2026 HSC Mathematics Standard 2 — Intuition Education Predicted Paper

Answers and marking notes

(Intuition Education — not part of the examination paper.)

Section I — correct responses

Brief reasons:

1. Cost per 100 g: 52.0c, 48.0c, 51.0c, 50.0c — the 750 g packet. B
2. $\binom{5}{2} = 10$ edges. D
3. $5a = 2b - c \Rightarrow b = \dfrac{5a + c}{2}$ (distractor B subtracts $c$ instead of adding; C halves only part of the expression — the flagged order-of-operations errors). A
4. $180° + 48° = 228°$ (distractors: $180° - 48° = 132°$, $90° + 48° = 138°$, $360° - 48° = 312°$). C
5. Relative frequency $\tfrac{36}{120} = 0.3$; $0.3 \times 300 = 90$. C
6. $t = \dfrac{k}{v}$ is a rectangular hyperbola. A
7. Monthly rate $= \tfrac{6\%}{12} = 0.5\%$, $n = 48$: $8000(1.005)^{48}$. D
8. Strong negative linear association: $r \approx -0.85$ ($-1$ would require perfectly collinear points). B
9. $50\% + 34\% = 84\%$. D
10. Absolute error $= 0.05$ cm; $\tfrac{0.05}{6.4} \times 100 = 0.78\%$. B
11. $z$-scores: English $1.25$, Maths $1.2$, Music $\tfrac{68-60}{5} = 1.6$, Science $0.83$. C
12. $\sin^{-1}(0.5) = 30°$; obtuse solution $180° - 30° = 150°$. D
13. A point well below the line at large $x$ drags the line down (gradient decreases) and weakens the linear fit ($r$ decreases). C
14. $\dfrac{40 \times 60}{8} = 300$. C
15. Only edges directed source → sink count: $6 + 5 = 11$ (the backward 4 is excluded). A

Question 16

Taxed items (GST-inclusive) $= 11 \times 4.50 = \$49.50$. GST-free items $= 87.45 - 49.50 = \$37.95$. (1 mark ×11 recognition; 1 mark answer. The ×10-vs-÷11 confusion scores 0 for $\$45.00$.)

Question 17

(a) Energy $= 0.9 \text{ kW} \times 5 \times 30 = 135$ kWh; cost $= 135 \times \$0.32 = \$43.20$. (1 mark W→kW and kWh; 1 mark cost in dollars.) (b) $21.60 \div 0.32 = 67.5$ kWh; $67.5 \div 0.9 = 75$ hours.

Question 18

(a) First branch $\tfrac{5}{8}$ R, $\tfrac{3}{8}$ B; second branches $\tfrac{4}{7}/\tfrac{3}{7}$ after R and $\tfrac{5}{7}/\tfrac{2}{7}$ after B. (b) $P(\text{same}) = \tfrac{5}{8}\cdot\tfrac{4}{7} + \tfrac{3}{8}\cdot\tfrac{2}{7} = \tfrac{20 + 6}{56} = \tfrac{26}{56} = \tfrac{13}{28}$. (1 mark both products; 1 mark sum. Unreduced $\tfrac{26}{56}$ accepted.)

Question 19

(a) Network on vertices $A$–$E$ with the eight weighted edges from the table. (b) MST edges: $DE(3), CD(4), AC(5), BC(6)$ — total $18$ km. (1 mark correct tree — it is unique; 1 mark length.) (c) With $AC$ closed: $A\to D\to E = 9 + 3 = 12$ km (beats $A\to B\to E = 15$). The shortest path does not use the MST's $A$–$C$ entry point — the designed trap.

Question 20

(a) $t = \dfrac{k}{n}$ with $k = 4 \times 18 = 72$, so $t = \dfrac{72}{n}$. (b) $t = \tfrac{72}{6} = 12$ hours. (c) $\tfrac{72}{n} \le 8 \Rightarrow n \ge 9$: 9 pumps.

Question 21

Tax $= 5092 + 0.325 \times (87{,}500 - 45{,}000) = 5092 + 13{,}812.50 = \$18{,}904.50$. Medicare levy $= 0.02 \times 87{,}500 = \$1{,}750$. Total $= \$20{,}654.50$. (1 mark correct bracket in dollars not cents; 1 mark levy; 1 mark total.)

Question 22

(a) Axis of symmetry: $p = \dfrac{-200}{2(-5)} = \$20$. (b) $R = -5(400) + 200(20) = \$2000$. (c) $-5p^2 + 200p = 1500 \Rightarrow p^2 - 40p + 300 = 0 \Rightarrow (p-10)(p-30) = 0$: $p = \$10$ or $\$30$.

Question 23

Arrival in Sydney time: 9:50 am + 13 h 40 min = 11:30 pm Tuesday. Time difference $10 - (-7) = 17$ hours (LA behind). $11{:}30$ pm Tuesday $- 17$ h $=$ 6:30 am Tuesday in Los Angeles. (1 mark arrival in Sydney time; 1 mark 17-hour difference in the right direction; 1 mark time AND day.)

Question 24

Daily rate $= \tfrac{18.25\%}{365} = 0.05\%$. Interest $= 840\left(1.0005^{20} - 1\right) = 840 \times 0.0100476 = \$8.44$. (1 mark daily rate and $n = 20$; 1 mark interest only, not the future value $\$848.44$.)

Question 25

(a) Gradient $= \tfrac{76 - 40}{8 - 2} = 6$; $40 = 6(2) + c \Rightarrow c = 28$: $y = 6x + 28$. (b) Each additional hour of revision is associated with an increase of 6 marks, on average. (Context, both variables, per-unit language required.) (c) $y = 6(5) + 28 = 58$ marks. (d) $x = 20$ gives $y = 148 > 100$, an impossible mark — 20 hours is far outside the data range, so extrapolation is unreliable.

Question 26

(a) $V = \pi r^2 h + \tfrac{1}{2}\cdot\tfrac{4}{3}\pi r^3$ with $r = 2$: $V = \pi(4)(5) + \tfrac{2}{3}\pi(8) = 20\pi + \tfrac{16\pi}{3} = \tfrac{76\pi}{3} = 79.587\ldots \approx 79.6 \text{ m}^3$. (1 mark diameter→radius; 1 mark halved sphere formula; 1 mark combined and rounded.) (b) $79.587 \text{ m}^3 \times 1000 = 79{,}587$ L $\approx$ 79,600 L (3 s.f.).

Question 27

(a) Quarterly rate $= \tfrac{3\%}{4} = 0.75\%$, $n = 16$ quarters → factor $16.932$. Deposit $= \dfrac{20{,}000}{16.932} = \$1{,}181$. (1 mark rate/period conversion and correct cell; 1 mark divide — not multiply — and round.) (b) Interest $= 20{,}000 - 16 \times 1{,}181.20 = \$1{,}101$ (accept $\$1{,}101$–$\$1{,}104$ depending on rounding of the deposit).

Question 28

(a) From the ogive at cumulative frequencies 15, 30 and 45: $Q_1 = 22$, median $= 28$, $Q_3 = 34$ minutes. (b) Box from 22 to 34 with median line at 28; whiskers to 12 and 48.

Question 29

$IQR = 34 - 22 = 12$; upper fence $= Q_3 + 1.5 \times IQR = 34 + 18 = 52$. Since $48 \le 52$, 48 minutes is not an outlier. (1 mark fence; 1 mark comparison WITH a stated conclusion.)

Question 30

(a) $N = 4 \times 1.5 = 6$, $H = 3$, $M = 68$: $BAC = \dfrac{10(6) - 7.5(3)}{6.8 \times 68} = \dfrac{37.5}{462.4} = 0.0811\ldots \approx 0.081$. (1 mark $N = 6$; 1 mark substitution and rounding.) (b) $\tfrac{0.081}{0.015} = 5.4$ h $= 5$ h $24$ min (0.4 h $= 24$ min, not 40 min — the flagged trap). (c) 11:30 pm $+ 5$ h $24$ min $=$ 4:54 am.

Question 31

(a) $48{,}000 - 4 \times 6{,}000 = \$24{,}000$. (b) $48{,}000(1 - 0.15)^4 = 48{,}000 \times 0.85^4 = 48{,}000 \times 0.52200625 = \$25{,}056.30$. (1 mark $S = V_0(1-r)^n$ with $r = 0.15$; 1 mark value.) (c) Declining balance, by $25{,}056.30 - 24{,}000 = \$1{,}056.30$.

Question 32

$16\%$ above 660 → $660$ is one standard deviation above the mean: $\mu + \sigma = 660$. $2.5\%$ below 540 → $540$ is two standard deviations below: $\mu - 2\sigma = 540$. Subtracting: $3\sigma = 120$, so $\sigma = 40$ hours and $\mu = 620$ hours. (1 mark each correct $z$-identification; 1 mark solving the pair.)

Question 33

(a) Scale $1:500$ → 1 cm represents 5 m. Offsets: $1.6 \times 5 = 8$ m, $2.4 \times 5 = 12$ m, $2.0 \times 5 = 10$ m; spacing $2 \times 5 = 10$ m. (Scale conversion BEFORE the rule.) (b) $A \approx \tfrac{10}{2}(8 + 12) + \tfrac{10}{2}(12 + 10) = 100 + 110 = 210 \text{ m}^2$. (1 mark per application.) (c) $V = 210 \times 0.15 = 31.5 \approx$ 32 m³ (2 s.f.).

Question 34

(a) $\cos 35° = \dfrac{24}{AC} \Rightarrow AC = \dfrac{24}{\cos 35°} = 29.2986\ldots \approx 29.3$ m. (b) Sine rule in $ACD$: $\dfrac{AD}{\sin 40°} = \dfrac{AC}{\sin 65°} \Rightarrow AD = \dfrac{29.3 \sin 40°}{\sin 65°} = 20.78\ldots \approx 20.8$ m. (1 mark sine rule with sides opposite the correct angles; 1 mark answer. Full-accuracy $AC$ gives the same rounded answer — no early-rounding penalty here, but working from $29.3$ or $29.2986$ both accepted.)

Question 35

(a) $R$: 0, 280, 560, 840, 1120; $C$: 400, 520, 640, 760, 880. (b) Two straight lines; $C$ has intercept 400, $R$ passes through the origin. (c) $400 + 12n = 28n \Rightarrow 16n = 400 \Rightarrow n = 25$ T-shirts (intersection of the graphs). (d) Profit $= 1120 - 880 = \$240$.

Question 36

(a) $z = \dfrac{4.9 - 4.2}{0.5} = 1.4$. (b) $P(Z < 1.4) = 0.9192$, so $P(\text{mass} > 4.9) = 1 - 0.9192 = 0.0808$. (1 mark table value; 1 mark complement.) (c) $250 \times 0.0808 = 20.2 \approx$ 20 echidnas (a whole-number count, not a probability).

Question 37

(a) Paths: $A$–$C$–$D$ $= 3+4+6 = 13$; $B$–$D$ $= 11$; $A$–$C$–$E$ $= 9$. Minimum completion time 13 days, critical path $A$–$C$–$D$. (1 mark time; 1 mark path.) (b) $D$ must start by day $13 - 6 = 7$, so $LST_B = 7 - 5 = 2$: float $= 2 - 0 = 2$ days. (c) The increase of 3 days exceeds $B$'s float of 2 days, so $B$ becomes critical: new critical path $B$–$D$, new minimum time $8 + 6 = 14$ days.

Question 38

(a) $8 + 7 = 15$ L/min. (b) The cut through $XT$ and $YT$ has capacity $5 + 9 = 14$; no flow can exceed the capacity of any cut, so maximum flow $\le 14$ (and 14 is achieved). (c) Increase $YT$ (the saturated edge in the minimum cut). Increasing it to 10 lifts the minimum cut, and the flow, to 15 (spare capacity exists on $S\to X\to Y$); increasing $XT$ alone also lies on the min cut — accept $XT$ or $YT$ with a min-cut justification; reject $SX$/$SY$/$XY$ (not in the minimum cut).

Question 39

(a) Both bearings are measured clockwise from north at $P$: $\angle XPY = 122° - 48° = 74°$. (b) $XY^2 = 25^2 + 32^2 - 2(25)(32)\cos 74° = 1649 - 1600(0.27564) = 1207.98$, so $XY = 34.8$ km. (1 mark cosine rule; 1 mark answer.) (c) $\dfrac{\sin \angle PXY}{32} = \dfrac{\sin 74°}{34.756} \Rightarrow \sin \angle PXY = 0.8850 \Rightarrow \angle PXY = 62°$. (d) Bearing of $P$ from $X$ is $48° + 180° = 228°$; bearing of $Y$ from $X$ = $228° - 62° = $ 166°. (The "of/from" reversal and the parallel-north-lines transfer are the marked skills.)

Question 40

(a) $2000 \times 4.452 = \$8{,}904$. (b) $3000 \times 8.111$ would fund $\$3{,}000$ every year for all 10 years; the withdrawals in years 1–5 are only $\$2{,}000$, so the years 6–10 stream is valued as the 10-year factor minus the 5-year factor: $3000 \times (8.111 - 4.452) = 3000 \times 3.659 = \$10{,}977$. (1 mark explanation as a difference of two present values; 1 mark evaluation.) (c) $8{,}904 + 10{,}977 = \$19{,}881$.