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

Variant 1 easy

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

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.

Your 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

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?