# Ten lesson starts and thirty routes Choose one route after the shared start. Each lesson is planned for about 45 minutes: 5–8 minutes to start, 20–25 minutes for the chosen route, and 8–12 minutes to compare a method and name a next step. The check lessons use their own timing. Paper and pencil are sufficient. Calculators are optional checking tools, except where a learner's agreed access plan requires one from the start. ## Day 21 · A mix and its whole **Start.** Draw two circles and three squares. Say, “There are two circles for every three squares. How many shapes are in one whole group?” Invite a learner to mark the whole five-shape group. Write circles:squares = 2:3, circles:all shapes = 2:5 and circles as a fraction of all shapes = 2/5. Ask which statement compares two parts and which compares a part with the whole. The misconception to notice is “2:3 means two-thirds of the whole.” Have the learner point to all five shapes before changing their notation. ### Route 21A · Build the whole Draw a repeating pattern with 3 triangles and 4 dots in each group. Make two groups. Label triangles:dots, triangles:all marks and the fraction of marks that are dots. Then add a third group and explain which relationships stayed the same. Keep the groups visible; do not simplify the total counts before showing where they came from. **Teacher next move:** expect 3:4, 3:7 and 4/7; with three groups, counts are 9 and 12, whole 21. Ask a learner who writes 4/3 to circle the denominator's whole. ### Route 21B · Find and repair a claim A fictional puzzle says: “A box has 4 plain tiles and 6 striped tiles. Plain tiles make up 4/6 of the box.” Draw the ten tiles, decide whether the sentence is correct, and write a corrected sentence. Find a different true comparison involving 4 and 6. Explain why both the ratio and the fraction need a named comparison. **Teacher next move:** plain:striped = 4:6 = 2:3; plain fraction of the whole = 4/10 = 2/5. A ratio of quantities is not wrong simply because it is not a fraction of the whole. ### Route 21C · Design a three-part pattern Design a paper border with 2 circles, 1 square and 3 triangles per group. Draw two groups. State the three-part ratio, the fraction of all marks that are triangles, and the ratio of non-triangles to triangles. Then design another group with six marks that has a different triangle fraction. Name what changed. **Teacher next move:** 2:1:3; triangles 3/6 = 1/2; non-triangles:triangles = 3:3 = 1:1. Accept any second design with six marks and a triangle count other than three, with its fraction explained. **Close:** “A ratio compares ___ with ___. My denominator means ___.” ## Day 22 · Scaling without changing the mix **Start.** Draw a bar in five equal parts: two lightly shaded, three dark. One group has total length 10 cm. Find the length of one part, then each colour: 2 cm, 4 cm and 6 cm. Double the total to 20 cm. Ask what doubles and what stays the same. Explain that actual length can be fractional, while a count of whole cards must stay whole. ### Route 22A · A double number line A paper pattern uses 4 dots for every 7 dashes. Make a double number line showing 0, 4, 8, 12 dots beside 0, 7, 14, 21 dashes. Add the point for 10 dots if partial lengths rather than whole marks were being modelled. Explain why 10 whole dots would not make an exact number of complete pattern groups. **Teacher next move:** the continuous paired amount is 17½ dash-units; the scale factor is 10/4 = 2½. For whole marks in complete groups, use a multiple of four dots. Do not round 17½ to 18 and call the ratio unchanged. ### Route 22B · Work backwards from a total A set of paper counters must have stars:circles = 3:5 and contain 48 counters. Find each count using equal parts. Decide whether 50 counters can have this exact ratio with whole counters. Explain your answer without drawing all fifty. **Teacher next move:** eight parts; 48/8 = 6, so 18 stars and 30 circles. Fifty would require 6¼ per part, yielding 18¾ and 31¼ counters; the exact whole-count condition is impossible. “Impossible under this condition” is a useful answer. ### Route 22C · Compare methods, then choose A length design has red:blue:grey = 1:2:3 and total 15 m on a fictional diagram. One learner finds one part first. Another scales the six-part base group by 2½. Show both methods and choose the one you would use for total 9 m. Explain why 1:2:3 and 2:4:6 describe the same mix but 2:3:4 does not. **Teacher next move:** for 15 m, 2½ m, 5 m, 7½ m; for 9 m, 1½ m, 3 m, 4½ m. Each component must be multiplied by the same factor. All quantities here are continuous lengths, not whole objects. **Close:** name the scale factor and the condition that permits or prevents a fractional answer. ## Day 23 · A rate carries two units **Start.** A fictional drawing tool makes 18 cm of line in 3 minutes at a constant rate. Write 18 cm / 3 min = 6 cm/min. Label both units. Predict 30 cm in 5 minutes and 12 cm in 2 minutes. Ask, “What assumption lets us use the same rate?” The assumption is constant output over the times used, with no start-up delay or stopping. ### Route 23A · Read a table and test constancy A fictional copying tool records: 2 minutes → 14 labels; 4 minutes → 28 labels; 6 minutes → 42 labels. Calculate labels/min and predict the output at 5 minutes. A second table is 2 → 14, 4 → 26, 6 → 36. Explain why one constant-rate model fits the first table but not the second. Avoid claiming either is measured performance. **Teacher next move:** first is 7 labels/min, giving 35 labels; second rates are 7, 6½ and 6 labels/min. A linear-looking increase is not automatically proportional. ### Route 23B · Put units on an unlabeled answer Three fictional calculations have lost their units: 240/4 = 60, 15/3 = 5 and 9/6 = 1½. Choose one context for each from these pairs: centimetres and minutes; dollars and metres; metres and seconds. Write the complete rate, describe what one unit of the denominator means, and reverse one rate with its units. Do not compare the three bare numbers as though they measure the same thing. **Teacher next move:** accept a consistent assignment. If 9 m in 6 s gives 1½ m/s, the inverse is 2/3 s/m. Division order changes the meaning, not just the appearance of the answer. ### Route 23C · Design a fair comparison Paper model A produces 45 marks in 9 minutes. Model B produces 32 marks in 8 minutes. Decide which has the greater constant rate. Find a time at which both predictions are whole numbers and calculate both outputs. Write one limitation of the comparison if A pauses after every ten marks. **Teacher next move:** A 5 marks/min; B 4 marks/min. At 2 minutes, for example, A 10 and B 8. The pause would require a different time model; the simple average given may not describe each interval. **Close:** say the answer with “per” and both named units. ## Day 24 · A useful rate, with conditions **Start.** Fictional paper strips are sold as A: 8 strips for $3.20 and B: 12 strips for $4.20. Find $/strip: $0.40 and $0.35. For exactly 24 strips with whole packs and no fees, A costs $9.60 and B costs $8.40. Now say B adds a $2 handling fee. Recalculate B's total as $10.40. Ask why the smaller unit price no longer decides the cheaper complete order. ### Route 24A · Whole packs and leftovers A fictional label offer sells 5 labels for $1.80; B sells 8 labels for $2.72. You need at least 18 labels. Packs cannot be split. Calculate each unit price, number of packs, cost and leftovers. Choose the cheaper sufficient order. State the assumption that prices include every fee. **Teacher next move:** A $0.36/label, four packs 20, $7.20, two spare; B $0.34/label, three packs 24, $8.16, six spare. A is cheaper for this need even though B's unit price is smaller. ### Route 24B · Find the point where a decision changes An imaginary diagram service charges A $2 per page with no fee, or B $1 per page plus a $6 start fee. Pages are whole positive numbers. Make a table for 4, 6 and 8 pages, find the equal-cost quantity, then write a decision rule for whole positive quantities. Show where the fee enters the calculation. **Teacher next move:** A 8/B 10 at 4; both 12 at 6; A 16/B 14 at 8. A cheaper for 1–5, equal 6, B cheaper ≥ 7. The proportional $/page rate is only part of B's total model. ### Route 24C · Ask for the missing condition One fictional card says “$0.30 per token”. Another says “$2.40 for a pack of 10”. You need 16 usable tokens. List what you must know before a fair decision: whether single tokens can be bought, whether packs split, whether fees exist and whether tokens are interchangeable. Then solve under these stated conditions: the first sells singles with no fee; the second sells whole packs with no fee; every token is usable. Explain how a different condition could change the answer. **Teacher next move:** singles 16 cost $4.80; two packs 20 cost $4.80, four spare. Equal stated cost; a learner can choose based on leftover usefulness. A new fee or maximum quantity would change the model. **Close:** name the useful comparison, then one condition the comparison needs. ## Day 25 · Fresh check E **Start.** Give the learner the complete E source and Parts 1–2 from [assessment/FIRST-CHECKS.md](assessment/FIRST-CHECKS.md). Say, “Show your quantities, units and decision so someone else can follow them. You can use a diagram, table or equations.” Do not work an equivalent example with these numbers first. Keep staff keys and all change/recheck cards out of the learner's first view. Allow 20 minutes for Parts 1–2. Preserve that first response. Reveal only E's change card from [staff/REVEAL-GUIDE.md](staff/REVEAL-GUIDE.md), then allow 10 minutes for Part 3. Keep both responses. Use up to 10 minutes to discuss the learner's own choice after collection. Extra processing time or a split session may be agreed; record the support and elapsed stages rather than describing the administration as identical. ### Route 25A · Diagram-led response Offer blank equal-part bars and a blank quantities table. The learner chooses and labels the representation, then completes every E part. Do not provide the number of parts, counts, unit prices, pack totals or recommendation. Record that blank representation support was available. ### Route 25B · Spoken reasoning with a record The learner explains each E part aloud, pointing to their own marks or counters. A recorder writes the learner's words and arithmetic, reads them back, and lets the learner correct them. Capture Parts 1–2 before the change card. Record prompts and transcription support. Spoken explanation here is actual learner evidence only if it occurs; a written plan is not a spoken performance. ### Route 25C · Compare two of your own methods The learner completes every E part and, for one calculation, supplies a second method of their own. They choose which method is easier to check and explain why. The comparison must be made after their first calculation, not supplied as a worked staff model. Keep the new condition withheld until the first response is captured. **Close:** use the E descriptive criteria and a specific reteach branch. Prior access to the source or key ends the fresh claim; choose an unused E recheck after teaching rather than recycling the same answer as fresh evidence. ## Day 26 · Signed changes and integer operations **Start.** Draw a number line. A fictional score starts at −3, rises by 8 and then falls by 4. Model −3 + 8 − 4 = 1 with two movements. Compare “subtract 4” with “add −4”. Then show three changes of −2: 3 × (−2) = −6. Explain multiplication signs with a pattern: 2 × (−3) = −6, 1 × (−3) = −3, 0 × (−3) = 0, so continuing the constant change gives (−1) × (−3) = 3. Division reverses multiplication: −12 ÷ 3 = −4 and −12 ÷ (−3) = 4. These are numerical models, not directions to incur a debt or use equipment. ### Route 26A · A log of changes A fictional level indicator starts at −5, changes by +9, then −7, then +2. Mark each stage on a number line and find the final level. Reorder the signed additions and check whether the final total changes. Explain why the intermediate levels can differ even when the final sum does not. **Teacher next move:** −5 → 4 → −3 → −1. Signed addends −5 + 9 − 7 + 2 total −1; ordering addition changes the path but not its sum. Do not extend this claim to arbitrary reordering of subtraction or division expressions. ### Route 26B · Use a pattern to justify a sign Complete a table for n × (−4) at n = 3, 2, 1, 0, −1, −2. Explain the change from one row to the next. Use multiplication to check −20 ÷ (−4) and 20 ÷ (−4). Then compare −2 × (3 − 7) with −2 × 3 − 7 and explain the role of the brackets. **Teacher next move:** −12, −8, −4, 0, 4, 8; quotients 5 and −5. Bracketed expression 8; unbracketed −13. A recalled sign rule is useful only when it is applied to the intended expression. ### Route 26C · Make a target, then diagnose Use the integers −6, 3 and 2 once each to make 0 using addition, multiplication, division and optional brackets. Show one valid expression. Diagnose this false claim: “−6 ÷ 3 + 2 is −6 ÷ 5, so it is not zero.” State the order of operations and check your expression by substitution. **Teacher next move:** −6 + 3 × 2 = 0 or −6 ÷ 3 + 2 = 0. The false combination changes the expression; −6 ÷ 3 + 2 = −2 + 2 = 0. Accept another valid use of all three integers with operations named. A learner's method must match the written expression. **Close:** explain one sign or bracket choice using a model, inverse operation or pattern. ### Check the operation you have not yet seen After the chosen Day 26 route, note which signs and operations the learner actually explained. Route A mainly shows signed addition; it cannot establish all four operations. Choose a short ordinary follow-up if another operation is uncertain: - **Signed addition/subtraction:** calculate 8 − 12 + 5 and show the changes on a number line. Key: 1. - **Multiplication:** explain why (−4) × (−3) = 12 using a sign pattern or reversal of a negative change. A named rule alone is a starting point for a question about why it works. - **Division:** calculate −28 ÷ 7 and −28 ÷ (−7), then check each with multiplication. Keys: −4 and 4; 7 × (−4) = −28 and (−7) × 4 = −28. - **Order and brackets:** compare 6 − 2 × (−3) with (6 − 2) × (−3). Keys: 12 and −12. Ask what the brackets keep together. Keep the chosen response and the support used. Offer another teaching example where needed. These are ordinary follow-ups with visible staff answers, not fresh checks or a requirement to complete every task. E and F do not establish full integer mastery. ## Day 27 · Fractions need a common whole **Start.** Draw two equally long bars. Mark 1/2 of one and 1/3 of the other. Partition both into sixths. Add 3/6 + 2/6 = 5/6. Ask why 1/2 + 1/3 is not 2/5. Fractions count equal-size parts of a named whole; adding the denominators would change that part size. ### Route 27A · Compare picture and exact notation Use equal whole bars to show 3/4 + 1/6. Find a common partition, calculate exactly, and say whether the sum is below or above one whole. Then show how much is missing from one whole. A ruler drawing can support the reasoning without pretending to measure an exact physical length. **Teacher next move:** 9/12 + 2/12 = 11/12; missing 1/12. Equal whole bars matter; unequal wholes cannot be combined as though the fractions refer to the same amount. ### Route 27B · Reconstruct a missing amount A paper length model uses 5/8 m in one section and an unknown length in a second. Together they use 1⅜ m. Find the second length, write an equation and check by adding. Give the answer as an exact fraction and a finite decimal. Explain the meaning of the metre unit in both forms. **Teacher next move:** 11/8 − 5/8 = 6/8 = 3/4 m = 0.75 m; check 5/8 + 6/8 = 11/8. Convert 1⅜ before subtracting; a mixed number is a sum, not a product of its written parts. ### Route 27C · Decide which calculation is efficient Calculate 7/10 + 3/5 − 1/2 using exact fractions. Show a second method using finite decimals. Choose which representation you find easier to check and explain why both methods have the same exact result here. Give one fraction for which a short rounded decimal would lose exactness. **Teacher next move:** 7/10 + 6/10 − 5/10 = 8/10 = 4/5 = 0.8. Finite decimals 0.7 + 0.6 − 0.5 are exact. For 1/3, 0.33 is an approximation, not an equal value. Full recurring-decimal notation is reserved for the later mapped sequence. **Close:** name the whole, the common part size and the exact answer. ## Day 28 · Groups, portions and rational operations **Start.** A paper diagram has four sections of length 3/4 m. Write 4 × 3/4 m = 3 m. Reverse the question: how many 3/4 m sections fit in 3 m? Write 3 m ÷ (3/4 m) = 4 sections. Contrast this with 3 m ÷ 4 = 3/4 m per section. The divisor and its units determine what the answer counts. Introduce cancellation in (3/5) × (10/9) = 2/3 by cancelling common factors before multiplication. ### Route 28A · Explain the size of a product On a unit square, shade 2/3 of its width and 3/4 of its height. Find the overlap as a fraction of the whole square. Calculate 2/3 × 3/4, simplify and explain why this product is smaller than either positive factor. Then contrast multiplication by 3/2, which enlarges a positive starting amount. **Teacher next move:** overlap 6/12 = 1/2; a factor between 0 and 1 scales down. “Multiplication always makes larger” needs conditions. Zero and negative factors are outside this particular area explanation. ### Route 28B · Count full pieces and keep the remainder A fictional length stock has 2½ m. Each complete section needs 3/8 m. Find the exact quotient, the number of complete sections, the used length and the remaining length. Explain why a quotient of 6⅔ does not mean seven complete sections. **Teacher next move:** (5/2) ÷ (3/8) = 20/3 = 6⅔; six complete sections use 18/8 = 9/4 m, remaining 1/4 m. The count of complete sections is discrete; the length is continuous. Rounding up would require 21/8 m, more than stock. ### Route 28C · Cancel before multiplying Compare two exact methods for (7/12) × (18/21), then calculate (5/6) ÷ (10/9). Explain which common factors can be cancelled and why cancelling across addition would be invalid. Give a small counterexample for cancelling in (2 + 3)/3. **Teacher next move:** product 1/2; quotient 3/4. Cancellation divides common factors of a product numerator and denominator. (2 + 3)/3 = 5/3, whereas 2 + 1 = 3 is not the same expression. **Close:** say whether the answer measures an amount, an amount per group or a count of groups. ## Day 29 · A calculation becomes a model **Start.** A fictional paper display has 8 segments, each 5/8 m long. Total length is 5 m. Stock comes in whole 2 m packs at $3 each. Calculate three packs, 6 m, $9 and 1 m leftover. Write the assumptions: every segment may use joined lengths; all stock is usable; no fees; prices fictional. If joins were forbidden, a total-length calculation alone would need a cutting plan. Ask why the numerical total does not prove every physical arrangement works. ### Route 29A · Formulate a question and check a limit A paper diagram uses six long sections of 3/4 m and four short sections of 1/2 m. Stock is sold in whole 1 m units at $1.60 each. Joins are allowed and every length can be used. Budget is $12. Write a useful question about whether the diagram can be supplied, choose the operations, calculate the exact needed length and whole-unit purchase, and interpret the result with leftovers and budget. Then review what would need checking if joins were forbidden. **Teacher next move:** 6 × 3/4 + 4 × 1/2 = 13/2 m = 6½ m; buy 7 m for $11.20, leftover 1/2 m, budget spare $0.80. Without joins, examine individual pieces and stock cuts rather than claiming the same calculation is sufficient. ### Route 29B · A rate with a rational duration A fictional plotting model draws at 3/5 m/min for 2½ minutes. Find the length. Compare with a second model drawing at 2/3 m/min for 2¼ minutes. Explain why their outputs can match even though their rates differ. Write the constant-rate/no-delay assumption and one change that would invalidate the simple prediction. **Teacher next move:** (3/5) × (5/2) = 3/2 m; (2/3) × (9/4) = 3/2 m. Equal output does not imply equal rate when duration differs. A start-up delay must be subtracted from productive time if that is the stated model. ### Route 29C · Build an efficient expression A fictional stock ledger begins with 4½ m, allocates three sections of 2/3 m, then returns 1/4 m of usable stock. Write one expression for the final usable amount. Calculate exactly and check using a second method. A note says “subtract every written number first.” Explain why that advice loses the meaning of the three equal allocations and the return. **Teacher next move:** 9/2 − 3 × 2/3 + 1/4 = 11/4 m = 2¾ m. The return adds stock; multiplication groups equal allocations. Reject a reordered expression only when its meaning or operation structure changes. **Close:** “My answer fits ___ because ___. My model assumes ___. I would check ___ next.” ## Day 30 · Fresh check F **Start.** Give the complete F source and Parts 1–2 in [assessment/FIRST-CHECKS.md](assessment/FIRST-CHECKS.md). Use the same collection routine as E: 20 minutes, preserve the first response, reveal only F's change card, then 10 minutes for Part 3. Keep the initial conditions visible during the review. Record approved time/support changes. Do not introduce the new fee or budget before the first capture. ### Route 30A · Make a representation that earns its space The learner selects a blank bar, table or equation space and completes every F part. They explain what the representation makes easier to see. Blank supports carry no ratios, totals, calculated lengths, unit prices or recommendation. Record the chosen support. ### Route 30B · Explain a model in a short conference The learner completes all F parts through a recorded mathematical conference or a scribed account. Ask only neutral prompts: “What does that quantity mean?”, “Which units belong here?” and “What changed?” Record each prompt and the learner's response. Do not supply the operation or direct the recommendation. Keep the staged first capture. ### Route 30C · Test your own model The learner completes all F parts and checks one result with a different exact calculation or representation. They identify one assumption they would test before using the model in a physical situation. The second method must test the learner's model rather than copy a staff answer. Reveal the change card after Parts 1–2 have been kept. **Close:** give feedback against F's criteria and select a particular next lesson or an unused F recheck. A learner may correctly change their recommendation, explain why neither option works, or retain a decision if supported by the new quantities. Judgement follows the calculation and conditions.