Free math worksheets

- What relationship do the two fractions in part a) have? - How can division by \(2\) be rewritten as multiplication? - Simplify the result in part b).

1. For a), \(\frac{5}{9}\times\frac{9}{5}=1\). 2. The two fractions are reciprocals, so every factor in the numerator cancels with a matching factor in the denominator. 3. For b), \(\frac{4}{5}\div2=\frac{4}{5}\times\frac{1}{2}=\frac{2}{5}\).

a) \(1\), because a nonzero number multiplied by its reciprocal equals \(1\). b) \(\frac{2}{5}\)

- Count the equal fourths that make up the shaded amount. - Treat one fourth as the size of each group. - The quotient tells how many groups of that size fit in the total.

1. The bar has \(3\) shaded fourths, so the shaded amount is \(\frac{3}{4}\). 2. Each group has size \(\frac{1}{4}\). 3. There are \(3\) such groups, so \(\frac{3}{4}\div\frac{1}{4}=3\).

\(\frac{3}{4}\div\frac{1}{4}=3\)

- Rewrite division by a whole number as multiplication by its reciprocal. - Simplify common factors before multiplying. - Rewrite the mixed number as an improper fraction first.

1. For a), \(\frac{8}{11}\div4=\frac{8}{11}\times\frac{1}{4}=\frac{2}{11}\). 2. For b), \(\frac{5}{6}\div3=\frac{5}{6}\times\frac{1}{3}=\frac{5}{18}\). 3. For c), rewrite \(1 \frac{2}{3}=\frac{5}{3}\). Then \(\frac{5}{3}\div5=\frac{5}{3}\times\frac{1}{5}=\frac{1}{3}\).

a) \(\frac{2}{11}\) b) \(\frac{5}{18}\) c) \(\frac{1}{3}\)

- Equal sharing is represented by division. - Rewrite division by \(4\) as multiplication by \(\frac{1}{4}\). - Once you know the amount for one crate, multiply by \(3\) for part b).

1. For a), divide the total amount by \(4\): \(\frac{3}{5}\div4=\frac{3}{5}\times\frac{1}{4}=\frac{3}{20}\) gallon per crate. 2. For b), multiply the amount for one crate by \(3\): \(3\times\frac{3}{20}=\frac{9}{20}\) gallon.

a) \(\frac{3}{20}\) gallon per crate b) \(\frac{9}{20}\) gallon

- Count the equal intervals from \(0\) to \(\frac{2}{3}\). - Rewrite \(\frac{2}{3}\) as an equivalent fraction whose numerator matches that interval count. - Count equal fraction steps from \(0\) to each marker and simplify.

1. There are \(4\) equal intervals from \(0\) to \(\frac{2}{3}\). 2. Rewrite \(\frac{2}{3}\) as \(\frac{4}{6}\). Therefore, each interval represents \(\frac{1}{6}\). 3. Thus, \(U=\frac{1}{6}\), \(V=\frac{2}{6}=\frac{1}{3}\), and \(W=\frac{3}{6}=\frac{1}{2}\).

\(U=\frac{1}{6}\), \(V=\frac{1}{3}\), \(W=\frac{1}{2}\)

- Interpret division by \(2\) as splitting into two equal shares. - Use the equal segments on the number line. - Multiply by the reciprocal of the whole-number divisor.

1. Equal sharing is represented by \(\frac{4}{7}\div2\). 2. Divide by \(2\): \(\frac{4}{7}\div2=\frac{4}{7}\times\frac{1}{2}=\frac{4}{14}=\frac{2}{7}\). 3. The two equal segments on the number line each have length \(\frac{2}{7}\) mile.

\(\frac{2}{7}\,\text{mi}\)

- Interpret the quotient as the number of equal bottle-size groups. - Count how the total is partitioned on the number line. - Rewrite division by a fraction as multiplication by its reciprocal.

1. The question asks how many groups of \(\frac{3}{8}\) are in \(3\), so compute \(3\div\frac{3}{8}\). 2. Multiply by the reciprocal: \(3\div\frac{3}{8}=3\times\frac{8}{3}=8\). 3. The eight equal segments on the number line confirm that \(8\) full bottles can be filled.

\(8\) bottles

- Ask how many groups of \(\frac{3}{4}\) gallon fit into \(6\) gallons. - Rewrite division by a fraction as multiplication by its reciprocal. - Check whether \(8\) groups of \(\frac{3}{4}\) gallon total \(6\) gallons.

1. The number of watering cans is the total amount divided by the amount in each can: \(6\div\frac{3}{4}\). 2. Divide by a fraction by multiplying by its reciprocal: \(6\times\frac{4}{3}=\frac{24}{3}=8\). 3. Therefore, the fertilizer is divided among \(8\) watering cans.

8 watering cans

- Replace division by a fraction with multiplication by its reciprocal. - Cancel common factors across numerators and denominators before multiplying. - Rewrite mixed numbers as improper fractions first.

1. For a), multiply by the reciprocal: \(\frac{21}{40}\times\frac{15}{14}\). Cancel common factors to get \(\frac{9}{16}\). 2. For b), rewrite the mixed numbers: \(\frac{45}{8}\div\frac{9}{4}=\frac{45}{8}\times\frac{4}{9}=\frac{5}{2}=2 \frac{1}{2}\). 3. For c), \(\frac{48}{125}\times\frac{25}{16}=\frac{3}{5}\) after canceling common factors.

a) \(\frac{9}{16}\) b) \(2 \frac{1}{2}\) c) \(\frac{3}{5}\)

- Think about what happens when a positive number is divided by a number greater than \(1\) or by a number less than \(1\). - For part b), use the relationship between a dividend, divisor, and quotient. - For part c), find the reciprocal of \(\frac{2}{3}\).

1. For a), \(I=\frac{1}{4}\), \(II=1\), and \(III=\frac{1}{8}\). Therefore, \(III<I<II\). 2. For b), the unknown divisor is \(\frac{3}{7}\div\frac{9}{14}=\frac{3}{7}\times\frac{14}{9}=\frac{2}{3}\). 3. For c), dividing by \(\frac{2}{3}\) is the same as multiplying by its reciprocal, \(\frac{3}{2}\). Since \(\frac{3}{2}=1.5\), the statement is true.

a) \(III<I<II\) b) \(\square=\frac{2}{3}\) c) True, because dividing by \(\frac{2}{3}\) is equivalent to multiplying by \(\frac{3}{2}=1.5\).

- Try a simple positive fraction and compare multiplying it by \(2\) with dividing it by \(2\). - How can multiplication by \(n\) be written in the numerator? - How can division by \(n\) be rewritten as multiplication by a reciprocal?

1. Multiplying by \(n\) gives \(\frac{a}{b}\times n=\frac{an}{b}\). Because \(n>1\) and the fraction is positive, the result is greater than \(\frac{a}{b}\). 2. Dividing by \(n\) gives \(\frac{a}{b}\div n=\frac{a}{b}\times\frac{1}{n}=\frac{a}{bn}\). Because \(0<\frac{1}{n}<1\), the result is less than \(\frac{a}{b}\). 3. Multiplication makes the positive value \(n\) times as large, while division makes it \(\frac{1}{n}\) as large.

Multiplying \(\frac{a}{b}\) by \(n>1\) can be represented by multiplying the numerator by \(n\), so the value increases. Dividing by \(n\) can be represented by multiplying the denominator by \(n\), so the value decreases.

- Find the multiplication and division results separately first. - Simplify both results before comparing them. - To find how many times as large one value is, divide the larger value by the smaller value.

1. For 1), \(\frac{2}{5}\times4=\frac{8}{5}=1 \frac{3}{5}\). 2. For 2), \(\frac{2}{5}\div4=\frac{2}{5}\times\frac{1}{4}=\frac{1}{10}\). 3. For 3), compare the results: \(\frac{8}{5}\div\frac{1}{10}=16\). Multiplying by \(4\) makes the original value four times as large, while dividing by \(4\) makes it one fourth as large, so the two results differ by a factor of \(4\times4=16\).

1) \(\frac{8}{5}=1 \frac{3}{5}\) 2) \(\frac{1}{10}\) 3) The product is \(16\) times as large as the quotient.

- Convert each mixed number to an improper fraction first. - To divide by a fraction, think about which multiplication gives an equivalent expression. - Look for common factors that can be simplified before multiplying.

1. Convert the mixed numbers to improper fractions: \(3\frac{3}{4}=\frac{15}{4}\), \(1\frac{1}{2}=\frac{3}{2}\), and \(2\frac{2}{3}=\frac{8}{3}\). 2. For a), \(\frac{15}{4}\div\frac{5}{8}=\frac{15}{4}\times\frac{8}{5}=6\), so the result is \(6\,\text{m}\). 3. For b), \(\frac{3}{2}\div\frac{3}{4}=\frac{3}{2}\times\frac{4}{3}=2\), so the result is \(2\,\text{kg}\). 4. For c), \(\frac{8}{3}\div\frac{4}{9}=\frac{8}{3}\times\frac{9}{4}=6\), so the result is \(6\,\text{h}\).

a) \(6\,\text{m}\) b) \(2\,\text{kg}\) c) \(6\,\text{h}\)

- Find the amount in one bottle for each group. - Equal sharing is represented by division. - Rewrite the mixed numbers as improper fractions before dividing.

1. Group A has \(4 \frac{1}{2}\div6=\frac{9}{2}\times\frac{1}{6}=\frac{3}{4}\) quart per bottle. 2. Group B has \(3 \frac{1}{5}\div4=\frac{16}{5}\times\frac{1}{4}=\frac{4}{5}\) quart per bottle. 3. Compare the amounts: \(\frac{3}{4}=\frac{15}{20}\) and \(\frac{4}{5}=\frac{16}{20}\). 4. Since \(\frac{16}{20}>\frac{15}{20}\), Group B has more water in each bottle.

Group B. It has \(\frac{4}{5}\) quart per bottle, compared with \(\frac{3}{4}\) quart per bottle for Group A.

- How is dividing by \(2\) related to multiplying by \(\frac{1}{2}\)? - Rewrite or simplify the numbers in \(C\) and \(D\) before dividing. - Look for expressions that become identical after rewriting.

1. \(A=\frac{3}{4}\div2=\frac{3}{8}\). 2. \(B=\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}\). This is equivalent to dividing by \(2\). 3. For \(C\), rewrite \(1 \frac{1}{2}=\frac{3}{2}\). Then \(\frac{3}{2}\div4=\frac{3}{8}\). 4. For \(D\), simplify \(\frac{6}{4}=\frac{3}{2}\). Then \(\frac{3}{2}\div4=\frac{3}{8}\). 5. All four expressions have the same value.

All four expressions, \(A\), \(B\), \(C\), and \(D\), equal \(\frac{3}{8}\).

- A division expression can be written as a fraction with the dividend in the numerator and the divisor in the denominator. - To divide by a fraction, multiply by its reciprocal. - Cancel common factors before multiplying.

1. For a), the complex fraction is \(\frac{\frac{14}{15}}{\frac{7}{10}}\). Multiply by the reciprocal: \(\frac{14}{15}\times\frac{10}{7}=\frac{4}{3}\). 2. For b), \(12\div5=\frac{12}{5}\), so the complex fraction is \(\frac{\frac{12}{5}}{\frac{18}{25}}\). Then \(\frac{12}{5}\times\frac{25}{18}=\frac{10}{3}\). 3. For c), the complex fraction is \(\frac{\frac{22}{7}}{11}\). Then \(\frac{22}{7}\times\frac{1}{11}=\frac{2}{7}\).

a) \(\frac{\frac{14}{15}}{\frac{7}{10}}=\frac{4}{3}\) b) \(\frac{\frac{12}{5}}{\frac{18}{25}}=\frac{10}{3}\) c) \(\frac{\frac{22}{7}}{11}=\frac{2}{7}\)

- Use the parentheses to decide which division happens first. - In a complex fraction, the main fraction bar represents the outermost division. - Compare the numerator and denominator of the main fraction in the two cases.

1. For a), \((12\div4)\div2\) becomes \(\frac{\frac{12}{4}}{2}\). The numerator of the main fraction is \(\frac{12}{4}\), and the denominator is \(2\). Its value is \(\frac{3}{2}\). 2. For b), \(12\div(4\div2)\) becomes \(\frac{12}{\frac{4}{2}}\). The numerator is \(12\), and the denominator is \(\frac{4}{2}\). Its value is \(6\). 3. The results differ because the grouping determines which division is represented by the main fraction bar.

a) \(\frac{\frac{12}{4}}{2}=\frac{3}{2}\); main numerator: \(\frac{12}{4}\), main denominator: \(2\) b) \(\frac{12}{\frac{4}{2}}=6\); main numerator: \(12\), main denominator: \(\frac{4}{2}\)

- What happens when a positive number is divided by a number less than \(1\)? - What happens when the divisor is greater than \(1\)? - Rewrite the mixed number as an improper fraction before dividing. - Divide by a fraction by multiplying by its reciprocal. - Simplify common factors before multiplying.

1. In (1), the divisor \(\frac{2}{3}\) is less than \(1\), so dividing by it makes a positive number larger. In (2), the divisor \(1 \frac{1}{3}\) is greater than \(1\), so dividing by it makes the number smaller. 2. Therefore, calculation (1) should give a result greater than \(\frac{4}{5}\). 3. For (1), \(\frac{4}{5}\div\frac{2}{3}=\frac{4}{5}\times\frac{3}{2}=\frac{6}{5}=1 \frac{1}{5}\). 4. For (2), rewrite \(1 \frac{1}{3}=\frac{4}{3}\). Then \(\frac{4}{5}\div\frac{4}{3}=\frac{4}{5}\times\frac{3}{4}=\frac{3}{5}\).

a) Calculation (1), because its divisor is less than \(1\). b) (1) \(\frac{6}{5}=1 \frac{1}{5}\); (2) \(\frac{3}{5}\)

- First find the unknown fraction using the inverse operation. - Divide by a fraction by multiplying by its reciprocal. - Compare your result with Sarah’s claim.

1. Let the unknown fraction be \(x\). Then \(x=\frac{5}{12}\div\frac{5}{6}=\frac{5}{12}\times\frac{6}{5}=\frac{1}{2}\). 2. Test the claim: \(\frac{1}{2}\div\frac{1}{4}=\frac{1}{2}\times4=2\). 3. Since the result is \(2\), not \(4\), Sarah is not correct.

No. The unknown fraction is \(\frac{1}{2}\), and \(\frac{1}{2}\div\frac{1}{4}=2\).

- Decide whether each quotient should be greater or less than \(\frac{3}{5}\). - Compare dividing by \(3\) with dividing by \(\frac{1}{3}\). - Think about “sharing into 3 groups” versus “how many one-third groups fit.”

1. For 1), \(\frac{3}{5}\div3=\frac{3}{5}\times\frac{1}{3}=\frac{1}{5}\). This can represent sharing \(\frac{3}{5}\) equally among \(3\) groups. 2. For 2), \(\frac{3}{5}\div\frac{1}{3}=\frac{3}{5}\times3=\frac{9}{5}=1 \frac{4}{5}\). This asks how many groups of size \(\frac{1}{3}\) fit into \(\frac{3}{5}\). 3. The first quotient is smaller than the starting value, while the second is greater.

1) \(\frac{1}{5}\) 2) \(\frac{9}{5}=1 \frac{4}{5}\) In 1), the amount is shared among \(3\) equal groups. In 2), the question is how many \(\frac{1}{3}\)-sized groups fit into \(\frac{3}{5}\).

- Rewrite division by a fraction as multiplication by its reciprocal. - Simplify equivalent fractions before comparing expressions. - Consider what multiplication or division by a positive number less than \(1\) does to a positive quantity.

1. Dividing by \(\frac{5}{2}\) is equivalent to multiplying by its reciprocal \(\frac{2}{5}\). Therefore, A and B are equal. 2. Simplify \(\frac{4}{10}\) to \(\frac{2}{5}\). Therefore, A and C are equal. 3. Thus, A, B, and C all have equal values. 4. A multiplies \(30\) by a positive fraction less than \(1\), so A is less than \(30\). D divides \(30\) by a positive fraction less than \(1\), which is equivalent to multiplying by \(\frac{5}{2} > 1\), so D is greater than \(30\). Therefore, D is greater than A.

a) A, B, and C are equal. b) D is greater than A because multiplication by \(\frac{2}{5}\) decreases \(30\), while division by \(\frac{2}{5}\) increases it.

- Count the equal intervals from \(0\) to \(\frac{3}{4}\). - Rewrite \(\frac{3}{4}\) as an equivalent fraction whose numerator matches that interval count. - Count equal fraction steps from \(0\) to each marker.

1. There are \(6\) equal intervals from \(0\) to \(\frac{3}{4}\). 2. Rewrite \(\frac{3}{4}\) as \(\frac{6}{8}\). Therefore, each of those six equal intervals represents \(\frac{1}{8}\). 3. Point \(A\) is \(3\) intervals from \(0\), so \(A=\frac{3}{8}\). 4. Point \(B\) is \(10\) intervals from \(0\), so \(B=\frac{10}{8}=\frac{5}{4}=1\frac{1}{4}\). 5. Point \(C\) is \(14\) intervals from \(0\), so \(C=\frac{14}{8}=\frac{7}{4}=1\frac{3}{4}\).

\(A=\frac{3}{8}\), \(B=\frac{5}{4}=1\frac{1}{4}\), \(C=\frac{7}{4}=1\frac{3}{4}\)

- Use the labeled \(\frac{1}{4}\) to determine the value of one small interval. - Express the marked values, \(\frac{2}{5}\), and \(\frac{1}{2}\) with a common denominator. - Your final ordered list must contain fraction values rather than the letters \(D\), \(E\), and \(F\).

1. There are \(5\) equal intervals from \(0\) to \(\frac{1}{4}\). Rewrite \(\frac{1}{4}\) as \(\frac{5}{20}\), so each small interval represents \(\frac{1}{20}\). 2. Therefore, \(D=\frac{3}{20}\), \(E=\frac{7}{20}\), and \(F=\frac{12}{20}=\frac{3}{5}\). 3. Rewrite the two additional fractions in twentieths: \(\frac{2}{5}=\frac{8}{20}\) and \(\frac{1}{2}=\frac{10}{20}\). 4. Comparing the numerators gives \(\frac{3}{20}<\frac{7}{20}<\frac{8}{20}<\frac{10}{20}<\frac{12}{20}\).

\(D=\frac{3}{20}\), \(E=\frac{7}{20}\), \(F=\frac{3}{5}\) Order: \(\frac{3}{20}<\frac{7}{20}<\frac{2}{5}<\frac{1}{2}<\frac{3}{5}\)

- Identify the known amount and the fraction of the full container it represents. - Divide the known weight by \(\frac{2}{3}\) to recover the whole. - Check that the full amount is greater than the weight of a partly filled container.

1. The known weight \(\frac{5}{8}\,\text{kg}\) represents \(\frac{2}{3}\) of the full amount. 2. Divide by the fraction of the whole: \(\frac{5}{8}\div\frac{2}{3}=\frac{5}{8}\times\frac{3}{2}=\frac{15}{16}\). 3. Because \(\frac{2}{3}<1\), the full-container weight should be greater than \(\frac{5}{8}\,\text{kg}\), and \(\frac{15}{16}>\frac{5}{8}\).

\(\frac{15}{16}\,\text{kg}\)

- Express the mixed number in eighths. - Each colored segment represents one equal ribbon length. - Connect the number of equal groups to a fraction-division equation.

1. Convert the total length: \(2\frac{1}{4}=\frac{9}{4}=\frac{18}{8}\). 2. The number line partitions \(\frac{18}{8}\) into equal groups of \(\frac{3}{8}\). 3. Compute \(\frac{18}{8}\div\frac{3}{8}=\frac{18}{8}\times\frac{8}{3}=6\).

\(6\) bookmarks

- The two shaded parts together are \(\frac{2}{3}\) of the unknown whole. - First determine the value of one of the three equal bar parts. - Then combine all three equal parts to recover the whole amount.

1. The shaded \(\frac{2}{3}\) of the bar has value \(\frac{3}{4}\) cup, so the whole amount is \(\frac{3}{4}\div\frac{2}{3}\). 2. One third of the whole is half of \(\frac{3}{4}\), which is \(\frac{3}{8}\) cup. 3. Three thirds make the whole, so the whole amount is \(3\times\frac{3}{8}=\frac{9}{8}=1\frac{1}{8}\) cups.

\(1\frac{1}{8}\,\text{cups}\)

- Identify what is unknown in each situation: the number of groups or the size of each group. - For part a, rewrite the total in sixths. - For part b, think about splitting the numerator's amount into four equal shares.

1. For a), the question asks how many groups of size \(\frac{1}{6}\) fit in \(\frac{2}{3}\). Since \(\frac{2}{3}=\frac{4}{6}\), \(\frac{2}{3}\div\frac{1}{6}=4\). 2. For b), the question asks for one of \(4\) equal shares: \(\frac{2}{3}\div4=\frac{1}{6}\) cup. 3. Part a is measurement division because the group size is known and the number of groups is unknown. Part b is partitive division because the number of groups is known and the size of each group is unknown.

a) \(\frac{2}{3}\div\frac{1}{6}=4\) servings b) \(\frac{2}{3}\div4=\frac{1}{6}\,\text{cup}\) per child Part a asks how many groups fit; part b asks the size of each equal share.

- Rewrite the mixed number as an improper fraction. - Evaluate the denominator of the large fraction before doing the final division. - To divide by a fraction, multiply by its reciprocal.

1. Rewrite the numerator: \(3 \frac{1}{2}=\frac{7}{2}\). 2. Evaluate the denominator: \(\frac{7}{4}\div\frac{5}{2}=\frac{7}{4}\times\frac{2}{5}=\frac{7}{10}\). 3. The complex fraction is now \(\frac{\frac{7}{2}}{\frac{7}{10}}\). Divide by multiplying by the reciprocal: \(\frac{7}{2}\times\frac{10}{7}=5\).

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

- For division by a whole number, rewrite the whole number as a fraction and use its reciprocal. - When dividing by a fraction, which fraction must be inverted? - Rewrite a mixed number as one improper fraction before dividing. - Compare each incorrect step with the division rule.

1. For a), Lucas multiplied by \(3\) instead of dividing by \(3\). Correctly, \(\frac{3}{7}\div3=\frac{3}{7}\times\frac{1}{3}=\frac{1}{7}\). 2. For b), Lucas multiplied by the divisor instead of its reciprocal. Correctly, \(\frac{4}{5}\div\frac{2}{3}=\frac{4}{5}\times\frac{3}{2}=\frac{6}{5}=1 \frac{1}{5}\). 3. For c), Lucas separated the whole-number and fractional parts of the mixed number. Rewrite \(2 \frac{1}{2}=\frac{5}{2}\), then \(\frac{5}{2}\div\frac{1}{4}=\frac{5}{2}\times4=10\).

a) Error: multiplied by \(3\) instead of dividing. Correct result: \(\frac{1}{7}\). b) Error: did not use the reciprocal of the divisor. Correct result: \(\frac{6}{5}=1 \frac{1}{5}\). c) Error: split the mixed number incorrectly. Correct result: \(10\).

- First find the fraction of the original charge that remains after both stages. - For part b), \(14\,\text{Wh}\) represents the fraction you found in part a). - To recover the whole from a known fractional part, divide by that fraction.

1. After the morning, \(1-\frac{5}{8}=\frac{3}{8}\) of the original charge remains. 2. During the break, \(\frac{2}{9}\) of that remaining charge is used, so \(1-\frac{2}{9}=\frac{7}{9}\) of it remains. 3. For a), \(\frac{3}{8}\times\frac{7}{9}=\frac{21}{72}=\frac{7}{24}\) of the original charge remains. 4. For b), let \(C\) be the original capacity. Since \(\frac{7}{24}C=14\), divide by \(\frac{7}{24}\): \(C=14\div\frac{7}{24}=14\times\frac{24}{7}=48\). The original capacity was \(48\,\text{Wh}\).

a) \(\frac{7}{24}\) b) \(48\,\text{Wh}\)

- Rewrite the division as a related multiplication statement. - Work backward using the dividend and quotient. - Verify by rebuilding the dividend.

1. The divisor satisfies \(d\times\frac{9}{8}=\frac{3}{4}\). 2. Divide \(\frac{3}{4}\) by \(\frac{9}{8}\): \(d=\frac{3}{4}\times\frac{8}{9}=\frac{24}{36}=\frac{2}{3}\). 3. Check: \(\frac{2}{3}\times\frac{9}{8}=\frac{18}{24}=\frac{3}{4}\).

\(d=\frac{2}{3}\)

- Compare each divisor with the dividend before calculating. - Think about whether a full group of the divisor can fit. - Compute exact fractions to check the prediction.

1. Since \(\frac{3}{10}<\frac{4}{5}\), more than one group fits, while \(\frac{9}{10}>\frac{4}{5}\), so less than one group fits. 2. \(\frac{4}{5}\div\frac{3}{10}=\frac{4}{5}\times\frac{10}{3}=\frac{8}{3}\), which is greater than \(1\). 3. \(\frac{4}{5}\div\frac{9}{10}=\frac{4}{5}\times\frac{10}{9}=\frac{8}{9}\), which is less than \(1\).

a) The first quotient is greater than \(1\); the second is less than \(1\). b) \(\frac{8}{3}\) and \(\frac{8}{9}\)

- Use the quotient to determine the number of complete bags. - Find how much trail mix those complete bags use. - Subtract the used amount from the original total.

1. Convert the total to \(\frac{11}{6}\,\text{kg}\). 2. Compute \(\frac{11}{6}\div\frac{2}{9}=\frac{11}{6}\times\frac{9}{2}=\frac{33}{4}=8\frac{1}{4}\). 3. The quotient means \(8\) full bags can be filled. 4. Eight bags use \(8\times\frac{2}{9}=\frac{16}{9}\,\text{kg}\). 5. The remainder is \(\frac{11}{6}-\frac{16}{9}=\frac{33}{18}-\frac{32}{18}=\frac{1}{18}\,\text{kg}\).

a) \(8\) full bags b) \(\frac{1}{18}\,\text{kg}\)

- Choose one measurable total and one equal group size. - Make sure both fractions use the same unit. - Check that the quotient answers “how many groups?”

1. One valid situation is: A track is \(\frac{2}{3}\,\text{mi}\) long and is divided into sections of \(\frac{1}{6}\,\text{mi}\) each. 2. The number of sections is \(\frac{2}{3}\div\frac{1}{6}=\frac{2}{3}\times6=4\). 3. Other contexts are valid if they represent four groups of size \(\frac{1}{6}\) within \(\frac{2}{3}\).

Example: A \(\frac{2}{3}\,\text{mi}\) track is divided into \(\frac{1}{6}\,\text{mi}\) sections. It contains \(4\) sections.