Practice · Compare two drain-outs to isolate the inflow · Sensim Math

Variant 1 easy

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $3$ cars every $5$ minutes, the lot becomes empty at 8:05 a.m. If cars leave at an average of $1$ cars every $3$ minutes, the lot becomes empty at 10:15 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $3$ cars per $5$ minutes, the lot is empty at 8:05 a.m.
Leaving $1$ cars per $3$ minutes, the lot is empty at 10:15 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:05 is

65

minutes; to 10:15 is

195

minutes. The leaving rates are

\frac{3}{5}

and

\frac{1}{3}

car per minute.

3 \div 5 = \frac{3}{5},\quad 1 \div 3 = \frac{1}{3}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 65r = \frac{3}{5} \times 65 = 39,\qquad a + 195r = \frac{1}{3} \times 195 = 65

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

130

minutes longer and

26

more cars leave. The starting cars are the same, so those

26

extra cars are exactly the arrivals during the extra

130

minutes.

130r = 65 - 39 = 26 \;\Rightarrow\; r = \frac{1}{5}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{5}

into the first equation.

a = 39 - 65 \times \frac{1}{5} = 39 - 13 = 26

One car arrives every

5

minutes, so

13

cars came in during the first

65

minutes.

Answer: 26 cars

4 · Reviewdoes it hold up?

Second scenario: $26 + 195 \times \frac{1}{5} = 26 + 39 = 65$ cars, and $\frac{1}{3} \times 195 = 65$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 39 - 65r$ ; substitute into the second, $39 - 65r + 195r = 65$ , so $130r = 26$ and $r = \frac{1}{5}$ , giving $a = 26$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $3$ cars per $5$ minutes and $1$ cars per $3$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 2 easy

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $9$ cars every $10$ minutes, the lot becomes empty at 9:00 a.m. If cars leave at an average of $7$ cars every $10$ minutes, the lot becomes empty at 10:20 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $9$ cars per $10$ minutes, the lot is empty at 9:00 a.m.
Leaving $7$ cars per $10$ minutes, the lot is empty at 10:20 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 9:00 is

120

minutes; to 10:20 is

200

minutes. The leaving rates are

\frac{9}{10}

and

\frac{7}{10}

car per minute.

9 \div 10 = \frac{9}{10},\quad 7 \div 10 = \frac{7}{10}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 120r = \frac{9}{10} \times 120 = 108,\qquad a + 200r = \frac{7}{10} \times 200 = 140

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

80

minutes longer and

32

more cars leave. The starting cars are the same, so those

32

extra cars are exactly the arrivals during the extra

80

minutes.

80r = 140 - 108 = 32 \;\Rightarrow\; r = \frac{2}{5}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{2}{5}

into the first equation.

a = 108 - 120 \times \frac{2}{5} = 108 - 48 = 60

2

cars arrive every

5

minutes, so

48

cars came in during the first

120

minutes.

Answer: 60 cars

4 · Reviewdoes it hold up?

Second scenario: $60 + 200 \times \frac{2}{5} = 60 + 80 = 140$ cars, and $\frac{7}{10} \times 200 = 140$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 108 - 120r$ ; substitute into the second, $108 - 120r + 200r = 140$ , so $80r = 32$ and $r = \frac{2}{5}$ , giving $a = 60$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $9$ cars per $10$ minutes and $7$ cars per $10$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 3 easy

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $4$ cars every $7$ minutes, the lot becomes empty at 9:20 a.m. If cars leave at an average of $3$ cars every $6$ minutes, the lot becomes empty at 10:00 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $4$ cars per $7$ minutes, the lot is empty at 9:20 a.m.
Leaving $3$ cars per $6$ minutes, the lot is empty at 10:00 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 9:20 is

140

minutes; to 10:00 is

180

minutes. The leaving rates are

\frac{4}{7}

and

\frac{1}{2}

car per minute.

4 \div 7 = \frac{4}{7},\quad 3 \div 6 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 140r = \frac{4}{7} \times 140 = 80,\qquad a + 180r = \frac{1}{2} \times 180 = 90

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

40

minutes longer and

10

more cars leave. The starting cars are the same, so those

10

extra cars are exactly the arrivals during the extra

40

minutes.

40r = 90 - 80 = 10 \;\Rightarrow\; r = \frac{1}{4}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{4}

into the first equation.

a = 80 - 140 \times \frac{1}{4} = 80 - 35 = 45

One car arrives every

4

minutes, so

35

cars came in during the first

140

minutes.

Answer: 45 cars

4 · Reviewdoes it hold up?

Second scenario: $45 + 180 \times \frac{1}{4} = 45 + 45 = 90$ cars, and $\frac{1}{2} \times 180 = 90$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 80 - 140r$ ; substitute into the second, $80 - 140r + 180r = 90$ , so $40r = 10$ and $r = \frac{1}{4}$ , giving $a = 45$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $4$ cars per $7$ minutes and $3$ cars per $6$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 4 easy

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $5$ cars every $8$ minutes, the lot becomes empty at 8:20 a.m. If cars leave at an average of $1$ cars every $2$ minutes, the lot becomes empty at 9:00 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $5$ cars per $8$ minutes, the lot is empty at 8:20 a.m.
Leaving $1$ cars per $2$ minutes, the lot is empty at 9:00 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:20 is

80

minutes; to 9:00 is

120

minutes. The leaving rates are

\frac{5}{8}

and

\frac{1}{2}

car per minute.

5 \div 8 = \frac{5}{8},\quad 1 \div 2 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 80r = \frac{5}{8} \times 80 = 50,\qquad a + 120r = \frac{1}{2} \times 120 = 60

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

40

minutes longer and

10

more cars leave. The starting cars are the same, so those

10

extra cars are exactly the arrivals during the extra

40

minutes.

40r = 60 - 50 = 10 \;\Rightarrow\; r = \frac{1}{4}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{4}

into the first equation.

a = 50 - 80 \times \frac{1}{4} = 50 - 20 = 30

One car arrives every

4

minutes, so

20

cars came in during the first

80

minutes.

Answer: 30 cars

4 · Reviewdoes it hold up?

Second scenario: $30 + 120 \times \frac{1}{4} = 30 + 30 = 60$ cars, and $\frac{1}{2} \times 120 = 60$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 50 - 80r$ ; substitute into the second, $50 - 80r + 120r = 60$ , so $40r = 10$ and $r = \frac{1}{4}$ , giving $a = 30$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $5$ cars per $8$ minutes and $1$ cars per $2$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 5 medium

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $3$ cars every $4$ minutes, the lot becomes empty at 8:12 a.m. If cars leave at an average of $1$ cars every $2$ minutes, the lot becomes empty at 9:24 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $3$ cars per $4$ minutes, the lot is empty at 8:12 a.m.
Leaving $1$ cars per $2$ minutes, the lot is empty at 9:24 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:12 is

72

minutes; to 9:24 is

144

minutes. The leaving rates are

\frac{3}{4}

and

\frac{1}{2}

car per minute.

3 \div 4 = \frac{3}{4},\quad 1 \div 2 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 72r = \frac{3}{4} \times 72 = 54,\qquad a + 144r = \frac{1}{2} \times 144 = 72

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

72

minutes longer and

18

more cars leave. The starting cars are the same, so those

18

extra cars are exactly the arrivals during the extra

72

minutes.

72r = 72 - 54 = 18 \;\Rightarrow\; r = \frac{1}{4}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{4}

into the first equation.

a = 54 - 72 \times \frac{1}{4} = 54 - 18 = 36

One car arrives every

4

minutes, so

18

cars came in during the first

72

minutes.

Answer: 36 cars

4 · Reviewdoes it hold up?

Second scenario: $36 + 144 \times \frac{1}{4} = 36 + 36 = 72$ cars, and $\frac{1}{2} \times 144 = 72$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 54 - 72r$ ; substitute into the second, $54 - 72r + 144r = 72$ , so $72r = 18$ and $r = \frac{1}{4}$ , giving $a = 36$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $3$ cars per $4$ minutes and $1$ cars per $2$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 6 medium

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $5$ cars every $7$ minutes, the lot becomes empty at 10:09 a.m. If cars leave at an average of $3$ cars every $5$ minutes, the lot becomes empty at 11:30 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $5$ cars per $7$ minutes, the lot is empty at 10:09 a.m.
Leaving $3$ cars per $5$ minutes, the lot is empty at 11:30 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 10:09 is

189

minutes; to 11:30 is

270

minutes. The leaving rates are

\frac{5}{7}

and

\frac{3}{5}

car per minute.

5 \div 7 = \frac{5}{7},\quad 3 \div 5 = \frac{3}{5}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 189r = \frac{5}{7} \times 189 = 135,\qquad a + 270r = \frac{3}{5} \times 270 = 162

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

81

minutes longer and

27

more cars leave. The starting cars are the same, so those

27

extra cars are exactly the arrivals during the extra

81

minutes.

81r = 162 - 135 = 27 \;\Rightarrow\; r = \frac{1}{3}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{3}

into the first equation.

a = 135 - 189 \times \frac{1}{3} = 135 - 63 = 72

One car arrives every

3

minutes, so

63

cars came in during the first

189

minutes.

Answer: 72 cars

4 · Reviewdoes it hold up?

Second scenario: $72 + 270 \times \frac{1}{3} = 72 + 90 = 162$ cars, and $\frac{3}{5} \times 270 = 162$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 135 - 189r$ ; substitute into the second, $135 - 189r + 270r = 162$ , so $81r = 27$ and $r = \frac{1}{3}$ , giving $a = 72$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $5$ cars per $7$ minutes and $3$ cars per $5$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 7 medium

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $2$ cars every $3$ minutes, the lot becomes empty at 8:00 a.m. If cars leave at an average of $1$ cars every $3$ minutes, the lot becomes empty at 10:30 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $2$ cars per $3$ minutes, the lot is empty at 8:00 a.m.
Leaving $1$ cars per $3$ minutes, the lot is empty at 10:30 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:00 is

60

minutes; to 10:30 is

210

minutes. The leaving rates are

\frac{2}{3}

and

\frac{1}{3}

car per minute.

2 \div 3 = \frac{2}{3},\quad 1 \div 3 = \frac{1}{3}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 60r = \frac{2}{3} \times 60 = 40,\qquad a + 210r = \frac{1}{3} \times 210 = 70

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

150

minutes longer and

30

more cars leave. The starting cars are the same, so those

30

extra cars are exactly the arrivals during the extra

150

minutes.

150r = 70 - 40 = 30 \;\Rightarrow\; r = \frac{1}{5}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{5}

into the first equation.

a = 40 - 60 \times \frac{1}{5} = 40 - 12 = 28

One car arrives every

5

minutes, so

12

cars came in during the first

60

minutes.

Answer: 28 cars

4 · Reviewdoes it hold up?

Second scenario: $28 + 210 \times \frac{1}{5} = 28 + 42 = 70$ cars, and $\frac{1}{3} \times 210 = 70$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 40 - 60r$ ; substitute into the second, $40 - 60r + 210r = 70$ , so $150r = 30$ and $r = \frac{1}{5}$ , giving $a = 28$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $2$ cars per $3$ minutes and $1$ cars per $3$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 8 medium

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $7$ cars every $9$ minutes, the lot becomes empty at 8:12 a.m. If cars leave at an average of $2$ cars every $4$ minutes, the lot becomes empty at 9:32 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $7$ cars per $9$ minutes, the lot is empty at 8:12 a.m.
Leaving $2$ cars per $4$ minutes, the lot is empty at 9:32 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:12 is

72

minutes; to 9:32 is

152

minutes. The leaving rates are

\frac{7}{9}

and

\frac{1}{2}

car per minute.

7 \div 9 = \frac{7}{9},\quad 2 \div 4 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 72r = \frac{7}{9} \times 72 = 56,\qquad a + 152r = \frac{1}{2} \times 152 = 76

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

80

minutes longer and

20

more cars leave. The starting cars are the same, so those

20

extra cars are exactly the arrivals during the extra

80

minutes.

80r = 76 - 56 = 20 \;\Rightarrow\; r = \frac{1}{4}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{4}

into the first equation.

a = 56 - 72 \times \frac{1}{4} = 56 - 18 = 38

One car arrives every

4

minutes, so

18

cars came in during the first

72

minutes.

Answer: 38 cars

4 · Reviewdoes it hold up?

Second scenario: $38 + 152 \times \frac{1}{4} = 38 + 38 = 76$ cars, and $\frac{1}{2} \times 152 = 76$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 56 - 72r$ ; substitute into the second, $56 - 72r + 152r = 76$ , so $80r = 20$ and $r = \frac{1}{4}$ , giving $a = 38$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $7$ cars per $9$ minutes and $2$ cars per $4$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 9 hard

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $7$ cars every $12$ minutes, the lot becomes empty at 9:00 a.m. If cars leave at an average of $6$ cars every $12$ minutes, the lot becomes empty at 9:40 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $7$ cars per $12$ minutes, the lot is empty at 9:00 a.m.
Leaving $6$ cars per $12$ minutes, the lot is empty at 9:40 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 9:00 is

120

minutes; to 9:40 is

160

minutes. The leaving rates are

\frac{7}{12}

and

\frac{1}{2}

car per minute.

7 \div 12 = \frac{7}{12},\quad 6 \div 12 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 120r = \frac{7}{12} \times 120 = 70,\qquad a + 160r = \frac{1}{2} \times 160 = 80

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

40

minutes longer and

10

more cars leave. The starting cars are the same, so those

10

extra cars are exactly the arrivals during the extra

40

minutes.

40r = 80 - 70 = 10 \;\Rightarrow\; r = \frac{1}{4}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{4}

into the first equation.

a = 70 - 120 \times \frac{1}{4} = 70 - 30 = 40

One car arrives every

4

minutes, so

30

cars came in during the first

120

minutes.

Answer: 40 cars

4 · Reviewdoes it hold up?

Second scenario: $40 + 160 \times \frac{1}{4} = 40 + 40 = 80$ cars, and $\frac{1}{2} \times 160 = 80$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 70 - 120r$ ; substitute into the second, $70 - 120r + 160r = 80$ , so $40r = 10$ and $r = \frac{1}{4}$ , giving $a = 40$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $7$ cars per $12$ minutes and $6$ cars per $12$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 10 hard

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $8$ cars every $9$ minutes, the lot becomes empty at 9:24 a.m. If cars leave at an average of $5$ cars every $6$ minutes, the lot becomes empty at 9:48 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $8$ cars per $9$ minutes, the lot is empty at 9:24 a.m.
Leaving $5$ cars per $6$ minutes, the lot is empty at 9:48 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 9:24 is

144

minutes; to 9:48 is

168

minutes. The leaving rates are

\frac{8}{9}

and

\frac{5}{6}

car per minute.

8 \div 9 = \frac{8}{9},\quad 5 \div 6 = \frac{5}{6}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 144r = \frac{8}{9} \times 144 = 128,\qquad a + 168r = \frac{5}{6} \times 168 = 140

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

24

minutes longer and

12

more cars leave. The starting cars are the same, so those

12

extra cars are exactly the arrivals during the extra

24

minutes.

24r = 140 - 128 = 12 \;\Rightarrow\; r = \frac{1}{2}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{2}

into the first equation.

a = 128 - 144 \times \frac{1}{2} = 128 - 72 = 56

One car arrives every

2

minutes, so

72

cars came in during the first

144

minutes.

Answer: 56 cars

4 · Reviewdoes it hold up?

Second scenario: $56 + 168 \times \frac{1}{2} = 56 + 84 = 140$ cars, and $\frac{5}{6} \times 168 = 140$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 128 - 144r$ ; substitute into the second, $128 - 144r + 168r = 140$ , so $24r = 12$ and $r = \frac{1}{2}$ , giving $a = 56$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $8$ cars per $9$ minutes and $5$ cars per $6$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 11 hard

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $2$ cars every $3$ minutes, the lot becomes empty at 8:30 a.m. If cars leave at an average of $6$ cars every $10$ minutes, the lot becomes empty at 8:50 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $2$ cars per $3$ minutes, the lot is empty at 8:30 a.m.
Leaving $6$ cars per $10$ minutes, the lot is empty at 8:50 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:30 is

90

minutes; to 8:50 is

110

minutes. The leaving rates are

\frac{2}{3}

and

\frac{3}{5}

car per minute.

2 \div 3 = \frac{2}{3},\quad 6 \div 10 = \frac{3}{5}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 90r = \frac{2}{3} \times 90 = 60,\qquad a + 110r = \frac{3}{5} \times 110 = 66

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

20

minutes longer and

6

more cars leave. The starting cars are the same, so those

6

extra cars are exactly the arrivals during the extra

20

minutes.

20r = 66 - 60 = 6 \;\Rightarrow\; r = \frac{3}{10}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{3}{10}

into the first equation.

a = 60 - 90 \times \frac{3}{10} = 60 - 27 = 33

3

cars arrive every

10

minutes, so

27

cars came in during the first

90

minutes.

Answer: 33 cars

4 · Reviewdoes it hold up?

Second scenario: $33 + 110 \times \frac{3}{10} = 33 + 33 = 66$ cars, and $\frac{3}{5} \times 110 = 66$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 60 - 90r$ ; substitute into the second, $60 - 90r + 110r = 66$ , so $20r = 6$ and $r = \frac{3}{10}$ , giving $a = 33$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $2$ cars per $3$ minutes and $6$ cars per $10$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Variant 12 hard

At 7:00 a.m., some cars are parked in a parking lot, and more cars keep arriving at a steady rate every minute. If cars leave at an average of $6$ cars every $7$ minutes, the lot becomes empty at 8:24 a.m. If cars leave at an average of $1$ cars every $2$ minutes, the lot becomes empty at 9:54 a.m. How many cars were parked in the lot at 7:00 a.m.?

Your answer

Show solution

1 · Understandwhat's really being asked

A parking lot starts with some cars at 7:00 a.m. and gains cars at a steady rate. Under two different leaving rates it empties at two different times. Find the starting number of cars.

Givens

Cars arrive at a constant rate per minute.
Leaving $6$ cars per $7$ minutes, the lot is empty at 8:24 a.m.
Leaving $1$ cars per $2$ minutes, the lot is empty at 9:54 a.m.

Unknowns

The number of cars at 7:00 a.m.

Constraints

Both scenarios start from the same lot at 7:00 a.m. with the same arrival rate.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #8 Analyze the Units#7 Identify Subproblems

Put everything per minute and count minutes from 7:00. In each scenario, (cars at start) + (cars that arrived) = (cars that left). Comparing the two scenarios removes the starting count and reveals the arrival rate.

3 · Execute4 carry out the plan

1Convert to minutes and per-minute rates

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

From 7:00 to 8:24 is

84

minutes; to 9:54 is

174

minutes. The leaving rates are

\frac{6}{7}

and

\frac{1}{2}

car per minute.

6 \div 7 = \frac{6}{7},\quad 1 \div 2 = \frac{1}{2}

Rates can only be compared once they share a unit.

2Count the cars that left

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

Let

a

be the starting cars and

r

the arrivals per minute. When the lot is empty, everything that was there or arrived has left.

a + 84r = \frac{6}{7} \times 84 = 72,\qquad a + 174r = \frac{1}{2} \times 174 = 87

Both equations say: start + arrivals = departures.

3Compare the two scenarios

#7 Identify Subproblems 7.EE.B.4

The second scenario lasts

90

minutes longer and

15

more cars leave. The starting cars are the same, so those

15

extra cars are exactly the arrivals during the extra

90

minutes.

90r = 87 - 72 = 15 \;\Rightarrow\; r = \frac{1}{6}

Subtracting one story from the other cancels

a

.

4Find the starting count

#13 Convert to Algebra 7.EE.B.3

Put

r = \frac{1}{6}

into the first equation.

a = 72 - 84 \times \frac{1}{6} = 72 - 14 = 58

One car arrives every

6

minutes, so

14

cars came in during the first

84

minutes.

Answer: 58 cars

4 · Reviewdoes it hold up?

Second scenario: $58 + 174 \times \frac{1}{6} = 58 + 29 = 87$ cars, and $\frac{1}{2} \times 174 = 87$ cars left. Both scenarios balance.

Another way: With one variable: the first scenario gives $a = 72 - 84r$ ; substitute into the second, $72 - 84r + 174r = 87$ , so $90r = 15$ and $r = \frac{1}{6}$ , giving $a = 58$ .

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the balance of cars in each scenario as an equation.
7.RP.A.1 Compute unit rates associated with ratios of fractions — Turning $6$ cars per $7$ minutes and $1$ cars per $2$ minutes into per-minute rates.
7.EE.B.3 Solve multi-step real-world problems posed with positive and negative rational numbers — Computing the starting count from the arrival rate.

💡Takeaway. Two versions of the same story differ only in what changed, and that difference reveals the hidden rate.

Compare two drain-outs to isolate the inflow · 12 practice problems

Generated variants — 12

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Givens

Unknowns

Constraints

Where next?