The trail with missing pieces

Start with what you know

You do not need the whole answer to find a useful place to begin.

Arjun fits a woodland sign panel while Li Wei explains an idea to pink coach CKRobi beside a boardwalk.
Before the discovery morning, the team checks its signs and plans. The story pictures set the scene; the problem cards give the exact clues.

Arjun is helping prepare a woodland visitor centre. There will be a trail to explore, a number puzzle and places to listen, sketch and make things. Some of the planning board is incomplete. Li Wei joins him, with CKRobi as their coach.

Arjun “There are missing pieces everywhere. I don’t know the whole plan.”

CKRobi “What could you settle without knowing the whole plan?”

Li Wei “Let’s check the clues. One part might already have enough information.”

CKSTEM Thinking CompassStuck? Find your next move.

You don’t need the whole route yet. Start with one useful move, then notice what it tells you.

What do you know for certain? What are you trying to find?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

Two blank signs

The complete loop Centre → Pond → Bridge → Centre is 1,600 metres long. Centre–Pond is 450 m, and Bridge–Centre is 650 m. The Pond–Bridge distance is missing.

A separate trail branches from Pond to Lookout. Its sign is also blank, and no distance clue is given for this branch. These trails have no other segments; the drawing is not to scale.

What distance belongs on the Pond–Bridge sign? Can the Pond–Lookout distance also be determined? Explain what the clues settle and what they leave open.

CPBL450650??

C: Centre · P: Pond · B: Bridge · L: Lookout
Numbers are metres. Loop total: 1,600 m.

Every line is a stated trail. The loop and the branch have separate blanks. The drawing is not to scale.
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A question to help me begin

Which smaller part has enough information, even though both signs are blank?

Show the answer and reasoning: the two signs

Answer: Pond–Bridge is 500 m. Pond–Lookout cannot be determined from these clues.

Look at the loop on its own. Its total and two of its three distances are known. There is only one missing part in that relationship.

\[\begin{aligned}450+650&=1{,}100,\\1{,}600-1{,}100&=500.\end{aligned}\]
CPBL450650500?

C: Centre · P: Pond · B: Bridge · L: Lookout
Numbers are metres. Loop total: 1,600 m.

The loop distance fixes Pond–Bridge. The separate branch stays unknown; its drawn length supplies no distance clue.

Li Wei “The loop gives us enough to fix one sign. The Lookout branch isn’t part of its total.”

CKRobi “How could you test whether that other distance has just one answer?”

Arjun “It could be two hundred metres or three hundred and fifty. Neither changes the loop. We need another clue.”

Check: \(450+500+650=1{,}600\) m. Both suggested branch lengths fit all the given facts, so choosing one would be a guess. The team can restore the 500 m sign now and leave the other one for a proper check.

Start with what you know

When a whole problem feels tangled, look for a useful part that can already be settled. Finish that part and ask what it unlocks.

Look locally. Which clue has enough information?

Ask why it is certain. Would a different answer also fit?

Keep uncertainty visible. What is still open?

This is a way to get moving without pretending to know more than you do. A step may give a number, a necessary connection or a smaller set of possible places.

Once you have a starting fact, Work forward helps you follow its useful consequences. Choosing a foothold and continuing from it are connected decisions.

Start where it is forced

At another display, visitors choose a route through six stations. A place with only one possible connection can give a starting fact even when the middle still offers choices.

The stations are Gate (G), Birch (B), Pool (P), Nest (N), Rest (R) and Exit (E). The only paths join G–B, B–P, B–N, P–N, P–R, N–R and R–E. Each path can be walked in either direction. There are no other paths.

Start at Gate, visit each of the six stations exactly once, and finish at Exit.

Which connection is forced at each end? Is the whole station order forced? Find every allowed order.

CKSTEM Thinking CompassPlan and check your reasoning.

You don’t need the whole route yet. Start with one useful move, then notice what it tells you.

What do you know for certain? What are you trying to find?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

GBPNRE

G: Gate · B: Birch · P: Pool · N: Nest · R: Rest · E: Exit

The seven lines are all the available paths. Positions and drawn lengths are not distance clues.
Show the answer and reasoning: the station routes

Answer: Gate–Birch and Rest–Exit are forced. The whole order is not unique. Exactly two orders work: G–B–P–N–R–E and G–B–N–P–R–E.

Gate has only one neighbour, Birch, so the route must begin G–B. Exit has only one neighbour, Rest, so it must end R–E. These conclusions use the complete path list, not how close the stations look in the drawing.

GBPNRE

G: Gate · B: Birch · P: Pool · N: Nest · R: Rest · E: Exit

Thick teal lines: forced end connections.

Allowed order 1

G → B → P → N → R → E

Allowed order 2

G → B → N → P → R → E

Both orders share the forced ends and use the middle stations in different orders.

The two middle stations still to visit are Pool and Nest. They can occur as P–N or N–P. Both orders have every required connection.

Arjun “Both ends are fixed. I thought that meant the whole route was fixed.”

CKRobi “What do the remaining choices tell you?”

Li Wei “We can put Pool before Nest or Nest before Pool. A certain first step doesn’t make every later step certain.”

Check: each route uses only listed paths and visits all six stations once, with the requested start and finish. After the forced ends, only the two orders of P and N remain, so the list is complete.

Place the largest first

Inside the centre, Arjun finds a wooden number display. Placing cards at random would mean many trials. Can one card tell us something certain before a full arrangement is found?

Use the six cards 2, 3, 4, 6, 7 and 9, once each, to fill the six spaces of a three-row pyramid.

Every space above the bottom row must contain the difference between the two spaces directly below it: subtract the smaller number from the larger. For example, the difference of 8 and 5 would be 3. The bottom spaces are the inputs.

In which row must 9 go, and why? Is its exact position forced? Give two complete valid arrangements with 9 in different positions.

CKSTEM Thinking CompassPlan and check your reasoning.

You don’t need the whole route yet. Start with one useful move, then notice what it tells you.

What do you know for certain? What are you trying to find?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

?
??
???
234679
Six spaces and six cards. Use each card once; an upper space records the difference of its adjacent lower pair.
Show the answer and reasoning: the number display

Answer: 9 must be in the bottom row. Its exact position is not forced. The two arrangements below put it at the left and in the middle.

All the cards are positive. The difference of two positive numbers is smaller than the larger one. To make a difference of 9, we would need a larger number below it—but no card is larger than 9. So 9 cannot occupy either upper row.

9 at the left
3
74
926
9 in the middle
4
73
296
Two complete arrangements satisfy every card and difference rule. They show why a forced row is not a forced position.
\[\begin{aligned}9-2&=7,&9-2&=7,\\6-2&=4,&9-6&=3,\\7-4&=3,&7-3&=4.\end{aligned}\]

Check: each arrangement uses exactly 2, 3, 4, 6, 7 and 9, once each. Each upper number is the difference of the adjacent pair below. The arrangements disagree about the position of 9, so its row is certain while its exact place remains open. Reflecting the first arrangement also puts 9 at the right.

The useful first step was a restriction, not a complete solution. This argument relies on positive inputs and this difference rule; a different rule needs a fresh check.

A meeting by the pond

The team must reach Birch by 10:20 and Pond by 10:40 that morning. It will walk Gate → Birch → Pond, with a 6-minute pause at Birch.

The Gate–Birch walking time is missing. An uninterrupted Gate → Birch → Lookout walk takes 23 minutes, and Birch–Lookout takes 9 minutes. Birch–Pond takes 8 minutes.

These walking times are fixed. The only pause is the six minutes at Birch, and there are no other delays.

What is the missing Gate–Birch walking time, and the latest time the team can leave Gate to meet both arrival limits?

Separate walking clue

Gate → Birch → Lookout

23 minutes, no pause

Birch → Lookout: 9 min

Required outing

Gate → Birch → Pond

Gate → Birch: ? min

Pause at Birch: 6 min

Birch → Pond: 8 min

Pond by 10:40
The two walks share their first leg but end at different destinations.
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Show the answer and reasoning: the meeting

Answer: Gate–Birch takes 14 minutes. Leave Gate by 10:06.

Li Wei “I can’t choose the departure yet, but the separate Lookout walk has only one missing part. Twenty-three minus nine gives fourteen minutes from Gate to Birch.”

Arjun “That settles the same first leg used by both arrival limits. It doesn’t tell us which deadline will control.”

Work backwards

Arjun “From Pond’s deadline I get a latest departure of 10:12: back eight minutes for the last walk, six for the pause and fourteen for the first walk.”

CKRobi “Does that departure meet every time condition?”

Li Wei “No. We would reach Birch at 10:26. Its 10:20 limit gives 10:06 at Gate. We need the earlier of the two departure limits.”

Work backwards · latest permitted times

10:40Pond deadline
− 8 min
10:32Leave Birch for Pond
− 6 min
10:26Birch arrival from Pond limit
− 14 min
10:12Gate limit from Pond
Pond alone allows a 10:12 departure. The separate Birch limit is stricter: 10:20 minus 14 minutes gives 10:06. Meet both by using the earlier limit.

Check forwards: leave at 10:06, reach Birch at 10:20, pause until 10:26, then walk eight minutes to Pond, arriving at 10:34. Both arrival limits hold. Also, \(14+9=23\) checks the separate Lookout clue. The 23-minute walk ends at Lookout; it is not the walking time for the Pond trip.

How the tools fit: Start with what you know finds a part that is already determined; working backwards uses that new fact to meet a deadline. Leaving later than 10:06 misses the Birch limit, even if Pond could still be reached before 10:40. A latest start must satisfy all the deadlines, not only the final one.

Revisit Work backwards to turn a required finish into an earlier deadline.

The discovery morning

The last board needs one complete outing. A group size changes which base fits and how long the activities take; a base changes the walking legs. Choosing those together by trial would create a lot of repeated work.

There are 84 wristbands. Keep 12 spare and divide all the rest equally among 3 groups, with one wristband per person.

One whole group needs a home base that is covered and has a seat for everyone. The choices are Oak: 30 seats, uncovered; Willow: 20 seats, covered; Birch: 28 seats, covered. The group cannot split between bases.

Starting at its selected base at 9:00 that morning, the group must visit Map, Print and Sound once each, in some order, and return to that base by 9:50. Map takes 10 minutes. Print takes 1 minute for each pair of people, with one pair taking its turn at a time. Sound takes 1 minute for each group of three people, with one trio taking its turn at a time. Each person participates once in each activity; there are no partial pairs or trios. Every activity begins when the group arrives.

The walking times below are fixed for this group and the same in both directions. Use the stated time for each leg between consecutive stops. There are no other stops, pauses or delays.

Find the group size, every suitable home base, and every activity order that gets the group back on time. Give the return time for each qualifying order.

Oak

30 seats · uncovered

To Map: 3 min
To Print: 5 min
To Sound: 9 min

Willow

20 seats · covered

To Map: 6 min
To Print: 4 min
To Sound: 7 min

Birch

28 seats · covered

To Map: 4 min
To Print: 8 min
To Sound: 6 min

Map

Activity: 10 min

Print

Activity: 1 min per pair

Sound

Activity: 1 min per trio

Walk between activities

Map–Print: 5 min

Map–Sound: 9 min

Print–Sound: 4 min

Outing window

Leave base: 9:00

Back by: 9:50

All walking times work in both directions. Activity durations are separate from walking durations. Base capacities include every person in the group.
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Show the answer and reasoning: the full plan

Answer: 24 people per group; Birch is the only suitable base. The qualifying orders are Map → Print → Sound and Sound → Print → Map. Both return to Birch at 9:49.

Arjun “I was about to try every base with every activity order. But the wristbands already settle the group size: seventy-two shared three ways gives twenty-four.”

Li Wei “That number chooses the seat test and fixes the activity times too. Twenty-four people make twelve pairs at Print and eight trios at Sound.”

CKRobi “Which choices can you settle before repeating any route calculations?”

Use logical reasoning

Combine the two restrictions on the same choice. Enough seats permits Oak or Birch. Cover permits Willow or Birch. The base must satisfy both, so only Birch remains.

Oak

Seats: enough

Cover: no

Reject
Willow

Seats: not enough

Cover: yes

Reject
Birch

Seats: enough

Cover: yes

Fits both
The intersection keeps only a choice allowed by both conditions. Having the most seats does not supply cover.

Arjun “Oak has the most seats. That seemed like the best choice.”

CKRobi “What must a choice satisfy before it belongs in your plan?”

Li Wei “Both conditions. Oak lacks cover and Willow lacks seats. Birch passes both. We still have to check the visit order.”

Make an organized list

The chosen group has 12 pairs and 8 trios, so the three activities take \(10+12+8=30\) minutes in any order. There are 50 minutes available, so walking must take no more than \(50-30=20\) minutes. List by the first activity, then try both orders of the remaining two.

Map → Print → Sound

Walk: 4 + 5 + 4 + 6 = 19 min

Activities: 30 min

Back at 9:49

✓ On time
Map → Sound → Print

Walk: 4 + 9 + 4 + 8 = 25 min

Activities: 30 min

Back at 9:55

× Too late
Print → Map → Sound

Walk: 8 + 5 + 9 + 6 = 28 min

Activities: 30 min

Back at 9:58

× Too late
Print → Sound → Map

Walk: 8 + 4 + 9 + 4 = 25 min

Activities: 30 min

Back at 9:55

× Too late
Sound → Map → Print

Walk: 6 + 9 + 5 + 8 = 28 min

Activities: 30 min

Back at 9:58

× Too late
Sound → Print → Map

Walk: 6 + 4 + 5 + 4 = 19 min

Activities: 30 min

Back at 9:49

✓ On time
Each first activity is held fixed while the other two are ordered both ways. All six orders appear once; only two finish by 9:50.

For Map → Print → Sound, walking takes \(4+5+4+6=19\) minutes. With the 30 minutes of activities, that is 49 minutes: back at 9:49. Sound → Print → Map also takes 19 walking minutes. The other orders need 25 or 28 walking minutes and return at 9:55 or 9:58.

Check: \(3\times24+12=84\) accounts for every wristband. Birch is covered and seats all 24 people. Each accepted plan begins and ends there, visits each activity once and finishes before 9:50. All six possible activity orders were checked, so no other one was missed.

How the tools fit: Start with what you know settles the group count first; combine the base conditions next; then cover every remaining activity order. A starting fact does not automatically solve the whole problem. It makes the next decision possible.

Revisit Make an organized list when you need every remaining order, rather than one successful plan.

A sign is a promise

Arjun fixes the distance that the clues support. The other blank stays on the team's checking list. A visitor deserves reliable directions, not a number chosen because it looks reasonable.

Some visitors may enjoy the sound station most; others may prefer sketching or making a print. A good shared plan leaves room for different ways to enjoy a place.

Your turn

Try these before opening the answers. Find a useful foothold—or explain why the clues do not settle one.

Tickets and a postcard

At a museum, 2 adult tickets and 3 child tickets cost $34 altogether. Every adult ticket costs $8, and all child tickets have the same price. A postcard has a separate unknown price and is not included in the $34.

What is the child ticket price? Can you also find the postcard price from these facts?

Ticket total

2 adults at $8 each

3 children at $ ? each

$34 altogether
Separate postcard

Price unknown

Not in the ticket total
The child tickets share one price. The postcard is a separate unknown.
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Show the answer and reasoning: the museum

Answer: A child ticket costs $6. The postcard price cannot be determined.

The adult tickets cost \(2\times8=16\) dollars. The three child tickets cost \(34-16=18\), so each costs \(18\div3=6\).

Check: \(2\times8+3\times6=34\). A $1 postcard or a $2 postcard would both fit these ticket facts. Its separate unknown price does not prevent us solving the ticket part.

Five cards in a line

Place cards 1, 2, 3, 4 and 5 once each in a row. Every card in an interior position—one with neighbours on both sides—must be greater than at least one of its two neighbours. End cards have no condition.

What can you be certain about where 1 goes? Is one exact place forced? Give two valid rows that support your answer.

CKSTEM Thinking CompassPlan and check your reasoning.

You don’t need the whole route yet. Start with one useful move, then notice what it tells you.

What do you know for certain? What are you trying to find?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

12345

Interior card: greater than at least one neighbour.
End cards: no condition.

The cards are shown as a collection, not a proposed row.
Show the answer and reasoning: the smallest card

Answer: 1 must be at an end, but which end is not forced. Valid rows are 1–2–3–4–5 and 5–4–3–2–1.

In an interior position, 1 would have to be greater than a neighbour. All other cards are larger, so that cannot happen. Either end remains possible.

1 — 2 — 3 — 4 — 5

5 — 4 — 3 — 2 — 1

Both rows satisfy the interior condition and put the smallest card at different ends.

Check: every interior card in each displayed row is greater than a neighbour, and all five cards occur once. The useful extreme is the smallest card this time; the same habit is to look for a value with a special restriction.

Two sticker piles

Two boxes together contain 14 stickers. Each box contains a positive whole number of stickers. Neither individual count is known, and there are no other clues.

Do these facts force one particular count for each box? Give two different allowed splits to support your answer.

CKSTEM Thinking CompassPlan and check your reasoning.

You don’t need the whole route yet. Start with one useful move, then notice what it tells you.

What do you know for certain? What are you trying to find?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

Show the answer and reasoning: the two boxes

Answer: No. The boxes could contain 6 and 8 stickers, or 5 and 9.

Check: each split has two positive whole counts totalling 14. Even without counting a swap as different, these are different splits. The total gives a relationship, but it has two unknown parts. We need more information to settle an individual count.

What useful step can you take first?

A problem can contain many blanks while one small part is ready to solve. Look for a nearly complete relationship, a place with only one choice or a value whose properties restrict its position.

Say exactly what your first step establishes. A forced connection need not force a whole route. A forced row need not force one position. If several answers still fit, keep them open.

Our next focus is Use logical reasoning: bring compatible clues together and rule out only what contradicts the stated conditions.

CKSTEM Thinking Framework

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