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A library of useful ideas

Use a related problem

A new problem may have a familiar shape—even when its story looks different.

Before story day, the team finds something useful among its old ideas.

Mina and Arjun are getting the community library ready for story day. Visitors can share a favourite story or teach a welcome in a language they know. There are reading tables to connect, a display to finish and book gifts to prepare. CKRobo is their coach.

Arjun “There are so many different jobs. Do we have to invent a new way for every one?”

CKRobo “What have you solved before that might help?”

Mina “Let’s look through our old models. Then check whether their rules fit.”

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.

A familiar problem in a different story

There are five different reading tables, A, B, C, D and E. Every pair of different tables will exchange a story card. Prepare one card for each pair.

A–B and B–A are the same exchange. A table does not exchange with itself.

An earlier four-table plan, A to D, already had 6 different pair cards. That complete set is available to reuse.

How many cards are needed? Explain how you know that every pair is included exactly once.

ATableBTableCTableDTableETable

Five different tables. Each exchange uses two different tables, with no direction.

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A question to help me begin

Have you solved another problem in which two different things were chosen together? What did the order mean there?

Show the answer and reasoning: the exchange cards

Answer: 10 cards are needed, one for each different pair.

Mina “The old six cards already cover A, B, C and D. Adding E does not change any of those exchanges.”

Arjun “I need only the four new exchanges EA, EB, EC and ED. Six plus four gives ten.”

CKRobo “Why does that reuse leave nothing out or count anything twice?”

The old set and the new exchanges cannot overlap: only the new ones contain E. Every pair either contains E or stays wholly among A to D. Reuse the known result and add only the new work.

Earlier seed problem	New table problem
One variety	One table
Two different varieties	Two different tables
Reversing makes the same pair	Reversing makes the same exchange
Four choices to pair	Five choices to pair

Match what the parts and rules mean. The number of choices has changed.

The complete list below is a check; it is not necessary to rebuild the old six from scratch.

A–BA–CA–DA–E

B–CB–DB–E

C–DC–E

D–E

Every row pairs its letter only with later letters. Row counts are 4, 3, 2 and 1; E has no later letter.

\[6+4=10.\]

Check: every pair has one earlier letter and one later letter. It appears in the earlier letter’s row and nowhere else. All five tables are covered. Listing the pairs directly is also a good solution; remembering the earlier problem gave us a plan to start.

Remember a problem you already know how to solve. Match its important parts to the new problem. Ask which result, method or helpful extra piece can carry over—and which conditions have changed.

Find a connection. What does each part stand for?

Check the fit. Are the rules, order and question still the same?

Adapt and test. Use what helps, then check the new answer against the new problem.

A similar story is only a clue. It is not proof that the same method works. A different-looking story can have the same useful structure.

If no useful connection comes to mind, try a small action from the givens: name one pair, sketch a boundary or follow one change. See what it reveals. You do not need the whole route before you begin.

The same tables, a changed rule

Each of the same five tables, A to E, now sends one message to every other table. Each message needs its own address label.

A→B and B→A are different messages. No table sends a message to itself.

How many address labels are needed? Which part of the exchange-card method can still help, and what must change?

A → Bsent by AB → Asent by B

Two different messages for these two tables. The sender and receiver have different roles.

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Show the answer and reasoning: the address labels

Answer: 20 labels. Keep the ten different table pairs, but prepare two directed messages for each pair.

For A and B there is a message from A to B and another from B to A. Every pair now contributes two, not one.

\[2\times10=20.\]

Row = sender · column = receiver
From ↓ / to →	A	B	C	D	E
A	—	A→B	A→C	A→D	A→E
B	B→A	—	B→C	B→D	B→E
C	C→A	C→B	—	C→D	C→E
D	D→A	D→B	D→C	—	D→E
E	E→A	E→B	E→C	E→D	—

Every non-diagonal cell is one message. A→B and B→A occupy different cells; the dash means no self-message.

Arjun “It looks like the old job, so I nearly kept ten.”

CKRobo “Which rule decides whether reversing the letters changes the choice?”

Mina “Who sends it matters now. The pair helps us organise the work, but we need both directions.”

Check: each of five senders needs four labels: \(5\times4=20\). The diagonal has none because sending to yourself is forbidden. The changed rule matters more than the familiar tables.

A window in the story frame

The flat front of Arjun’s display frame has an outer rectangle 10 cm wide and 8 cm high. A centred rectangular window, 6 cm wide and 4 cm high, is cut out.

He paints only the front face of the frame. The window is empty; the sides and back are not painted.

What area will he paint, in square centimetres? Explain how a rectangle problem he already knows can help.

Only the shaded front face receives paint. The centred white window is a hole.

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Show the answer and reasoning: the painted frame

Answer: Arjun paints 56 cm².

A solid rectangular poster is familiar: width times height gives its area. Imagine filling the window temporarily. That helpful extra piece turns the awkward shape into a known rectangle problem.

Each small square is 1 cm². The completed rectangle has eighty squares; twenty-four lie in the window, leaving fifty-six to paint.

Find the filled rectangle, the window and then the painted part, in square centimetres:

\[\begin{aligned}10\times8&=80,\\6\times4&=24,\\80-24&=56.\end{aligned}\]

Check: the unit squares in the window are removed, not painted. Alternatively, the two full-width top and bottom strips contain \(2\times10\times2=40\) square centimetres. The two side strips between them contain \(2\times2\times4=16\), with no corner counted twice. Together they give 56 cm².

The library job looked different from a solid poster. Adding an imaginary piece made the known result useful; remembering a shape’s name alone would not have solved it.

A circle of reading stations

There are eight different stations, A to H, in a circle, in that order. Neighbours are consecutive stations around the circle; H and A are neighbours too.

Prepare one link card for each pair of stations that are not neighbours. Reversing a pair makes the same link. No station links to itself.

How many link cards are needed? Find a way to cover every allowed pair without counting a neighbour pair.

The labels give the order around the circle. Read neighbours from that order, including the closing H-to-A pair.

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A question to help me begin

Which earlier pair problem has matching rules? What is different about the allowed pairs this time?

Show the answer and reasoning: the circle links

Answer: 20 link cards are needed.

\[10+5+6+7=28.\]

Count the complement

There are fewer unwanted pairs than wanted ones. Remove the eight neighbour pairs from the whole set of 28. Each edge of the circle names a different unwanted pair.

× = neighbour pair to remove. All unmarked pairs are allowed links.

A–B ×A–CA–DA–EA–FA–GA–H ×

B–C ×B–DB–EB–FB–GB–H

C–D ×C–EC–FC–GC–H

D–E ×D–FD–GD–H

E–F ×E–GE–H

F–G ×F–H

G–H ×

There are 28 pairs altogether. Exactly eight have a ×, leaving twenty allowed pairs.

\[28-8=20.\]

Mina “The old pair method gives the whole set. Then I can remove AB, BC, CD, DE, EF, FG, GH and HA.”

CKRobo “How can you check that the subtraction removes exactly the forbidden links?”

Arjun “Each station has two neighbours, but each neighbour pair touches two stations. That gives eight different edges. We included the edge that closes the circle.”

Check: each station has five non-neighbours. That counts every allowed pair at both ends, so \(8\times5\div2=20\). A direct list also works; removing eight pairs is the shorter list here.

The two tools have different jobs: reuse a known method to count the whole set, then remove exactly its unwanted part.

Will the book gifts work?

Two crates, East and West, began with the same unknown whole number of books. No other books were added or removed.

East: reserve 12 books first, then donate one quarter of the books that remain. Exactly 27 books are now available.

West: donate one third of the original crate first, then reserve 6 books. All the fractions here are exact whole-book amounts; nothing is rounded.

How many books did each crate begin with? Can West’s available books make 6 packs of 4 books and 3 individual book gifts? If not, how many books is it short? After setting aside the three individual gifts, what is the greatest number of complete four-book packs West can make, and how many available books then remain?

East · changes in their original order

?Original books

Reserve 12

?After reserving

Donate ¼ of remainder

27Available books

West · changes in their original order

?Same original count

Donate ⅓ of original

?After donation

Reserve 6

?Available books

The two starting amounts are equal. Only East’s final available amount is known.

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Your answer

Enter each uniquely requested quantity. The checker does not assess the structural match or reasoning about operation order.

Break it into steps

Step 1 of 9

How many books form one of East’s three remaining equal quarters?

A question to help me begin

What does the known number count? Can a problem you have solved with equal parts help you recover an earlier amount?

Show the answer and reasoning: the book gifts

Answer: Each crate began with 48 books. West has 26 available, so no: the 27-book gift plan is 1 book short. After reserving the three gifts, West can make 5 full packs, with 3 available books left.

Use a related problem: Mina recalls the muffin stall. One quarter of a batch was sold, eight muffins were added, and 26 were left. Removing the extra eight left 18 in three equal parts; one part was six, so the original batch was 24.

Work backwards

Start from East’s known finish. Its 27 available books are three equal quarters of the amount after reserving twelve. One quarter is \(27\div3=9\), so the four quarters make \(4\times9=36\). Now restore the twelve reserved earlier: \(36+12=48\).

Kept9books

Donated9books

Work backwards · start from East’s known finish

27Available books

÷ 3, then × 4

36Before donation

Restore 12

48Original books

The four equal parts rebuild the thirty-six books after reserving. Restoring the earlier twelve recovers the original forty-eight.

Check East: \(48-12=36\); a quarter of 36 is 9; \(36-9=27\). The quarter was taken from 36, not from the original 48.

Work forward

Recovering East’s beginning does not yet answer the West question. The crates began equally, so West also began with 48. Use that newly known amount to work out what happens next, in West’s own order.

Work forward · follow West’s different journey

48Original books

Donate 16

32After donation

Reserve 6

26Available books

Each box counts books still in that journey. The donated sixteen are removed, not the number remaining after donation.

\[\begin{aligned}48\div3&=16\text{ donated},\\48-16&=32\text{ remaining},\\32-6&=26\text{ available}.\end{aligned}\]

Arjun “Copying East’s finish would give twenty-seven, but West donated from the original amount. Following West’s own stages gives twenty-six.”

Mina “Then the three individual gifts leave twenty-three books for packs. Five packs use twenty, leaving three; six packs would need twenty-four. So five is the most.”

CKRobo “Does that allocation account for every available book?”

The original plan needs \(6\times4+3=27\) books, so it is one short. The revised allocation checks as \(5\times4+3+3=26\): packs, individual gifts and unused available books.

Check West: its 16 donated, 6 reserved and 26 available books total 48. The requested six packs and three gifts need 27. The answer concerns that exact plan.

Use a related problem supplies the useful equal-parts idea. Work backwards recovers a shared beginning from East. Work forward carries it through a different journey to decide what West can do. Each tool answers a different obstacle.

A place for more than one story

The displays are ready. Mina leaves space for visitors to add their own welcome if they wish; Arjun makes sure a quiet reading corner is ready too. A shared place can make room for different voices.

The team’s old ideas also have different jobs. A useful connection does not make two problems identical. It tells you where to begin looking.

Your turn

Think through each new problem before opening its answer. Choose what you can reuse and what needs checking. No writing is needed.

Two ribbons, with a new permission

A bookmark tie uses two ribbons. There are six colours, A to F, with enough ribbon of every colour. The two ribbons may have the same colour.

Only the chosen colours matter: A–B and B–A are the same style. Ribbons of the same colour are interchangeable.

How many different two-ribbon styles are possible? Explain how you have included every style once.

AColourBColourCColourDColourEColourFColour

Choose two ribbons. A colour may be chosen twice; changing their order does not create a new style.

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Show the answer and reasoning: the ribbon styles

Answer: 21 styles.

The old different-pair method gives \(6\times5\div2=15\) two-colour styles. The changed permission adds AA, BB, CC, DD, EE and FF: six same-colour styles.

A–AA–BA–CA–DA–EA–F

B–BB–CB–DB–EB–F

C–CC–DC–EC–F

D–DD–ED–F

E–EE–F

F–F

Every row begins with its same-colour pair, then includes the later colours. Same-colour pairs are not silently discarded.

\[15+6=21.\]

Check: every style either has different colours or the same colour; those cases cannot overlap. The rows show \(6+5+4+3+2+1=21\), with each unordered style appearing once. Reusing only the old answer pattern would miss the six repeats.

A frame, a different question

A new frame has an outer rectangle 10 cm wide and 8 cm high, with a centred inner window 8 cm wide and 6 cm high.

A thin trim follows every outer edge and every inner edge once. There is no overlap or cutting waste.

What total length of trim, in centimetres, is needed? Which familiar rectangle method helps?

Both the outer boundary and the inner window boundary need trim. The shaded face is not the requested quantity.

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Show the answer and reasoning: the trim

Answer: 64 cm of trim.

Length follows the boundaries, so use the known method of adding a rectangle’s four sides. Outer boundary: \(2(10+8)=36\) cm. Inner boundary: \(2(8+6)=28\) cm. Both need trim.

Outer boundary

10 + 8 + 10 + 8

36 cm

Inner boundary

8 + 6 + 8 + 6

28 cm

Add the two separate boundary lengths once each: 36 cm + 28 cm = 64 cm.

\[36+28=64\text{ cm}.\]

Check: two outer widths, two outer heights, two inner widths and two inner heights are each included once. Square centimetres would measure paint, not trim. A familiar frame picture does not decide which quantity the question asks for.

Equal starts, different journeys

Two game kits, A and B, each begin with 40 counters.

A: reserve 8 counters first, then give away one quarter of those that remain. B: give away one quarter of the original 40 first, then reserve 8.

All fractions are exact whole-counter amounts. No other counters are added or removed.

How many counters remain available in each kit? Does the equal beginning guarantee an equal finish here?

Kit A · original order

40Original counters

Reserve 8

?After reserving

Give ¼ of remainder

?Available counters

Kit B · original order

40Original counters

Give ¼ of original

?After giving

Reserve 8

?Available counters

The starts match; the order and the amount from which the quarter is taken differ.

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Show the answer and reasoning: the two kits

Answer: A has 24 counters available; B has 22. No, the equal beginning does not guarantee the same finish here.

The beginning is known, so Work forward. Keep the useful habit of tracking what each amount counts, but follow each kit’s actual order.

Work forward · kit A

40Original counters

Reserve 8

32After reserving

Give 8

24Available counters

Work forward · kit B

40Original counters

Give 10

30After giving

Reserve 8

22Available counters

Kit A gives away a quarter of 32. Kit B gives away a quarter of 40. Both conservation checks return to forty.

For A, \(40-8=32\), then a quarter of 32 is 8, leaving \(32-8=24\). For B, a quarter of 40 is 10, leaving 30; reserving 8 then leaves \(30-8=22\).

Check: A’s 8 reserved, 8 given away and 24 available total 40. B’s 10 given away, 8 reserved and 22 available also total 40. The fraction was taken from different amounts.

Which parts of the old method still apply?

Keep a useful question ready: What have I solved before, and why could it help here? Match the parts, check the rules and adapt the method. A familiar answer is not a substitute for a new check.

Sometimes the useful piece is a way of organising pairs, an extra piece that completes a shape, or a way of finding one equal part. Keep the reasoning that fits; change what the new problem requires.

Our next focus is Work forward: use what you know to find one useful new fact, then see what it unlocks.