← Objects in a line leave one fewer gap than there are objects · Objects versus Gaps (Fencepost Counting)

Objects in a line leave one fewer gap than there are objects · 12 practice problems

6.NS.B.34.OA.A.3

Generated variants — 12

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

Variant 1 easy answer: 2.06 m

A path is 10.3 m10.3\ \text{m} long on each side, and 1212 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 10.3 m long on each side gets 12 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 10.3 m long.
  • There are 12 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
12÷2=612 \div 2 = 6
6 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 6 sculptures in a row make 5 spaces between them.
6−1=56 - 1 = 5
5 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
10.3÷5=2.0610.3 \div 5 = 2.06
They stand 2.06 m apart.
Answer: 2.06 m
4 · Reviewdoes it hold up?

5 gaps of 2.06 m come back to 10.3 m, the length of one side.

Another way: Dividing by 6 instead of 5 is the usual slip. It spaces the row as if there were 6 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 2 easy answer: 2.25 m

A path is 22.5 m22.5\ \text{m} long on each side, and 2222 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 22.5 m long on each side gets 22 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 22.5 m long.
  • There are 22 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
22÷2=1122 \div 2 = 11
11 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 11 sculptures in a row make 10 spaces between them.
11−1=1011 - 1 = 10
10 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
22.5÷10=2.2522.5 \div 10 = 2.25
They stand 2.25 m apart.
Answer: 2.25 m
4 · Reviewdoes it hold up?

10 gaps of 2.25 m come back to 22.5 m, the length of one side.

Another way: Dividing by 11 instead of 10 is the usual slip. It spaces the row as if there were 11 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 3 easy answer: 6.96 m

A path is 34.8 m34.8\ \text{m} long on each side, and 1212 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 34.8 m long on each side gets 12 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 34.8 m long.
  • There are 12 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
12÷2=612 \div 2 = 6
6 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 6 sculptures in a row make 5 spaces between them.
6−1=56 - 1 = 5
5 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
34.8÷5=6.9634.8 \div 5 = 6.96
They stand 6.96 m apart.
Answer: 6.96 m
4 · Reviewdoes it hold up?

5 gaps of 6.96 m come back to 34.8 m, the length of one side.

Another way: Dividing by 6 instead of 5 is the usual slip. It spaces the row as if there were 6 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 4 easy answer: 2.56 m

A path is 38.4 m38.4\ \text{m} long on each side, and 3232 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 38.4 m long on each side gets 32 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 38.4 m long.
  • There are 32 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
32÷2=1632 \div 2 = 16
16 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 16 sculptures in a row make 15 spaces between them.
16−1=1516 - 1 = 15
15 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
38.4÷15=2.5638.4 \div 15 = 2.56
They stand 2.56 m apart.
Answer: 2.56 m
4 · Reviewdoes it hold up?

15 gaps of 2.56 m come back to 38.4 m, the length of one side.

Another way: Dividing by 16 instead of 15 is the usual slip. It spaces the row as if there were 16 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 5 medium answer: 9.18 m

A path is 45.9 m45.9\ \text{m} long on each side, and 1212 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 45.9 m long on each side gets 12 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 45.9 m long.
  • There are 12 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
12÷2=612 \div 2 = 6
6 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 6 sculptures in a row make 5 spaces between them.
6−1=56 - 1 = 5
5 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
45.9÷5=9.1845.9 \div 5 = 9.18
They stand 9.18 m apart.
Answer: 9.18 m
4 · Reviewdoes it hold up?

5 gaps of 9.18 m come back to 45.9 m, the length of one side.

Another way: Dividing by 6 instead of 5 is the usual slip. It spaces the row as if there were 6 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 6 medium answer: 2.2 m

A path is 50.6 m50.6\ \text{m} long on each side, and 4848 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 50.6 m long on each side gets 48 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 50.6 m long.
  • There are 48 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
48÷2=2448 \div 2 = 24
24 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 24 sculptures in a row make 23 spaces between them.
24−1=2324 - 1 = 23
23 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
50.6÷23=2.250.6 \div 23 = 2.2
They stand 2.2 m apart.
Answer: 2.2 m
4 · Reviewdoes it hold up?

23 gaps of 2.2 m come back to 50.6 m, the length of one side.

Another way: Dividing by 24 instead of 23 is the usual slip. It spaces the row as if there were 24 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 7 medium answer: 5.81 m

A path is 58.1 m58.1\ \text{m} long on each side, and 2222 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 58.1 m long on each side gets 22 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 58.1 m long.
  • There are 22 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
22÷2=1122 \div 2 = 11
11 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 11 sculptures in a row make 10 spaces between them.
11−1=1011 - 1 = 10
10 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
58.1÷10=5.8158.1 \div 10 = 5.81
They stand 5.81 m apart.
Answer: 5.81 m
4 · Reviewdoes it hold up?

10 gaps of 5.81 m come back to 58.1 m, the length of one side.

Another way: Dividing by 11 instead of 10 is the usual slip. It spaces the row as if there were 11 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 8 medium answer: 1.25 m

A path is 42.5 m42.5\ \text{m} long on each side, and 7070 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 42.5 m long on each side gets 70 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 42.5 m long.
  • There are 70 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
70÷2=3570 \div 2 = 35
35 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 35 sculptures in a row make 34 spaces between them.
35−1=3435 - 1 = 34
34 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
42.5÷34=1.2542.5 \div 34 = 1.25
They stand 1.25 m apart.
Answer: 1.25 m
4 · Reviewdoes it hold up?

34 gaps of 1.25 m come back to 42.5 m, the length of one side.

Another way: Dividing by 35 instead of 34 is the usual slip. It spaces the row as if there were 35 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 9 hard answer: 14.16 m

A path is 70.8 m70.8\ \text{m} long on each side, and 1212 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 70.8 m long on each side gets 12 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 70.8 m long.
  • There are 12 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
12÷2=612 \div 2 = 6
6 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 6 sculptures in a row make 5 spaces between them.
6−1=56 - 1 = 5
5 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
70.8÷5=14.1670.8 \div 5 = 14.16
They stand 14.16 m apart.
Answer: 14.16 m
4 · Reviewdoes it hold up?

5 gaps of 14.16 m come back to 70.8 m, the length of one side.

Another way: Dividing by 6 instead of 5 is the usual slip. It spaces the row as if there were 6 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 10 hard answer: 3.95 m

A path is 79 m79\ \text{m} long on each side, and 4242 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 79 m long on each side gets 42 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 79 m long.
  • There are 42 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
42÷2=2142 \div 2 = 21
21 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 21 sculptures in a row make 20 spaces between them.
21−1=2021 - 1 = 20
20 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
79÷20=3.9579 \div 20 = 3.95
They stand 3.95 m apart.
Answer: 3.95 m
4 · Reviewdoes it hold up?

20 gaps of 3.95 m come back to 79 m, the length of one side.

Another way: Dividing by 21 instead of 20 is the usual slip. It spaces the row as if there were 21 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 11 hard answer: 44.4 m

A path is 88.8 m88.8\ \text{m} long on each side, and 66 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 88.8 m long on each side gets 6 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 88.8 m long.
  • There are 6 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
6÷2=36 \div 2 = 3
3 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 3 sculptures in a row make 2 spaces between them.
3−1=23 - 1 = 2
2 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
88.8÷2=44.488.8 \div 2 = 44.4
They stand 44.4 m apart.
Answer: 44.4 m
4 · Reviewdoes it hold up?

2 gaps of 44.4 m come back to 88.8 m, the length of one side.

Another way: Dividing by 3 instead of 2 is the usual slip. It spaces the row as if there were 3 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.
Variant 12 hard answer: 0.5 m

A path is 26.5 m26.5\ \text{m} long on each side, and 108108 sculptures are to be placed along both sides at equal intervals, including one at each end. How many meters apart are the sculptures? (Ignore the thickness of the sculptures.)

Show solution
1 · Understandwhat's really being asked

A path 26.5 m long on each side gets 108 sculptures spread along both sides, one at each end. We want the distance between neighbours.

Givens
  • Each side of the path is 26.5 m long.
  • There are 108 sculptures in all.
  • They go along both sides, with one at each end.
Unknowns
  • The distance between two neighbouring sculptures.
Constraints
  • The intervals are all equal, and the sculptures have no thickness.
2 · Planchoose the strategy

#1 Draw a Diagram · also uses: #7 Identify Subproblems#8 Analyze the Units

The division at the end is easy; the counting before it is not. Split the sculptures between the two sides first, then count gaps rather than sculptures -- a row with one at each end has one fewer gap than it has objects.

3 · Execute3 carry out the plan

1Share the sculptures between the sides

#7 Identify Subproblems 4.OA.A.3
Both sides are the same length and treated the same way.
108÷2=54108 \div 2 = 54
54 sculptures on each side.

2Count the gaps, not the sculptures

#1 Draw a Diagram 4.OA.A.3
With one at each end, 54 sculptures in a row make 53 spaces between them.
54−1=5354 - 1 = 53
53 gaps to share the length between.

3Divide the length by the gaps

#8 Analyze the Units 6.NS.B.3
The gaps are equal, so each one is the same fraction of the side.
26.5÷53=0.526.5 \div 53 = 0.5
They stand 0.5 m apart.
Answer: 0.5 m
4 · Reviewdoes it hold up?

53 gaps of 0.5 m come back to 26.5 m, the length of one side.

Another way: Dividing by 54 instead of 53 is the usual slip. It spaces the row as if there were 54 gaps, which would leave one hanging off the far end with no sculpture after it.

Standardsmin grade 6
  • 6.NS.B.3 Fluently add, subtract, multiply, divide multi-digit decimals — Dividing a decimal length by a whole number of gaps.
  • 4.OA.A.3 Solve multi-step word problems using four operations with whole numbers — Working out how many sculptures and how many gaps there are.
💡Takeaway. Count the spaces between things, not the things. With one at each end there is always one space fewer.