← Find the whole from a fractional part · Part-Whole Fraction Reasoning

Find the whole from a fractional part · 12 practice problems

5.NF.B.45.NF.B.6

Generated variants — 12

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 easy answer: 110 liters

From the water in a tank, $\dfrac{1}{5}$ of all the water was used. Then $\dfrac{7}{8}$ of what remained was poured out, leaving $11$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 1 of every 5 parts of its water, then loses 7 of every 8 parts of what is still there. After both, 11 liters remain. We need how much it held to begin with.

Givens

$\frac{1}{5}$ of the original water was used first.
$\frac{7}{8}$ of the remainder was poured out next.
11 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{1}{5}

leaves the rest of the whole.

1 - \frac{1}{5} = \frac{4}{5}

\frac{4}{5}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{7}{8}

of what remains keeps

\frac{1}{8}

of it -- and that is

\frac{1}{8}

\frac{4}{5}

of the original, so multiply.

\frac{4}{5} \times \frac{1}{8} = \frac{1}{10}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 11 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{10}

of the original equals 11 liters. Dividing 11 by 1 gives one part, and there are 10 of them in the whole.

11 \div 1 = 11,\quad 11 \times 10 = 110

The tank started with 110 liters.

Answer: 110 liters

4 · Reviewdoes it hold up?

Run it forwards: 1/5 of 110 is 22, leaving 88; pouring 7/8 of that leaves 11. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{1}{5}$ , then shading $\frac{7}{8}$ of only the unshaded part shows why the answer is 110.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 11 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 2 easy answer: 84 liters

From the water in a tank, $\dfrac{2}{7}$ of all the water was used. Then $\dfrac{4}{5}$ of what remained was poured out, leaving $12$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 2 of every 7 parts of its water, then loses 4 of every 5 parts of what is still there. After both, 12 liters remain. We need how much it held to begin with.

Givens

$\frac{2}{7}$ of the original water was used first.
$\frac{4}{5}$ of the remainder was poured out next.
12 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{2}{7}

leaves the rest of the whole.

1 - \frac{2}{7} = \frac{5}{7}

\frac{5}{7}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{4}{5}

of what remains keeps

\frac{1}{5}

of it -- and that is

\frac{1}{5}

\frac{5}{7}

of the original, so multiply.

\frac{5}{7} \times \frac{1}{5} = \frac{1}{7}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 12 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{7}

of the original equals 12 liters. Dividing 12 by 1 gives one part, and there are 7 of them in the whole.

12 \div 1 = 12,\quad 12 \times 7 = 84

The tank started with 84 liters.

Answer: 84 liters

4 · Reviewdoes it hold up?

Run it forwards: 2/7 of 84 is 24, leaving 60; pouring 4/5 of that leaves 12. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{2}{7}$ , then shading $\frac{4}{5}$ of only the unshaded part shows why the answer is 84.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 12 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 3 easy answer: 140 liters

From the water in a tank, $\dfrac{2}{5}$ of all the water was used. Then $\dfrac{5}{6}$ of what remained was poured out, leaving $14$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 2 of every 5 parts of its water, then loses 5 of every 6 parts of what is still there. After both, 14 liters remain. We need how much it held to begin with.

Givens

$\frac{2}{5}$ of the original water was used first.
$\frac{5}{6}$ of the remainder was poured out next.
14 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{2}{5}

leaves the rest of the whole.

1 - \frac{2}{5} = \frac{3}{5}

\frac{3}{5}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{5}{6}

of what remains keeps

\frac{1}{6}

of it -- and that is

\frac{1}{6}

\frac{3}{5}

of the original, so multiply.

\frac{3}{5} \times \frac{1}{6} = \frac{1}{10}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 14 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{10}

of the original equals 14 liters. Dividing 14 by 1 gives one part, and there are 10 of them in the whole.

14 \div 1 = 14,\quad 14 \times 10 = 140

The tank started with 140 liters.

Answer: 140 liters

4 · Reviewdoes it hold up?

Run it forwards: 2/5 of 140 is 56, leaving 84; pouring 5/6 of that leaves 14. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{2}{5}$ , then shading $\frac{5}{6}$ of only the unshaded part shows why the answer is 140.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 14 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 4 easy answer: 60 liters

From the water in a tank, $\dfrac{1}{4}$ of all the water was used. Then $\dfrac{2}{3}$ of what remained was poured out, leaving $15$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 1 of every 4 parts of its water, then loses 2 of every 3 parts of what is still there. After both, 15 liters remain. We need how much it held to begin with.

Givens

$\frac{1}{4}$ of the original water was used first.
$\frac{2}{3}$ of the remainder was poured out next.
15 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{1}{4}

leaves the rest of the whole.

1 - \frac{1}{4} = \frac{3}{4}

\frac{3}{4}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{2}{3}

of what remains keeps

\frac{1}{3}

of it -- and that is

\frac{1}{3}

\frac{3}{4}

of the original, so multiply.

\frac{3}{4} \times \frac{1}{3} = \frac{1}{4}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 15 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{4}

of the original equals 15 liters. Dividing 15 by 1 gives one part, and there are 4 of them in the whole.

15 \div 1 = 15,\quad 15 \times 4 = 60

The tank started with 60 liters.

Answer: 60 liters

4 · Reviewdoes it hold up?

Run it forwards: 1/4 of 60 is 15, leaving 45; pouring 2/3 of that leaves 15. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{1}{4}$ , then shading $\frac{2}{3}$ of only the unshaded part shows why the answer is 60.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 15 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 5 medium answer: 108 liters

From the water in a tank, $\dfrac{1}{3}$ of all the water was used. Then $\dfrac{3}{4}$ of what remained was poured out, leaving $18$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 1 of every 3 parts of its water, then loses 3 of every 4 parts of what is still there. After both, 18 liters remain. We need how much it held to begin with.

Givens

$\frac{1}{3}$ of the original water was used first.
$\frac{3}{4}$ of the remainder was poured out next.
18 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{1}{3}

leaves the rest of the whole.

1 - \frac{1}{3} = \frac{2}{3}

\frac{2}{3}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{3}{4}

of what remains keeps

\frac{1}{4}

of it -- and that is

\frac{1}{4}

\frac{2}{3}

of the original, so multiply.

\frac{2}{3} \times \frac{1}{4} = \frac{1}{6}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 18 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{6}

of the original equals 18 liters. Dividing 18 by 1 gives one part, and there are 6 of them in the whole.

18 \div 1 = 18,\quad 18 \times 6 = 108

The tank started with 108 liters.

Answer: 108 liters

4 · Reviewdoes it hold up?

Run it forwards: 1/3 of 108 is 36, leaving 72; pouring 3/4 of that leaves 18. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{1}{3}$ , then shading $\frac{3}{4}$ of only the unshaded part shows why the answer is 108.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 18 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 6 medium answer: 80 liters

From the water in a tank, $\dfrac{3}{8}$ of all the water was used. Then $\dfrac{3}{5}$ of what remained was poured out, leaving $20$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 3 of every 8 parts of its water, then loses 3 of every 5 parts of what is still there. After both, 20 liters remain. We need how much it held to begin with.

Givens

$\frac{3}{8}$ of the original water was used first.
$\frac{3}{5}$ of the remainder was poured out next.
20 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{3}{8}

leaves the rest of the whole.

1 - \frac{3}{8} = \frac{5}{8}

\frac{5}{8}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{3}{5}

of what remains keeps

\frac{2}{5}

of it -- and that is

\frac{2}{5}

\frac{5}{8}

of the original, so multiply.

\frac{5}{8} \times \frac{2}{5} = \frac{1}{4}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 20 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{4}

of the original equals 20 liters. Dividing 20 by 1 gives one part, and there are 4 of them in the whole.

20 \div 1 = 20,\quad 20 \times 4 = 80

The tank started with 80 liters.

Answer: 80 liters

4 · Reviewdoes it hold up?

Run it forwards: 3/8 of 80 is 30, leaving 50; pouring 3/5 of that leaves 20. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{3}{8}$ , then shading $\frac{3}{5}$ of only the unshaded part shows why the answer is 80.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 20 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 7 medium answer: 54 liters

From the water in a tank, $\dfrac{1}{6}$ of all the water was used. Then $\dfrac{5}{9}$ of what remained was poured out, leaving $20$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 1 of every 6 parts of its water, then loses 5 of every 9 parts of what is still there. After both, 20 liters remain. We need how much it held to begin with.

Givens

$\frac{1}{6}$ of the original water was used first.
$\frac{5}{9}$ of the remainder was poured out next.
20 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{1}{6}

leaves the rest of the whole.

1 - \frac{1}{6} = \frac{5}{6}

\frac{5}{6}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{5}{9}

of what remains keeps

\frac{4}{9}

of it -- and that is

\frac{4}{9}

\frac{5}{6}

of the original, so multiply.

\frac{5}{6} \times \frac{4}{9} = \frac{10}{27}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 20 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{10}{27}

of the original equals 20 liters. Dividing 20 by 10 gives one part, and there are 27 of them in the whole.

20 \div 10 = 2,\quad 2 \times 27 = 54

The tank started with 54 liters.

Answer: 54 liters

4 · Reviewdoes it hold up?

Run it forwards: 1/6 of 54 is 9, leaving 45; pouring 5/9 of that leaves 20. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{1}{6}$ , then shading $\frac{5}{9}$ of only the unshaded part shows why the answer is 54.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 20 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 8 medium answer: 88 liters

From the water in a tank, $\dfrac{5}{8}$ of all the water was used. Then $\dfrac{1}{3}$ of what remained was poured out, leaving $22$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 5 of every 8 parts of its water, then loses 1 of every 3 parts of what is still there. After both, 22 liters remain. We need how much it held to begin with.

Givens

$\frac{5}{8}$ of the original water was used first.
$\frac{1}{3}$ of the remainder was poured out next.
22 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{5}{8}

leaves the rest of the whole.

1 - \frac{5}{8} = \frac{3}{8}

\frac{3}{8}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{1}{3}

of what remains keeps

\frac{2}{3}

of it -- and that is

\frac{2}{3}

\frac{3}{8}

of the original, so multiply.

\frac{3}{8} \times \frac{2}{3} = \frac{1}{4}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 22 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{4}

of the original equals 22 liters. Dividing 22 by 1 gives one part, and there are 4 of them in the whole.

22 \div 1 = 22,\quad 22 \times 4 = 88

The tank started with 88 liters.

Answer: 88 liters

4 · Reviewdoes it hold up?

Run it forwards: 5/8 of 88 is 55, leaving 33; pouring 1/3 of that leaves 22. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{5}{8}$ , then shading $\frac{1}{3}$ of only the unshaded part shows why the answer is 88.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 22 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 9 hard answer: 70 liters

From the water in a tank, $\dfrac{3}{7}$ of all the water was used. Then $\dfrac{2}{5}$ of what remained was poured out, leaving $24$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 3 of every 7 parts of its water, then loses 2 of every 5 parts of what is still there. After both, 24 liters remain. We need how much it held to begin with.

Givens

$\frac{3}{7}$ of the original water was used first.
$\frac{2}{5}$ of the remainder was poured out next.
24 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{3}{7}

leaves the rest of the whole.

1 - \frac{3}{7} = \frac{4}{7}

\frac{4}{7}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{2}{5}

of what remains keeps

\frac{3}{5}

of it -- and that is

\frac{3}{5}

\frac{4}{7}

of the original, so multiply.

\frac{4}{7} \times \frac{3}{5} = \frac{12}{35}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 24 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{12}{35}

of the original equals 24 liters. Dividing 24 by 12 gives one part, and there are 35 of them in the whole.

24 \div 12 = 2,\quad 2 \times 35 = 70

The tank started with 70 liters.

Answer: 70 liters

4 · Reviewdoes it hold up?

Run it forwards: 3/7 of 70 is 30, leaving 40; pouring 2/5 of that leaves 24. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{3}{7}$ , then shading $\frac{2}{5}$ of only the unshaded part shows why the answer is 70.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 24 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 10 hard answer: 50 liters

From the water in a tank, $\dfrac{3}{10}$ of all the water was used. Then $\dfrac{2}{7}$ of what remained was poured out, leaving $25$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 3 of every 10 parts of its water, then loses 2 of every 7 parts of what is still there. After both, 25 liters remain. We need how much it held to begin with.

Givens

$\frac{3}{10}$ of the original water was used first.
$\frac{2}{7}$ of the remainder was poured out next.
25 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{3}{10}

leaves the rest of the whole.

1 - \frac{3}{10} = \frac{7}{10}

\frac{7}{10}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{2}{7}

of what remains keeps

\frac{5}{7}

of it -- and that is

\frac{5}{7}

\frac{7}{10}

of the original, so multiply.

\frac{7}{10} \times \frac{5}{7} = \frac{1}{2}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 25 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{2}

of the original equals 25 liters. Dividing 25 by 1 gives one part, and there are 2 of them in the whole.

25 \div 1 = 25,\quad 25 \times 2 = 50

The tank started with 50 liters.

Answer: 50 liters

4 · Reviewdoes it hold up?

Run it forwards: 3/10 of 50 is 15, leaving 35; pouring 2/7 of that leaves 25. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{3}{10}$ , then shading $\frac{2}{7}$ of only the unshaded part shows why the answer is 50.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 25 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 11 hard answer: 90 liters

From the water in a tank, $\dfrac{2}{9}$ of all the water was used. Then $\dfrac{4}{7}$ of what remained was poured out, leaving $30$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 2 of every 9 parts of its water, then loses 4 of every 7 parts of what is still there. After both, 30 liters remain. We need how much it held to begin with.

Givens

$\frac{2}{9}$ of the original water was used first.
$\frac{4}{7}$ of the remainder was poured out next.
30 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{2}{9}

leaves the rest of the whole.

1 - \frac{2}{9} = \frac{7}{9}

\frac{7}{9}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{4}{7}

of what remains keeps

\frac{3}{7}

of it -- and that is

\frac{3}{7}

\frac{7}{9}

of the original, so multiply.

\frac{7}{9} \times \frac{3}{7} = \frac{1}{3}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 30 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{1}{3}

of the original equals 30 liters. Dividing 30 by 1 gives one part, and there are 3 of them in the whole.

30 \div 1 = 30,\quad 30 \times 3 = 90

The tank started with 90 liters.

Answer: 90 liters

4 · Reviewdoes it hold up?

Run it forwards: 2/9 of 90 is 20, leaving 70; pouring 4/7 of that leaves 30. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{2}{9}$ , then shading $\frac{4}{7}$ of only the unshaded part shows why the answer is 90.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 30 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.

Variant 12 hard answer: 88 liters

From the water in a tank, $\dfrac{4}{11}$ of all the water was used. Then $\dfrac{3}{8}$ of what remained was poured out, leaving $35$ liters of water.

How many liters of water were in the tank at the start?

Show solution

1 · Understandwhat's really being asked

A tank loses 4 of every 11 parts of its water, then loses 3 of every 8 parts of what is still there. After both, 35 liters remain. We need how much it held to begin with.

Givens

$\frac{4}{11}$ of the original water was used first.
$\frac{3}{8}$ of the remainder was poured out next.
35 liters were left at the end.

Unknowns

The number of liters at the start.

Constraints

The second fraction is taken from what remained, not from the original amount.

2 · Planchoose the strategy

#11 Work Backwards · also uses: #7 Identify Subproblems

Track what fraction of the ORIGINAL survives each step. Two survival fractions multiply into one, and then the final amount can be scaled back up to the whole.

3 · Execute3 carry out the plan

1What fraction is left after the first step

#7 Identify Subproblems 5.NF.B.4

Using

\frac{4}{11}

leaves the rest of the whole.

1 - \frac{4}{11} = \frac{7}{11}

\frac{7}{11}

of the original water is still in the tank.

2What fraction is left after the second step

#7 Identify Subproblems 5.NF.B.4

Pouring out

\frac{3}{8}

of what remains keeps

\frac{5}{8}

of it -- and that is

\frac{5}{8}

\frac{7}{11}

of the original, so multiply.

\frac{7}{11} \times \frac{5}{8} = \frac{35}{88}

A fraction of a fraction is found by multiplying them.

3Match the leftover to 35 liters and work backwards

#11 Work Backwards 5.NF.B.6

\frac{35}{88}

of the original equals 35 liters. Dividing 35 by 35 gives one part, and there are 88 of them in the whole.

35 \div 35 = 1,\quad 1 \times 88 = 88

The tank started with 88 liters.

Answer: 88 liters

4 · Reviewdoes it hold up?

Run it forwards: 4/11 of 88 is 32, leaving 56; pouring 3/8 of that leaves 35. It matches.

Another way: Taking the second fraction from the original instead of the remainder is the trap here; drawing one bar, shading $\frac{4}{11}$ , then shading $\frac{3}{8}$ of only the unshaded part shows why the answer is 88.

Standardsmin grade 5

5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction — Multiplying the two survival fractions to get the fraction of the original that is left.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Scaling 35 liters back up to the whole tank.

💡Takeaway. When a share is taken from what is left, not from the original, multiply the leftover fractions instead of subtracting them.