}{6}\). - Then compare the numerators to order the fractions.

1. One whole is \(\frac{6}{6}\). 2. Five sixths is \(1\) sixth below one, while eight sixths is \(2\) sixths above one. 3. Therefore, \(\frac{5}{6}\) is closer to \(1\). 4. Since \(5<8\), \(\frac{5}{6}<\frac{8}{6}\).

\(\frac{5}{6}<\frac{8}{6}\), and \(\frac{5}{6}\) is closer to \(1\).

- Treat the final comparison and the amount of change as two separate questions. - The shared denominator means every counted part has the same size. - Compare ending numerators, then compare how much each numerator changed.

1. The final numerators are \(5\) and \(6\), so \(\frac{6}{8}>\frac{5}{8}\). 2. Fraction A gains \(5-2=3\) eighth-size parts. 3. Fraction B gains \(6-4=2\) eighth-size parts. 4. Fraction B ends greater, but fraction A increases by more parts.

Fraction B ends greater: \(\frac{6}{8}>\frac{5}{8}\). Fraction A increases by more, gaining \(3\) eighths instead of \(2\) eighths.

- Keep the denominator fixed because the bar still has eight sections. - Update only the number of lit sections at each change. - Order the resulting fractions by their numerators.

1. The starting fraction is \(\frac{5}{8}\). 2. After \(2\) more sections light up, the fraction is \(\frac{7}{8}\). 3. After \(1\) section turns off, the fraction is \(\frac{6}{8}\). 4. Since \(5<6<7\), \(\frac{5}{8}<\frac{6}{8}<\frac{7}{8}\).

The stages are \(\frac{5}{8}\), \(\frac{7}{8}\), and \(\frac{6}{8}\). From least to greatest: \(\frac{5}{8}<\frac{6}{8}<\frac{7}{8}\).

- Count the “yes” cards in both groups. - Put that count over the \(8\) cards in all. - Write one half and one whole as eighths before comparing.

1. Two groups contain \(3+3=6\) yes cards. 2. There are \(8\) cards in all, so the yes fraction is \(\frac{6}{8}\). 3. One half is \(\frac{4}{8}\), and one whole is \(\frac{8}{8}\). 4. Therefore, \(\frac{1}{2}<\frac{6}{8}<1\).

The yes fraction is \(\frac{6}{8}\), and \(\frac{1}{2}<\frac{6}{8}<1\).

- Find the uncompleted part of the first bar. - Take half of that remaining count. - Add it to the original completed count before comparing.

1. The first bar has \(8-4=4\) eighths remaining. 2. Half of the \(4\) remaining eighths is \(2\) eighths. 3. The second bar shows \(4+2=6\) eighths, or \(\frac{6}{8}\). 4. Since \(6>4\), \(\frac{6}{8}>\frac{4}{8}\).

The second bar shows \(\frac{6}{8}\), and \(\frac{6}{8}>\frac{4}{8}\).

- Fraction A has \(2\) shaded sixths. - Give Fraction B \(2\) unshaded sixths, then count its shaded sixths. - Compare the numerators because both denominators are \(6\).

1. Fraction A has \(2\) shaded sixths. 2. Fraction B therefore has \(2\) unshaded sixths. 3. Out of \(6\) parts, that leaves \(6-2=4\) shaded sixths, so B is \(\frac{4}{6}\). 4. Since \(4>2\), \(\frac{4}{6}>\frac{2}{6}\).

Fraction B is \(\frac{4}{6}\), and \(\frac{4}{6}>\frac{2}{6}\).

- All three fractions are measured in sixths. - Order the numerators first. - Compare the differences between neighboring numerators to find the closest pair.

1. Ordering the numerators gives \(4<5<7\), so \(\frac{4}{6}<\frac{5}{6}<\frac{7}{6}\). 2. The first neighboring pair differs by \(5-4=1\) sixth-size part. 3. The second neighboring pair differs by \(7-5=2\) sixth-size parts. 4. Therefore, \(\frac{4}{6}\) and \(\frac{5}{6}\) are the closest pair.

\(\frac{4}{6}<\frac{5}{6}<\frac{7}{6}\). The first pair is \(1\) sixth apart, the second pair is \(2\) sixths apart, and \(\frac{4}{6}\) with \(\frac{5}{6}\) is closest.

- Compare the original equal-part counts first. - Apply the same numerator change to both fractions. - Check whether the new numerator order matches the original order.

1. Initially, \(3<5\), so \(\frac{3}{4}<\frac{5}{4}\). 2. Adding \(2\) fourths changes A to \(\frac{5}{4}\). 3. Adding \(2\) fourths changes B to \(\frac{7}{4}\). 4. Since \(5<7\), \(\frac{5}{4}<\frac{7}{4}\), so the order does not change.

Before: \(\frac{3}{4}<\frac{5}{4}\). After: \(\frac{5}{4}<\frac{7}{4}\). The order stays the same.

- Count the equal spaces between consecutive whole numbers. - Count third-size spaces from \(0\) to each marked point. - With the same denominator, compare the numerators and the number of spaces between the points.

1. Each whole unit is divided into \(3\) equal spaces, so each small space represents \(\frac{1}{3}\). 2. Point \(A\) is \(4\) third-size spaces from \(0\), so \(A=\frac{4}{3}\). 3. Point \(B\) is \(5\) third-size spaces from \(0\), so \(B=\frac{5}{3}\). 4. Since \(5>4\), \(\frac{5}{3}>\frac{4}{3}\). The points are \(1\) third-size space apart.

\(A=\frac{4}{3}\), \(B=\frac{5}{3}\), and \(\frac{4}{3}<\frac{5}{3}\). They are \(1\) third-size space apart.

- Count how many half-size spaces make one whole on the line. - Name the shown point by counting half-size spaces from \(0\). - Moving one half-size space left changes the numerator by one while keeping the denominator fixed.

1. Each whole is divided into \(2\) equal spaces, so each small space is one half. 2. Point \(X\) is \(5\) half-size spaces from \(0\), so \(X=\frac{5}{2}\). 3. One half-size space to the left is \(4\) halves from \(0\), so \(Y=\frac{4}{2}\). 4. Therefore, \(\frac{4}{2}<\frac{5}{2}\), and \(\frac{4}{2}=2\).

\(X=\frac{5}{2}\), \(Y=\frac{4}{2}\), and \(\frac{4}{2}<\frac{5}{2}\). Also, \(Y=2\).

- Work backward from the equal ending fractions. - Keep the total number of lights in each panel fixed while changing how many are on. - Compare the original numerators only after finding the missing one.

1. Panel B gains \(2\) lights that are on and ends at \(\frac{5}{8}\). 2. Its starting numerator was \(5-2=3\), so it started at \(\frac{3}{8}\). 3. Both starting fractions use eighths, and \(7>3\). 4. Therefore, \(\frac{7}{8}>\frac{3}{8}\).

Panel B started at \(\frac{3}{8}\). The starting comparison is \(\frac{7}{8}>\frac{3}{8}\).

- First compare the two fraction numbers using their common denominator. - For each banner, determine the length of one fourth from that banner's own total length. - Compare the two painted lengths with the order of the fraction numbers. - In the explanation, consider why the size of the reference whole matters.

1. a) The fractions have the same denominator. Three fourths contains more fourth-size fraction parts than two fourths, so \(\frac{3}{4} > \frac{2}{4}\). 2. b) Banner A is divided into fourths, so each fourth is \(8 \div 4 = 2\,\text{in.}\). Three fourths is \(3 \times 2 = 6\,\text{in.}\). 3. Banner B is divided into fourths, so each fourth is \(12 \div 4 = 3\,\text{in.}\). Two fourths is \(2 \times 3 = 6\,\text{in.}\). 4. c) The fraction numbers compare shares of a same-size whole, but the banners are different lengths. Here the painted physical lengths are equal even though \(\frac{3}{4} > \frac{2}{4}\).

a) \(\frac{3}{4} > \frac{2}{4}\) b) Banner A: \(6\,\text{in.}\); Banner B: \(6\,\text{in.}\) c) The wholes are different lengths. The same-denominator fraction comparison tells which fraction number is greater, but it does not by itself compare physical painted lengths from different-size wholes.