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

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

- Think about the number of quarter-cups in the total. - The two fractions have the same-sized units. - Which number tells how many groups fit?

1. Each portion is one fourth of a cup. 2. The total has three fourths, so it contains \(3\) portions. 3. \(\frac{3}{4}\div\frac{1}{4}=3\).

\(3\) portions; \(\frac{3}{4}\div\frac{1}{4}=3\).

- First consider how much of the plank one half-meter length would cover. - Compare the length remaining with the size of one half-meter. - Express the leftover part as a fraction of the half-meter length, not as a fraction of the whole plank.

1. The number of half-meter lengths is \(\frac{3}{4}\div\frac{1}{2}\). 2. One half-meter fits, leaving \(\frac{1}{4}\,\text{m}\). 3. The remaining quarter-meter is half of a half-meter, so there are \(1\frac{1}{2}\) half-meter lengths.

\(1\frac{1}{2}\) half-meter lengths.

- Imagine the total divided into equal sixth-sized parts. - Set aside one complete group and examine what remains. - Compare the remaining parts with the number of parts in a full group.

1. Both fractions are measured in sixths. One full group uses \(\frac{3}{6}\), leaving \(\frac{2}{6}\). 2. The remainder is \(\frac{2}{3}\) of a full group, because a group contains three sixths and two sixths remain. 3. Therefore, \(\frac{5}{6}\div\frac{3}{6}=\frac{5}{3}=1\frac{2}{3}\) groups.

\(1\frac{2}{3}\) groups: one full group and two-thirds of another.

- Compare the size of the total with the size of one group. - Think about one whole group of the stated size. - What part of that group does the total make?

1. One third is half of two thirds. 2. Therefore \(\frac{1}{3}\) is half of one group of \(\frac{2}{3}\). 3. The quotient is \(\frac{1}{2}\), which is less than \(1\).

\(\frac{1}{2}<1\). One third is half of a two-thirds group, so only half a group fits.

- 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

- 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.

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

- Could you express the total water as a number of eighth-liters? - How does the amount needed for one bottle relate to the total amount? - Check your bottle count by multiplying it by the amount in each bottle.

1. Divide the total amount of water by the amount in one bottle: \(3\div\frac{3}{8}\). 2. Multiply by the reciprocal: \(3\times\frac{8}{3}=8\). 3. Exactly \(8\) full bottles can be filled.

\(8\) full bottles.

- Which fraction tells you the size of one group? - Before multiplying, estimate whether more or fewer than two groups fit. - The check should recover the original amount from the group count and group size.

1. The reciprocal of \(\frac{2}{7}\) is \(\frac{7}{2}\). 2. Multiply: \(\frac{3}{5}\times\frac{7}{2}=\frac{21}{10}=2\frac{1}{10}\). 3. Check: \(\frac{21}{10}\times\frac{2}{7}=\frac{3}{5}\).

\(\frac{3}{5}\div\frac{2}{7}=\frac{21}{10}=2\frac{1}{10}\). Check: \(\frac{21}{10}\times\frac{2}{7}=\frac{3}{5}\).

- Rewrite the division as multiplication by a reciprocal. - Look for matching factors in the numerator and denominator of the product. - Reduce the final fraction.

1. Multiply by the reciprocal: \(\frac{5}{6}\times\frac{9}{5}\). 2. The two factors of \(5\) cancel, giving \(\frac{9}{6}=\frac{3}{2}\). 3. The quotient is \(1\frac{1}{2}\).

\(\frac{5}{6}\div\frac{5}{9}=\frac{5}{6}\times\frac{9}{5}=\frac{9}{6}=\frac{3}{2}=1\frac{1}{2}\). The factors of \(5\) cancel.

- Ask which number of equal groups produces the stated total. - Use the relationship between multiplication and division. - Check your divisor by substituting it in the original statement.

1. The missing divisor equals the total divided by the given quotient. 2. \(\frac{2}{5}\div\frac{4}{3}=\frac{2}{5}\times\frac{3}{4}=\frac{3}{10}\). 3. Check: \(\frac{2}{5}\div\frac{3}{10}=\frac{4}{3}\).

\(\frac{3}{10}\). Check: \(\frac{2}{5}\div\frac{3}{10}=\frac{2}{5}\times\frac{10}{3}=\frac{4}{3}\).

- Compare the batter available with one muffin's amount. - What is the most batter you can use for whole muffins without running out? - Subtract the amount you actually use to find the leftover.

1. \(\frac{3}{4}\div\frac{1}{3}=\frac{9}{4}=2\frac{1}{4}\), so there is enough for two whole muffins. 2. Two muffins use \(\frac{2}{3}\) cup. 3. The leftover is \(\frac{3}{4}-\frac{2}{3}=\frac{9}{12}-\frac{8}{12}=\frac{1}{12}\) cup.

\(2\) whole muffins; \(\frac{1}{12}\) cup of batter left.

- How can you determine the number of complete quarter-meter pieces that fit? - Find how much tape the full pieces use together. - What calculation gives the amount of tape not used?

1. The total length divided by one piece's length is \(\frac{7}{8}\div\frac{1}{4}=3\frac{1}{2}\). Only three full pieces can be cut. 2. Those pieces use \(3\times\frac{1}{4}=\frac{3}{4}\,\text{m}\). 3. The tape left over is \(\frac{7}{8}-\frac{3}{4}=\frac{1}{8}\,\text{m}\), too short for a fourth full piece.

a) \(3\) full pieces. b) \(\frac{1}{8}\,\text{m}\) of tape.

- 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.

- 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 \(\frac{15}{16}>\frac{5}{8}\).

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

- Which operation tells how many equal-length pieces fit into the roll? - How could you rewrite the mixed number before calculating? - Check your count by multiplying it by the ribbon needed for one bookmark.

1. Write the total as an improper fraction: \(2\frac{1}{4}=\frac{9}{4}=\frac{18}{8}\). 2. Divide by the ribbon for one bookmark: \(\frac{18}{8}\div\frac{3}{8}=6\). 3. The ribbon makes \(6\) bookmarks with none left over.

\(6\) bookmarks.

- How many equal bar parts are shaded? - The shaded amount is known. First find what one part represents. - Use the number of equal parts in the whole bar.

1. The diagram shows \(2\) of \(3\) equal parts shaded, or \(\frac{2}{3}\) of the whole. 2. One part is half of \(\frac{3}{4}\) cup, or \(\frac{3}{8}\) cup. 3. All three parts represent \(3\times\frac{3}{8}=\frac{9}{8}=1\frac{1}{8}\) 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.