Practice · A fraction of a quantity equals whole times fraction · Sensim Math

Variant 1 easy answer:

\frac{5}{6}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{4}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{2}{3}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 4 of A on top. C holds 2 over 3 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{4}$ of (A).
(C) is $\frac{2}{3}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 12

12 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{4}

of A, which is

\frac{5}{4}

of A altogether.

12 + 12 \times \frac{1}{4} = 15

(B) holds 15.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{2}{3}

of B -- of B, not of A.

15 \times \frac{2}{3} = 10

(C) holds 10.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{10}{12} = \frac{5}{6}

(C) is

\frac{5}{6}

of (A).

Answer:

\frac{5}{6}

4 · Reviewdoes it hold up?

Try a different starting amount, say 24: B becomes 30 and C becomes 20, and the ratio is still $\frac{5}{6}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{5}{4}$ times $\frac{2}{3}$ is $\frac{5}{6}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 2 easy answer:

\frac{9}{10}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{5}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{3}{4}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 5 of A on top. C holds 3 over 4 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{5}$ of (A).
(C) is $\frac{3}{4}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 20

20 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{5}

of A, which is

\frac{6}{5}

of A altogether.

20 + 20 \times \frac{1}{5} = 24

(B) holds 24.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{3}{4}

of B -- of B, not of A.

24 \times \frac{3}{4} = 18

(C) holds 18.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{18}{20} = \frac{9}{10}

(C) is

\frac{9}{10}

of (A).

Answer:

\frac{9}{10}

4 · Reviewdoes it hold up?

Try a different starting amount, say 40: B becomes 48 and C becomes 36, and the ratio is still $\frac{9}{10}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{6}{5}$ times $\frac{3}{4}$ is $\frac{9}{10}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 3 easy answer:

\frac{4}{3}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{2}{3}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{4}{5}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 2 over 3 of A on top. C holds 4 over 5 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{2}{3}$ of (A).
(C) is $\frac{4}{5}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 15

15 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{2}{3}

of A, which is

\frac{5}{3}

of A altogether.

15 + 15 \times \frac{2}{3} = 25

(B) holds 25.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{4}{5}

of B -- of B, not of A.

25 \times \frac{4}{5} = 20

(C) holds 20.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{20}{15} = \frac{4}{3}

(C) is

\frac{4}{3}

of (A).

Answer:

\frac{4}{3}

4 · Reviewdoes it hold up?

Try a different starting amount, say 30: B becomes 50 and C becomes 40, and the ratio is still $\frac{4}{3}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{5}{3}$ times $\frac{4}{5}$ is $\frac{4}{3}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 4 easy answer:

\frac{6}{5}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{2}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{4}{5}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 2 of A on top. C holds 4 over 5 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{2}$ of (A).
(C) is $\frac{4}{5}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 10

10 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{2}

of A, which is

\frac{3}{2}

of A altogether.

10 + 10 \times \frac{1}{2} = 15

(B) holds 15.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{4}{5}

of B -- of B, not of A.

15 \times \frac{4}{5} = 12

(C) holds 12.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{12}{10} = \frac{6}{5}

(C) is

\frac{6}{5}

of (A).

Answer:

\frac{6}{5}

4 · Reviewdoes it hold up?

Try a different starting amount, say 20: B becomes 30 and C becomes 24, and the ratio is still $\frac{6}{5}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{3}{2}$ times $\frac{4}{5}$ is $\frac{6}{5}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 5 medium answer:

\frac{7}{8}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{6}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{3}{4}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 6 of A on top. C holds 3 over 4 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{6}$ of (A).
(C) is $\frac{3}{4}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 24

24 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{6}

of A, which is

\frac{7}{6}

of A altogether.

24 + 24 \times \frac{1}{6} = 28

(B) holds 28.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{3}{4}

of B -- of B, not of A.

28 \times \frac{3}{4} = 21

(C) holds 21.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{21}{24} = \frac{7}{8}

(C) is

\frac{7}{8}

of (A).

Answer:

\frac{7}{8}

4 · Reviewdoes it hold up?

Try a different starting amount, say 48: B becomes 56 and C becomes 42, and the ratio is still $\frac{7}{8}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{7}{6}$ times $\frac{3}{4}$ is $\frac{7}{8}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 6 medium answer:

\frac{10}{9}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{3}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{5}{6}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 3 of A on top. C holds 5 over 6 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{3}$ of (A).
(C) is $\frac{5}{6}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 18

18 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{3}

of A, which is

\frac{4}{3}

of A altogether.

18 + 18 \times \frac{1}{3} = 24

(B) holds 24.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{5}{6}

of B -- of B, not of A.

24 \times \frac{5}{6} = 20

(C) holds 20.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{20}{18} = \frac{10}{9}

(C) is

\frac{10}{9}

of (A).

Answer:

\frac{10}{9}

4 · Reviewdoes it hold up?

Try a different starting amount, say 36: B becomes 48 and C becomes 40, and the ratio is still $\frac{10}{9}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{4}{3}$ times $\frac{5}{6}$ is $\frac{10}{9}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 7 medium answer:

\frac{5}{7}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{3}{7}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{1}{2}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 3 over 7 of A on top. C holds 1 over 2 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{3}{7}$ of (A).
(C) is $\frac{1}{2}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 14

14 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{3}{7}

of A, which is

\frac{10}{7}

of A altogether.

14 + 14 \times \frac{3}{7} = 20

(B) holds 20.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{1}{2}

of B -- of B, not of A.

20 \times \frac{1}{2} = 10

(C) holds 10.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{10}{14} = \frac{5}{7}

(C) is

\frac{5}{7}

of (A).

Answer:

\frac{5}{7}

4 · Reviewdoes it hold up?

Try a different starting amount, say 28: B becomes 40 and C becomes 20, and the ratio is still $\frac{5}{7}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{10}{7}$ times $\frac{1}{2}$ is $\frac{5}{7}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 8 medium answer:

\frac{1}{2}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{3}{4}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{2}{7}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 3 over 4 of A on top. C holds 2 over 7 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{3}{4}$ of (A).
(C) is $\frac{2}{7}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 28

28 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{3}{4}

of A, which is

\frac{7}{4}

of A altogether.

28 + 28 \times \frac{3}{4} = 49

(B) holds 49.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{2}{7}

of B -- of B, not of A.

49 \times \frac{2}{7} = 14

(C) holds 14.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{14}{28} = \frac{1}{2}

(C) is

\frac{1}{2}

of (A).

Answer:

\frac{1}{2}

4 · Reviewdoes it hold up?

Try a different starting amount, say 56: B becomes 98 and C becomes 28, and the ratio is still $\frac{1}{2}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{7}{4}$ times $\frac{2}{7}$ is $\frac{1}{2}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 9 hard answer:

\frac{27}{40}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{4}{5}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{3}{8}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 4 over 5 of A on top. C holds 3 over 8 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{4}{5}$ of (A).
(C) is $\frac{3}{8}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 40

40 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{4}{5}

of A, which is

\frac{9}{5}

of A altogether.

40 + 40 \times \frac{4}{5} = 72

(B) holds 72.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{3}{8}

of B -- of B, not of A.

72 \times \frac{3}{8} = 27

(C) holds 27.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{27}{40} = \frac{27}{40}

(C) is

\frac{27}{40}

of (A).

Answer:

\frac{27}{40}

4 · Reviewdoes it hold up?

Try a different starting amount, say 80: B becomes 144 and C becomes 54, and the ratio is still $\frac{27}{40}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{9}{5}$ times $\frac{3}{8}$ is $\frac{27}{40}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 10 hard answer:

\frac{49}{40}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{2}{5}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{7}{8}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 2 over 5 of A on top. C holds 7 over 8 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{2}{5}$ of (A).
(C) is $\frac{7}{8}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 40

40 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{2}{5}

of A, which is

\frac{7}{5}

of A altogether.

40 + 40 \times \frac{2}{5} = 56

(B) holds 56.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{7}{8}

of B -- of B, not of A.

56 \times \frac{7}{8} = 49

(C) holds 49.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{49}{40} = \frac{49}{40}

(C) is

\frac{49}{40}

of (A).

Answer:

\frac{49}{40}

4 · Reviewdoes it hold up?

Try a different starting amount, say 80: B becomes 112 and C becomes 98, and the ratio is still $\frac{49}{40}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{7}{5}$ times $\frac{7}{8}$ is $\frac{49}{40}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 11 hard answer:

\frac{55}{54}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{2}{9}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{5}{6}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 2 over 9 of A on top. C holds 5 over 6 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{2}{9}$ of (A).
(C) is $\frac{5}{6}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 54

54 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{2}{9}

of A, which is

\frac{11}{9}

of A altogether.

54 + 54 \times \frac{2}{9} = 66

(B) holds 66.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{5}{6}

of B -- of B, not of A.

66 \times \frac{5}{6} = 55

(C) holds 55.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{55}{54} = \frac{55}{54}

(C) is

\frac{55}{54}

of (A).

Answer:

\frac{55}{54}

4 · Reviewdoes it hold up?

Try a different starting amount, say 108: B becomes 132 and C becomes 110, and the ratio is still $\frac{55}{54}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{11}{9}$ times $\frac{5}{6}$ is $\frac{55}{54}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

Variant 12 hard answer:

\frac{1}{2}

There are three containers (A), (B), and (C) holding water. The amount of water in (B) is $\dfrac{1}{8}$ of (A) more than the amount in (A), and the amount of water in (C) is $\dfrac{4}{9}$ of the amount in (B). The amount of water in (C) is how many times the amount in (A)? Find the answer.

Show solution

1 · Understandwhat's really being asked

Three containers. B holds what A holds plus 1 over 8 of A on top. C holds 4 over 9 of what B holds. We need how many times A's amount C is.

Givens

(B) is (A) plus $\frac{1}{8}$ of (A).
(C) is $\frac{4}{9}$ of (B).
No actual amount is given for any container.

Unknowns

How many times (A)'s amount (C) is.

Constraints

The answer is a ratio, so it cannot depend on the actual amount in (A).

2 · Planchoose the strategy

#9 Solve an Easier Related Problem · also uses: #7 Identify Subproblems

Since the answer is a ratio, any amount for (A) gives the same result. Pick one that both fractions divide evenly, work forwards through B and C, and compare.

3 · Execute4 carry out the plan

1Choose an easy amount for (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Pick an amount both fractions divide evenly, so no step lands on a part-cup.

A = 72

72 cups keeps every step whole.

2Find (B) from (A)

#7 Identify Subproblems 5.NF.B.4

B is A plus

\frac{1}{8}

of A, which is

\frac{9}{8}

of A altogether.

72 + 72 \times \frac{1}{8} = 81

(B) holds 81.

3Find (C) from (B)

#7 Identify Subproblems 5.NF.B.4

C is

\frac{4}{9}

of B -- of B, not of A.

81 \times \frac{4}{9} = 36

(C) holds 36.

4Compare (C) to (A)

#9 Solve an Easier Related Problem 5.NF.B.6

Write C over A and reduce.

\frac{36}{72} = \frac{1}{2}

(C) is

\frac{1}{2}

of (A).

Answer:

\frac{1}{2}

4 · Reviewdoes it hold up?

Try a different starting amount, say 144: B becomes 162 and C becomes 72, and the ratio is still $\frac{1}{2}$ . The choice of A really does not matter.

Another way: Multiply the two steps directly: $\frac{9}{8}$ times $\frac{4}{9}$ is $\frac{1}{2}$ , no numbers needed -- but picking an amount is what makes it believable.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Taking a fraction of each container's amount in turn.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Choosing a workable amount and reading the final ratio.

💡Takeaway. When only ratios are given, pick a friendly amount to start -- the answer comes out the same whichever you choose.

A fraction of a quantity equals whole times fraction · 12 practice problems

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