# Twenty solved contexts · staff

These are complete staff models for discussion after a learner has tried a related idea. They are not fresh check sources or reserved rechecks. Every situation and price is fictional. Each calculation models only the stated conditions. Paper diagrams replace any need for real purchases, equipment, travel, food preparation or household measurements.

## W01 · A puzzle border

**Question.** A border repeats 5 dots and 3 triangles. What fraction of its marks are triangles? How many of each are in four groups?

**Solution.** One whole group has 5 + 3 = 8 marks, so triangles are 3/8 of the whole. Four groups multiply both counts by four: 20 dots and 12 triangles, 32 marks altogether. Check: 12/32 = 3/8. The ratio dots:triangles is 5:3; it does not make the triangle fraction 3/5.

## W02 · Three colours of counters

**Question.** A paper counter set has pale:dark:plain = 2:3:4 and total 54. Find every count and the fraction that is dark.

**Solution.** There are 2 + 3 + 4 = 9 equal parts. One part is 54/9 = 6 counters. Counts are 12, 18 and 24. Their total is 54. Dark fraction is 18/54 = 1/3. Whole counters work because all three counts are integers.

## W03 · A ribbon on a diagram

**Question.** A drawn ribbon has light:dark length = 3:7 and total length 2½ m. Find both lengths.

**Solution.** Ten parts share 5/2 m, so each part is 1/4 m. Light length is 3/4 m and dark length is 7/4 m = 1¾ m. Check: 3/4 + 7/4 = 10/4 = 5/2 m. Fractional lengths are allowed; this is not a count of complete objects.

## W04 · An impossible exact set

**Question.** Can 35 whole cards have circles:squares = 4:5 exactly?

**Solution.** Nine parts would each contain 35/9 cards. The circle count would be 140/9 = 15 5/9 and the square count 175/9 = 19 4/9. Neither is a whole count. Therefore the exact set is impossible with whole cards. Nearby valid totals include 27 and 36; choosing 36 gives 16 circles and 20 squares. Rounding individual counts would change the ratio.

## W05 · A garden-path walking model

**Question.** A fictional garden-path plan models a walker covering 28 m in 4 minutes at a constant rate without stops. What is the rate and predicted distance in 6½ minutes? Work on paper; no walk or timing trial is needed.

**Solution.** 28 m / 4 min = 7 m/min. For 13/2 min, distance is 7 × 13/2 = 91/2 m = 45½ m. The minute units cancel. This is a model prediction, not an observed person's pace. A stop would change the productive duration.

## W06 · Sorting bookmarks

**Question.** Two fictional library-sorting models process 63 bookmarks in 9 minutes and 48 bookmarks in 8 minutes. Compare their constant rates and predict how many each would sort in 3 minutes. No person is being timed or compared.

**Solution.** First: 63/9 = 7 bookmarks/min; second: 48/8 = 6 bookmarks/min. The first has a greater constant rate. Three-minute predictions are 21 and 18 bookmarks. A pause, varying item difficulty or set-up time would need a different model.

## W07 · A toy-train diagram and its inverse rate

**Question.** A fictional toy-train diagram models 15 m in 6 seconds at constant speed. State metres per second and seconds per metre. This is a paper quantity model, not an instruction to operate a train or measure a speed.

**Solution.** 15/6 = 5/2 m/s. The inverse is 6/15 = 2/5 s/m. One says how far per second; the other says how long per metre. Their units and meanings differ. Neither establishes how a real toy runs around a bend or stops.

## W08 · Notebook labels: a smaller unit price can cost more

**Question.** Fictional notebook-label offer A has whole packs of 6 labels for $2.10. Offer B has whole packs of 10 for $3.20. At least 13 usable labels are needed. Which sufficient order costs less, with no fees?

**Solution.** A's unit price is $0.35/label; B's is $0.32/label. A needs ceiling(13/6) = 3 packs: 18 labels, $6.30, five spare. B needs ceiling(13/10) = 2 packs: 20 labels, $6.40, seven spare. A costs $0.10 less for this bounded order. A unit-price comparison alone misses the pack condition.

## W09 · Poster copies with an order fee

**Question.** Two fictional shops quote copies of a community poster. A charges $1.50/page, no fee. B charges $1/page plus $4 per order. Copies have the same stated quality and every quoted fee is included. Find the equal-cost whole-page quantity and compare orders of 6 and 10 pages.

**Solution.** Equality is 1.5 n = n + 4, so 0.5 n = 4 and n = 8 pages. At six pages A costs $9 and B $10. At ten pages A costs $15 and B $14. The rate units are $/page, but B's fee is $/order. The fee is added once, not once per page.

## W10 · A rate with a cap

**Question.** A fictional stamp model produces 9 marks/min but can run for at most 4 minutes in one session. Another produces 7 marks/min for at most 6 minutes. Can either complete 40 marks in one session at its constant rate?

**Solution.** First maximum is 9 × 4 = 36 marks, so it cannot complete 40 under this limit. Second maximum is 7 × 6 = 42 marks; time for 40 is 40/7 min = 5 5/7 min, within six minutes. A faster rate does not necessarily meet the constrained need. Counts at the end are whole; the modelled duration may be fractional.

## W11 · A fictional temperature log

**Question.** A fictional weather log begins at −8 °C, rises by 13 °C and then falls by 6 °C. Find each stage and the final temperature. Use the supplied log rather than measuring or exposing anyone to a temperature.

**Solution.** −8 + 13 = 5 °C. Then 5 − 6 = −1 °C. A number-line model moves thirteen units right, then six left. The combined expression is −8 + 13 − 6 = −1. The changes and absolute readings are labelled in degrees Celsius here; they are supplied fictional values.

## W12 · A score penalty and its reversal

**Question.** A fictional game applies four penalties of −3 points. What is the total change? Removing those same four penalties reverses what change?

**Solution.** Four penalties give 4 × (−3) = −12 points. Reversing the complete change adds +12 points: −(−12) = 12. Equivalently, (−4) × (−3) = 12 describes reversing four negative changes. Verify with 12 + (−12) = 0. This interpretation supports the sign pattern; it does not imply that all negative numbers are penalties.

## W13 · A signed height change on a diagram

**Question.** A fictional height diagram models a decrease of 18 m across 6 equal intervals. State the signed change per interval. How many changes of −3 m make a total of −18 m? This is a diagram, not a route to travel.

**Solution.** −18 m / 6 intervals = −3 m/interval. For the second question, (−18 m)/(−3 m) = 6 changes. Check: 6 × (−3 m) = −18 m. The divisor's sign and its meaning distinguish a signed amount per interval from a count of changes.

## W14 · Two fictional score rules

**Question.** One fictional game's round adjustment is “subtract five from two, then multiply that result by negative three”: −3 × (2 − 5). Another rule says “multiply negative three by two, then subtract five”: −3 × 2 − 5. Compare the score adjustments in points.

**Solution.** First calculate inside the brackets: 2 − 5 = −3; then −3 × (−3) = 9. Without brackets, multiply first: −3 × 2 − 5 = −6 − 5 = −11. The expressions ask different questions. They are not two notations for the same operation sequence.

## W15 · Ribbon in a parcel diagram

**Question.** Two fictional parcel diagrams allocate 2/3 m of ribbon to one parcel and 3/8 m to another. What total length is allocated? No actual wrapping or purchase is needed.

**Solution.** A common denominator is 24. Convert 2/3 = 16/24 and 3/8 = 9/24. Total is 25/24 m = 1 1/24 m. Check against estimation: both parts are positive, and the sum is a little above one metre. Do not add denominators: 5/11 would be smaller than 2/3 alone.

## W16 · A repair-tape ledger

**Question.** A fictional repair-tape ledger begins with 2¼ m of usable tape and allocates 5/6 m to a model repair. What usable length remains? Solve the ledger on paper.

**Solution.** 9/4 − 5/6 = 27/12 − 10/12 = 17/12 m = 1 5/12 m. Adding the allocation back checks the result: 17/12 + 10/12 = 27/12 = 9/4 m. A short decimal display for 17/12 would be an approximation; the fraction is exact.

## W17 · A rug region on a room plan

**Question.** A rectangular room plan marks a rug covering 3/5 of the room's width and 5/8 of its length. What fraction of the whole room area does this rectangular rug cover? No room dimensions or purchase are needed.

**Solution.** (3/5) × (5/8) = 15/40 = 3/8. Cancelling the common factor five before multiplying also gives 3/8. Both dimensions are fractions of their respective whole dimensions; the result is a dimensionless fraction of the whole area, not a length in metres.

## W18 · Complete cord sections

**Question.** A fictional craft-cord ledger has one usable 3¾ m length; each full section needs 2/5 m. Cutting loss is zero in this model. Find the exact quotient, full sections and leftover length. Use a drawing rather than cutting cord.

**Solution.** (15/4) ÷ (2/5) = (15/4) × (5/2) = 75/8 = 9⅜ section-lengths. Nine complete sections use 9 × 2/5 = 18/5 m. Leftover is 15/4 − 18/5 = 75/20 − 72/20 = 3/20 m. A tenth section needs four-tenths of a metre, more than the remainder. The exact quotient must be interpreted before choosing a whole count.

## W19 · Two watering models

**Question.** Two fictional garden-watering models deliver 4/7 L/min for 3½ minutes and 5/8 L/min for 3⅕ minutes. Compare the modelled volumes with constant rates and no start-up delay. No water or equipment is used.

**Solution.** First: (4/7) × (7/2) = 2 L. Second: (5/8) × (16/5) = 2 L. They deliver the same modelled volume over different durations. Their rates are different. Cancelling factors keeps the calculation exact and efficient. These rates are not actual product specifications or watering advice.

## W20 · Bunting that fits a fictional budget

**Question.** A fictional bunting diagram needs five sections of 3/4 m and three of 2/3 m. Stock is sold in whole 1 m units for $1.40/unit. Every part of the stock is usable and joins are allowed. Budget is $9. Can the order be supplied?

**Solution.** Needed length is 5 × 3/4 + 3 × 2/3 = 15/4 + 2 = 23/4 m = 5¾ m. Buy six whole metre units: cost 6 × $1.40 = $8.40. Leftover is 6 − 23/4 = 1/4 m; budget spare is $0.60. This model fits the stated budget. If joins were forbidden, a separate cutting feasibility check would be needed; total usable length alone would not establish that every section can be cut.