Primary Mathematics: Intervals, Fence Posts and Cuts | Worked Learning Guide

BTT Mathematics / Primary Mathematics Learning Hub

Count the gaps before counting the objects. Fence posts, trees, cuts, bell sounds, pages and repeated events often look like different problems. Their shared difficulty is deciding whether the calculation counts a boundary, an interval, a physical object or an occurrence.

Place four counters in a straight row and look at the spaces between neighbouring counters. There are three spaces, not four. Now join the last counter back to the first in a loop. There are four spaces. The counters did not change; the boundary condition changed.

This small difference explains why a correct division can still lead to a wrong answer. A learner divides a 48 m path by a 6 m spacing and obtains eight. The arithmetic is correct. However, eight is the number of intervals. When both ends require a post, the number of posts is nine.

This worked guide develops interval reasoning from concrete marks and gaps to multi-step word problems. The later extensions include shared corners, maximum spacing, circular arrangements and repeated events. They are teaching extensions, not a claim that every example is required at every Primary level. All physical situations are simplified mathematical models: posts have negligible width, lengths are measured along the stated route, and cutting rules are explicitly specified.

Marks and gaps · Boundary conditions · Cuts and pieces · Repeated events · 24 questions · Worked solutions

1. An interval is the space between two neighbouring marks

Label marks along a straight path A, B, C and D. The intervals are AB, BC and CD. Every interval has a start and an end, but neighbouring intervals share a mark. Counting two endpoints for every interval would count the internal marks twice.

A-----B-----C-----D
   1     2     3

There are four marks and three intervals. To extend the row, add one new interval and one new mark at the far end. The first mark is needed before the first interval can begin. That is why an open chain with n intervals and marks at both ends contains n + 1 marks.

Worked example: equal spacing along a path

A 48 m straight path has posts at both ends and every 6 m between them. First count intervals: 48 ÷ 6 = 8. Then include the initial boundary mark: 8 + 1 = 9 posts.

The positions provide an independent check: 0, 6, 12, 18, 24, 30, 36, 42 and 48 m. Listing positions is slower for a long path but extremely useful while learning the structure. It shows exactly where the extra post comes from.

Reverse the question

Eleven posts stand in a straight row, equally spaced 2 m apart. Both end posts mark the route’s endpoints. There are ten intervals, so the distance from the first post to the last is 10 × 2 = 20 m.

Multiplying eleven by two would give every post its own extra interval, including one beyond the final post. The error is not multiplication; it is attaching the spacing to the wrong count.

2. The endpoints decide whether to add or subtract one

Keep the same 48 m route divided into eight intervals of 6 m. With posts at both endpoints, there are nine posts. With a post at exactly one endpoint and at every internal division, there are eight. With neither endpoint included but every internal division marked, there are seven.

The same eight intervals can produce three different post counts

Endpoint rule	Post positions	Count
Both included	0 through 48 m, every 6 m	9
Start included, end excluded	0 through 42 m, every 6 m	8
Both excluded	6 through 42 m, every 6 m	7

These rules assume that the route boundaries lie on the same regular spacing pattern. They are not instructions to add or subtract one whenever the word “post” appears. Begin by drawing a short version and marking which endpoints belong to the count.

Existing objects versus new objects

A 30 m path needs posts every 3 m, including both ends. The complete arrangement needs 30 ÷ 3 + 1 = 11 posts. If both end posts already exist, only 11 − 2 = 9 new posts are needed.

The requested quantity is new posts, not total posts. This distinction often appears after the main calculation and is easy to miss. Write a final sentence that names the quantity: “Nine additional posts are needed.” That sentence checks interpretation as well as arithmetic.

Inside a range is not the same as including its boundaries

Suppose lamps are located every 4 m along a 24 m model path, at positions 0, 4, 8, 12, 16, 20 and 24 m. A gate extends from 8 m to 16 m, and lamps at the two gate ends must remain. Only the lamp at 12 m is strictly inside the gate and must be removed.

Dividing the 8 m gate length by 4 m gives two intervals. It does not give the number of internal lamps. Those two intervals have one internal dividing mark. After removing that mark, six of the original seven lamps remain.

3. A closed route has no extra first post

On a loop, every post begins one interval and ends another. There is no unpaired endpoint. A 48 m closed route with equally spaced posts every 6 m contains eight intervals and eight distinct posts.

If we list distances travelled from a chosen starting post, we reach 0, 6, 12, 18, 24, 30, 36, 42 and 48 m. The locations labelled 0 m and 48 m are the same physical post. The list records nine arrival descriptions but only eight distinct locations.

This is the first important bridge to repeated-event problems: a physical place may be counted once while visits to that place are counted many times. Always ask what counts as a different item.

Worked example: posts around a rectangle

A rectangular route has sides 18 m and 12 m. Posts are placed every 3 m around the boundary, including all corners. Its perimeter is 2 × (18 + 12) = 60 m. Because it is a closed route, the distinct post count is 60 ÷ 3 = 20.

Check by sides. Each 18 m side has seven posts including its two corners. Each 12 m side has five. Adding gives 7 + 5 + 7 + 5 = 24, but each of four corners was counted twice. Subtract one duplicate copy of each corner: 24 − 4 = 20.

The corner requirement is compatible because both side lengths are multiples of 3 m. If a side were not divisible by the exact spacing, the stated arrangement might be impossible. A perimeter divisible by the spacing is not by itself enough when every corner must be a post.

Shared endpoints in joined routes

Two open arms meet at one endpoint. The first arm contains nine posts and the second seven, with the shared junction included in both counts. There are 9 + 7 − 1 = 15 distinct posts, not sixteen.

If routes intersect at more than one marked point, account for each shared point separately. If three routes meet at one common post and the post appears in all three counts, remove two duplicate copies, not three. It must remain counted once.

4. Exact spacing and maximum spacing are different conditions

A 25 m path cannot have posts at both ends with every consecutive gap exactly 6 m. Exact gaps of 6 m would make the total length a whole multiple of six. But 25 ÷ 6 is not whole.

Placing posts at 0, 6, 12, 18 and 24 m leaves the endpoint at 25 m unmarked. Adding a post there creates a final gap of 1 m, breaking the “exactly 6 m” condition. The construction is useful for a different question, but it is not a solution to this one.

Worked example: no gap may exceed 6 m

Change the rule: both ends of the 25 m path need posts, and no gap may be longer than 6 m. Four gaps can cover at most 4 × 6 = 24 m, which is too short. At least five gaps are necessary. Five gaps of 5 m work, giving six posts.

This is a minimum proof with two parts. Four gaps cannot work; five gaps can work. Merely rounding 25 ÷ 6 upward would give the correct gap count, but the reasoning explains why rounding upward is appropriate.

The placements are not unique. Gaps of 6, 6, 6, 6 and 1 m also obey the maximum-gap rule. The question asks for the minimum number of posts, not necessarily equal gaps. Distinguish the best count from a unique arrangement.

Do not round an impossible exact-spacing answer

For exact spacing, a non-whole interval count signals incompatibility. For maximum spacing, it signals the need for another interval. The same division result leads to different actions because the conditions differ. This is why a memorised “always round up” rule is unsafe.

5. Separate components keep separate endpoints

Three disjoint open paths each contain six intervals and have marks at both ends. Each path requires seven marks, so the total is twenty-one. Adding the eighteen intervals and then adding just one would incorrectly join three separate paths into a single chain.

For r disjoint open chains, each with marked endpoints, the total mark count equals the total interval count plus r. This formula is limited to separate chains. It should not be applied blindly to branched networks, where junctions and sharing need their own accounting.

A useful small-case test is to draw two one-interval paths. They contain four endpoint marks. Merging one pair of endpoints joins the paths into a two-interval chain with three marks. The number of intervals is unchanged; the number of distinct endpoints changes.

6. Count what one cut changes

For cutting questions in this section, a cut passes completely through one existing straight piece and separates it into two non-empty pieces. Pieces are not stacked, and a single cut does not pass through several different pieces. Under these rules, every cut increases the piece count by exactly one.

Starting with one rod, one cut produces two pieces, two cuts produce three, and seven cuts produce eight. To obtain p pieces from one straight rod, p − 1 cuts are needed. The formula comes from the change per cut, not from the shape of a familiar diagram.

Worked example: several rods

Five separate rods are each divided into four pieces. Each rod needs three cuts, so the total is 5 × 3 = 15 cuts. There are twenty final pieces, but five pieces existed before any cutting. The increase is 20 − 5 = 15.

This gives a second check: final pieces = initial pieces + cuts. With four initial pieces and nine permitted cuts, there will be thirteen pieces. It does not matter which pieces are selected at each stage, provided every cut obeys the one-piece rule.

Why stacking changes the model

If a cut passes through a stack of three rods, it can create three extra pieces at once. The one-extra-piece rule no longer describes that operation. A question permitting stacking must state or allow a different interpretation of “one cut”.

Do not silently use a physical shortcut to answer a mathematical model with different rules. Likewise, do not insist on the one-piece formula when the problem explicitly permits simultaneous cuts. The operation definition belongs to the problem.

A closed wire ring has no initial open endpoints

Cutting a closed wire ring once opens it into one continuous length. It does not yet create two separate lengths. Cutting at two distinct points gives two lengths; cutting at six distinct positions gives six separate open lengths.

Thus one closed ring cut into six separate open lengths requires six cuts, not five. The first cut changes the topology of the object from closed to open; only later cuts increase the number of pieces. This is another boundary-condition difference, like comparing a row of posts with a loop.

For the special case of one open length from a closed ring, one opening cut is needed. If the task merely asks to keep one closed piece, zero cuts suffice. Naming the desired final object avoids an apparent contradiction.

7. Count elapsed intervals between repeated events

A bell sounds at equally spaced instants. The elapsed time from the beginning of the first sound to the beginning of the seventh contains six intervals. It does not contain seven, because no interval before the first sound is included.

If that elapsed time is 18 seconds, each interval lasts 18 ÷ 6 = 3 seconds. From the first sound to the twelfth there are eleven intervals, so the elapsed time is 11 × 3 = 33 seconds.

We measure onset-to-onset times here. If a different problem includes the duration of each sound, that additional information may matter. A long bell ring is not the same object as the gap between consecutive ring beginnings.

Worked example: begin in the middle of the sequence

The time from the fourth beep to the eleventh beep is 28 seconds. Count intervals by index difference: 11 − 4 = 7. Each interval is 28 ÷ 7 = 4 seconds.

The time from the first beep to the twentieth is then nineteen intervals, or 76 seconds. Counting the seven named positions 4, 5, 6, 7, 8, 9 and 10 can help, because each begins one interval leading to the next beep.

Repeated clock-time events

A device beeps at 9:10 a.m. and every twelve minutes afterward. How many beeps occur from 9:10 a.m. to 10:10 a.m., including both endpoints? The elapsed time is sixty minutes, giving five intervals. Include the initial beep to obtain six events.

The times are 9:10, 9:22, 9:34, 9:46, 9:58 and 10:10. Excluding both endpoints leaves four events. Excluding only the final endpoint leaves five. The arithmetic and the inclusion rule must both be right.

For longer timetable calculations, the Time, Money and Schedules guide develops clock-time conversion separately. Here, the focus is the number of repeated occurrences inside the interval.

8. Repeated activities have breaks only where stated

Seven activity rounds each last four minutes. Between consecutive rounds there is a two-minute break, but there is no break before the first or after the last. The activity time is 7 × 4 = 28 minutes. There are six intervening breaks, giving another 6 × 2 = 12 minutes. Total time is forty minutes.

Multiplying seven by six would attach a break to every round, including the last. That answers a different schedule in which the final break is included. Draw the sequence activity–break–activity for a two-round case before expanding it.

Pages, labels and inclusive counts

Pages numbered 48 through 73 inclusive contain 73 − 48 + 1 = 26 page numbers. Subtraction gives the number of steps between labels; including both labelled positions requires one more.

Check with pages 48 and 49 alone. Their difference is one, but there are two page numbers. This small example reveals the endpoint issue without requiring a long list. Do not automatically use the same rule for “the number of pages strictly between 48 and 73”, which would exclude both endpoint pages.

Physical posts versus repeated visits

A runner starts at a marked line and completes a lap every forty seconds. By 120 seconds the runner has completed three laps and has been at the starting line at times 0, 40, 80 and 120 seconds. That is four occurrences, including the initial one, but only one physical starting line.

Counting unique places, completed intervals and recorded visits produces three different answers. The learner must identify which kind the question requests before deciding whether to add one.

9. Practice: 24 original questions

Assume negligible marker width and exact stated route lengths. Unless changed explicitly, marks are regularly spaced, and cuts separate one existing piece at a time without stacking.

Questions 1–8: Marks, gaps and shared endpoints

1. A 48 m straight path has posts every 6 m, including both endpoints. How many posts are there?
2. Eleven posts stand in a straight row, with 2 m between neighbours. Find the distance from the first to the last.
3. On the path in Question 1, include the starting post but exclude the finishing post. How many posts remain?
4. On the same path, include neither endpoint but retain all internal posts. Find the count.
5. A closed route is 48 m long with posts every 6 m. How many distinct posts are there?
6. A rectangular route has sides 18 m and 12 m, with posts every 3 m including every corner. Count distinct posts.
7. A 30 m straight path needs posts every 3 m including both ends. The two endpoint posts already exist. How many new posts are required?
8. Two open arms share one endpoint. Their post counts are nine and seven, each including the shared junction. Find the distinct total.

Questions 9–16: Exact conditions and cuts

9. Can a 25 m straight path have posts at both ends with every consecutive gap exactly 6 m? Explain.
10. For a 25 m straight path with both ends marked and every gap at most 6 m, find the minimum number of posts.
11. Lamps stand at 0, 4, 8, 12, 16, 20 and 24 m. A gate runs from 8 m to 16 m. Retain the lamps at the gate endpoints but remove all lamps strictly inside it. How many lamps remain?
12. Three disjoint open paths each have six intervals with all endpoints marked. Find the total number of marks.
13. One straight rod is divided into eight pieces. How many cuts are needed?
14. Five separate rods are each divided into four pieces. No stacking is allowed. Find the total cuts.
15. Four separate pieces undergo nine one-piece cuts. How many pieces result?
16. One closed wire ring is cut into six separate open lengths at distinct positions. How many cuts are required?

Questions 17–24: Repeated events and interpretation

17. Equally spaced bell onsets take 18 seconds from the first to the seventh. Find the interval between consecutive onsets.
18. Using Question 17, find the elapsed time from the first onset to the twelfth.
19. From the fourth beep to the eleventh takes 28 seconds. Find the interval between consecutive beeps.
20. A device beeps at 9:10 a.m. and every twelve minutes. Count events through 10:10 a.m., including both endpoints.
21. Using Question 20, count events strictly after 9:10 a.m. and strictly before 10:10 a.m.
22. Seven four-minute activities have two-minute breaks only between consecutive activities. Find the total duration.
23. Count the page numbers from 48 through 73 inclusive.
24. A runner starts at a marked line at time zero and completes each lap in forty seconds. By 120 seconds, how many laps are completed, how many start-line occurrences have been recorded including time zero, and how many physical starting lines are involved?

10. Worked solutions

Solutions 1–8

1. Nine posts. The division 48 ÷ 6 = 8 counts intervals. An open chain with both ends marked needs one more post than intervals: 8 + 1 = 9.

2. 20 m. Eleven posts create ten gaps. Multiply the number of gaps by their common length: 10 × 2 = 20 m. No extra gap follows the last post.

3. Eight posts. Remove the finishing post from the nine-post arrangement. The retained positions are 0 through 42 m in steps of 6 m.

4. Seven posts. Remove both endpoint posts from the original nine. The internal positions are 6, 12, 18, 24, 30, 36 and 42 m.

5. Eight posts. A closed route has as many distinct posts as intervals. Returning to the starting position after 48 m does not create a new post.

6. Twenty posts. Perimeter is 60 m, giving twenty 3 m intervals on a closed boundary. Alternatively add side counts 7 + 5 + 7 + 5 and subtract the four duplicated corners.

7. Nine new posts. The finished arrangement needs 30 ÷ 3 + 1 = 11 posts. Two already exist, leaving 11 − 2 = 9 to add.

8. Fifteen distinct posts. Adding the two arm totals counts the shared endpoint twice. Subtract its duplicate copy: 9 + 7 − 1 = 15.

Solutions 9–16

9. No. A whole number of exact 6 m gaps can cover only a multiple of 6 m. Twenty-five is not such a multiple. A final shorter gap would violate the stated rule.

10. Six posts. Four gaps of at most 6 m cover at most 24 m, too little. Five gaps are necessary, and five equal gaps of 5 m achieve the length. Five gaps need six endpoint-inclusive posts.

11. Six lamps remain. Only the 12 m lamp is strictly inside the gate. The 8 m and 16 m lamps are retained. Remove one from the original seven.

12. Twenty-one marks. Each six-interval open path has seven marks. The paths are disjoint, so none are shared: 3 × 7 = 21.

13. Seven cuts. One piece exists initially. Each permitted cut adds one piece. To reach eight pieces, the increase must be seven.

14. Fifteen cuts. Each rod requires 4 − 1 = 3 cuts. Five rods require 5 × 3 = 15. This matches twenty final pieces minus five initial pieces.

15. Thirteen pieces. Begin with four and add one for each of nine cuts: 4 + 9 = 13. The result depends on the no-stacking, one-piece rule.

16. Six cuts. The first cut opens the closed ring without producing a second piece. Five more cuts create five additional pieces, leaving six separate open lengths.

Solutions 17–24

17. Three seconds. Seven onsets contain six onset-to-onset intervals. Divide 18 by 6, not by 7.

18. Thirty-three seconds. The first-to-twelfth span has eleven intervals. At three seconds each, the total is 11 × 3 = 33 seconds.

19. Four seconds. Index difference 11 − 4 = 7 gives the interval count. Then 28 ÷ 7 = 4 seconds.

20. Six events. Sixty minutes contain five twelve-minute intervals. Including the initial event gives six. Listing 9:10, 9:22, 9:34, 9:46, 9:58 and 10:10 checks the count.

21. Four events. Remove both endpoint events from the six-event inclusive list. The four internal times are 9:22, 9:34, 9:46 and 9:58.

22. Forty minutes. Activities take 7 × 4 = 28 minutes. Six intervening breaks take 6 × 2 = 12 minutes. Total 28 + 12 = 40.

23. Twenty-six page numbers. There are 73 − 48 = 25 steps between the endpoint labels. Include both labels to obtain 25 + 1 = 26.

24. Three laps, four occurrences, one physical starting line. Each forty-second interval completes one lap. The recorded starting-line times are 0, 40, 80 and 120 seconds, all at the same location.

11. Transfer: change one condition at a time

A rectangular model boundary measures 24 m by 18 m. Posts stand every 6 m, including corners. A gate is later opened along the 24 m side from the 6 m post to the 18 m post. Retain both gate-end posts and remove posts strictly between them. Find the initial post total and the remaining total.

Solution: the perimeter is 2 × (24 + 18) = 84 m. A closed boundary with exact 6 m spacing has fourteen distinct posts. Between the gate endpoints, the only internal post is at 12 m. Remove that one to leave thirteen. The gate length contains two intervals but only one internal post.

Now change the task from physical posts to repeated signals. A signal first occurs at 6:40 a.m. and repeats every eighteen minutes through 8:10 a.m., including both endpoints. The elapsed time is ninety minutes, so there are five intervals and six signals. Their times are 6:40, 6:58, 7:16, 7:34, 7:52 and 8:10.

The two questions use related counting but different boundaries. The rectangular route is closed and counts unique posts. The time range is open-ended as a sequence of occurrences and explicitly includes its initial and final events. Do not transfer the final formula without transferring its assumptions.

12. A practical diagnostic routine for parents and tutors

When an answer is one too large or one too small, do not immediately tell the child to add or subtract one. Ask the child to label the counted objects and draw the smallest meaningful example. Three posts with two spaces often reveal the issue more effectively than repeating a long formula.

Then separate the three decisions: calculate how many intervals fit, decide which boundaries are included, and adjust for shared or already-existing objects. A child may perform the first step correctly while missing the second. Repeating division practice would not address that error.

For cutting problems, ask what one allowed cut does. If it separates one piece into two, record the increase of one. If the model starts with a ring, discuss the opening cut separately. If stacking is permitted, stop using the one-piece rule. The operation, not the everyday word “cut”, determines the count.

Use a changed example rather than an identical answer

After teaching an endpoint-inclusive row, try an endpoint-exclusive row with the same length and spacing. After teaching a loop, ask for the number of recorded visits when a walker returns to the starting point. These small changes test whether the learner understands the boundary condition rather than remembers a number.

Later, return to one open-path problem, one closed-route problem, one cutting problem and one repeated-event problem without notes. Ask for a sentence naming the answer quantity before the calculation. “I am counting intervals” and “I am counting distinct posts” should lead to visibly different reasoning.

Check without repeating the same error

Listing actual positions is one check. Rebuilding the length from gaps is another. For cuts, compare final pieces minus initial pieces. For events, compare the last index minus the first index. Repeating the original division may confirm the arithmetic while leaving the endpoint mistake untouched.

A successful check therefore asks a different question about the same arrangement. If nine posts are claimed to be 6 m apart, their eight gaps should span 48 m. If eight posts are claimed on the same open path with both ends included, their seven gaps span only 42 m. The physical or temporal interpretation exposes the error.

Scope and related learning routes

The worked relationships in this article are derived directly from explicitly stated counting models. No real construction specification, transport timetable or safety standard is being prescribed. For official subject documents, use the MOE Primary curriculum and syllabus page alongside school instructions.

For broader enumeration, read Systematic Listing and Logical Counting. For what remains unchanged when the representation changes, use Invariants and Unchanged Quantities.

Continue the worked-learning collection

Balance Scales and Weighing Puzzles develops equality and complete branching strategies. Counting Rectangles, Squares and Triangles extends boundary selection to exhaustive shape counting. Digit Sums and Arithmetic Error Checks examines what different checks can and cannot establish.

Return to the BTT Primary Mathematics Learning Hub. Before the next calculation, name the unit being counted. A gap is not a post, a cut is not a final piece, and another visit does not create another physical place.