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Year 5 / Mathematics / Term 1 / Weeks 03 04 / Teacher

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Teacher copy · worked answers, feedback and next movesYear 5 Maths · T1 W3–4 · Teacher · Teacher key

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Teacher copy · prompts and answer keys

Teacher copy: This page may include teaching prompts or answer keys. Answer keys in this public library can be viewed by anyone. Give learners a clean prompt, use checks as formative evidence, and change a case locally when prior access matters.

Keep this file separate from learner cards and fresh checks. Record actual task, child's first answer, method, access route, adult prompts, revision and next move. A mathematical idea shown through tactile cards, AAC, speech, handwriting or typing has the same status when the target is number reasoning. A correct calculator display without a chosen operation/scale check is incomplete evidence of this fortnight's target. A missing answer can mean access or time failed; mark not yet observed, offer a usable route and gather another sample before inferring a gap.

Worked answer bank · 30 core learner cards

Day A B C
11 52×18=52×(20−2)=1,040−104=936; 50×20=1,000 is a high rough anchor. 73×14=730+292=1,022; 70×15=1,050 is nearby, so 100 is implausible. 208×5=1,040; 200×5=1,000, with another 8×5=40.
12 235×7=1,400+210+35=1,645; more than 200×7=1,400. 602×4=2,400+0+8=2,408; zero tens retained. 481×6=2,400+480+6=2,886; near 480×6=2,880.
13 215×14=2,150+860=3,010; ten sets contribute 2,150. 76×32=2,280+152=2,432; 75×32=2,400 is nearby. 142×26=2,840+852=3,692; twenty strips contribute 2,840.
14 396×9=400×9−4×9=3,600−36=3,564. 125×24=2,500+500=3,000; also 125×8×3=1,000×3. 318×19=318×20−318=6,360−318=6,042; near 300×20=6,000, not 60,000.
15, after check 106×13=1,060+318=1,378. 72×28=72×30−72×2=2,160−144=2,016; also 1,440+576. 309×6=1,800+0+54=1,854.
16 12×$34=$408; under $420 by $12. 19×$27=$513; under $530 by $17. 11×$42=$462; under $480 by $18.
17 215×9=1,935; inverse 1,935÷9=215, near 200×9=1,800. 324×7=2,268; inverse 2,268÷7=324, near 320×7=2,240. 507×6=3,042; inverse 3,042÷6=507, near 500×6=3,000.
18 18×39=702; unknown 39, substitute or use 18×(40−1)=720−18. 32×26=832; unknown 32, 832÷26=32. 45×28=1,260; unknown 28, 45×(20+8)=900+360.
19 30×24=720 and 45×16=720; Plan 1 fits at most 32 bundles. 40×21=840 and 30×28=840; Plan 2 fits at most 35 envelopes. 36×26=936 and 24×39=936; Plan 1 fits at least 30 stations.
20, after check 57×19=57×20−57=1,083; near 1,140. 24×35=720+120=840; also 12×70=840 by halving/doubling. 108×11=1,080+108=1,188; inverse 1,188÷11=108, near 1,100.

Feedback that changes the next lesson: If a child loses a zero in ×20, physically put 20 counters in two rows of ten and write number×(2×10). If a result is one-tenth size, first compare it with a lower bound such as 128×20 before repeating the algorithm. If the exact product is right but an estimate is missing, model rounding before the calculation with a fresh pair. If a constraint is ignored, make a two-column table: arithmetic total and stated limit. A different correct strategy should be accepted and discussed, not converted into the teacher's method by default.

Optional extension answers · 20 routes

Day A B
11 39×21=819; 39×20=780 < 819 < 40×21=840. 62×15=930, near 60×15=900; 93 misses a factor of ten.
12 504×3=1,512; 54×3=162; a zero in 504 separates hundreds from ones. 706×4=2,800+0+24=2,824.
13 84×20=1,680, not 168; 84×7=588; total 2,268. 46×32=1,380+92=1,472; (40+6)×32=1,280+192=1,472.
14 214×19=4,280−214=4,066; near 4,000. 25×48=1,200; 25×4×12=100×12, or 1,250−50.
15 Wrong partial 132×4 is 528, not 428; total 1,848, not 1,748. Example 119×20=2,380, near 2,400; other justified two-digit multipliers may fit.
16 14×$29=$406; $14 remains from $420. $456÷12=$38 each; the source gives no real vendor price.
17 306×8=2,448; 2,448÷8=306, 2,448÷306=8; near 2,400. 390×7=2,730; thus 2,730÷7=390 is correct but 390×7=2,370 is false. Correct the stated product to 2,730.
18 ?=50, since 50×18=900; 5×18 is only 90. ?=42, since 24×42=1,008 (960+48).
19 Both totals 600. “At most 15 bundles” favours 12×50; “at least 18 groups” favours 20×30. These are invented possible criteria, not given. Both totals 420. No pack prices were supplied, so cheaper is unknown.
20 Open design: independently check the child's chosen factors, both partial products and the deliberate error against exact arithmetic. Open design: verify both products really match and the declared limit selects the child's named plan; a deliberately unresolved tie is valid only if stated.

Day 15 · Fresh Check A key

Estimate 200×15≈3,000 is a sensible benchmark; so is 210×14=2,940. Exact 213×14 = 213×10 + 213×4 = 2,130+852 = 2,982. The proposed 29,820 is ten times too large; even 213×20=4,260 is well below it. Accept another valid efficient method with its value/place shown. Do not require one particular estimate, but ask whether its rounding direction makes sense. A pupil who says 2,982 without method may know it; ask for a partial product or reverse check on a new item before claiming strategy security.

Construct 2 · independent and explained 1 · partial or prompted 0 · not yet observed after usable access
Benchmark Nearby benchmark with reason and scale Nearby value but weak reason, or correct after prompt No defensible size sense yet
Partial products/exact 2,130+852=2,982 or valid equivalent, places intact Method mostly right with arithmetic/place slip, repairs after prompt Group value or operation not established
Plausibility Rejects 29,820 using benchmark, bound or factor of ten Rejects but gives vague reason Accepts tenfold total without a check

Next move: If ten pages become 213 instead of 2,130, teach ×10 with place cards then retest using a different 3-digit×two-digit example. If a learner's method is good but adding 2,130+852 slips, give a short addition check separately; do not erase multiplication understanding. If access/time prevented response, repeat with a new source after adapting the access route; mark this item not yet observed.

Day 20 · Fresh Check B key

One useful estimate is 180×24≈4,320, intentionally above exact because 180 exceeds 178; another is 180×25=4,500, a looser upper anchor. Exact 178×24 = 178×20 + 178×4 = 3,560+712 = 4,272. Inverse 4,272÷24=178 or two-way check 24×(180−2)=4,320−48=4,272. The plan fits the 4,500-label count limit with 228 spaces left. Price, actual shelf safety and existence of labels are unknown; “cheapest” has no evidence.

Construct 2 · independent and explained 1 · partial or prompted 0 · not yet observed after usable access
Estimate Useful rounded product and correct direction/scale Reasonable rounded value with incomplete explanation No reliable benchmark yet
Product Exact 4,272 with inspectable tens/ones or equivalent Valid structure, one arithmetic slip repaired with prompt Group/place structure not evident
Independent check Inverse or different route gives 178/4,272 and is explained Copies a fact or checks after cue No check yet
Interpretation 4,272≤4,500, 228 spare; refuses unsupported price claim Correct fit with weak evidence limit Constraint or price boundary misunderstood

Next move: If 178×20 becomes 356, compare it to 178×2×10 and use the partial-products mat. If the result is exact but the recommendation ignores the 4,500 limit, highlight only the constraint sentence and ask the child to compare. If a child asserts “cheap,” give two invented price cards with different values and show why price data is required; never ask about family purchases. If inverse division is inaccessible, let the learner verify with 24×(180−2) and revisit division in a later separate lesson.

Decision after these two weeks

These checks sample one fortnight of multiplication and estimation. They do not prove all of AC9M5N06/08/09 or the Year 5 achievement standard. Collate classwork, alternate representations and later transfer with the school's local assessment rules. A 0 is an invitation to find out why evidence is missing, not a diagnosis. Keep named work outside the public repository.

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