Free math worksheets

- The whole-number factor tells how many equal parts make the whole. - Use the model to name one of those equal parts. - Verify by multiplying the term count by the unit fraction.

1. Twelve equal unit fractions must combine to one whole. 2. One whole divided into \(12\) equal parts gives the unit fraction \(\frac{1}{12}\). 3. Check: \(12\times\frac{1}{12}=\frac{12}{12}=1\).

\(\frac{1}{12}\)

- Think about what “zero groups” means. - Compare this situation with having zero groups of whole objects. - Decide whether any eighths are present.

1. The expression is \(0\times\frac{7}{8}\). 2. Zero groups contain no fraction parts, so \(0\times\frac{7}{8}=0\). 3. Noah is not correct.

No. \(0\times\frac{7}{8}=0\).

- You can write the whole number as a fraction with denominator \(1\). - Look for a common factor between the whole number and the denominator before multiplying.

1. For a), simplify \(18\div6=3\). Then \(\frac{5}{6}\times18=5\times3=15\). 2. For b), simplify \(24\div8=3\). Then \(24\times\frac{3}{8}=3\times3=9\). 3. For c), simplify \(35\div7=5\). Then \(\frac{2}{7}\times35=2\times5=10\).

a) \(15\) b) \(9\) c) \(10\)

- Think about which number you divide by to find one-half, one-third, or another unit fraction. - What does the denominator tell you to do? - You may be able to work with \(6\) first and then use the two zeros.

1. \(\frac{1}{2} \times 600 = 300\). 2. \(\frac{1}{3} \times 600 = 200\). 3. \(\frac{1}{4} \times 600 = 150\). 4. \(\frac{1}{5} \times 600 = 120\). 5. \(\frac{1}{6} \times 600 = 100\).

a) \(300\) b) \(200\) c) \(150\) d) \(120\) e) \(100\)

- What happens to the result when the denominator doubles? - If the denominator doubles, try halving a result you already know. - Think about division as the inverse of multiplication.

1. Find each fraction by dividing \(240\) by its denominator. 2. a) \(240 \div 2 = 120\), \(240 \div 4 = 60\), and \(240 \div 8 = 30\). 3. b) \(240 \div 3 = 80\), \(240 \div 6 = 40\), and \(240 \div 12 = 20\).

a) \(\frac{1}{2}\) is \(120\), \(\frac{1}{4}\) is \(60\), and \(\frac{1}{8}\) is \(30\). b) \(\frac{1}{3}\) is \(80\), \(\frac{1}{6}\) is \(40\), and \(\frac{1}{12}\) is \(20\).

- Imagine dividing the same pizza into \(4\) equal slices and then into \(5\) equal slices. Which slices are larger? - Work one division at a time. - Look for a pattern as the denominator increases.

1. a) \(1000 \div 5 = 200\), \(1000 \div 10 = 100\), and \(1000 \div 20 = 50\). 2. b) \(1000 \div 4 = 250\), and \(1000 \div 8 = 125\). 3. c) Since \(200 < 250\), \(\frac{1}{5}\) of the number is less than \(\frac{1}{4}\) of the number. 4. When the same whole is divided into more equal parts, each part is smaller.

a) \(\frac{1}{5}\) is \(200\), \(\frac{1}{10}\) is \(100\), and \(\frac{1}{20}\) is \(50\). b) \(\frac{1}{4}\) is \(250\), and \(\frac{1}{8}\) is \(125\). c) \(\frac{1}{5}\) is less than \(\frac{1}{4}\) because dividing the same whole into more equal parts makes each part smaller.

- First find the value of one equal part, such as one-tenth. - Think about how the whole number is divided into equal groups. - After finding one part, how can you find several parts?

1. a) Find one-tenth: \(4000 \div 10 = 400\). Then find three-tenths: \(3 \times 400 = 1200\). 2. b) Find one-fifth: \(1500 \div 5 = 300\). Then find four-fifths: \(4 \times 300 = 1200\).

a) \(1200\) b) \(1200\)

- Find the value of each expression separately. - Use the denominator to decide what division to perform. - Compare the three calculated values at the end.

1. Find A: \(150 \div 3 = 50\). 2. Find B: \(200 \div 5 = 40\). 3. Find C: \(180 \div 4 = 45\). 4. Order the results: \(40 < 45 < 50\). 5. Match the letters to the values: B, C, A.

A) \(\frac{1}{3}\) of \(150\) is \(50\). B) \(\frac{1}{5}\) of \(200\) is \(40\). C) \(\frac{1}{4}\) of \(180\) is \(45\). From least to greatest: B, C, A, because \(40<45<50\).

- Use the denominator to think about how many equal groups make the whole. - What fraction completes two-thirds to make one whole?

1. Find one-third of the dots: \(54 \div 3 = 18\). 2. The fraction not shaded is the part that completes \(\frac{2}{3}\) to one whole, so it is \(\frac{1}{3}\). 3. Therefore, \(18\) dots are unshaded.

\(18\) dots are unshaded, and the unshaded fraction is \(\frac{1}{3}\).

- Count how many equal fraction models are shown. - Read the shaded fraction in one model. - The whole-number factor tells how many equal addends to write.

1. Each model shows \(\frac{2}{5}\), and there are \(4\) equal groups, so the multiplication expression is \(4\times\frac{2}{5}\). 2. Repeated addition is \(\frac{2}{5}+\frac{2}{5}+\frac{2}{5}+\frac{2}{5}\). 3. Altogether there are \(8\) fifths: \(\frac{8}{5}=1\frac{3}{5}\).

a) \(4\times\frac{2}{5}\) b) \(\frac{2}{5}+\frac{2}{5}+\frac{2}{5}+\frac{2}{5}\) c) \(1\frac{3}{5}\)

- Compare the new group count with the known group count. - Add one additional copy of the fraction. - Simplify the resulting improper fraction.

1. Four groups are one more group than three groups. 2. Add one more \(\frac{5}{8}\): \(\frac{15}{8}+\frac{5}{8}=\frac{20}{8}\). 3. Simplify \(\frac{20}{8}=\frac{5}{2}=2\frac{1}{2}\).

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

- Interpret the multiplication as several equal copies of the same unit fraction. - Ask whether making more copies changes the size of each fractional part. - Use repeated addition to check the product.

1. The product means five copies of one sixth. 2. As repeated addition, \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}\). 3. Multiplication changes the number of sixth-sized parts being counted; it does not change the size of each part. 4. Therefore, \(5\times\frac{1}{6}=\frac{5}{6}\), not \(\frac{5}{30}\).

Maya changed the denominator even though the parts are still sixths. \(\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{5}{6}\), so \(5\times\frac{1}{6}=\frac{5}{6}\).

- Find the exact product in the example. - Compare the product with the whole-number factor. - One counterexample is enough to disprove an “always” claim.

1. Compute the product: \(3\times\frac{1}{4}=\frac{3}{4}\). 2. The result \(\frac{3}{4}\) is less than the whole-number factor \(3\). 3. Therefore, the claim is false.

The claim is false because \(3\times\frac{1}{4}=\frac{3}{4}<3\).

- Compare the new whole-number factor with the known factor. - Apply the same multiplication to the known product. - Simplify and convert the result to a mixed number.

1. Eight groups are twice as many as four groups. 2. Double the known product: \(2\times\frac{12}{10}=\frac{24}{10}\). 3. Simplify \(\frac{24}{10}=\frac{12}{5}=2\frac{2}{5}\).

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

- Count how many fifth-sized parts each product represents. - In the first product, multiply the number of groups by the number of fifths in each group. - Equal numbers of fifths have equal values.

1. Three groups of four fifths contain \(3\times4=12\) fifths, so \(3\times\frac{4}{5}=\frac{12}{5}\). 2. Twelve groups of one fifth also contain \(12\) fifths, so \(12\times\frac{1}{5}=\frac{12}{5}\). 3. Therefore, the products are equal.

Yes. Both products equal \(\frac{12}{5}=2\frac{2}{5}\).

- A factor pair uses two whole numbers whose product is \(12\). - Start with the smallest possible factor and work upward without skipping possibilities. - Decide whether changing only the order changes which two factors are used.

1. A factor pair is a pair of whole numbers whose product is the target number. 2. \(1\times12=12\), \(2\times6=12\), and \(3\times4=12\), so the factor pairs are \(1\) and \(12\), \(2\) and \(6\), and \(3\) and \(4\). 3. Reversing a multiplication, such as \(12\times1\), uses the same two factors and therefore does not create a new factor pair.

\(1\) and \(12\); \(2\) and \(6\); \(3\) and \(4\). Reversing the order does not make a new factor pair.

- Use the grid itself to count the equal rows and the number of cells in each row. - Read the shaded fraction in one row before writing the multiplication expression. - Regroup the total number of eighths into wholes and a remaining fraction.

1. The model has \(4\) rows, and each row has \(\frac{3}{8}\) shaded, so it represents \(4\times\frac{3}{8}\). 2. Across all rows there are \(4\times3=12\) shaded eighth-sized cells. 3. The total is \(\frac{12}{8}=1\frac{4}{8}=1\frac{1}{2}\).

a) \(4\times\frac{3}{8}\) b) \(12\) eighth-sized cells c) \(1\frac{1}{2}\)

- Convert to the requested smaller unit before calculating. - Divide by the denominator and multiply by the numerator. - Finding one unit fraction first can make the calculation easier.

1. For a), \(240\div8=30\), and \(30\times3=90\). The result is \(90\,\text{kg}\). 2. For b), \(4\,\text{km}=4000\,\text{m}\). Then \(4000\div20=200\), and \(200\times7=1400\). The result is \(1400\,\text{m}\). 3. For c), \(2\,\text{h}=120\,\text{min}\). Then \(120\div12=10\), and \(10\times11=110\). The result is \(110\,\text{min}\).

a) \(90\,\text{kg}\) b) \(1400\,\text{m}\) c) \(110\,\text{min}\)

- Find the numerical value of each fractional amount. - Divide by the denominator, then multiply by the numerator. - Compare the two final measurements.

1. For Value A, \(420\div6=70\), and \(70\times5=350\). Thus, Value A is \(350\,\text{m}\). 2. For Value B, \(400\div8=50\), and \(50\times7=350\). Thus, Value B is \(350\,\text{m}\). 3. The values are equal.

The values are equal; both are \(350\,\text{m}\).

- Use the stated base fraction for each note before applying the dot rule. - Compare repeated copies of the resulting fractional durations. - Rewrite the dotted half note's value in eighths for part b).

1. A dotted quarter note has value \(\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\). Two have value \(2\times\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\). 2. Three undotted quarter notes have value \(3\times\frac{1}{4}=\frac{3}{4}\), so the durations are equal. 3. A dotted half note has value \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}=\frac{6}{8}\). Therefore, it has the duration of six eighth notes.

a) Both durations equal \(\frac{3}{4}\). b) \(6\) eighth notes

- Find the two values separately. - Record the intermediate results so you can compare the final values. - Use multiplication facts to help with the division.

1. Find Calculation A: \(3200 \div 8 = 400\), and \(5 \times 400 = 2000\). 2. Find Calculation B: \(2700 \div 3 = 900\), and \(2 \times 900 = 1800\). 3. Since \(2000 > 1800\), Calculation A has the greater result.

Calculation A, \(2000\), is greater than Calculation B, \(1800\).

- Find the value on the left side and then the value on the right side. - To find a fraction such as two-fifths, divide by the denominator and multiply by the numerator. - Compare the two final values in each part.

1. a) \(\frac{2}{5} \times 200 = 80\), and \(\frac{1}{2} \times 180 = 90\). Therefore, \(80 < 90\). 2. b) \(\frac{3}{4} \times 120 = 90\), and \(\frac{2}{3} \times 150 = 100\). Therefore, \(90 < 100\). 3. c) \(\frac{5}{10} \times 400 = 200\), and \(\frac{1}{4} \times 800 = 200\). Therefore, \(200 = 200\).

a) \(<\) b) \(<\) c) \(=\)

- Use the grid itself to determine the total number of squares. - Determine how many squares would represent one-half of that total. - Compare the target number of blue squares with the number already shaded.

1. The grid has \(4\) rows and \(5\) columns, so it contains \(4\times5=20\) squares. 2. One-half of \(20\) squares is \(10\) squares. 3. The model already has \(6\) blue squares, so \(10-6=4\) more squares must be shaded.

\(4\) more squares

- Use the given decomposition to split the product into two easier products. - Find each partial product separately. - Combine the two results.

1. Multiply each part by \(6\): \(6\times\frac{1}{2}=3\) and \(6\times\frac{1}{4}=\frac{6}{4}=1\frac{1}{2}\). 2. Add the results: \(3+1\frac{1}{2}=4\frac{1}{2}\).

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

- Think of the product as four equal groups of sixth-sized parts. - The total contains \(28\) sixths. - Determine how many sixths must be in each equal group.

1. Four equal groups must contain \(28\) sixth-sized parts altogether. 2. Split the \(28\) sixths equally among the \(4\) groups: \(28\div4=7\). 3. Each group is therefore \(\frac{7}{6}\), so the missing numerator is \(7\).

\(7\)

- Compute both products in exact fraction form. - Rewrite the whole-number product using fifths. - For part b), subtract the smaller result from the larger one.

1. \(7\times\frac{2}{5}=\frac{14}{5}=2\frac{4}{5}\). 2. \(4\times\frac{3}{4}=3=\frac{15}{5}\). 3. Since \(\frac{15}{5}>\frac{14}{5}\), \(4\times\frac{3}{4}\) is greater. 4. The difference is \(\frac{15}{5}-\frac{14}{5}=\frac{1}{5}\).

a) \(4\times\frac{3}{4}\) b) \(\frac{1}{5}\)

- Imagine the total if every bottle were completely full. - Combine the equal missing pieces from all eight bottles. - Remove the combined missing amount from the full-bottle total.

1. Eight completely full bottles would contain \(8\) bottles’ worth of water. 2. Their missing parts total \(8\times\frac{1}{8}=1\) bottle’s worth. 3. Subtract the missing amount: \(8-1=7\).

\(7\) bottles’ worth of water

- Express the mixed-number total entirely in fifths. - Compare the numerator contributed by one packet with the total numerator. - Find the whole-number factor that changes \(4\) fifths into \(16\) fifths.

1. Rewrite the total mass: \(3\frac{1}{5}=\frac{16}{5}\) pounds. 2. The equation becomes \(p\times\frac{4}{5}=\frac{16}{5}\). 3. Each group contributes four fifth-sized parts. Since \(4\times4=16\), \(p=4\).

\(p=4\), so there are \(4\) packets.

- For part a), compare the fraction factor with \(\frac{1}{2}\) before multiplying exactly. - For part b), multiply the numerator by the whole-number factor. - Convert the improper result to a mixed number.

1. Since \(\frac{5}{12}<\frac{1}{2}\), seven groups of \(\frac{5}{12}\) are less than seven groups of \(\frac{1}{2}\), which equal \(3\frac{1}{2}\). 2. The exact product is \(7\times\frac{5}{12}=\frac{35}{12}\). 3. Rewrite \(\frac{35}{12}=2\frac{11}{12}\).

a) Less than \(3\frac{1}{2}\) b) \(2\frac{11}{12}\)

- Compare \(\frac{3}{4}\) with \(\frac{1}{4}\). - Decide how many copies of the known product are needed. - Combine those copies and convert the result to a mixed number.

1. The fraction \(\frac{3}{4}\) is three copies of \(\frac{1}{4}\). 2. Therefore, \(6\times\frac{3}{4}\) is three times \(6\times\frac{1}{4}\). 3. Since \(6\times\frac{1}{4}=\frac{6}{4}\), three times that amount is \(\frac{18}{4}=4\frac{2}{4}=4\frac{1}{2}\).

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

- For part a), write each product as a number of eighths. - Compare each numerator with \(8\), and check whether it is a multiple of \(8\). - For part b), choose the least \(n\) listed in each category.

1. The products are \(\frac{3}{8},\frac{6}{8},\frac{9}{8},\frac{12}{8},\frac{15}{8},\frac{18}{8},\frac{21}{8},\frac{24}{8}\). 2. For \(n=1,2\), the products are proper fractions. For \(n=3,4,5,6,7\), they are mixed numbers. For \(n=8\), the product is a whole number. 3. The first values are \(n=1\) for a proper fraction, \(n=3\) for a mixed number, and \(n=8\) for a whole number.

a) Proper fraction: \(n=1,2\); mixed number: \(n=3,4,5,6,7\); whole number: \(n=8\) b) \(n=1\), \(n=3\), and \(n=8\), respectively

- Express the product as a single fraction. - Determine what must be true for the numerator to make a whole number of fourths. - Check every multiple of \(4\) in the allowed range.

1. The product is \(\frac{6a}{8}=\frac{3a}{4}\). 2. For the result to be a whole number, \(3a\) must be divisible by \(4\). 3. Because \(3\) is not divisible by \(2\), the factor of \(4\) must come from \(a\). Among \(1\) through \(7\), only \(a=4\) is divisible by \(4\). 4. It gives \(6\times\frac{4}{8}=3\). Therefore, \(a=4\) is the only solution.

\(a=4\)

- What part of the two multiplication expressions is the same? - Imagine subtracting the smaller amount from the larger amount inside each equal group. - Simplify the fraction that remains in one group before multiplying.

1. Both products have \(5\) equal groups, so subtract the fraction in each group: \(\frac{7}{8}-\frac{3}{8}=\frac{4}{8}\). 2. The expression becomes \(5\times\frac{4}{8}=\frac{20}{8}=2\frac{1}{2}\).

\(2\frac{1}{2}\). The shared factor means the fractions can be subtracted within each of the \(5\) equal groups before multiplying.

- For part a), use the model to determine one of the six equal shares. - Rename that share using fourths and eighths without changing its value. - For part b), multiply each equivalent share by the same group count.

1. Six equal groups share a total of \(3\), so each group is \(3\div6=\frac{1}{2}\). 2. Equivalent forms are \(\frac{1}{2}=\frac{2}{4}=\frac{4}{8}\). 3. The equations are \(6\times\frac{1}{2}=3\), \(6\times\frac{2}{4}=3\), and \(6\times\frac{4}{8}=3\).

a) \(\frac{1}{2},\frac{2}{4},\frac{4}{8}\) b) \(6\times\frac{1}{2}=3\); \(6\times\frac{2}{4}=3\); \(6\times\frac{4}{8}=3\)

- The products change by the same amount each time \(n\) increases by \(1\). - Find the products just below and just above \(2\). - Compare their exact distances from \(2\).

1. The products increase by \(\frac{5}{12}\) each time \(n\) increases by \(1\). 2. The products nearest \(2\) occur for \(n=4\) and \(n=5\): \(4\times\frac{5}{12}=\frac{20}{12}=1\frac{2}{3}\) and \(5\times\frac{5}{12}=\frac{25}{12}=2\frac{1}{12}\). 3. Their distances from \(2\) are \(\frac{1}{3}\) and \(\frac{1}{12}\), respectively. 4. Therefore, \(n=5\) gives the closest product.

\(5\times\frac{5}{12}=2\frac{1}{12}\), which is \(\frac{1}{12}\) from \(2\)

- Replace the factor \(9\) with the stated sum. - Find the amount contributed by each group of equal addends. - Check that the two partial group counts still total \(9\).

1. Split the factor: \(9\times\frac{5}{6}=6\times\frac{5}{6}+3\times\frac{5}{6}\). 2. The partial products are \(5\) and \(\frac{15}{6}=2\frac{1}{2}\). 3. Their sum is \(7\frac{1}{2}\). The split works because the same \(9\) equal groups were separated into groups of \(6\) and \(3\).

\(6\times\frac{5}{6}+3\times\frac{5}{6}=5+2\frac{1}{2}=7\frac{1}{2}\). Splitting the factor preserves the total because the two partial group counts still add to \(9\).

- Read the number of rows and the shaded fraction in one row from the grid. - Count the total number of eighth-sized shaded parts across all rows. - Compare the resulting mixed number with Nora's claim.

1. The grid has \(6\) equal rows, and \(\frac{3}{8}\) of each row is shaded, so it represents \(6\times\frac{3}{8}\). 2. Six groups contain \(18\) eighth-sized parts: \(6\times\frac{3}{8}=\frac{18}{8}\). 3. \(\frac{18}{8}=2\frac{2}{8}=2\frac{1}{4}\). 4. Nora is not correct because the shaded amount is one fourth more than \(2\) wholes.

\(6\times\frac{3}{8}=2\frac{1}{4}\). Nora is not correct.

- Identify what is identical in all three expressions. - Compare the amount in one group before considering all seven groups. - Ask whether repeating three ordered positive amounts equally can reverse their order.

1. Each expression has the same number of equal groups: \(7\). 2. The amount in one group is ordered \(\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\). 3. Repeating each positive amount the same number of times preserves that order.

\(7\times\frac{1}{3}<7\times\frac{1}{2}<7\times\frac{2}{3}\). The order is guaranteed because all three positive fractions are repeated the same number of times.

- Write the product as one fraction. - Determine what factor must be supplied to cancel the denominator \(8\). - List the multiples of that factor in the allowed range. - Check the neighboring multiples just outside the range to confirm completeness.

1. The product is \(\frac{3n}{8}\). 2. Since \(3\) has no factor of \(2\), \(n\) must supply all three factors of \(2\) in the denominator \(8\). Therefore, \(n\) must be a multiple of \(8\). 3. The multiples of \(8\) from \(1\) through \(24\) are \(8,16,24\). 4. The next multiples outside the range are \(0\) and \(32\), so the list contains every allowed value. The valid values increase by \(8\).

\(n=8,16,24\); the valid values are multiples of \(8\)

- First read the number of equal groups and the fraction in each group from the model. - Translate each description into a multiplication expression. - Compare the numbers of fifth-sized parts in the totals.

1. The model shows \(4\) equal groups of \(\frac{2}{5}\), so its total is \(\frac{8}{5}\). 2. A matches the model directly. 3. B gives \(2\times\frac{4}{5}=\frac{8}{5}\), so it has the same total. 4. C represents \(4\times5=20\), so it does not have the same value. 5. D gives \(8\times\frac{1}{5}=\frac{8}{5}\), so it also matches.

A, B, and D

- Read the shaded fraction in one row of each model. - Find the change in a single row before considering all the rows. - The same change repeats once for every row in the grid.

1. Each row in model a) has \(\frac{2}{8}\) shaded, while each row in model b) has \(\frac{5}{8}\) shaded. 2. The increase in one row is \(\frac{3}{8}\). 3. There are \(6\) rows, so the total increase is \(6\times\frac{3}{8}=\frac{18}{8}=2\frac{1}{4}\). 4. The same per-row increase occurs in all six rows, so it is repeated six times.

The total increases by \(2\frac{1}{4}\).

- Compute enough candidate products to decide which ones do not exceed \(3\). - Among allowed products, the larger product is closer to the upper limit. - Find the exact gap between the chosen product and \(3\).

1. Compute the candidate products: \(3\times\frac{5}{8}=\frac{15}{8}=1\frac{7}{8}\), \(4\times\frac{5}{8}=\frac{20}{8}=2\frac{1}{2}\), \(5\times\frac{5}{8}=\frac{25}{8}=3\frac{1}{8}\), and \(6\times\frac{5}{8}=\frac{30}{8}=3\frac{3}{4}\). 2. The products for \(n=5\) and \(n=6\) are greater than \(3\), so they are not allowed. 3. Of the remaining products, \(2\frac{1}{2}\) is closer to \(3\) than \(1\frac{7}{8}\). 4. Its distance from \(3\) is \(\frac{1}{2}\).

\(n=4\); the product is \(2\frac{1}{2}\), and its distance from \(3\) is \(\frac{1}{2}\)

- Determine the total number of eighth-yard parts in all eleven pieces. - Group those eighths into as many complete sets of eight as possible. - Report both the number of whole yards and the leftover fraction.

1. Eleven pieces contain \(11\times\frac{3}{8}=\frac{33}{8}\) yards. 2. Four complete yards use \(\frac{32}{8}\). 3. The remaining length is \(\frac{1}{8}\) yard.

\(4\) complete yards with \(\frac{1}{8}\) yard remaining

- Keep both the number of equal amounts and the size of each amount unchanged. - Make the meaning of the five equal amounts different in the two situations. - In each situation, identify what quantity the product measures.

1. One equal-groups example is five containers holding \(\frac{2}{5}\) liter each; the product represents total volume. 2. One movement example is five jumps of \(\frac{2}{5}\) meter each; the product represents total distance. 3. In either structure, \(5\times\frac{2}{5}=\frac{10}{5}=2\).

One valid pair of situations is: Equal groups: Five containers each hold \(\frac{2}{5}\,\text{L}\). Together they hold \(2\,\text{L}\). Equal moves: Elena makes five jumps of \(\frac{2}{5}\,\text{m}\) each. The total distance is \(2\,\text{m}\).

- Read the shaded tenths in one strip. - Pair the ten equal parts two at a time and determine what fraction name those pairs create. - Use the equivalent fraction for one group to find the total across five groups.

1. Each strip has \(6\) shaded tenths. Pairing the tenths gives \(3\) shaded fifths, so \(\frac{6}{10}=\frac{3}{5}\). 2. Relabeling equal-sized amounts does not change the value of any group. 3. Across five groups, the total is \(5\times\frac{3}{5}=3\).

\(\frac{6}{10}=\frac{3}{5}\), and the total of the five groups is \(3\).

- Compare the two whole-number factors before calculating either complete product. - Determine how many copies of the unknown fraction remain after subtraction. - Rename the required difference in tenths and split it equally among those copies.

1. The first product has two more equal groups than the second, so their difference is two copies of the unknown fraction. 2. Rename \(\frac{4}{5}\) as \(\frac{8}{10}\). Two equal copies must total \(\frac{8}{10}\), so each copy is \(\frac{4}{10}\). 3. The missing numerator is \(4\). 4. The products are \(6\times\frac{4}{10}=\frac{24}{10}\) and \(4\times\frac{4}{10}=\frac{16}{10}\); their difference is \(\frac{8}{10}=\frac{4}{5}\).

\(4\)

- For each product, count how many tenths appear in one group and how many groups there are. - Compare the two whole-number multiplication facts that count the tenths. - Describe what changes in the grouping and what does not change in the total.

1. Three groups of four tenths contain \(3\times4=12\) tenths, so \(3\times\frac{4}{10}=\frac{12}{10}\). 2. Four groups of three tenths contain \(4\times3=12\) tenths, so \(4\times\frac{3}{10}=\frac{12}{10}\). 3. Swapping the group count and the number of tenths in each group changes the arrangement but not the total number of tenths, because \(3\times4\) and \(4\times3\) count the same \(12\) parts.

a) Both products equal \(\frac{12}{10}\). b) Swapping the group count and the number of tenths per group keeps the same total number of tenths.

- For each fraction, let the denominator determine equal groups of the full collection. - Use the numerator to select the needed number of groups. - Compare the two selected counts after finding both.

1. \(\frac{4}{7}\times35\): divide \(35\) into \(7\) groups of \(5\), then take \(4\) groups, giving \(20\). 2. \(\frac{3}{5}\times35\): divide \(35\) into \(5\) groups of \(7\), then take \(3\) groups, giving \(21\). 3. The second selection is larger by \(21-20=1\) object.

\(\frac{3}{5}\) of the collection is larger: \(21\) objects versus \(20\), a difference of \(1\) object.

- First use the selected groups to determine the size of one equal group. - Extend that group size to the full number of equal groups. - Check that the stated fraction of your whole collection gives the known selected amount.

1. Seven equal groups contain \(21\) objects, so one group contains \(21\div7=3\) objects. 2. All \(15\) groups contain \(15\times3=45\) objects. 3. Therefore \(\frac{7}{15}\times45=21\), so the complete collection has \(45\) objects.

The complete collection has \(45\) objects. The fraction connection is \(\frac{7}{15}\times45=21\): dividing the whole into \(15\) equal groups gives \(3\) objects per group, and selecting \(7\) groups gives \(21\).

- Convert each product to a mixed number and record only what remains after the whole-number part. - Look for where a whole number appears and the cycle begins again. - Relate \(11\) to the position numbers in one complete cycle.

1. The products have fractional parts \(\frac{3}{4},\frac{1}{2},\frac{1}{4},0,\frac{3}{4},\frac{1}{2},\frac{1}{4},0\). 2. The four-term pattern repeats because every four groups add \(3\) whole units. 3. Since \(11\) has the same position in the cycle as \(3\), the fractional part is \(\frac{1}{4}\).

\(\frac{3}{4},\frac{1}{2},\frac{1}{4},0,\frac{3}{4},\frac{1}{2},\frac{1}{4},0\); the pattern repeats every four products. For \(n=11\), the fractional part is \(\frac{1}{4}\).