Paper #12 maths section (no calculator): 9/17 in about 19 min against a 16-min guide, no blanks. Right: solving a linear equation with brackets 3/3 with every M-line written, area of a rectangle with a corner cut 2/2, substitution with (−4)² 2/2, error interval 2/2. Zero: 5 + 2 × 4² circled 21 (4² evaluated as 8 — squaring as doubling, same as 3² = 6 on 14 Oct); estimate rounded to 6, 29 and 1 rather than 1 s.f. (0/2); without-replacement probability with denominators 7 and 7 and one order only (0/3); inequality 15 − 4x > 3 with 3 − 15 taken as +18 (0/2) — though the divide-by-negative flip was written and done correctly under time, first time seen. Errors are three content slips in number (a square, 1 s.f., negatives) and one untaught-looking probability method; the algebra was clean.
Priority: Squares and negatives inside bigger sums under time; then without-replacement probability as a tree Start with: Six squares in order-of-operations sums, timed, with the square written as its own line Teach using: Tree diagram for Q4-type probability; written check line for estimates and inequalities Avoid: Reteaching linear equations — A17 was clean Check whether: 4², 3², 5² come out as 16, 9, 25 inside a sum under time; 0.52 rounds to 0.5 at 1 s.f. Success: A timed set with no square-as-doubling, 1 s.f. roundings written, and a without-replacement item with 8 then 7 denominators
| Topic | Name | Change | Evidence recorded |
|---|---|---|---|
| A17 | Developing → Developing | paper #12 Q3: 3/3, loss=none — 6x−15=4x+7, 6x=4x+22, 2x=22, x=11, every line valid, 2 min, first time inside a timed section | |
| A4 | Secure → Secure | paper #12 Q3: 3/3, loss=none — 3(2x−5) expanded to 6x−15 correctly, both terms multiplied, under time | |
| N1-N3 | Developing → Developing | paper #12 Q1: 0/1, loss=knowledge — order of operations right (power, ×, +) but 4² taken as 8; paper #12 Q5: the 0/2 came from 3−15 written as +18. Two negatives/powers arithmetic slips in one timed section. | |
| N6-N7 | Developing → Developing | paper #12 Q1: 0/1, loss=knowledge — wrote 4*4 then evaluated it as 8 (squaring as doubling) under time; order of operations itself was right | |
| N14 | Secure → Secure | paper #12 Q2: 0/2, loss=knowledge — rounded first and wrote the rounded values before the sum, but to 6, 29 and 1 (not 1 s.f.: 30 and 0.5 missing), so no M1; 174 given | |
| P8 | Secure → Secure | paper #12 Q4: 0/3, loss=knowledge — without replacement: 7*7=49 and 5*3=15 → 15/49; denominators 7 and 7, one order only, no tree drawn | |
| P6-P7 | Secure → Secure | paper #12 Q4: 0/3, loss=knowledge — only red-then-blue counted; blue-then-red not added; no branch written | |
| A22 | Developing → Developing | paper #12 Q5: 0/2, loss=knowledge (negatives, N1-N3) — −4x>18 from 3−15 taken as +18; x<−4.5 followed. The flip when dividing by −4 was written and done correctly under time — the loose end's flip-under-time now seen, right. Mark lost to the constant, not the inequality method. | |
| G14-G16 | Developing → Developing | paper #12 Q6: 2/2, loss=none — 9*6=54, 54−4=50, unit written, 1 min | |
| A1-A3 | Secure → Secure | paper #12 Q7: 2/2, loss=none — (−4)²=16, 48−10=38, sign handled; page timed 6 min but she had moved around the paper, treat as a guide | |
| N15 | Secure → Secure | paper #12 Q8: 2/2, loss=none — 335 ⩽ n < 345, both bounds and signs, question's letter n used first time (loose end closed on the facts) |