Practice · Same way: lap gap; opposite way: lap sum · Sensim Math

Variant 1 easy

A and B decide to walk around a lake. While A walks $100$ m, B walks $300$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $50$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.2$ times B's original speed, and $5$ minutes later there are still $7.7$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 1:3. Walking the same way, B laps A after 50 minutes. Then B turns around at 1.2 times the old speed; after 5 more minutes they are still 7.7 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $100$ m, B walks $300$ m.
Same direction from the same start: first meeting after $50$ minutes.
Then B reverses direction at $1.2$ times B's original speed.
After $5$ more minutes, $7.7$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

100:300 = 1:3

, so let A walk

x

m per minute and B walk

3x

m per minute. Let the lap be

y

m.

\text{A}: x,\quad \text{B}: 3x,\quad 1.2 \times 3x = 3.6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

50 \times 3x - 50 \times x = y,\quad 100x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

5

minutes they have covered

5x + 18x

and

7700

m is still left.

5 \times x + 5 \times 3.6x = y - 7700,\quad 23x = y - 7700

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 100x

into the second equation.

23x = 100x - 7700,\quad 77x = 7700,\quad x = 100,\quad y = 100 \times 100 = 10000

The extra

77x

in the first stage is exactly the missing

7700

m.

5Convert to kilometers

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

10000

m is

10

km.

10000 \text{ m} = 10 \text{ km}

The question asks for kilometers.

Answer:

10

km

4 · Reviewdoes it hold up?

A walks $100$ m per minute and B $300$ m per minute. In $50$ minutes B gains $200 \times 50 = 10000$ m, one lap. Afterward B walks $360$ m per minute; in $5$ minutes they cover $500 + 1800 = 2300$ m, leaving $10000 - 2300 = 7700$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $100x$ (from the chase) and also $23x + 7700$ (from the approach). The difference $77x$ must be $7700$ m, so $x = 100$ and the lap is $100 \times 100 = 10000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 2 medium

A and B decide to walk around a lake. While A walks $300$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $2$ hours. After that, B starts walking in the opposite direction at a constant speed $1.5$ times B's original speed, and $10$ minutes later there are still $8.1$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:5. Walking the same way, B laps A after 120 minutes. Then B turns around at 1.5 times the old speed; after 10 more minutes they are still 8.1 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $500$ m.
Same direction from the same start: first meeting after $2$ hours.
Then B reverses direction at $1.5$ times B's original speed.
After $10$ more minutes, $8.1$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:500 = 3:5

, so let A walk

3x

m per minute and B walk

5x

m per minute. Let the lap be

y

m. Note that

2

hours is

120

minutes.

\text{A}: 3x,\quad \text{B}: 5x,\quad 1.5 \times 5x = 7.5x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

120 \times 5x - 120 \times 3x = y,\quad 240x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

10

minutes they have covered

30x + 75x

and

8100

m is still left.

10 \times 3x + 10 \times 7.5x = y - 8100,\quad 105x = y - 8100

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 240x

into the second equation.

105x = 240x - 8100,\quad 135x = 8100,\quad x = 60,\quad y = 240 \times 60 = 14400

The extra

135x

in the first stage is exactly the missing

8100

m.

5Convert to kilometers

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

14400

m is

14.4

km.

14400 \text{ m} = 14.4 \text{ km}

The question asks for kilometers.

Answer:

14.4

km

4 · Reviewdoes it hold up?

A walks $180$ m per minute and B $300$ m per minute. In $120$ minutes B gains $120 \times 120 = 14400$ m, one lap. Afterward B walks $450$ m per minute; in $10$ minutes they cover $1800 + 4500 = 6300$ m, leaving $14400 - 6300 = 8100$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $240x$ (from the chase) and also $105x + 8100$ (from the approach). The difference $135x$ must be $8100$ m, so $x = 60$ and the lap is $240 \times 60 = 14400$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 3 easy

A and B decide to walk around a lake. While A walks $300$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour. After that, B starts walking in the opposite direction at a constant speed $1.4$ times B's original speed, and $5$ minutes later there are still $5.6$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:5. Walking the same way, B laps A after 60 minutes. Then B turns around at 1.4 times the old speed; after 5 more minutes they are still 5.6 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $500$ m.
Same direction from the same start: first meeting after $1$ hour.
Then B reverses direction at $1.4$ times B's original speed.
After $5$ more minutes, $5.6$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:500 = 3:5

, so let A walk

3x

m per minute and B walk

5x

m per minute. Let the lap be

y

m. Note that

1

hour is

60

minutes.

\text{A}: 3x,\quad \text{B}: 5x,\quad 1.4 \times 5x = 7x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

60 \times 5x - 60 \times 3x = y,\quad 120x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

5

minutes they have covered

15x + 35x

and

5600

m is still left.

5 \times 3x + 5 \times 7x = y - 5600,\quad 50x = y - 5600

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 120x

into the second equation.

50x = 120x - 5600,\quad 70x = 5600,\quad x = 80,\quad y = 120 \times 80 = 9600

The extra

70x

in the first stage is exactly the missing

5600

m.

5Convert to kilometers

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

9600

m is

9.6

km.

9600 \text{ m} = 9.6 \text{ km}

The question asks for kilometers.

Answer:

9.6

km

4 · Reviewdoes it hold up?

A walks $240$ m per minute and B $400$ m per minute. In $60$ minutes B gains $160 \times 60 = 9600$ m, one lap. Afterward B walks $560$ m per minute; in $5$ minutes they cover $1200 + 2800 = 4000$ m, leaving $9600 - 4000 = 5600$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $120x$ (from the chase) and also $50x + 5600$ (from the approach). The difference $70x$ must be $5600$ m, so $x = 80$ and the lap is $120 \times 80 = 9600$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 4 easy

A and B decide to walk around a lake. While A walks $200$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $50$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.4$ times B's original speed, and $10$ minutes later there are still $3.6$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 2:5. Walking the same way, B laps A after 50 minutes. Then B turns around at 1.4 times the old speed; after 10 more minutes they are still 3.6 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $200$ m, B walks $500$ m.
Same direction from the same start: first meeting after $50$ minutes.
Then B reverses direction at $1.4$ times B's original speed.
After $10$ more minutes, $3.6$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

200:500 = 2:5

, so let A walk

2x

m per minute and B walk

5x

m per minute. Let the lap be

y

m.

\text{A}: 2x,\quad \text{B}: 5x,\quad 1.4 \times 5x = 7x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

50 \times 5x - 50 \times 2x = y,\quad 150x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

10

minutes they have covered

20x + 70x

and

3600

m is still left.

10 \times 2x + 10 \times 7x = y - 3600,\quad 90x = y - 3600

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 150x

into the second equation.

90x = 150x - 3600,\quad 60x = 3600,\quad x = 60,\quad y = 150 \times 60 = 9000

The extra

60x

in the first stage is exactly the missing

3600

m.

5Convert to kilometers

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

9000

m is

9

km.

9000 \text{ m} = 9 \text{ km}

The question asks for kilometers.

Answer:

9

km

4 · Reviewdoes it hold up?

A walks $120$ m per minute and B $300$ m per minute. In $50$ minutes B gains $180 \times 50 = 9000$ m, one lap. Afterward B walks $420$ m per minute; in $10$ minutes they cover $1200 + 4200 = 5400$ m, leaving $9000 - 5400 = 3600$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $150x$ (from the chase) and also $90x + 3600$ (from the approach). The difference $60x$ must be $3600$ m, so $x = 60$ and the lap is $150 \times 60 = 9000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 5 easy

A and B decide to walk around a lake. While A walks $300$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $40$ minutes. After that, B starts walking in the opposite direction at a constant speed $0.8$ times B's original speed, and $5$ minutes later there are still $1.8$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:5. Walking the same way, B laps A after 40 minutes. Then B turns around at 0.8 times the old speed; after 5 more minutes they are still 1.8 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $500$ m.
Same direction from the same start: first meeting after $40$ minutes.
Then B reverses direction at $0.8$ times B's original speed.
After $5$ more minutes, $1.8$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:500 = 3:5

, so let A walk

3x

m per minute and B walk

5x

m per minute. Let the lap be

y

m.

\text{A}: 3x,\quad \text{B}: 5x,\quad 0.8 \times 5x = 4x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

40 \times 5x - 40 \times 3x = y,\quad 80x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

5

minutes they have covered

15x + 20x

and

1800

m is still left.

5 \times 3x + 5 \times 4x = y - 1800,\quad 35x = y - 1800

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 80x

into the second equation.

35x = 80x - 1800,\quad 45x = 1800,\quad x = 40,\quad y = 80 \times 40 = 3200

The extra

45x

in the first stage is exactly the missing

1800

m.

5Convert to kilometers

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

3200

m is

3.2

km.

3200 \text{ m} = 3.2 \text{ km}

The question asks for kilometers.

Answer:

3.2

km

4 · Reviewdoes it hold up?

A walks $120$ m per minute and B $200$ m per minute. In $40$ minutes B gains $80 \times 40 = 3200$ m, one lap. Afterward B walks $160$ m per minute; in $5$ minutes they cover $600 + 800 = 1400$ m, leaving $3200 - 1400 = 1800$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $80x$ (from the chase) and also $35x + 1800$ (from the approach). The difference $45x$ must be $1800$ m, so $x = 40$ and the lap is $80 \times 40 = 3200$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 6 medium

A and B decide to walk around a lake. While A walks $200$ m, B walks $300$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $20$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.2$ times B's original speed, and $10$ minutes later there are still $3.6$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 2:3. Walking the same way, B laps A after 80 minutes. Then B turns around at 1.2 times the old speed; after 10 more minutes they are still 3.6 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $200$ m, B walks $300$ m.
Same direction from the same start: first meeting after $1$ hour $20$ minutes.
Then B reverses direction at $1.2$ times B's original speed.
After $10$ more minutes, $3.6$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

200:300 = 2:3

, so let A walk

2x

m per minute and B walk

3x

m per minute. Let the lap be

y

m. Note that

1

hour

20

minutes is

80

minutes.

\text{A}: 2x,\quad \text{B}: 3x,\quad 1.2 \times 3x = 3.6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

80 \times 3x - 80 \times 2x = y,\quad 80x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

10

minutes they have covered

20x + 36x

and

3600

m is still left.

10 \times 2x + 10 \times 3.6x = y - 3600,\quad 56x = y - 3600

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 80x

into the second equation.

56x = 80x - 3600,\quad 24x = 3600,\quad x = 150,\quad y = 80 \times 150 = 12000

The extra

24x

in the first stage is exactly the missing

3600

m.

5Convert to kilometers

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

12000

m is

12

km.

12000 \text{ m} = 12 \text{ km}

The question asks for kilometers.

Answer:

12

km

4 · Reviewdoes it hold up?

A walks $300$ m per minute and B $450$ m per minute. In $80$ minutes B gains $150 \times 80 = 12000$ m, one lap. Afterward B walks $540$ m per minute; in $10$ minutes they cover $3000 + 5400 = 8400$ m, leaving $12000 - 8400 = 3600$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $80x$ (from the chase) and also $56x + 3600$ (from the approach). The difference $24x$ must be $3600$ m, so $x = 150$ and the lap is $80 \times 150 = 12000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 7 medium

A and B decide to walk around a lake. While A walks $300$ m, B walks $400$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $30$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.1$ times B's original speed, and $5$ minutes later there are still $5.3$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:4. Walking the same way, B laps A after 90 minutes. Then B turns around at 1.1 times the old speed; after 5 more minutes they are still 5.3 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $400$ m.
Same direction from the same start: first meeting after $1$ hour $30$ minutes.
Then B reverses direction at $1.1$ times B's original speed.
After $5$ more minutes, $5.3$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:400 = 3:4

, so let A walk

3x

m per minute and B walk

4x

m per minute. Let the lap be

y

m. Note that

1

hour

30

minutes is

90

minutes.

\text{A}: 3x,\quad \text{B}: 4x,\quad 1.1 \times 4x = 4.4x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

90 \times 4x - 90 \times 3x = y,\quad 90x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

5

minutes they have covered

15x + 22x

and

5300

m is still left.

5 \times 3x + 5 \times 4.4x = y - 5300,\quad 37x = y - 5300

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 90x

into the second equation.

37x = 90x - 5300,\quad 53x = 5300,\quad x = 100,\quad y = 90 \times 100 = 9000

The extra

53x

in the first stage is exactly the missing

5300

m.

5Convert to kilometers

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

9000

m is

9

km.

9000 \text{ m} = 9 \text{ km}

The question asks for kilometers.

Answer:

9

km

4 · Reviewdoes it hold up?

A walks $300$ m per minute and B $400$ m per minute. In $90$ minutes B gains $100 \times 90 = 9000$ m, one lap. Afterward B walks $440$ m per minute; in $5$ minutes they cover $1500 + 2200 = 3700$ m, leaving $9000 - 3700 = 5300$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $90x$ (from the chase) and also $37x + 5300$ (from the approach). The difference $53x$ must be $5300$ m, so $x = 100$ and the lap is $90 \times 100 = 9000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 8 medium

A and B decide to walk around a lake. While A walks $300$ m, B walks $400$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $30$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.5$ times B's original speed, and $5$ minutes later there are still $3.6$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:4. Walking the same way, B laps A after 90 minutes. Then B turns around at 1.5 times the old speed; after 5 more minutes they are still 3.6 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $400$ m.
Same direction from the same start: first meeting after $1$ hour $30$ minutes.
Then B reverses direction at $1.5$ times B's original speed.
After $5$ more minutes, $3.6$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:400 = 3:4

, so let A walk

3x

m per minute and B walk

4x

m per minute. Let the lap be

y

m. Note that

1

hour

30

minutes is

90

minutes.

\text{A}: 3x,\quad \text{B}: 4x,\quad 1.5 \times 4x = 6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

90 \times 4x - 90 \times 3x = y,\quad 90x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

5

minutes they have covered

15x + 30x

and

3600

m is still left.

5 \times 3x + 5 \times 6x = y - 3600,\quad 45x = y - 3600

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 90x

into the second equation.

45x = 90x - 3600,\quad 45x = 3600,\quad x = 80,\quad y = 90 \times 80 = 7200

The extra

45x

in the first stage is exactly the missing

3600

m.

5Convert to kilometers

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

7200

m is

7.2

km.

7200 \text{ m} = 7.2 \text{ km}

The question asks for kilometers.

Answer:

7.2

km

4 · Reviewdoes it hold up?

A walks $240$ m per minute and B $320$ m per minute. In $90$ minutes B gains $80 \times 90 = 7200$ m, one lap. Afterward B walks $480$ m per minute; in $5$ minutes they cover $1200 + 2400 = 3600$ m, leaving $7200 - 3600 = 3600$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $90x$ (from the chase) and also $45x + 3600$ (from the approach). The difference $45x$ must be $3600$ m, so $x = 80$ and the lap is $90 \times 80 = 7200$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 9 hard

A and B decide to walk around a lake. While A walks $400$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $40$ minutes. After that, B starts walking in the opposite direction at a constant speed $0.8$ times B's original speed, and $10$ minutes later there are still $2$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 4:5. Walking the same way, B laps A after 100 minutes. Then B turns around at 0.8 times the old speed; after 10 more minutes they are still 2 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $400$ m, B walks $500$ m.
Same direction from the same start: first meeting after $1$ hour $40$ minutes.
Then B reverses direction at $0.8$ times B's original speed.
After $10$ more minutes, $2$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

400:500 = 4:5

, so let A walk

4x

m per minute and B walk

5x

m per minute. Let the lap be

y

m. Note that

1

hour

40

minutes is

100

minutes.

\text{A}: 4x,\quad \text{B}: 5x,\quad 0.8 \times 5x = 4x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

100 \times 5x - 100 \times 4x = y,\quad 100x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

10

minutes they have covered

40x + 40x

and

2000

m is still left.

10 \times 4x + 10 \times 4x = y - 2000,\quad 80x = y - 2000

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 100x

into the second equation.

80x = 100x - 2000,\quad 20x = 2000,\quad x = 100,\quad y = 100 \times 100 = 10000

The extra

20x

in the first stage is exactly the missing

2000

m.

5Convert to kilometers

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

10000

m is

10

km.

10000 \text{ m} = 10 \text{ km}

The question asks for kilometers.

Answer:

10

km

4 · Reviewdoes it hold up?

A walks $400$ m per minute and B $500$ m per minute. In $100$ minutes B gains $100 \times 100 = 10000$ m, one lap. Afterward B walks $400$ m per minute; in $10$ minutes they cover $4000 + 4000 = 8000$ m, leaving $10000 - 8000 = 2000$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $100x$ (from the chase) and also $80x + 2000$ (from the approach). The difference $20x$ must be $2000$ m, so $x = 100$ and the lap is $100 \times 100 = 10000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 10 hard

A and B decide to walk around a lake. While A walks $200$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $20$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.2$ times B's original speed, and $15$ minutes later there are still $2.4$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 2:5. Walking the same way, B laps A after 80 minutes. Then B turns around at 1.2 times the old speed; after 15 more minutes they are still 2.4 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $200$ m, B walks $500$ m.
Same direction from the same start: first meeting after $1$ hour $20$ minutes.
Then B reverses direction at $1.2$ times B's original speed.
After $15$ more minutes, $2.4$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

200:500 = 2:5

, so let A walk

2x

m per minute and B walk

5x

m per minute. Let the lap be

y

m. Note that

1

hour

20

minutes is

80

minutes.

\text{A}: 2x,\quad \text{B}: 5x,\quad 1.2 \times 5x = 6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

80 \times 5x - 80 \times 2x = y,\quad 240x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

15

minutes they have covered

30x + 90x

and

2400

m is still left.

15 \times 2x + 15 \times 6x = y - 2400,\quad 120x = y - 2400

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 240x

into the second equation.

120x = 240x - 2400,\quad 120x = 2400,\quad x = 20,\quad y = 240 \times 20 = 4800

The extra

120x

in the first stage is exactly the missing

2400

m.

5Convert to kilometers

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

4800

m is

4.8

km.

4800 \text{ m} = 4.8 \text{ km}

The question asks for kilometers.

Answer:

4.8

km

4 · Reviewdoes it hold up?

A walks $40$ m per minute and B $100$ m per minute. In $80$ minutes B gains $60 \times 80 = 4800$ m, one lap. Afterward B walks $120$ m per minute; in $15$ minutes they cover $600 + 1800 = 2400$ m, leaving $4800 - 2400 = 2400$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $240x$ (from the chase) and also $120x + 2400$ (from the approach). The difference $120x$ must be $2400$ m, so $x = 20$ and the lap is $240 \times 20 = 4800$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 11 hard

A and B decide to walk around a lake. While A walks $300$ m, B walks $500$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $50$ minutes. After that, B starts walking in the opposite direction at a constant speed $1.2$ times B's original speed, and $20$ minutes later there are still $1.6$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 3:5. Walking the same way, B laps A after 110 minutes. Then B turns around at 1.2 times the old speed; after 20 more minutes they are still 1.6 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $300$ m, B walks $500$ m.
Same direction from the same start: first meeting after $1$ hour $50$ minutes.
Then B reverses direction at $1.2$ times B's original speed.
After $20$ more minutes, $1.6$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

300:500 = 3:5

, so let A walk

3x

m per minute and B walk

5x

m per minute. Let the lap be

y

m. Note that

1

hour

50

minutes is

110

minutes.

\text{A}: 3x,\quad \text{B}: 5x,\quad 1.2 \times 5x = 6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

110 \times 5x - 110 \times 3x = y,\quad 220x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

20

minutes they have covered

60x + 120x

and

1600

m is still left.

20 \times 3x + 20 \times 6x = y - 1600,\quad 180x = y - 1600

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 220x

into the second equation.

180x = 220x - 1600,\quad 40x = 1600,\quad x = 40,\quad y = 220 \times 40 = 8800

The extra

40x

in the first stage is exactly the missing

1600

m.

5Convert to kilometers

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

8800

m is

8.8

km.

8800 \text{ m} = 8.8 \text{ km}

The question asks for kilometers.

Answer:

8.8

km

4 · Reviewdoes it hold up?

A walks $120$ m per minute and B $200$ m per minute. In $110$ minutes B gains $80 \times 110 = 8800$ m, one lap. Afterward B walks $240$ m per minute; in $20$ minutes they cover $2400 + 4800 = 7200$ m, leaving $8800 - 7200 = 1600$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $220x$ (from the chase) and also $180x + 1600$ (from the approach). The difference $40x$ must be $1600$ m, so $x = 40$ and the lap is $220 \times 40 = 8800$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Variant 12 hard

A and B decide to walk around a lake. While A walks $400$ m, B walks $700$ m. If they start from the same point at the same time and walk around in the same direction at constant speeds, they meet for the first time after $1$ hour $20$ minutes. After that, B starts walking in the opposite direction at a constant speed $0.8$ times B's original speed, and $15$ minutes later there are still $4.8$ km left until the two meet again. How many kilometers is the distance around the lake?

Your answer

Show solution

1 · Understandwhat's really being asked

A and B walk around a lake with speeds in the ratio 4:7. Walking the same way, B laps A after 80 minutes. Then B turns around at 0.8 times the old speed; after 15 more minutes they are still 4.8 km apart along the path toward each other. Find the distance around the lake.

Givens

While A walks $400$ m, B walks $700$ m.
Same direction from the same start: first meeting after $1$ hour $20$ minutes.
Then B reverses direction at $0.8$ times B's original speed.
After $15$ more minutes, $4.8$ km remain before they meet again.

Unknowns

The distance around the lake, in km.

Constraints

Speeds are constant within each stage.

2 · Planchoose the strategy

#13 Convert to Algebra · also uses: #1 Draw a Diagram#8 Analyze the Units

Draw the lake as a circle. Walking the same way, the faster walker must gain one full lap to meet the slower one; walking toward each other from the meeting point, the two distances must add up to one full lap. Each picture gives one equation.

3 · Execute5 carry out the plan

1Name the speeds and the lap

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

Speeds are in the ratio

400:700 = 4:7

, so let A walk

4x

m per minute and B walk

7x

m per minute. Let the lap be

y

m. Note that

1

hour

20

minutes is

80

minutes.

\text{A}: 4x,\quad \text{B}: 7x,\quad 0.8 \times 7x = 5.6x

A ratio of speeds lets one letter stand for both.

2Same direction: B gains one lap

#1 Draw a Diagram 8.EE.C.8

Starting together and walking the same way, they next meet when B is exactly one lap ahead.

80 \times 7x - 80 \times 4x = y,\quad 240x = y

Same direction means the distances differ by one lap.

3Opposite directions: together they cover the lap

#1 Draw a Diagram 8.EE.C.8

From the meeting point they walk toward each other around the lake, so they meet again when their distances add to one lap. After

15

minutes they have covered

60x + 84x

and

4800

m is still left.

15 \times 4x + 15 \times 5.6x = y - 4800,\quad 144x = y - 4800

Opposite directions means the distances add to one lap.

4Solve

#13 Convert to Algebra 8.EE.C.8

Substitute

y = 240x

into the second equation.

144x = 240x - 4800,\quad 96x = 4800,\quad x = 50,\quad y = 240 \times 50 = 12000

The extra

96x

in the first stage is exactly the missing

4800

m.

5Convert to kilometers

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

12000

m is

12

km.

12000 \text{ m} = 12 \text{ km}

The question asks for kilometers.

Answer:

12

km

4 · Reviewdoes it hold up?

A walks $200$ m per minute and B $350$ m per minute. In $80$ minutes B gains $150 \times 80 = 12000$ m, one lap. Afterward B walks $280$ m per minute; in $15$ minutes they cover $3000 + 4200 = 7200$ m, leaving $12000 - 7200 = 4800$ m, as given.

Another way: Compare the two stages per unit $x$ : one lap is $240x$ (from the chase) and also $144x + 4800$ (from the approach). The difference $96x$ must be $4800$ m, so $x = 50$ and the lap is $240 \times 50 = 12000$ m.

Standardsmin grade 8

8.EE.C.8 Analyze and solve pairs of simultaneous linear equations — Writing and solving one equation for each stage of the walk.
6.RP.A.3 Use ratio and rate reasoning to solve problems — Using the speed ratio and converting meters to kilometers.

💡Takeaway. On a loop, chasing means the gap closes by one lap; walking toward each other means the two distances add to one lap.

Same way: lap gap; opposite way: lap sum · 12 practice problems

Generated variants — 12

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