Free math worksheets

- Count all equal parts or objects to find the denominator. - Count the shaded parts or objects to find the numerator. - Read the numerator as a counting number and use the fraction name for the denominator.

1. In a), \(3\) of \(8\) equal sectors are shaded, so the fraction is \(\frac{3}{8}\), or three eighths. 2. In b), \(2\) of \(3\) equal parts are shaded, so the fraction is \(\frac{2}{3}\), or two thirds. 3. In c), \(5\) of \(6\) objects are shaded, so the fraction is \(\frac{5}{6}\), or five sixths.

a) \(\frac{3}{8}\); three eighths b) \(\frac{2}{3}\); two thirds c) \(\frac{5}{6}\); five sixths

- How many equal slices make the whole pizza? - A fraction tells how many equal parts of a whole are being described. - How many slices remain after Lucas and Sarah eat theirs? - Can any fraction be written in an equivalent simpler form?

1. The whole pizza has \(8\) equal slices. 2. Lucas eats \(2\) of the \(8\) slices, so his fraction is \(\frac{2}{8} = \frac{1}{4}\). 3. Sarah eats \(1\) of the \(8\) slices, so her fraction is \(\frac{1}{8}\). 4. The parents receive \(8 - 2 - 1 = 5\) slices, so their fraction is \(\frac{5}{8}\).

Lucas: \(\frac{1}{4}\) (or \(\frac{2}{8}\)) Sarah: \(\frac{1}{8}\) Parents: \(\frac{5}{8}\)

- How many equal sections make the whole bed? That number is the denominator. - Count how many sections each flower type covers. - After accounting for the red and blue flowers, how many sections remain for the yellow flowers? - Compare the fractions for the red and yellow flowers.

1. The whole bed has \(4\) equal sections. 2. The blue flowers cover \(2\) of the \(4\) sections, so they cover \(\frac{2}{4} = \frac{1}{2}\). 3. The yellow flowers cover \(4 - 1 - 2 = 1\) section, so they cover \(\frac{1}{4}\). 4. The red roses and yellow sunflowers each cover \(1\) of the \(4\) equal sections, so their fractions are equal.

a) \(\frac{2}{4}\), or \(\frac{1}{2}\) b) \(\frac{1}{4}\) c) The fractions are equal because each flower type covers \(\frac{1}{4}\) of the bed.

- Count the total number of equal parts in the whole cake. - How many parts are shaded yellow? - Use the number of selected parts as the numerator and the total number of parts as the denominator. - How many unshaded slices are needed to complete the whole cake?

1. Count the yellow slices: \(5\) slices remain. 2. The cake has \(8\) equal slices, so the fraction remaining is \(\frac{5}{8}\). 3. The number eaten is \(8 - 5 = 3\), so the fraction eaten is \(\frac{3}{8}\).

a) \(5\) slices b) \(\frac{5}{8}\) c) \(\frac{3}{8}\)

- Count the total number of equal boards. - Count the blue-shaded boards, then find how many are unshaded. - Write the number of unshaded boards over the total number of boards.

1. The fence has \(8\) equal boards in all. 2. The diagram shows \(3\) painted boards, so \(8-3=5\) boards remain unpainted. 3. Therefore, the fraction that still needs to be painted is \(\frac{5}{8}\).

\(\frac{5}{8}\) of the fence still needs to be painted.

- Identify the labeled sector that covers half of the circle. - Find the labeled sector that is half as large as the one-half sector. - Compare the two smallest labeled sectors with the one-fourth sector.

1. Sector A covers one half of the circle, so it represents the peanuts. 2. Sector B covers one fourth of the circle, so it represents the cashews. 3. Sectors C and D each cover one eighth of the circle. They represent the almonds and raisins in either order.

Peanuts: A, \(\frac{1}{2}\) Cashews: B, \(\frac{1}{4}\) Almonds and raisins: C and D in either order, \(\frac{1}{8}\) each

- Count all the sections to find the denominator. - The number of shaded sections gives the numerator for the shaded fraction. - For the unshaded fraction, determine how many sections remain.

1. Count all the sections: the strip has \(6\) equal sections. 2. Count the shaded sections: \(2\) sections are shaded. 3. The shaded fraction is \(\frac{2}{6} = \frac{1}{3}\). 4. The number of unshaded sections is \(6 - 2 = 4\). 5. The unshaded fraction is \(\frac{4}{6} = \frac{2}{3}\).

a) \(6\) b) \(2\) c) \(\frac{2}{6}\), or \(\frac{1}{3}\) d) \(\frac{4}{6}\), or \(\frac{2}{3}\)

- Count how many equal slices make the whole pizza. - Compare the shaded slices with the unshaded slices. - Use the total number of equal slices as the denominator for both fractions. - Check that the two parts together account for every slice.

1. Three of the \(8\) slices have pepperoni, so \(8 - 3 = 5\) slices have only cheese. 2. The pepperoni fraction is \(\frac{3}{8}\). 3. The cheese-only fraction is \(\frac{5}{8}\). 4. Together the two parts make all \(8\) slices: \(\frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1\).

a) \(5\) slices b) \(\frac{3}{8}\) c) \(\frac{5}{8}\) d) \(\frac{3}{8}\) and \(\frac{5}{8}\)

- For each model, first count how many equal parts or objects make the whole. - Then count how many of those parts are shaded. - Put the shaded count over the total count. - Check each model separately because its denominator may be different.

1. In a), \(1\) of \(3\) equal parts is shaded, so the fraction is \(\frac{1}{3}\). 2. In b), \(3\) of \(8\) equal squares are shaded, so the fraction is \(\frac{3}{8}\). 3. In c), \(5\) of \(6\) equal sections are shaded, so the fraction is \(\frac{5}{6}\). 4. In d), \(3\) of \(4\) objects are shaded, so the fraction is \(\frac{3}{4}\).

a) \(\frac{1}{3}\) b) \(\frac{3}{8}\) c) \(\frac{5}{6}\) d) \(\frac{3}{4}\)

- Count the equal parts in the whole rectangle and the parts that are shaded. - Think about what the denominator tells you about the whole. - Compare the shaded count with the total number of parts to find the unshaded fraction.

1. The model has \(8\) equal squares, and \(5\) are shaded, so the shaded fraction is \(\frac{5}{8}\). 2. The numerator is \(5\), and the denominator is \(8\). 3. The denominator tells the number of equal parts in the whole rectangle. 4. There are \(8 - 5 = 3\) unshaded squares, so the unshaded fraction is \(\frac{3}{8}\).

a) Numerator: \(5\); denominator: \(8\) b) The denominator represents the \(8\) equal parts in the whole rectangle. c) \(\frac{3}{8}\)

- Count all equal parts in each figure to determine its denominator. - Count the shaded parts for the shaded numerator. - The unshaded numerator comes from the equal parts that remain. - Check that the shaded and unshaded counts together make the whole figure.

1. Figure a) has \(8\) equal parts, with \(3\) shaded. The shaded fraction is \(\frac{3}{8}\), and the unshaded fraction is \(\frac{5}{8}\). 2. Figure b) has \(4\) equal parts, with \(1\) shaded. The shaded fraction is \(\frac{1}{4}\), and the unshaded fraction is \(\frac{3}{4}\). 3. Figure c) has \(6\) equal parts, with \(4\) shaded. The shaded fraction is \(\frac{4}{6}\), and the unshaded fraction is \(\frac{2}{6}\). 4. Figure d) has \(6\) equal parts, with \(5\) shaded. The shaded fraction is \(\frac{5}{6}\), and the unshaded fraction is \(\frac{1}{6}\).

a) shaded: \(\frac{3}{8}\); unshaded: \(\frac{5}{8}\) b) shaded: \(\frac{1}{4}\); unshaded: \(\frac{3}{4}\) c) shaded: \(\frac{4}{6}\); unshaded: \(\frac{2}{6}\) d) shaded: \(\frac{5}{6}\); unshaded: \(\frac{1}{6}\)

- Use the bar to determine how many equal sections make the whole. - The red section tells you the number of dots in one equal part. - Think about how many equal groups of \(18\) are needed to make the whole.

1. The bar has \(4\) equal sections, so the red section is one-fourth of the whole. 2. If one section represents \(18\) dots, the whole set has four groups of \(18\) dots. 3. \(4 \times 18 = 72\), so the whole set has \(72\) dots.

There are \(72\) dots. The whole has \(4\) equal sections, so \(4 \times 18 = 72\).

- Add the slice values to find the whole. - Find one-half of the whole. - Match that value to a slice.

1. The total is \(12 + 6 + 6 = 24\). 2. One-half of \(24\) is \(24 \div 2 = 12\). 3. Reading has a value of \(12\), so it takes up one-half of the chart.

Reading takes up exactly one-half of the pie chart.

- Add all four values to find the whole. - Find one-half of the whole. - Look for two equal slice values.

1. The total is \(4 + 8 + 4 + 16 = 32\). 2. One-half of \(32\) is \(16\), so walking takes up one-half of the chart. 3. Train and bike each have a value of \(4\), so their slices are equal.

1) Walking takes up one-half of the chart. 2) Train and bike have equal-sized slices.

- Compare the slice sizes. - Look for two equal smaller slices. - Use the whole circle to identify halves and fourths.

1. The water slice is as large as the other two slices combined, so it takes up one-half of the circle. 2. The juice and milk slices are equal and split the other half into two equal parts. 3. Therefore, juice and milk each take up one-fourth of the circle.

1) Water takes up one-half. 2) Juice and milk each take up one-fourth.

- Think about what the word “half” means. - Compare the widths of the two pieces. - Decide whether the two parts are equal in size.

1. Halves are \(2\) equal parts of one whole. 2. The card has \(2\) pieces, but they are not the same size. 3. Therefore, the pieces are not halves of the card.

No. Halves must be \(2\) equal parts of the same whole.

- Treat the full loop as the whole. - Count the equal time parts in that whole. - Describe the longer sound as repeated copies of one beat.

1. The loop is partitioned into \(8\) equal beats. 2. One beat is \(\frac{1}{8}\) of the loop. 3. The held note uses \(3\) beats, so it is made of \(3\) copies of \(\frac{1}{8}\).

The clap is \(\frac{1}{8}\) of the loop. The held note contains \(3\) copies of \(\frac{1}{8}\).

- Identify the whole set before naming either fraction. - Count equal groups for the pair question. - Count equal individual objects for the lantern question.

1. The whole set is partitioned into \(4\) equal pairs, so one pair is \(\frac{1}{4}\) of the set. 2. The whole set also contains \(8\) individual lanterns, so one lantern is \(\frac{1}{8}\) of the set. 3. The fraction name depends on which equal parts are being counted.

One pair is \(\frac{1}{4}\) of the lanterns. One lantern is \(\frac{1}{8}\) of the lanterns.

- Keep the original strip as the whole for every part. - First name one of the six equal original pieces. - For the block, count how many original sixth-size pieces are taped together. - In part c), compare one original piece with the two-piece block rather than changing the whole.

1. One original piece is one of \(6\) equal parts, so it is \(\frac{1}{6}\) of the strip. 2. The block contains \(2\) of those sixth-size pieces, so it is \(\frac{2}{6}\) of the strip. 3. The block is not one of the original sixth-size unit-fraction pieces because it contains two of them, not one.

a) \(\frac{1}{6}\) b) \(\frac{2}{6}\) c) No. The block is made of two original \(\frac{1}{6}\)-size pieces.

- Count the equal parts in each badge. - One part is shaded in each whole. - Equal parts can have different shapes as long as they have equal area within their whole.

1. Each badge is its own whole and is divided into \(4\) equal-area parts. 2. One part is shaded in each badge, so each shaded amount is \(\frac{1}{4}\) of its badge. 3. Equal parts do not need the same shape; they must have equal size within the whole.

Each badge has \(\frac{1}{4}\) shaded. The part shapes may differ because the fraction name depends on equal size and part count.

- Keep the original object as the whole even when part of it is hidden. - Recall how many equal panels made the display before one was covered. - Distinguish the number visible from the number in the whole.

1. Covering a panel does not change the original whole or its partition. 2. The original display still has \(3\) equal panels. 3. One panel is \(\frac{1}{3}\) of the original display. 4. The two visible panels are not halves of the original display; together they are \(\frac{2}{3}\).

No. One panel is \(\frac{1}{3}\) of the original display, and the two visible panels together are \(\frac{2}{3}\).

- Count the equal parts and shaded parts in diagram a). - Do the same count in diagram b) instead of judging by which way the rectangle points. - Decide whether a turn changes the size of the whole or any of its parts.

1. In diagram a), the rectangle is divided into \(4\) equal parts and \(3\) are shaded, so it represents \(\frac{3}{4}\). 2. In diagram b), the same rectangle is still divided into the same \(4\) equal parts and the same \(3\) parts are shaded, so it also represents \(\frac{3}{4}\). 3. A quarter-turn changes only the direction of the paper. It does not change the whole, the equal parts, or which parts are shaded. 4. Therefore, both diagrams represent \(\frac{3}{4}\).

Both diagrams represent \(\frac{3}{4}\). Turning the paper changes its direction but not the whole, its \(4\) equal parts, or the \(3\) shaded parts.

- Identify the quantity that was divided equally. - Separate physical shape from the measured amount. - Use one part over the total number of equal weight parts.

1. The whole loaf's weight is divided into \(4\) equal amounts. 2. One piece is \(1\) of those \(4\) equal weight parts. 3. One piece is \(\frac{1}{4}\) of the loaf's weight. 4. Equal weight, not identical shape, makes the pieces equal parts in this problem.

One piece is \(\frac{1}{4}\) of the loaf's weight.

- Identify the equal groups made from the whole set. - A unit fraction names one entire equal group. - Share the total card count equally among the stacks.

1. The whole set is divided into \(6\) equal stacks. 2. One of those stacks is \(\frac{1}{6}\) of the cards. 3. \(24 \div 6=4\), so each stack contains \(4\) cards. 4. One card is only part of a stack, so it is smaller than \(\frac{1}{6}\) of the set.

One stack is \(\frac{1}{6}\) of all the cards, and it contains \(4\) cards.

- Identify which markings actually form boundaries of the whole. - Count equal regions, not objects printed inside them. - Match the denominator to the true partition of the logo.

1. The logo's area is partitioned into \(4\) equal sectors. 2. The stars are decorations inside the sectors, not boundaries that create equal parts of the logo. 3. One sector is therefore one of \(4\) equal parts. 4. One sector is \(\frac{1}{4}\) of the logo.

One sector is \(\frac{1}{4}\) of the logo. The stars do not create additional parts of the whole.

- Count the number of visible sections in each rectangle. - The denominator tells more than how many pieces there are; it also requires something about their sizes. - Compare the widths of all four sections in each diagram.

1. In diagram a), the \(4\) sections have equal widths, so they are equal parts of the same whole. 2. Each section in diagram a) is therefore \(\frac{1}{4}\) of the rectangle. 3. In diagram b), the \(4\) sections have different widths. 4. A denominator of \(4\) requires \(4\) equal parts, so diagram b) cannot name every section as one fourth.

Diagram a). Its \(4\) sections are equal. Diagram b) has unequal sections, so its sections cannot all be called \(\frac{1}{4}\).

- Read the number of equal slices and the number of orange slices from the picture. - Think about how many of the equal slices make one-half of the pizza. - For the last part, compare the number eaten with the total number of slices.

1. The pizza has \(8\) equal slices, and \(3\) are orange, so the eaten fraction is \(\frac{3}{8}\). 2. Half of \(8\) equal slices is \(4\) slices. Since \(3\) have already been eaten, \(4 - 3 = 1\) more slice must be eaten. 3. If \(5\) slices have been eaten, \(8 - 5 = 3\) slices remain. The remaining fraction is \(\frac{3}{8}\).

a) \(\frac{3}{8}\) b) \(1\) more slice c) \(\frac{3}{8}\)

- Count the visible pieces, then compare their sizes. - A fraction partition requires every part to be equal. - Use the original equal partition to name an unsplit section.

1. The strip now has \(6\) visible pieces, but the pieces are not all equal in size. 2. Each end piece is half the size of an unsplit middle section. 3. The original \(4\) equal sections still show that each unsplit section is \(\frac{1}{4}\) of the strip. 4. Unequal visible pieces cannot all be sixths.

Mia is incorrect because the \(6\) visible pieces are not equal. Each unsplit section is \(\frac{1}{4}\) of the original strip.

- One section is one of \(6\) equal parts of the route. - Think about how many sections make \(1\) mile when \(6\) sections make \(3\) miles. - One of two equal sections in a mile is half a mile.

1. One of \(6\) equal sections is \(\frac{1}{6}\) of the entire route. 2. Six equal sections make \(3\) miles, so two sections make \(1\) mile. 3. One section is half of that mile, so each section is \(\frac{1}{2}\,\text{mile}\) long.

One section is \(\frac{1}{6}\) of the route and is \(\frac{1}{2}\,\text{mile}\) long.

- Track how the second partition affects both original halves. - Count all equal smallest parts in the whole rectangle. - Keep the original half visible as a group of the new parts.

1. Each of the \(2\) halves is split into \(3\) equal parts. 2. The whole rectangle therefore has \(2 \times 3=6\) equal smallest parts. 3. One smallest part is \(\frac{1}{6}\) of the original rectangle. 4. One original half contains \(3\) smallest parts.

One smallest part is \(\frac{1}{6}\) of the rectangle, and \(3\) smallest parts make one original half.

- Check both the number and size of the regions in each plan. - Count the cells created by the row-and-column partition. - A denominator can count only equal parts of the whole.

1. Plan A makes \(8\) equal strips, so each strip is \(\frac{1}{8}\) of the square. 2. Plan B makes \(2 \times 4=8\) equal cells, so each cell is \(\frac{1}{8}\). 3. Plan C has \(8\) regions, but the regions are not equal. 4. Therefore, only Plans A and B make eighths.

Plans A and B make eighths. Plan C does not because its eight regions are unequal.

- Keep the \(2\)-inch strip the same each time. - Count how many copies of it fit in each longer strip. - Use each copy count as the denominator.

1. A \(6\)-inch whole contains \(3\) equal \(2\)-inch parts, so the strip is \(\frac{1}{3}\) of that whole. 2. An \(8\)-inch whole contains \(4\) equal \(2\)-inch parts, so the strip is \(\frac{1}{4}\). 3. A \(12\)-inch whole contains \(6\) equal \(2\)-inch parts, so the strip is \(\frac{1}{6}\).

The \(2\)-inch strip is \(\frac{1}{3}\) of a \(6\)-inch whole, \(\frac{1}{4}\) of an \(8\)-inch whole, and \(\frac{1}{6}\) of a \(12\)-inch whole.

- Track the first partition and the second partition separately. - Count the total number of smallest equal groups in the whole set. - Share the total card count among those smallest groups.

1. One of \(3\) equal rows is \(\frac{1}{3}\) of the set. 2. Dividing each row into \(2\) equal teams creates \(3 \times 2=6\) equal teams in the whole set. 3. One team is \(\frac{1}{6}\) of the set. 4. \(18 \div 6=3\), so one team contains \(3\) cards.

One row is \(\frac{1}{3}\) of the set. One team is \(\frac{1}{6}\) of the set and contains \(3\) cards.

- Count how many equal-width grid steps span the whole rectangle. - Think about how wide each part must be if four parts are to be equal. - The partition lines should repeat that equal width from the left edge.

1. Count \(8\) equal cell-width steps across the rectangle. 2. Four equal parts must each be \(8\div4=2\) cell-width steps wide. 3. Starting at the left edge, the partition lines belong after \(2\), \(4\), and \(6\) cell-width steps. 4. The rectangle is then divided into \(4\) equal parts, so each part represents \(\frac{1}{4}\) of the whole.

Draw the partition lines after \(2\), \(4\), and \(6\) cell-width steps from the left edge. Each part is \(\frac{1}{4}\) of the whole.

- The same section value can represent different fractions when the chart totals are different. - Find the total represented by each chart. - Write Section A as a fraction of each chart's own total before comparing the slices.

1. The total in a) is \(2 + 1 + 1 = 4\), so Section A is \(\frac{2}{4}\) of that whole. 2. The total in b) is \(2 + 4 + 2 = 8\), so Section A is \(\frac{2}{8}\) of that whole. 3. In a), Section A covers one half of the chart. In b), it covers one fourth of the chart. 4. Therefore, the slices are not the same size even though both sections have the same numerical value.

No. Section A is \(\frac{2}{4}\) of chart a) but \(\frac{2}{8}\) of chart b), so its slice is larger in chart a).

- First use one small card as the whole. - Then use the two-card rectangle as the whole. - Count how many tile-size parts fit in each whole.

1. The tile is \(\frac{1}{2}\) of one small card. 2. The larger card contains \(2\) small cards, so it contains \(4\) tile-size parts. 3. The same tile is \(1\) of those \(4\) equal parts, or \(\frac{1}{4}\) of the larger card. 4. The fraction name changes because the whole changes.

The tile is \(\frac{1}{4}\) of the larger card. It is also \(\frac{1}{2}\) of one small card because the two questions use different wholes.