Appendix A — Answer Key, Chapter 6: Problem Solving with Fractions

SOL 6.CE.1 d–e · Covers textbook Chapter 6 and the companion workbook. Item numbers match the textbook; workbook items that repeat textbook problems share the same answers, and workbook-only items are keyed at the end. Reasoning answers show an acceptable response, not the only wording. Every answer is in simplest form.

Lesson 6.1 — Estimating with Fractions

Guided practice

1. a) $0$ b) $\tfrac{1}{2}$ c) $1$ d) $\tfrac{1}{2}$
2. a) $3$ b) $7$ c) $3$
3. $\tfrac{5}{6} \approx 1$ and $\tfrac{1}{8} \approx 0$, so the sum is about $1$. (Exact: $\tfrac{23}{24}$.)
4. $8 - 2 = 6$. (Exact: $5\tfrac{2}{3}$.)
5. Adding a positive number to $\tfrac{1}{2}$ must give more than $\tfrac{1}{2}$, but $\tfrac{2}{5}$ is less than $\tfrac{1}{2}$. The correct sum is $\tfrac{5}{6}$.

Independent practice

6. a) $0$ b) $\tfrac{1}{2}$ c) $1$ d) $\tfrac{1}{2}$
7. a) $1 + 1 = 2$ (exact $1\tfrac{19}{24}$) b) $1 - 0 = 1$ (exact $\tfrac{49}{60}$) c) $5 + 3 = 8$ (exact $8\tfrac{1}{24}$)
8. a) $\tfrac{1}{2} \times 10 = 5$ (exact $4\tfrac{7}{8}$) b) $3 \times 6 = 18$ (exact $18\tfrac{23}{64}$)
9. $18 \div 3 = 6$. (Exact: $\tfrac{142}{23} = 6\tfrac{4}{23}$.)
10. Half of $12$ is $6$, and $7$ is only one more than $6$, so $\tfrac{7}{12}$ sits just above $\tfrac{1}{2}$. It is nowhere near $\tfrac{12}{12}$, so rounding to $1$ would nearly double the value.
11. Estimate: $3 + 2 = 5$ cups, which is exactly the size of the bag, so the estimate is too close to decide. Exact: $2\tfrac{3}{4} + 1\tfrac{7}{8} = 2\tfrac{6}{8} + 1\tfrac{7}{8} = 3\tfrac{13}{8} = 4\tfrac{5}{8}$ cups. She has enough, with $5 - 4\tfrac{5}{8} = \tfrac{3}{8}$ cup left over.
12. $\tfrac{5}{8}$ is less than 1, so the product must be less than $24$. Estimating with $\tfrac{1}{2} \times 24 = 12$ suggests an answer near 12. Exact: $\tfrac{5}{8} \times 24 = \tfrac{120}{8} = 15$.

Exit ticket 6.1

1. $\tfrac{1}{2}$ (half of 7 is 3.5, and the numerator 3 is close to that)
2. $7 + 3 = 10$. (Exact: $9\tfrac{23}{24}$.)
3. Use $27$: $\tfrac{1}{3} \times 27 = 9$. (Exact: $\tfrac{26}{3} = 8\tfrac{2}{3}$.)
4. Both $\tfrac{1}{2}$ and $\tfrac{1}{3}$ are positive, so the sum has to be greater than $\tfrac{1}{2}$; an estimate of about $\tfrac{1}{2} + \tfrac{1}{2} = 1$ shows the answer should be near 1. Since $\tfrac{2}{5} < \tfrac{1}{2}$, the answer cannot be right. The correct sum is $\tfrac{5}{6}$.

Lesson 6.2 — Addition and Subtraction Problems in Context

Guided practice

1. $\tfrac{3}{8} + \tfrac{1}{8} = \tfrac{4}{8} = \tfrac{1}{2}$ mile
2. $\tfrac{10}{12} - \tfrac{3}{12} = \tfrac{7}{12}$
3. $2\tfrac{2}{10} + 3\tfrac{3}{10} = 5\tfrac{5}{10} = 5\tfrac{1}{2}$ inches
4. $6 - 2\tfrac{3}{7} = 5\tfrac{7}{7} - 2\tfrac{3}{7} = 3\tfrac{4}{7}$
5. Estimate: $5 + 1 = 6$. Exact: $4\tfrac{21}{24} + 1\tfrac{4}{24} = 5\tfrac{25}{24} = 6\tfrac{1}{24}$, which is very close to the estimate.

Independent practice

6. $\tfrac{7}{10} + \tfrac{2}{10} = \tfrac{9}{10}$ gallon
7. $\tfrac{11}{12} - \tfrac{4}{12} = \tfrac{7}{12}$
8. $3\tfrac{9}{12} + 2\tfrac{10}{12} = 5\tfrac{19}{12} = 6\tfrac{7}{12}$
9. $9\tfrac{1}{6} - 4\tfrac{4}{6} = 8\tfrac{7}{6} - 4\tfrac{4}{6} = 4\tfrac{3}{6} = 4\tfrac{1}{2}$
10. LCD 12: $2\tfrac{4}{12} + 1\tfrac{6}{12} + \tfrac{9}{12} = 3\tfrac{19}{12} = 4\tfrac{7}{12}$ feet
11. Step 1: $5\tfrac{1}{8} - 1\tfrac{3}{4} = 5\tfrac{1}{8} - 1\tfrac{6}{8} = 4\tfrac{9}{8} - 1\tfrac{6}{8} = 3\tfrac{3}{8}$ gallons. Step 2: $3\tfrac{3}{8} + 2\tfrac{4}{8} = 5\tfrac{7}{8}$ gallons.
12. The student added numerators and denominators, which does not work because the pieces are different sizes. An estimate shows the sum must be more than $\tfrac{2}{3}$, but $\tfrac{1}{3}$ is less. Correct: $\tfrac{2}{3} + \tfrac{1}{6} = \tfrac{4}{6} + \tfrac{1}{6} = \tfrac{5}{6}$.

Exit ticket 6.2

1. $\tfrac{5}{12} + \tfrac{3}{12} = \tfrac{8}{12} = \tfrac{2}{3}$
2. $4\tfrac{1}{6} - 1\tfrac{5}{6} = 3\tfrac{7}{6} - 1\tfrac{5}{6} = 2\tfrac{2}{6} = 2\tfrac{1}{3}$
3. $3\tfrac{4}{8} - 1\tfrac{7}{8} = 2\tfrac{12}{8} - 1\tfrac{7}{8} = 1\tfrac{5}{8}$ yards
4. Rewrite both fractions with a common denominator and compare them. If the fraction being subtracted is larger than the fraction it is subtracted from, trade one whole for its equivalent fraction and add it to the top fraction, reducing the whole number by one.

Lesson 6.3 — Multiplication and Division Problems in Context

Guided practice

1. $\tfrac{1}{2} \times \tfrac{5}{6} = \tfrac{5}{12}$
2. $\tfrac{2}{5} \times 10 = \tfrac{20}{5} = 4$ students
3. $8 \div \tfrac{2}{3} = 8 \times \tfrac{3}{2} = 12$ scoops
4. $\tfrac{3}{4} \div 3 = \tfrac{3}{4} \times \tfrac{1}{3} = \tfrac{3}{12} = \tfrac{1}{4}$ pound each
5. $\tfrac{9}{4} \times \tfrac{4}{3} = \tfrac{36}{12} = 3$

Independent practice

6. $\tfrac{3}{8} \times \tfrac{2}{3} = \tfrac{6}{24} = \tfrac{1}{4}$
7. $5 \div \tfrac{5}{6} = 5 \times \tfrac{6}{5} = 6$ laps
8. $\tfrac{10}{3} \times \tfrac{21}{10} = \tfrac{210}{30} = 7$
9. $\tfrac{14}{3} \div \tfrac{7}{6} = \tfrac{14}{3} \times \tfrac{6}{7} = \tfrac{84}{21} = 4$
10. $\tfrac{5}{6} \div 4 = \tfrac{5}{6} \times \tfrac{1}{4} = \tfrac{5}{24}$ pound
11. $2\tfrac{1}{2} \times 1\tfrac{3}{5} = \tfrac{5}{2} \times \tfrac{8}{5} = \tfrac{40}{10} = 4$ square yards. Multiplying is right because area covers rows of equal length: the bed is $1\tfrac{3}{5}$ yards wide repeated across $2\tfrac{1}{2}$ yards of length.
12. Dividing by $\tfrac{2}{3}$ asks how many two-thirds fit inside 12. Since each group is smaller than 1, more than 12 groups fit. Exact: $12 \times \tfrac{3}{2} = 18$.

Exit ticket 6.3

1. $\tfrac{3}{5} \times \tfrac{2}{9} = \tfrac{6}{45} = \tfrac{2}{15}$
2. $9 \div \tfrac{3}{4} = 9 \times \tfrac{4}{3} = 12$
3. $\tfrac{5}{2} \div \tfrac{5}{8} = \tfrac{5}{2} \times \tfrac{8}{5} = \tfrac{40}{10} = 4$ pizzas
4. Taking $\tfrac{3}{4}$ of something means taking only part of it, and $\tfrac{3}{4}$ is less than 1 whole. So $\tfrac{3}{4} \times 8 = 6$, which is less than 8.

Lesson 6.4 — Multistep Problems and Justifying Your Answer

Guided practice

1. Used: $\tfrac{1}{4} + \tfrac{1}{3} = \tfrac{3}{12} + \tfrac{4}{12} = \tfrac{7}{12}$ gallon. Left: $\tfrac{7}{8} - \tfrac{7}{12} = \tfrac{21}{24} - \tfrac{14}{24} = \tfrac{7}{24}$ gallon.
2. $2\tfrac{1}{2} + 1\tfrac{3}{4} = 2\tfrac{2}{4} + 1\tfrac{3}{4} = 3\tfrac{5}{4} = 4\tfrac{1}{4}$; doubled, $8\tfrac{1}{2}$.
3. Ripe: $\tfrac{2}{3} \times 6 = 4$. Not ripe: $6 - 4 = 2$ melons.
4. Estimate: $10 - 3 - 3 = 4$. Exact: $9\tfrac{7}{8} - 3\tfrac{2}{8} = 6\tfrac{5}{8}$, then $6\tfrac{5}{8} - 2\tfrac{4}{8} = 4\tfrac{1}{8}$.
5. Less than 5. $\tfrac{1}{2} \times 9\tfrac{1}{3} = \tfrac{1}{2} \times \tfrac{28}{3} = \tfrac{28}{6} = \tfrac{14}{3} = 4\tfrac{2}{3}$, and $4\tfrac{2}{3} < 5$. (Half of 10 would be 5, and $9\tfrac{1}{3}$ is less than 10.)

Independent practice

6. $3\tfrac{3}{4} - 1\tfrac{1}{2} = 3\tfrac{3}{4} - 1\tfrac{2}{4} = 2\tfrac{1}{4}$ liters. Then $\tfrac{9}{4} \div 3 = \tfrac{9}{4} \times \tfrac{1}{3} = \tfrac{9}{12} = \tfrac{3}{4}$ liter per glass.
7. Hiked: $2\tfrac{3}{4} + 3\tfrac{1}{2} = 2\tfrac{3}{4} + 3\tfrac{2}{4} = 5\tfrac{5}{4} = 6\tfrac{1}{4}$ miles. Left: $8\tfrac{4}{12} - 6\tfrac{3}{12} = 2\tfrac{1}{12}$ miles.
8. Used: $\tfrac{2}{3} \times \tfrac{21}{4} = \tfrac{42}{12} = \tfrac{7}{2} = 3\tfrac{1}{2}$ pounds. Each container: $\tfrac{7}{2} \div 2 = \tfrac{7}{4} = 1\tfrac{3}{4}$ pounds.
9. Used: $4 \times 1\tfrac{3}{8} = 4 \times \tfrac{11}{8} = \tfrac{44}{8} = 5\tfrac{1}{2}$ yards. Left: $10 - 5\tfrac{1}{2} = 4\tfrac{1}{2}$ yards.
10. Estimate: $8 + 5 - 4 = 9$. Exact, LCD 12: $7\tfrac{10}{12} + 4\tfrac{6}{12} = 11\tfrac{16}{12} = 12\tfrac{4}{12}$; then $12\tfrac{4}{12} - 3\tfrac{9}{12} = 11\tfrac{16}{12} - 3\tfrac{9}{12} = 8\tfrac{7}{12}$.
11. Time used: $1\tfrac{1}{4} + \tfrac{2}{3} = 1\tfrac{3}{12} + \tfrac{8}{12} = 1\tfrac{11}{12}$ hours. Time left: $4\tfrac{6}{12} - 1\tfrac{11}{12} = 3\tfrac{18}{12} - 1\tfrac{11}{12} = 2\tfrac{7}{12}$ hours. Each chore: $\tfrac{31}{12} \div 2 = \tfrac{31}{24} = 1\tfrac{7}{24}$ hours. Justification: two chores at $1\tfrac{7}{24}$ hours total $\tfrac{62}{24} = 2\tfrac{7}{12}$ hours, which matches the time left, and roughly $1\tfrac{1}{4}$ hours each is reasonable for about $2\tfrac{1}{2}$ hours split in two.
12. $\tfrac{3}{4}$ is less than 1, so the product must be less than $8\tfrac{2}{3}$; an estimate of $\tfrac{3}{4} \times 9$ is under 7, so 26 is far too large. Rosa most likely multiplied by 3 and forgot to divide by 4. Correct: $\tfrac{3}{4} \times \tfrac{26}{3} = \tfrac{78}{12} = \tfrac{13}{2} = 6\tfrac{1}{2}$.

Exit ticket 6.4

1. LCD 6: $6\tfrac{3}{6} - 2\tfrac{2}{6} = 4\tfrac{1}{6}$; then $4\tfrac{1}{6} - 1\tfrac{3}{6} = 3\tfrac{7}{6} - 1\tfrac{3}{6} = 2\tfrac{4}{6} = 2\tfrac{2}{3}$.
2. Each piece: $\tfrac{9}{2} \div 6 = \tfrac{9}{12} = \tfrac{3}{4}$ pound. Two pieces: $\tfrac{3}{2} = 1\tfrac{1}{2}$ pounds.
3. Estimate: $6 + 2 = 8$. Exact: $5\tfrac{21}{24} + 2\tfrac{2}{24} = 7\tfrac{23}{24}$.
4. To justify an answer is to give a reason it is correct, not just to state it. For a subtraction, you can add the answer back to the amount subtracted and check that you get the original amount, or compare the answer to your estimate.

Chapter 6 Review

Part A — Estimating and reasonableness (6.CE.1d, 6.CE.1e)

1. a) $0$ b) $\tfrac{1}{2}$ c) $1$
2. $8 - 3 = 5$. (Exact: $\tfrac{42}{8} = 5\tfrac{1}{4}$.)
3. $\tfrac{1}{4} \times 20 = 5$. (Exact: $\tfrac{79}{16} = 4\tfrac{15}{16}$.)
4. Not reasonable. Both fractions are at least $\tfrac{1}{2}$, so the sum must be more than 1, but $\tfrac{8}{11}$ is less than 1. Correct: $\tfrac{18}{30} + \tfrac{25}{30} = \tfrac{43}{30} = 1\tfrac{13}{30}$.

Part B — Addition and subtraction in context (6.CE.1d)

5. LCD 24: $\tfrac{15}{24} + \tfrac{4}{24} = \tfrac{19}{24}$
6. LCD 12: $7\tfrac{3}{12} - 3\tfrac{8}{12} = 6\tfrac{15}{12} - 3\tfrac{8}{12} = 3\tfrac{7}{12}$
7. Combined: $12\tfrac{3}{4} + 9\tfrac{2}{4} = 21\tfrac{5}{4} = 22\tfrac{1}{4}$ pounds. Difference: $12\tfrac{3}{4} - 9\tfrac{2}{4} = 3\tfrac{1}{4}$ pounds.
8. LCD 12: $2\tfrac{10}{12} + 1\tfrac{9}{12} = 3\tfrac{19}{12} = 4\tfrac{7}{12}$ inches

Part C — Multiplication and division in context (6.CE.1e)

9. $\tfrac{12}{72} = \tfrac{1}{6}$
10. $\tfrac{21}{4} \times \tfrac{4}{3} = \tfrac{84}{12} = 7$
11. $\tfrac{3}{5} \times \tfrac{5}{2} = \tfrac{15}{10} = \tfrac{3}{2} = 1\tfrac{1}{2}$ hours
12. $\tfrac{21}{2} \div \tfrac{3}{4} = \tfrac{21}{2} \times \tfrac{4}{3} = \tfrac{84}{6} = 14$ bows

Part D — Multistep problems and justification (6.CE.1d, 6.CE.1e)

13. Inside the parentheses: $1\tfrac{2}{6} + 2\tfrac{5}{6} = 3\tfrac{7}{6} = 4\tfrac{1}{6}$. Then $9\tfrac{3}{6} - 4\tfrac{1}{6} = 5\tfrac{2}{6} = 5\tfrac{1}{3}$.
14. $\tfrac{2}{3} \times \tfrac{27}{4} = \tfrac{54}{12} = 4\tfrac{1}{2}$; then $4\tfrac{1}{2} - 1\tfrac{1}{2} = 3$.
15. They are equal. $\tfrac{3}{4} \times 8 = 6$ and $\tfrac{2}{3} \times 9 = 6$. Justification: in each case the denominator divides the whole number evenly, giving parts of 2 and 3 respectively, and three parts of 2 equal two parts of 3.
16. Estimate: $5 - 3 = 2$, so the answer should be near 2, not near $3\tfrac{1}{2}$. The student most likely subtracted the whole numbers and then handled the fractions backward. Correct: $5\tfrac{1}{4} - 2\tfrac{3}{4} = 4\tfrac{5}{4} - 2\tfrac{3}{4} = 2\tfrac{2}{4} = 2\tfrac{1}{2}$. Check: $2\tfrac{1}{2} + 2\tfrac{3}{4} = 4\tfrac{5}{4} = 5\tfrac{1}{4}$.

Workbook-only items

Page 2, benchmark table. $\tfrac{2}{9} \to 0$; $\tfrac{5}{9} \to \tfrac{1}{2}$; $\tfrac{8}{9} \to 1$; $\tfrac{1}{12} \to 0$; $\tfrac{7}{12} \to \tfrac{1}{2}$; $\tfrac{11}{12} \to 1$

Page 2, rounding table. $5\tfrac{2}{9} \to 5$; $7\tfrac{3}{4} \to 8$; $10\tfrac{1}{2} \to 11$; $3\tfrac{1}{12} \to 3$

Page 2, two rules. Multiplying by a number less than 1 makes the result smaller. Dividing by a number less than 1 makes the result larger.

Page 3, estimate table. $\tfrac{3}{4} + \tfrac{7}{8} \approx 1 + 1 = 2$ (exact $1\tfrac{5}{8}$). $\tfrac{11}{12} - \tfrac{1}{8} \approx 1 - 0 = 1$ (exact $\tfrac{19}{24}$). $6\tfrac{1}{8} + 2\tfrac{11}{12} \approx 6 + 3 = 9$ (exact $9\tfrac{1}{24}$). $9\tfrac{7}{8} - 4\tfrac{1}{6} \approx 10 - 4 = 6$ (exact $5\tfrac{17}{24}$). $\tfrac{1}{3} \times 14\tfrac{3}{4} \approx \tfrac{1}{3} \times 15 = 5$ (exact $4\tfrac{11}{12}$). $20\tfrac{1}{4} \div \tfrac{3}{4} \approx 21 \div \tfrac{3}{4} = 28$ (exact $27$).

Page 3, reasonable or not. $\tfrac{5}{6} + \tfrac{3}{4} = \tfrac{8}{10}$: not reasonable; both addends are more than $\tfrac{1}{2}$, so the sum exceeds 1. Correct: $\tfrac{10}{12} + \tfrac{9}{12} = \tfrac{19}{12} = 1\tfrac{7}{12}$. $\tfrac{3}{4} \times 12 = 9$: reasonable and correct. $8 \div \tfrac{1}{2} = 4$: not reasonable; dividing by a number less than 1 makes the result larger. Correct: $16$.

Page 3, apply it. Estimate $3 + 2 = 5$; too close to call from the estimate alone. Exact total $4\tfrac{5}{8}$ cups, so yes, she has enough, with $\tfrac{3}{8}$ cup left.

Page 5, compute. a) $\tfrac{7}{10}$ b) $\tfrac{2}{8} = \tfrac{1}{4}$ c) $\tfrac{5}{6}$ d) $\tfrac{9}{12} - \tfrac{2}{12} = \tfrac{7}{12}$ e) $\tfrac{4}{10} + \tfrac{5}{10} = \tfrac{9}{10}$ f) $\tfrac{7}{8} - \tfrac{4}{8} = \tfrac{3}{8}$

Page 5, word problems. 1. $\tfrac{9}{10}$ gallon. 2. $4\tfrac{7}{12}$ feet.

Page 6, compute and circle. a) $6\tfrac{4}{8} = 6\tfrac{1}{2}$ b) $4\tfrac{1}{3}$ c) $3\tfrac{5}{6} + 1\tfrac{3}{6} = 4\tfrac{8}{6} = 5\tfrac{1}{3}$ d) $6\tfrac{6}{5} - 2\tfrac{3}{5} = 4\tfrac{3}{5}$ e) $7\tfrac{12}{12} - 3\tfrac{5}{12} = 4\tfrac{7}{12}$ f) $5\tfrac{15}{12} - 2\tfrac{10}{12} = 3\tfrac{5}{12}$. Regrouping was needed in d, e, and f.

Page 6, apply it. Step 1: $3\tfrac{3}{8}$ gallons. Step 2: $5\tfrac{7}{8}$ gallons.

Page 6, explain. Write both fractions with a common denominator. If the fraction you are subtracting is greater than the fraction on top, you must regroup one whole first.

Page 8, match the operation. M; D; D; M

Page 8, compute. a) $\tfrac{3}{15} = \tfrac{1}{5}$ b) $\tfrac{10}{30} = \tfrac{1}{3}$ c) $\tfrac{2}{3} \times 6 = 4$ d) $\tfrac{3}{8}$ e) $\tfrac{3}{2} \times \tfrac{8}{3} = 4$ f) $\tfrac{9}{2} \times \tfrac{2}{3} = 3$

Page 9, word problems. 1. $\tfrac{7}{2} \div \tfrac{1}{4} = \tfrac{7}{2} \times 4 = 14$ scoops. 2. $\tfrac{2}{5} \times 45 = 18$ minutes. 3. $\tfrac{27}{4} \div 9 = \tfrac{27}{36} = \tfrac{3}{4}$ pound per box. 4. $6$ pieces. 5. $4$ square yards.

Page 9, explain. Each group is only $\tfrac{2}{3}$ of a unit, so more than 12 of them fit inside 12. The quotient is $18$.

Page 11, routine. Understand: how much flour is left. Estimate: about 2 to 3 cups. Step 1: $2\tfrac{5}{12}$ cups used. Step 2: $2\tfrac{5}{6}$ cups left. Justify: $2\tfrac{5}{6} + 2\tfrac{5}{12} = 5\tfrac{1}{4}$, the starting amount, and the result matches the estimate.

Page 11, your turn. 1. Poured out $2 \times \tfrac{3}{4} = 1\tfrac{1}{2}$ L; left $2\tfrac{1}{2} - 1\tfrac{1}{2} = 1$ L. 2. LCD 12: $1\tfrac{4}{12} + \tfrac{9}{12} + 1\tfrac{6}{12} = 2\tfrac{19}{12} = 3\tfrac{7}{12}$ hours. 3. $\tfrac{3}{4} \times \tfrac{16}{3} = \tfrac{48}{12} = 4$; then $4 - 1\tfrac{1}{4} = 2\tfrac{3}{4}$.

Page 12, error hunt. $\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{2}{6}$: false, correct $\tfrac{3}{4}$. $3\tfrac{1}{2} - 1\tfrac{3}{4} = 2\tfrac{1}{4}$: false, correct $1\tfrac{3}{4}$. $\tfrac{2}{3} \times \tfrac{3}{4} = \tfrac{1}{2}$: true. $5 \div \tfrac{1}{2} = 2\tfrac{1}{2}$: false, correct $10$. $2\tfrac{1}{4} + 1\tfrac{1}{2} = 3\tfrac{3}{4}$: true.

Page 12, apply it. Time used $1\tfrac{11}{12}$ hours; time left $2\tfrac{7}{12}$ hours; each chore $1\tfrac{7}{24}$ hours.

Page 12, explain. Since $\tfrac{3}{4} < 1$, the product must be less than $8\tfrac{2}{3}$, and an estimate gives about $6\tfrac{1}{2}$. The correct product is $6\tfrac{1}{2}$.