← 3-2 · Posts versus gaps on lines and loops · Objects versus Gaps (Fencepost Counting)

Posts versus gaps on lines and loops · 12 practice problems

3.OA.A.33.MD.D.8

Generated variants — 12

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

Variant 1 answer: 396 ft

Around the edge of a round pond, $18$ flags are set up at equal intervals of $22\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

18 flags stand at equal 22-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

18 flags are placed at equal intervals.
Each gap between neighboring flags is 22 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 22 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 18 flags there are 18 gaps.

\text{number of gaps} = \text{number of flags} = 18

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 18 gaps is 22 feet, so the distance is 18 times 22.

18 \times 22 = 396

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 396 ft

Review

About 18 gaps of about 22 feet is roughly 396 feet, so our exact 396 feet is reasonable.

Compute 18 x 22 directly: 17 x 22 = 374 plus 22 gives 396 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 2 answer: 280 ft

Along one side of a straight road, $9$ trees are planted in a straight row at equal intervals of $35\ \text{ft}$ apart. What is the distance from the first tree to the last tree, in feet? (Ignore the thickness of the trees.)

Show solution

Understand

9 trees stand at equal 35-foot gaps along one side of a straight road. Find the distance from the first tree to the last tree.

Givens

9 trees are placed at equal intervals.
Each gap between neighboring trees is 35 feet.
The trees are planted in a straight row (an open path).

Unknowns

The distance from the first tree to the last tree, in feet.

Constraints

The path is open, so the two end posts have no gap beyond them.
Trees thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few trees to see how the number of gaps relates to the number of trees. Then the total distance is the number of gaps times 35 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a short row with, say, 4 trees: there are only 3 gaps because the row has two loose ends. So on a straight line the number of gaps is one fewer than the number of trees. With 9 trees there are 8 gaps.

\text{number of gaps} = 9 - 1 = 8

An open row has two free ends, so it has one fewer gap than posts.

#1 Draw a Diagram 3.MD.D.8

Each of the 8 gaps is 35 feet, so the distance is 8 times 35.

8 \times 35 = 280

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 280 ft

Review

About 8 gaps of about 35 feet is roughly 280 feet, so our exact 280 feet is reasonable.

Compute 8 x 35 directly: 7 x 35 = 245 plus 35 gives 280 feet, confirming the line rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 3 answer: 300 ft

Along one side of a straight road, $21$ trees are planted in a straight row at equal intervals of $15\ \text{ft}$ apart. What is the distance from the first tree to the last tree, in feet? (Ignore the thickness of the trees.)

Show solution

Understand

21 trees stand at equal 15-foot gaps along one side of a straight road. Find the distance from the first tree to the last tree.

Givens

21 trees are placed at equal intervals.
Each gap between neighboring trees is 15 feet.
The trees are planted in a straight row (an open path).

Unknowns

The distance from the first tree to the last tree, in feet.

Constraints

The path is open, so the two end posts have no gap beyond them.
Trees thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few trees to see how the number of gaps relates to the number of trees. Then the total distance is the number of gaps times 15 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a short row with, say, 4 trees: there are only 3 gaps because the row has two loose ends. So on a straight line the number of gaps is one fewer than the number of trees. With 21 trees there are 20 gaps.

\text{number of gaps} = 21 - 1 = 20

An open row has two free ends, so it has one fewer gap than posts.

#1 Draw a Diagram 3.MD.D.8

Each of the 20 gaps is 15 feet, so the distance is 20 times 15.

20 \times 15 = 300

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 300 ft

Review

About 20 gaps of about 15 feet is roughly 300 feet, so our exact 300 feet is reasonable.

Compute 20 x 15 directly: 19 x 15 = 285 plus 15 gives 300 feet, confirming the line rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 4 answer: 360 ft

Around the edge of a round pond, $24$ flags are set up at equal intervals of $15\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

24 flags stand at equal 15-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

24 flags are placed at equal intervals.
Each gap between neighboring flags is 15 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 15 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 24 flags there are 24 gaps.

\text{number of gaps} = \text{number of flags} = 24

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 24 gaps is 15 feet, so the distance is 24 times 15.

24 \times 15 = 360

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 360 ft

Review

About 24 gaps of about 15 feet is roughly 360 feet, so our exact 360 feet is reasonable.

Compute 24 x 15 directly: 23 x 15 = 345 plus 15 gives 360 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 5 answer: 300 ft

Around the edge of a round pond, $15$ flags are set up at equal intervals of $20\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

15 flags stand at equal 20-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

15 flags are placed at equal intervals.
Each gap between neighboring flags is 20 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 20 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 15 flags there are 15 gaps.

\text{number of gaps} = \text{number of flags} = 15

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 15 gaps is 20 feet, so the distance is 15 times 20.

15 \times 20 = 300

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 300 ft

Review

About 15 gaps of about 20 feet is roughly 300 feet, so our exact 300 feet is reasonable.

Compute 15 x 20 directly: 14 x 20 = 280 plus 20 gives 300 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 6 answer: 558 ft

Around the edge of a round pond, $31$ flags are set up at equal intervals of $18\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

31 flags stand at equal 18-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

31 flags are placed at equal intervals.
Each gap between neighboring flags is 18 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 18 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 31 flags there are 31 gaps.

\text{number of gaps} = \text{number of flags} = 31

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 31 gaps is 18 feet, so the distance is 31 times 18.

31 \times 18 = 558

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 558 ft

Review

About 31 gaps of about 18 feet is roughly 558 feet, so our exact 558 feet is reasonable.

Compute 31 x 18 directly: 30 x 18 = 540 plus 18 gives 558 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 7 answer: 300 ft

Along one side of a straight road, $11$ trees are planted in a straight row at equal intervals of $30\ \text{ft}$ apart. What is the distance from the first tree to the last tree, in feet? (Ignore the thickness of the trees.)

Show solution

Understand

11 trees stand at equal 30-foot gaps along one side of a straight road. Find the distance from the first tree to the last tree.

Givens

11 trees are placed at equal intervals.
Each gap between neighboring trees is 30 feet.
The trees are planted in a straight row (an open path).

Unknowns

The distance from the first tree to the last tree, in feet.

Constraints

The path is open, so the two end posts have no gap beyond them.
Trees thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few trees to see how the number of gaps relates to the number of trees. Then the total distance is the number of gaps times 30 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a short row with, say, 4 trees: there are only 3 gaps because the row has two loose ends. So on a straight line the number of gaps is one fewer than the number of trees. With 11 trees there are 10 gaps.

\text{number of gaps} = 11 - 1 = 10

An open row has two free ends, so it has one fewer gap than posts.

#1 Draw a Diagram 3.MD.D.8

Each of the 10 gaps is 30 feet, so the distance is 10 times 30.

10 \times 30 = 300

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 300 ft

Review

About 10 gaps of about 30 feet is roughly 300 feet, so our exact 300 feet is reasonable.

Compute 10 x 30 directly: 9 x 30 = 270 plus 30 gives 300 feet, confirming the line rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 8 answer: 300 ft

Around the edge of a round pond, $12$ flags are set up at equal intervals of $25\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

12 flags stand at equal 25-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

12 flags are placed at equal intervals.
Each gap between neighboring flags is 25 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 25 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 12 flags there are 12 gaps.

\text{number of gaps} = \text{number of flags} = 12

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 12 gaps is 25 feet, so the distance is 12 times 25.

12 \times 25 = 300

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 300 ft

Review

About 12 gaps of about 25 feet is roughly 300 feet, so our exact 300 feet is reasonable.

Compute 12 x 25 directly: 11 x 25 = 275 plus 25 gives 300 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 9 answer: 480 ft

Around the edge of a round pond, $40$ flags are set up at equal intervals of $12\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

40 flags stand at equal 12-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

40 flags are placed at equal intervals.
Each gap between neighboring flags is 12 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 12 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 40 flags there are 40 gaps.

\text{number of gaps} = \text{number of flags} = 40

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 40 gaps is 12 feet, so the distance is 40 times 12.

40 \times 12 = 480

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 480 ft

Review

About 40 gaps of about 12 feet is roughly 480 feet, so our exact 480 feet is reasonable.

Compute 40 x 12 directly: 39 x 12 = 468 plus 12 gives 480 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 10 answer: 375 ft

Along one side of a straight road, $16$ trees are planted in a straight row at equal intervals of $25\ \text{ft}$ apart. What is the distance from the first tree to the last tree, in feet? (Ignore the thickness of the trees.)

Show solution

Understand

16 trees stand at equal 25-foot gaps along one side of a straight road. Find the distance from the first tree to the last tree.

Givens

16 trees are placed at equal intervals.
Each gap between neighboring trees is 25 feet.
The trees are planted in a straight row (an open path).

Unknowns

The distance from the first tree to the last tree, in feet.

Constraints

The path is open, so the two end posts have no gap beyond them.
Trees thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few trees to see how the number of gaps relates to the number of trees. Then the total distance is the number of gaps times 25 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a short row with, say, 4 trees: there are only 3 gaps because the row has two loose ends. So on a straight line the number of gaps is one fewer than the number of trees. With 16 trees there are 15 gaps.

\text{number of gaps} = 16 - 1 = 15

An open row has two free ends, so it has one fewer gap than posts.

#1 Draw a Diagram 3.MD.D.8

Each of the 15 gaps is 25 feet, so the distance is 15 times 25.

15 \times 25 = 375

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 375 ft

Review

About 15 gaps of about 25 feet is roughly 375 feet, so our exact 375 feet is reasonable.

Compute 15 x 25 directly: 14 x 25 = 350 plus 25 gives 375 feet, confirming the line rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 11 answer: 260 ft

Around the edge of a round pond, $26$ flags are set up at equal intervals of $10\ \text{ft}$ apart. What is the distance around the pond, in feet? (Ignore the thickness of the flags.)

Show solution

Understand

26 flags stand at equal 10-foot gaps along the edge of a round pond. Find the distance around the pond.

Givens

26 flags are placed at equal intervals.
Each gap between neighboring flags is 10 feet.
The flags are arranged around a closed loop (the pond edge).

Unknowns

The distance around the pond, in feet.

Constraints

The path is a closed loop, so the gaps wrap back to the start.
Flags thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few flags to see how the number of gaps relates to the number of flags. Then the total distance is the number of gaps times 10 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a small ring with, say, 4 flags: there are 4 gaps because the last flag connects back to the first. So on a closed loop the number of gaps equals the number of flags. With 26 flags there are 26 gaps.

\text{number of gaps} = \text{number of flags} = 26

On a circle there is no loose end, so each flag starts exactly one gap.

#1 Draw a Diagram 3.MD.D.8

Each of the 26 gaps is 10 feet, so the distance is 26 times 10.

26 \times 10 = 260

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 260 ft

Review

About 26 gaps of about 10 feet is roughly 260 feet, so our exact 260 feet is reasonable.

Compute 26 x 10 directly: 25 x 10 = 250 plus 10 gives 260 feet, confirming the loop rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!

Variant 12 answer: 216 ft

Along one side of a straight road, $13$ trees are planted in a straight row at equal intervals of $18\ \text{ft}$ apart. What is the distance from the first tree to the last tree, in feet? (Ignore the thickness of the trees.)

Show solution

Understand

13 trees stand at equal 18-foot gaps along one side of a straight road. Find the distance from the first tree to the last tree.

Givens

13 trees are placed at equal intervals.
Each gap between neighboring trees is 18 feet.
The trees are planted in a straight row (an open path).

Unknowns

The distance from the first tree to the last tree, in feet.

Constraints

The path is open, so the two end posts have no gap beyond them.
Trees thickness is ignored.

Plan

#1 Draw a Diagram · also uses: #9 Solve an Easier Related Problem

Sketch a few trees to see how the number of gaps relates to the number of trees. Then the total distance is the number of gaps times 18 feet.

Execute

#9 Solve an Easier Related Problem 3.OA.A.3

Draw a short row with, say, 4 trees: there are only 3 gaps because the row has two loose ends. So on a straight line the number of gaps is one fewer than the number of trees. With 13 trees there are 12 gaps.

\text{number of gaps} = 13 - 1 = 12

An open row has two free ends, so it has one fewer gap than posts.

#1 Draw a Diagram 3.MD.D.8

Each of the 12 gaps is 18 feet, so the distance is 12 times 18.

12 \times 18 = 216

Adding up equal gaps along the path is just multiplication, and the total is the distance.

Answer: 216 ft

Review

About 12 gaps of about 18 feet is roughly 216 feet, so our exact 216 feet is reasonable.

Compute 12 x 18 directly: 11 x 18 = 198 plus 18 gives 216 feet, confirming the line rule.

Standards · min grade 3

3.OA.A.3 Solve multiplication and division word problems within 100 — Relating gaps to posts and setting up the multiplication.
3.MD.D.8 Solve real-world problems involving perimeters of polygons — Finding the total distance along the path.

💡 On a loop, gaps equal posts; on a line, gaps are one fewer -- draw a tiny sketch and Grade 3 multiplication does the rest!