← Express smaller units as larger using fractions · Multiplicative Comparison and Unit Rate

Express smaller units as larger using fractions · 12 practice problems

5.NF.A.15.MD.A.15.NF.B.6

Generated variants — 12

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

Variant 1 easy answer: 12 L

There are two faucets that pour out water at a steady rate of $\dfrac{1}{4}\text{ L}$ and $3\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 12 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 1 over 4 litres a minute and the other at 3 and 1 over 2 litres a minute. They run for 3 minutes and 12 seconds. We need the total collected.

Givens

First tap: $\frac{1}{4}$ L per minute.
Second tap: $3\frac{1}{2}$ L per minute.
Both run for 3 minutes 12 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{1}{4} + \frac{7}{2} = \frac{15}{4}

Together they deliver

\frac{15}{4}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 12 seconds is 12 sixtieths of a minute.

3 + \frac{12}{60} = 3 + \frac{1}{5} = \frac{16}{5}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{15}{4} \times \frac{16}{5} = 12

They collect 12 L.

Answer: 12 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 3 L a minute for a bit over 3 minutes, so a total near 12 L is the right size.

Another way: Work each tap separately over the $\frac{16}{5}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 12 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 2 easy answer: 9 L

There are two faucets that pour out water at a steady rate of $\dfrac{1}{5}\text{ L}$ and $2\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 20 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 1 over 5 litres a minute and the other at 2 and 1 over 2 litres a minute. They run for 3 minutes and 20 seconds. We need the total collected.

Givens

First tap: $\frac{1}{5}$ L per minute.
Second tap: $2\frac{1}{2}$ L per minute.
Both run for 3 minutes 20 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{1}{5} + \frac{5}{2} = \frac{27}{10}

Together they deliver

\frac{27}{10}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 20 seconds is 20 sixtieths of a minute.

3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{27}{10} \times \frac{10}{3} = 9

They collect 9 L.

Answer: 9 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 2 L a minute for a bit over 3 minutes, so a total near 9 L is the right size.

Another way: Work each tap separately over the $\frac{10}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 20 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 3 easy answer: 14 L

There are two faucets that pour out water at a steady rate of $\dfrac{7}{8}\text{ L}$ and $1\dfrac{3}{4}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 5 minutes 20 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 7 over 8 litres a minute and the other at 1 and 3 over 4 litres a minute. They run for 5 minutes and 20 seconds. We need the total collected.

Givens

First tap: $\frac{7}{8}$ L per minute.
Second tap: $1\frac{3}{4}$ L per minute.
Both run for 5 minutes 20 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{7}{8} + \frac{7}{4} = \frac{21}{8}

Together they deliver

\frac{21}{8}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 20 seconds is 20 sixtieths of a minute.

5 + \frac{20}{60} = 5 + \frac{1}{3} = \frac{16}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{21}{8} \times \frac{16}{3} = 14

They collect 14 L.

Answer: 14 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 2 L a minute for a bit over 5 minutes, so a total near 14 L is the right size.

Another way: Work each tap separately over the $\frac{16}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 20 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 4 medium answer: 13 L

There are two faucets that pour out water at a steady rate of $\dfrac{2}{5}\text{ L}$ and $3\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 20 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 2 over 5 litres a minute and the other at 3 and 1 over 2 litres a minute. They run for 3 minutes and 20 seconds. We need the total collected.

Givens

First tap: $\frac{2}{5}$ L per minute.
Second tap: $3\frac{1}{2}$ L per minute.
Both run for 3 minutes 20 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{2}{5} + \frac{7}{2} = \frac{39}{10}

Together they deliver

\frac{39}{10}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 20 seconds is 20 sixtieths of a minute.

3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{39}{10} \times \frac{10}{3} = 13

They collect 13 L.

Answer: 13 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 3 L a minute for a bit over 3 minutes, so a total near 13 L is the right size.

Another way: Work each tap separately over the $\frac{10}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 20 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 5 easy answer: 12 L

There are two faucets that pour out water at a steady rate of $\dfrac{3}{4}\text{ L}$ and $1\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 5 minutes 20 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 3 over 4 litres a minute and the other at 1 and 1 over 2 litres a minute. They run for 5 minutes and 20 seconds. We need the total collected.

Givens

First tap: $\frac{3}{4}$ L per minute.
Second tap: $1\frac{1}{2}$ L per minute.
Both run for 5 minutes 20 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{3}{4} + \frac{3}{2} = \frac{9}{4}

Together they deliver

\frac{9}{4}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 20 seconds is 20 sixtieths of a minute.

5 + \frac{20}{60} = 5 + \frac{1}{3} = \frac{16}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{9}{4} \times \frac{16}{3} = 12

They collect 12 L.

Answer: 12 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 2 L a minute for a bit over 5 minutes, so a total near 12 L is the right size.

Another way: Work each tap separately over the $\frac{16}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 20 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 6 medium answer: 11 L

There are two faucets that pour out water at a steady rate of $\dfrac{4}{5}\text{ L}$ and $2\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 20 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 4 over 5 litres a minute and the other at 2 and 1 over 2 litres a minute. They run for 3 minutes and 20 seconds. We need the total collected.

Givens

First tap: $\frac{4}{5}$ L per minute.
Second tap: $2\frac{1}{2}$ L per minute.
Both run for 3 minutes 20 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{4}{5} + \frac{5}{2} = \frac{33}{10}

Together they deliver

\frac{33}{10}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 20 seconds is 20 sixtieths of a minute.

3 + \frac{20}{60} = 3 + \frac{1}{3} = \frac{10}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{33}{10} \times \frac{10}{3} = 11

They collect 11 L.

Answer: 11 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 3 L a minute for a bit over 3 minutes, so a total near 11 L is the right size.

Another way: Work each tap separately over the $\frac{10}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 20 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 7 medium answer: 9 L

There are two faucets that pour out water at a steady rate of $\dfrac{1}{3}\text{ L}$ and $1\dfrac{2}{3}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 4 minutes 30 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 1 over 3 litres a minute and the other at 1 and 2 over 3 litres a minute. They run for 4 minutes and 30 seconds. We need the total collected.

Givens

First tap: $\frac{1}{3}$ L per minute.
Second tap: $1\frac{2}{3}$ L per minute.
Both run for 4 minutes 30 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

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

Together they deliver 2 L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 30 seconds is 30 sixtieths of a minute.

4 + \frac{30}{60} = 4 + \frac{1}{2} = \frac{9}{2}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

2 \times \frac{9}{2} = 9

They collect 9 L.

Answer: 9 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 2 L a minute for a bit over 4 minutes, so a total near 9 L is the right size.

Another way: Work each tap separately over the $\frac{9}{2}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 30 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 8 hard answer: 15 L

There are two faucets that pour out water at a steady rate of $\dfrac{2}{3}\text{ L}$ and $3\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 36 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 2 over 3 litres a minute and the other at 3 and 1 over 2 litres a minute. They run for 3 minutes and 36 seconds. We need the total collected.

Givens

First tap: $\frac{2}{3}$ L per minute.
Second tap: $3\frac{1}{2}$ L per minute.
Both run for 3 minutes 36 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{2}{3} + \frac{7}{2} = \frac{25}{6}

Together they deliver

\frac{25}{6}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 36 seconds is 36 sixtieths of a minute.

3 + \frac{36}{60} = 3 + \frac{3}{5} = \frac{18}{5}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{25}{6} \times \frac{18}{5} = 15

They collect 15 L.

Answer: 15 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 4 L a minute for a bit over 3 minutes, so a total near 15 L is the right size.

Another way: Work each tap separately over the $\frac{18}{5}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 36 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 9 hard answer: 12 L

There are two faucets that pour out water at a steady rate of $\dfrac{5}{6}\text{ L}$ and $2\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 36 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 5 over 6 litres a minute and the other at 2 and 1 over 2 litres a minute. They run for 3 minutes and 36 seconds. We need the total collected.

Givens

First tap: $\frac{5}{6}$ L per minute.
Second tap: $2\frac{1}{2}$ L per minute.
Both run for 3 minutes 36 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{5}{6} + \frac{5}{2} = \frac{10}{3}

Together they deliver

\frac{10}{3}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 36 seconds is 36 sixtieths of a minute.

3 + \frac{36}{60} = 3 + \frac{3}{5} = \frac{18}{5}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{10}{3} \times \frac{18}{5} = 12

They collect 12 L.

Answer: 12 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 3 L a minute for a bit over 3 minutes, so a total near 12 L is the right size.

Another way: Work each tap separately over the $\frac{18}{5}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 36 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 10 medium answer: 15 L

There are two faucets that pour out water at a steady rate of $\dfrac{1}{2}\text{ L}$ and $3\dfrac{2}{3}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 3 minutes 36 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 1 over 2 litres a minute and the other at 3 and 2 over 3 litres a minute. They run for 3 minutes and 36 seconds. We need the total collected.

Givens

First tap: $\frac{1}{2}$ L per minute.
Second tap: $3\frac{2}{3}$ L per minute.
Both run for 3 minutes 36 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{1}{2} + \frac{11}{3} = \frac{25}{6}

Together they deliver

\frac{25}{6}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 36 seconds is 36 sixtieths of a minute.

3 + \frac{36}{60} = 3 + \frac{3}{5} = \frac{18}{5}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{25}{6} \times \frac{18}{5} = 15

They collect 15 L.

Answer: 15 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 4 L a minute for a bit over 3 minutes, so a total near 15 L is the right size.

Another way: Work each tap separately over the $\frac{18}{5}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 36 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 11 hard answer: 14 L

There are two faucets that pour out water at a steady rate of $\dfrac{3}{5}\text{ L}$ and $1\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 6 minutes 40 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 3 over 5 litres a minute and the other at 1 and 1 over 2 litres a minute. They run for 6 minutes and 40 seconds. We need the total collected.

Givens

First tap: $\frac{3}{5}$ L per minute.
Second tap: $1\frac{1}{2}$ L per minute.
Both run for 6 minutes 40 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{3}{5} + \frac{3}{2} = \frac{21}{10}

Together they deliver

\frac{21}{10}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 40 seconds is 40 sixtieths of a minute.

6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{21}{10} \times \frac{20}{3} = 14

They collect 14 L.

Answer: 14 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 2 L a minute for a bit over 6 minutes, so a total near 14 L is the right size.

Another way: Work each tap separately over the $\frac{20}{3}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 40 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.

Variant 12 hard answer: 8 L

There are two faucets that pour out water at a steady rate of $\dfrac{1}{6}\text{ L}$ and $1\dfrac{1}{2}\text{ L}$ per minute. If you turn on both faucets at the same time and collect water for 4 minutes 48 seconds, how many liters of water can you collect in all?

Show solution

1 · Understandwhat's really being asked

Two taps run at the same time, one at 1 over 6 litres a minute and the other at 1 and 1 over 2 litres a minute. They run for 4 minutes and 48 seconds. We need the total collected.

Givens

First tap: $\frac{1}{6}$ L per minute.
Second tap: $1\frac{1}{2}$ L per minute.
Both run for 4 minutes 48 seconds.

Unknowns

The total litres collected.

Constraints

The rates are per minute, so the time must be in minutes too.

2 · Planchoose the strategy

#8 Analyze the Units · also uses: #7 Identify Subproblems

Two taps into one container add up, so combine the rates first. Then make the time match those units -- litres-per-minute times minutes gives litres, but only if the seconds are converted first.

3 · Execute3 carry out the plan

1Combine the two per-minute rates

#7 Identify Subproblems 5.NF.A.1

Both pour into the same container, so their rates add. Give them a common denominator first.

\frac{1}{6} + \frac{3}{2} = \frac{5}{3}

Together they deliver

\frac{5}{3}

L each minute.

2Write the time as minutes

#8 Analyze the Units 5.MD.A.1

A minute is 60 seconds, so 48 seconds is 48 sixtieths of a minute.

4 + \frac{48}{60} = 4 + \frac{4}{5} = \frac{24}{5}\ \text{min}

Now the time is measured in the same unit the rate uses.

3Multiply rate by time

#8 Analyze the Units 5.NF.B.6

Litres per minute times minutes leaves litres -- the minutes cancel.

\frac{5}{3} \times \frac{24}{5} = 8

They collect 8 L.

Answer: 8 L

4 · Reviewdoes it hold up?

Round to check: the taps together give a bit over 1 L a minute for a bit over 4 minutes, so a total near 8 L is the right size.

Another way: Work each tap separately over the $\frac{24}{5}$ minutes and add the two totals; combining the rates first just saves one multiplication.

Standardsmin grade 5

5.NF.A.1 Add and subtract fractions with unlike denominators — Adding the two per-minute rates over a common denominator.
5.MD.A.1 Convert among different-sized standard measurement units within a given system — Turning 48 seconds into a fraction of a minute.
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers — Multiplying the combined rate by the time in minutes.

💡Takeaway. A rate tells you which unit the time must be in. Convert the seconds first, and the multiplication takes care of itself.