Practice · Subtract the full-time pump's share first · Sensim Math

Variant 1 easy

Pump A alone can empty a swimming pool in $4$ hours, and pump B alone can empty it in $3$ hours. Pump B starts pumping water out of the pool at 2 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 4 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 2 p.m. to 4 p.m.; pump A joins at some point and runs until 4 p.m. The pool is exactly empty at 4 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $4$ hours.
Pump B empties the pool in $3$ hours.
B starts at 2 p.m., A joins later, and the pool is empty at 4 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 4 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $2$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{4}

of it per hour and pump B empties

\frac{1}{3}

per hour.

\text{A}: \frac{1}{4},\qquad \text{B}: \frac{1}{3}

A pump that needs

3

hours does

\frac{1}{3}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 2 p.m. to 4 p.m., which is

2

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{4}t = 1 - \frac{2}{3} = \frac{1}{3} \;\Rightarrow\; t = \frac{4}{3}

A only has

\frac{1}{3}

of the pool left to do, so it runs for less than the full

2

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{4}{3}

hours is

1

hour and

\frac{1}{3} \times 60 = 20

minutes. A stopped at 4 p.m., so count back.

4{:}00 - 1{:}20 = 2{:}40

A was turned on at 2:40 p.m.

Answer: 2:40 p.m.

4 · Reviewdoes it hold up?

B alone from 2:00 to 2:40 is $\frac{2}{3}$ hours, doing $\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$ . Together for $\frac{4}{3}$ hours they do $\frac{4}{3} \times \left(\frac{1}{3} + \frac{1}{4}\right) = \frac{4}{3} \times \frac{7}{12} = \frac{7}{9}$ . The total is $\frac{9}{9} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{3}x + \left(\frac{1}{3} + \frac{1}{4}\right)(2 - x) = 1$ , which gives $x = \frac{2}{3}$ , i.e. $40$ minutes after 2 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 2 easy

Pump A alone can empty a swimming pool in $3$ hours, and pump B alone can empty it in $5$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 4 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 4 p.m.; pump A joins at some point and runs until 4 p.m. The pool is exactly empty at 4 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $3$ hours.
Pump B empties the pool in $5$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 4 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 4 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $3$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{3}

of it per hour and pump B empties

\frac{1}{5}

per hour.

\text{A}: \frac{1}{3},\qquad \text{B}: \frac{1}{5}

A pump that needs

5

hours does

\frac{1}{5}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 4 p.m., which is

3

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{3}t = 1 - \frac{3}{5} = \frac{2}{5} \;\Rightarrow\; t = \frac{6}{5}

A only has

\frac{2}{5}

of the pool left to do, so it runs for less than the full

3

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{6}{5}

hours is

1

hour and

\frac{1}{5} \times 60 = 12

minutes. A stopped at 4 p.m., so count back.

4{:}00 - 1{:}12 = 2{:}48

A was turned on at 2:48 p.m.

Answer: 2:48 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 2:48 is $\frac{9}{5}$ hours, doing $\frac{9}{5} \times \frac{1}{5} = \frac{9}{25}$ . Together for $\frac{6}{5}$ hours they do $\frac{6}{5} \times \left(\frac{1}{5} + \frac{1}{3}\right) = \frac{6}{5} \times \frac{8}{15} = \frac{16}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{5}x + \left(\frac{1}{5} + \frac{1}{3}\right)(3 - x) = 1$ , which gives $x = \frac{9}{5}$ , i.e. $1$ hour $48$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 3 medium

Pump A alone can empty a swimming pool in $5$ hours, and pump B alone can empty it in $6$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 4 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 4 p.m.; pump A joins at some point and runs until 4 p.m. The pool is exactly empty at 4 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $5$ hours.
Pump B empties the pool in $6$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 4 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 4 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $3$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{5}

of it per hour and pump B empties

\frac{1}{6}

per hour.

\text{A}: \frac{1}{5},\qquad \text{B}: \frac{1}{6}

A pump that needs

6

hours does

\frac{1}{6}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 4 p.m., which is

3

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{5}t = 1 - \frac{1}{2} = \frac{1}{2} \;\Rightarrow\; t = \frac{5}{2}

A only has

\frac{1}{2}

of the pool left to do, so it runs for less than the full

3

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{5}{2}

hours is

2

hours and

\frac{1}{2} \times 60 = 30

minutes. A stopped at 4 p.m., so count back.

4{:}00 - 2{:}30 = 1{:}30

A was turned on at 1:30 p.m.

Answer: 1:30 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 1:30 is $\frac{1}{2}$ hours, doing $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ . Together for $\frac{5}{2}$ hours they do $\frac{5}{2} \times \left(\frac{1}{6} + \frac{1}{5}\right) = \frac{5}{2} \times \frac{11}{30} = \frac{11}{12}$ . The total is $\frac{12}{12} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{6}x + \left(\frac{1}{6} + \frac{1}{5}\right)(3 - x) = 1$ , which gives $x = \frac{1}{2}$ , i.e. $30$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 4 easy

Pump A alone can empty a swimming pool in $6$ hours, and pump B alone can empty it in $5$ hours. Pump B starts pumping water out of the pool at 2 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 6 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 2 p.m. to 6 p.m.; pump A joins at some point and runs until 6 p.m. The pool is exactly empty at 6 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $6$ hours.
Pump B empties the pool in $5$ hours.
B starts at 2 p.m., A joins later, and the pool is empty at 6 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 6 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $4$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{6}

of it per hour and pump B empties

\frac{1}{5}

per hour.

\text{A}: \frac{1}{6},\qquad \text{B}: \frac{1}{5}

A pump that needs

5

hours does

\frac{1}{5}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 2 p.m. to 6 p.m., which is

4

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{6}t = 1 - \frac{4}{5} = \frac{1}{5} \;\Rightarrow\; t = \frac{6}{5}

A only has

\frac{1}{5}

of the pool left to do, so it runs for less than the full

4

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{6}{5}

hours is

1

hour and

\frac{1}{5} \times 60 = 12

minutes. A stopped at 6 p.m., so count back.

6{:}00 - 1{:}12 = 4{:}48

A was turned on at 4:48 p.m.

Answer: 4:48 p.m.

4 · Reviewdoes it hold up?

B alone from 2:00 to 4:48 is $\frac{14}{5}$ hours, doing $\frac{14}{5} \times \frac{1}{5} = \frac{14}{25}$ . Together for $\frac{6}{5}$ hours they do $\frac{6}{5} \times \left(\frac{1}{5} + \frac{1}{6}\right) = \frac{6}{5} \times \frac{11}{30} = \frac{11}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{5}x + \left(\frac{1}{5} + \frac{1}{6}\right)(4 - x) = 1$ , which gives $x = \frac{14}{5}$ , i.e. $2$ hours $48$ minutes after 2 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 5 easy

Pump A alone can empty a swimming pool in $6$ hours, and pump B alone can empty it in $5$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 4 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 4 p.m.; pump A joins at some point and runs until 4 p.m. The pool is exactly empty at 4 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $6$ hours.
Pump B empties the pool in $5$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 4 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 4 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $3$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{6}

of it per hour and pump B empties

\frac{1}{5}

per hour.

\text{A}: \frac{1}{6},\qquad \text{B}: \frac{1}{5}

A pump that needs

5

hours does

\frac{1}{5}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 4 p.m., which is

3

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{6}t = 1 - \frac{3}{5} = \frac{2}{5} \;\Rightarrow\; t = \frac{12}{5}

A only has

\frac{2}{5}

of the pool left to do, so it runs for less than the full

3

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{12}{5}

hours is

2

hours and

\frac{2}{5} \times 60 = 24

minutes. A stopped at 4 p.m., so count back.

4{:}00 - 2{:}24 = 1{:}36

A was turned on at 1:36 p.m.

Answer: 1:36 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 1:36 is $\frac{3}{5}$ hours, doing $\frac{3}{5} \times \frac{1}{5} = \frac{3}{25}$ . Together for $\frac{12}{5}$ hours they do $\frac{12}{5} \times \left(\frac{1}{5} + \frac{1}{6}\right) = \frac{12}{5} \times \frac{11}{30} = \frac{22}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{5}x + \left(\frac{1}{5} + \frac{1}{6}\right)(3 - x) = 1$ , which gives $x = \frac{3}{5}$ , i.e. $36$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 6 medium

Pump A alone can empty a swimming pool in $8$ hours, and pump B alone can empty it in $5$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 5 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 5 p.m.; pump A joins at some point and runs until 5 p.m. The pool is exactly empty at 5 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $8$ hours.
Pump B empties the pool in $5$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 5 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 5 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $4$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{8}

of it per hour and pump B empties

\frac{1}{5}

per hour.

\text{A}: \frac{1}{8},\qquad \text{B}: \frac{1}{5}

A pump that needs

5

hours does

\frac{1}{5}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 5 p.m., which is

4

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{8}t = 1 - \frac{4}{5} = \frac{1}{5} \;\Rightarrow\; t = \frac{8}{5}

A only has

\frac{1}{5}

of the pool left to do, so it runs for less than the full

4

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{8}{5}

hours is

1

hour and

\frac{3}{5} \times 60 = 36

minutes. A stopped at 5 p.m., so count back.

5{:}00 - 1{:}36 = 3{:}24

A was turned on at 3:24 p.m.

Answer: 3:24 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 3:24 is $\frac{12}{5}$ hours, doing $\frac{12}{5} \times \frac{1}{5} = \frac{12}{25}$ . Together for $\frac{8}{5}$ hours they do $\frac{8}{5} \times \left(\frac{1}{5} + \frac{1}{8}\right) = \frac{8}{5} \times \frac{13}{40} = \frac{13}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{5}x + \left(\frac{1}{5} + \frac{1}{8}\right)(4 - x) = 1$ , which gives $x = \frac{12}{5}$ , i.e. $2$ hours $24$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 7 medium

Pump A alone can empty a swimming pool in $4$ hours, and pump B alone can empty it in $9$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 4 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 4 p.m.; pump A joins at some point and runs until 4 p.m. The pool is exactly empty at 4 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $4$ hours.
Pump B empties the pool in $9$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 4 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 4 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $3$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{4}

of it per hour and pump B empties

\frac{1}{9}

per hour.

\text{A}: \frac{1}{4},\qquad \text{B}: \frac{1}{9}

A pump that needs

9

hours does

\frac{1}{9}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 4 p.m., which is

3

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{4}t = 1 - \frac{1}{3} = \frac{2}{3} \;\Rightarrow\; t = \frac{8}{3}

A only has

\frac{2}{3}

of the pool left to do, so it runs for less than the full

3

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{8}{3}

hours is

2

hours and

\frac{2}{3} \times 60 = 40

minutes. A stopped at 4 p.m., so count back.

4{:}00 - 2{:}40 = 1{:}20

A was turned on at 1:20 p.m.

Answer: 1:20 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 1:20 is $\frac{1}{3}$ hours, doing $\frac{1}{3} \times \frac{1}{9} = \frac{1}{27}$ . Together for $\frac{8}{3}$ hours they do $\frac{8}{3} \times \left(\frac{1}{9} + \frac{1}{4}\right) = \frac{8}{3} \times \frac{13}{36} = \frac{26}{27}$ . The total is $\frac{27}{27} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{9}x + \left(\frac{1}{9} + \frac{1}{4}\right)(3 - x) = 1$ , which gives $x = \frac{1}{3}$ , i.e. $20$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 8 medium

Pump A alone can empty a swimming pool in $6$ hours, and pump B alone can empty it in $9$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 5 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 5 p.m.; pump A joins at some point and runs until 5 p.m. The pool is exactly empty at 5 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $6$ hours.
Pump B empties the pool in $9$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 5 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 5 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $4$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{6}

of it per hour and pump B empties

\frac{1}{9}

per hour.

\text{A}: \frac{1}{6},\qquad \text{B}: \frac{1}{9}

A pump that needs

9

hours does

\frac{1}{9}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 5 p.m., which is

4

hours.

4 \times \frac{1}{9} = \frac{4}{9}

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{6}t = 1 - \frac{4}{9} = \frac{5}{9} \;\Rightarrow\; t = \frac{10}{3}

A only has

\frac{5}{9}

of the pool left to do, so it runs for less than the full

4

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{10}{3}

hours is

3

hours and

\frac{1}{3} \times 60 = 20

minutes. A stopped at 5 p.m., so count back.

5{:}00 - 3{:}20 = 1{:}40

A was turned on at 1:40 p.m.

Answer: 1:40 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 1:40 is $\frac{2}{3}$ hours, doing $\frac{2}{3} \times \frac{1}{9} = \frac{2}{27}$ . Together for $\frac{10}{3}$ hours they do $\frac{10}{3} \times \left(\frac{1}{9} + \frac{1}{6}\right) = \frac{10}{3} \times \frac{5}{18} = \frac{25}{27}$ . The total is $\frac{27}{27} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{9}x + \left(\frac{1}{9} + \frac{1}{6}\right)(4 - x) = 1$ , which gives $x = \frac{2}{3}$ , i.e. $40$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 9 hard

Pump A alone can empty a swimming pool in $4$ hours, and pump B alone can empty it in $10$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 5 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 5 p.m.; pump A joins at some point and runs until 5 p.m. The pool is exactly empty at 5 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $4$ hours.
Pump B empties the pool in $10$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 5 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 5 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $4$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{4}

of it per hour and pump B empties

\frac{1}{10}

per hour.

\text{A}: \frac{1}{4},\qquad \text{B}: \frac{1}{10}

A pump that needs

10

hours does

\frac{1}{10}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 5 p.m., which is

4

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{4}t = 1 - \frac{2}{5} = \frac{3}{5} \;\Rightarrow\; t = \frac{12}{5}

A only has

\frac{3}{5}

of the pool left to do, so it runs for less than the full

4

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{12}{5}

hours is

2

hours and

\frac{2}{5} \times 60 = 24

minutes. A stopped at 5 p.m., so count back.

5{:}00 - 2{:}24 = 2{:}36

A was turned on at 2:36 p.m.

Answer: 2:36 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 2:36 is $\frac{8}{5}$ hours, doing $\frac{8}{5} \times \frac{1}{10} = \frac{4}{25}$ . Together for $\frac{12}{5}$ hours they do $\frac{12}{5} \times \left(\frac{1}{10} + \frac{1}{4}\right) = \frac{12}{5} \times \frac{7}{20} = \frac{21}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{10}x + \left(\frac{1}{10} + \frac{1}{4}\right)(4 - x) = 1$ , which gives $x = \frac{8}{5}$ , i.e. $1$ hour $36$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 10 hard

Pump A alone can empty a swimming pool in $3$ hours, and pump B alone can empty it in $10$ hours. Pump B starts pumping water out of the pool at 1 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 7 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 1 p.m. to 7 p.m.; pump A joins at some point and runs until 7 p.m. The pool is exactly empty at 7 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $3$ hours.
Pump B empties the pool in $10$ hours.
B starts at 1 p.m., A joins later, and the pool is empty at 7 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 7 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $6$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{3}

of it per hour and pump B empties

\frac{1}{10}

per hour.

\text{A}: \frac{1}{3},\qquad \text{B}: \frac{1}{10}

A pump that needs

10

hours does

\frac{1}{10}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 1 p.m. to 7 p.m., which is

6

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{3}t = 1 - \frac{3}{5} = \frac{2}{5} \;\Rightarrow\; t = \frac{6}{5}

A only has

\frac{2}{5}

of the pool left to do, so it runs for less than the full

6

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{6}{5}

hours is

1

hour and

\frac{1}{5} \times 60 = 12

minutes. A stopped at 7 p.m., so count back.

7{:}00 - 1{:}12 = 5{:}48

A was turned on at 5:48 p.m.

Answer: 5:48 p.m.

4 · Reviewdoes it hold up?

B alone from 1:00 to 5:48 is $\frac{24}{5}$ hours, doing $\frac{24}{5} \times \frac{1}{10} = \frac{12}{25}$ . Together for $\frac{6}{5}$ hours they do $\frac{6}{5} \times \left(\frac{1}{10} + \frac{1}{3}\right) = \frac{6}{5} \times \frac{13}{30} = \frac{13}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{10}x + \left(\frac{1}{10} + \frac{1}{3}\right)(6 - x) = 1$ , which gives $x = \frac{24}{5}$ , i.e. $4$ hours $48$ minutes after 1 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 11 hard

Pump A alone can empty a swimming pool in $6$ hours, and pump B alone can empty it in $10$ hours. Pump B starts pumping water out of the pool at 2 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 6 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 2 p.m. to 6 p.m.; pump A joins at some point and runs until 6 p.m. The pool is exactly empty at 6 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $6$ hours.
Pump B empties the pool in $10$ hours.
B starts at 2 p.m., A joins later, and the pool is empty at 6 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 6 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $4$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{6}

of it per hour and pump B empties

\frac{1}{10}

per hour.

\text{A}: \frac{1}{6},\qquad \text{B}: \frac{1}{10}

A pump that needs

10

hours does

\frac{1}{10}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 2 p.m. to 6 p.m., which is

4

hours.

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

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{6}t = 1 - \frac{2}{5} = \frac{3}{5} \;\Rightarrow\; t = \frac{18}{5}

A only has

\frac{3}{5}

of the pool left to do, so it runs for less than the full

4

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{18}{5}

hours is

3

hours and

\frac{3}{5} \times 60 = 36

minutes. A stopped at 6 p.m., so count back.

6{:}00 - 3{:}36 = 2{:}24

A was turned on at 2:24 p.m.

Answer: 2:24 p.m.

4 · Reviewdoes it hold up?

B alone from 2:00 to 2:24 is $\frac{2}{5}$ hours, doing $\frac{2}{5} \times \frac{1}{10} = \frac{1}{25}$ . Together for $\frac{18}{5}$ hours they do $\frac{18}{5} \times \left(\frac{1}{10} + \frac{1}{6}\right) = \frac{18}{5} \times \frac{4}{15} = \frac{24}{25}$ . The total is $\frac{25}{25} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{10}x + \left(\frac{1}{10} + \frac{1}{6}\right)(4 - x) = 1$ , which gives $x = \frac{2}{5}$ , i.e. $24$ minutes after 2 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Variant 12 hard

Pump A alone can empty a swimming pool in $3$ hours, and pump B alone can empty it in $12$ hours. Pump B starts pumping water out of the pool at 2 p.m. Partway through, pump A is turned on as well, and together they finish emptying the pool at 7 p.m. At what time was pump A turned on?

Your answer (check it yourself)

Show solution

1 · Understandwhat's really being asked

Pump B runs from 2 p.m. to 7 p.m.; pump A joins at some point and runs until 7 p.m. The pool is exactly empty at 7 p.m. Find when A was switched on.

Givens

Pump A empties the pool in $3$ hours.
Pump B empties the pool in $12$ hours.
B starts at 2 p.m., A joins later, and the pool is empty at 7 p.m.

Unknowns

The time pump A was turned on.

Constraints

Each pump works at a steady rate.
Both pumps stop at 7 p.m.

2 · Planchoose the strategy

#16 Count the Complement · also uses: #13 Convert to Algebra#8 Analyze the Units

Call the whole pool $1$ . Instead of splitting the afternoon into a B-only part and a both-pumps part, change focus: B ran the entire $5$ hours, so its share is known at once. Whatever is left was A's work, which tells how long A ran.

3 · Execute4 carry out the plan

1Write each pump's hourly rate

#8 Analyze the Units 7.RP.A.1

With the whole pool as

1

, pump A empties

\frac{1}{3}

of it per hour and pump B empties

\frac{1}{12}

per hour.

\text{A}: \frac{1}{3},\qquad \text{B}: \frac{1}{12}

A pump that needs

12

hours does

\frac{1}{12}

of the job each hour.

2B's share over the whole afternoon

#16 Count the Complement 7.EE.B.3

B ran from 2 p.m. to 7 p.m., which is

5

hours.

5 \times \frac{1}{12} = \frac{5}{12}

Looking at B alone avoids splitting the time into two phases.

3Find how long A ran

#13 Convert to Algebra 7.EE.B.4

A had to empty the rest of the pool. Let A run for

t

hours.

\frac{1}{3}t = 1 - \frac{5}{12} = \frac{7}{12} \;\Rightarrow\; t = \frac{7}{4}

A only has

\frac{7}{12}

of the pool left to do, so it runs for less than the full

5

hours.

4Turn hours into a clock time

#8 Analyze the Units 7.EE.B.3

\frac{7}{4}

hours is

1

hour and

\frac{3}{4} \times 60 = 45

minutes. A stopped at 7 p.m., so count back.

7{:}00 - 1{:}45 = 5{:}15

A was turned on at 5:15 p.m.

Answer: 5:15 p.m.

4 · Reviewdoes it hold up?

B alone from 2:00 to 5:15 is $\frac{13}{4}$ hours, doing $\frac{13}{4} \times \frac{1}{12} = \frac{13}{48}$ . Together for $\frac{7}{4}$ hours they do $\frac{7}{4} \times \left(\frac{1}{12} + \frac{1}{3}\right) = \frac{7}{4} \times \frac{5}{12} = \frac{35}{48}$ . The total is $\frac{48}{48} = 1$ .

Another way: Let $x$ be the hours B works alone. Then $\frac{1}{12}x + \left(\frac{1}{12} + \frac{1}{3}\right)(5 - x) = 1$ , which gives $x = \frac{13}{4}$ , i.e. $3$ hours $15$ minutes after 2 p.m.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Setting A's share equal to the work that remained.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Writing each pump's work per hour as a unit rate.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Converting fractional hours to minutes and a clock time.

💡Takeaway. Look at the worker who never stopped first; the rest of the job belongs to the one who joined late.

Subtract the full-time pump's share first · 12 practice problems

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