Practice · Find the time first, not the zigzags · Sensim Math

Variant 1 easy

Two people, A and B, start $1\ \text{km}$ apart and walk toward each other, A at $15\ \text{m}$ per minute and B at $10\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $100\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $1\ \text{km}$ gap from opposite ends while a dog shuttles between them at $100\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $1\ \text{km}$ apart.
A walks $15\ \text{m}$ per minute; B walks $10\ \text{m}$ per minute.
The dog runs back and forth between them at $100\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

1\ \text{km} = 1000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

15x + 10x = 1000 \;\Rightarrow\; 25x = 1000 \;\Rightarrow\; x = 40

Together they close

25\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

40

minutes at

100\ \text{m}

per minute, no matter how many times it turns.

100 \times 40 = 4000\ \text{m} = 4\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 4 km

4 · Reviewdoes it hold up?

The dog is $4$ times as fast as the walkers' combined closing speed ( $100$ vs. $25$ m per minute), so it should cover $4$ times the gap: $4 \times 1 = 4\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{1000}{110} = \frac{100}{11}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $40$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $15x + 10x = 1000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 2 easy

Two people, A and B, start $3\ \text{km}$ apart and walk toward each other, A at $20\ \text{m}$ per minute and B at $10\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $150\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $3\ \text{km}$ gap from opposite ends while a dog shuttles between them at $150\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $3\ \text{km}$ apart.
A walks $20\ \text{m}$ per minute; B walks $10\ \text{m}$ per minute.
The dog runs back and forth between them at $150\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

3\ \text{km} = 3000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

20x + 10x = 3000 \;\Rightarrow\; 30x = 3000 \;\Rightarrow\; x = 100

Together they close

30\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

100

minutes at

150\ \text{m}

per minute, no matter how many times it turns.

150 \times 100 = 15000\ \text{m} = 15\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 15 km

4 · Reviewdoes it hold up?

The dog is $5$ times as fast as the walkers' combined closing speed ( $150$ vs. $30$ m per minute), so it should cover $5$ times the gap: $5 \times 3 = 15\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{3000}{160} = \frac{75}{4}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $100$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $20x + 10x = 3000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 3 easy

Two people, A and B, start $4\ \text{km}$ apart and walk toward each other, A at $30\ \text{m}$ per minute and B at $20\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $150\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $4\ \text{km}$ gap from opposite ends while a dog shuttles between them at $150\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $4\ \text{km}$ apart.
A walks $30\ \text{m}$ per minute; B walks $20\ \text{m}$ per minute.
The dog runs back and forth between them at $150\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

4\ \text{km} = 4000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

30x + 20x = 4000 \;\Rightarrow\; 50x = 4000 \;\Rightarrow\; x = 80

Together they close

50\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

80

minutes at

150\ \text{m}

per minute, no matter how many times it turns.

150 \times 80 = 12000\ \text{m} = 12\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 12 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $150$ vs. $50$ m per minute), so it should cover $3$ times the gap: $3 \times 4 = 12\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{4000}{170} = \frac{400}{17}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $80$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $30x + 20x = 4000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 4 easy

Two people, A and B, start $2\ \text{km}$ apart and walk toward each other, A at $25\ \text{m}$ per minute and B at $15\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $160\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $2\ \text{km}$ gap from opposite ends while a dog shuttles between them at $160\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $2\ \text{km}$ apart.
A walks $25\ \text{m}$ per minute; B walks $15\ \text{m}$ per minute.
The dog runs back and forth between them at $160\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

2\ \text{km} = 2000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

25x + 15x = 2000 \;\Rightarrow\; 40x = 2000 \;\Rightarrow\; x = 50

Together they close

40\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

160\ \text{m}

per minute, no matter how many times it turns.

160 \times 50 = 8000\ \text{m} = 8\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 8 km

4 · Reviewdoes it hold up?

The dog is $4$ times as fast as the walkers' combined closing speed ( $160$ vs. $40$ m per minute), so it should cover $4$ times the gap: $4 \times 2 = 8\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{2000}{175} = \frac{80}{7}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $25x + 15x = 2000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 5 medium

Two people, A and B, start $3\ \text{km}$ apart and walk toward each other, A at $40\ \text{m}$ per minute and B at $20\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $180\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $3\ \text{km}$ gap from opposite ends while a dog shuttles between them at $180\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $3\ \text{km}$ apart.
A walks $40\ \text{m}$ per minute; B walks $20\ \text{m}$ per minute.
The dog runs back and forth between them at $180\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

3\ \text{km} = 3000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

40x + 20x = 3000 \;\Rightarrow\; 60x = 3000 \;\Rightarrow\; x = 50

Together they close

60\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

180\ \text{m}

per minute, no matter how many times it turns.

180 \times 50 = 9000\ \text{m} = 9\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 9 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $180$ vs. $60$ m per minute), so it should cover $3$ times the gap: $3 \times 3 = 9\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{3000}{200} = 15$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $40x + 20x = 3000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 6 medium

Two people, A and B, start $2\ \text{km}$ apart and walk toward each other, A at $30\ \text{m}$ per minute and B at $20\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $200\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $2\ \text{km}$ gap from opposite ends while a dog shuttles between them at $200\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $2\ \text{km}$ apart.
A walks $30\ \text{m}$ per minute; B walks $20\ \text{m}$ per minute.
The dog runs back and forth between them at $200\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

2\ \text{km} = 2000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

30x + 20x = 2000 \;\Rightarrow\; 50x = 2000 \;\Rightarrow\; x = 40

Together they close

50\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

40

minutes at

200\ \text{m}

per minute, no matter how many times it turns.

200 \times 40 = 8000\ \text{m} = 8\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 8 km

4 · Reviewdoes it hold up?

The dog is $4$ times as fast as the walkers' combined closing speed ( $200$ vs. $50$ m per minute), so it should cover $4$ times the gap: $4 \times 2 = 8\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{2000}{220} = \frac{100}{11}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $40$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $30x + 20x = 2000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 7 medium

Two people, A and B, start $6\ \text{km}$ apart and walk toward each other, A at $45\ \text{m}$ per minute and B at $30\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $225\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $6\ \text{km}$ gap from opposite ends while a dog shuttles between them at $225\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $6\ \text{km}$ apart.
A walks $45\ \text{m}$ per minute; B walks $30\ \text{m}$ per minute.
The dog runs back and forth between them at $225\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

6\ \text{km} = 6000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

45x + 30x = 6000 \;\Rightarrow\; 75x = 6000 \;\Rightarrow\; x = 80

Together they close

75\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

80

minutes at

225\ \text{m}

per minute, no matter how many times it turns.

225 \times 80 = 18000\ \text{m} = 18\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 18 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $225$ vs. $75$ m per minute), so it should cover $3$ times the gap: $3 \times 6 = 18\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{6000}{255} = \frac{400}{17}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $80$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $45x + 30x = 6000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 8 medium

Two people, A and B, start $4\ \text{km}$ apart and walk toward each other, A at $50\ \text{m}$ per minute and B at $30\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $240\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $4\ \text{km}$ gap from opposite ends while a dog shuttles between them at $240\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $4\ \text{km}$ apart.
A walks $50\ \text{m}$ per minute; B walks $30\ \text{m}$ per minute.
The dog runs back and forth between them at $240\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

4\ \text{km} = 4000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

50x + 30x = 4000 \;\Rightarrow\; 80x = 4000 \;\Rightarrow\; x = 50

Together they close

80\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

240\ \text{m}

per minute, no matter how many times it turns.

240 \times 50 = 12000\ \text{m} = 12\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 12 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $240$ vs. $80$ m per minute), so it should cover $3$ times the gap: $3 \times 4 = 12\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{4000}{270} = \frac{400}{27}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $50x + 30x = 4000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 9 hard

Two people, A and B, start $3\ \text{km}$ apart and walk toward each other, A at $35\ \text{m}$ per minute and B at $25\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $240\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $3\ \text{km}$ gap from opposite ends while a dog shuttles between them at $240\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $3\ \text{km}$ apart.
A walks $35\ \text{m}$ per minute; B walks $25\ \text{m}$ per minute.
The dog runs back and forth between them at $240\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

3\ \text{km} = 3000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

35x + 25x = 3000 \;\Rightarrow\; 60x = 3000 \;\Rightarrow\; x = 50

Together they close

60\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

240\ \text{m}

per minute, no matter how many times it turns.

240 \times 50 = 12000\ \text{m} = 12\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 12 km

4 · Reviewdoes it hold up?

The dog is $4$ times as fast as the walkers' combined closing speed ( $240$ vs. $60$ m per minute), so it should cover $4$ times the gap: $4 \times 3 = 12\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{3000}{265} = \frac{600}{53}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $35x + 25x = 3000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 10 hard

Two people, A and B, start $2\ \text{km}$ apart and walk toward each other, A at $45\ \text{m}$ per minute and B at $35\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $240\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $2\ \text{km}$ gap from opposite ends while a dog shuttles between them at $240\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $2\ \text{km}$ apart.
A walks $45\ \text{m}$ per minute; B walks $35\ \text{m}$ per minute.
The dog runs back and forth between them at $240\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

2\ \text{km} = 2000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

45x + 35x = 2000 \;\Rightarrow\; 80x = 2000 \;\Rightarrow\; x = 25

Together they close

80\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

25

minutes at

240\ \text{m}

per minute, no matter how many times it turns.

240 \times 25 = 6000\ \text{m} = 6\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 6 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $240$ vs. $80$ m per minute), so it should cover $3$ times the gap: $3 \times 2 = 6\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{2000}{275} = \frac{80}{11}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $25$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $45x + 35x = 2000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 11 hard

Two people, A and B, start $5\ \text{km}$ apart and walk toward each other, A at $60\ \text{m}$ per minute and B at $40\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $300\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $5\ \text{km}$ gap from opposite ends while a dog shuttles between them at $300\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $5\ \text{km}$ apart.
A walks $60\ \text{m}$ per minute; B walks $40\ \text{m}$ per minute.
The dog runs back and forth between them at $300\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

5\ \text{km} = 5000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

60x + 40x = 5000 \;\Rightarrow\; 100x = 5000 \;\Rightarrow\; x = 50

Together they close

100\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

300\ \text{m}

per minute, no matter how many times it turns.

300 \times 50 = 15000\ \text{m} = 15\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 15 km

4 · Reviewdoes it hold up?

The dog is $3$ times as fast as the walkers' combined closing speed ( $300$ vs. $100$ m per minute), so it should cover $3$ times the gap: $3 \times 5 = 15\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{5000}{340} = \frac{250}{17}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $60x + 40x = 5000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Variant 12 hard

Two people, A and B, start $5\ \text{km}$ apart and walk toward each other, A at $70\ \text{m}$ per minute and B at $30\ \text{m}$ per minute. At the same moment, a dog starts out with A and runs toward B at $400\ \text{m}$ per minute; when it reaches B it turns back toward A, and it keeps running back and forth this way. How many kilometers has the dog run by the time A and B meet?

Your answer

Show solution

1 · Understandwhat's really being asked

Two walkers close a $5\ \text{km}$ gap from opposite ends while a dog shuttles between them at $400\ \text{m}$ per minute. Find the dog's total distance when the walkers meet.

Givens

A and B start $5\ \text{km}$ apart.
A walks $70\ \text{m}$ per minute; B walks $30\ \text{m}$ per minute.
The dog runs back and forth between them at $400\ \text{m}$ per minute, starting with A.

Unknowns

The total distance the dog runs, in kilometers.

Constraints

The dog runs nonstop until A and B meet.

2 · Planchoose the strategy

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

Adding up the dog's back-and-forth legs would mean infinitely many shorter and shorter trips. Change focus: the dog runs for exactly as long as the walkers take to meet, and at a constant speed. So find that time, then multiply.

3 · Execute3 carry out the plan

1Use the same unit

#8 Analyze the Units 6.RP.A.3

The speeds are in meters per minute, so write the gap in meters.

5\ \text{km} = 5000\ \text{m}

Mixing kilometers and meters would throw the answer off by a factor of

1000

.

2Find when A and B meet

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

Let them meet after

x

minutes. Walking toward each other, their distances add up to the whole gap.

70x + 30x = 5000 \;\Rightarrow\; 100x = 5000 \;\Rightarrow\; x = 50

Together they close

100\ \text{m}

every minute.

3Use the time for the dog

#16 Count the Complement 6.RP.A.3

The dog runs the whole

50

minutes at

400\ \text{m}

per minute, no matter how many times it turns.

400 \times 50 = 20000\ \text{m} = 20\ \text{km}

Turning around changes the dog's direction, not its speed or running time.

Answer: 20 km

4 · Reviewdoes it hold up?

The dog is $4$ times as fast as the walkers' combined closing speed ( $400$ vs. $100$ m per minute), so it should cover $4$ times the gap: $4 \times 5 = 20\ \text{km}$ .

Another way: Compute the first legs to see why the direct sum is hard: the dog meets B after $\frac{5000}{430} = \frac{500}{43}$ minutes, then turns back, and the legs keep shrinking forever. Their total time still adds up to the $50$ minutes found above.

Standardsmin grade 7

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Writing the meeting condition as $70x + 30x = 5000$.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Converting units and using distance $=$ speed $\times$ time for the dog.

💡Takeaway. When the path is messy but the clock is simple, find the time first.

Find the time first, not the zigzags · 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?