Mathematics Compulsory | DSE Core Math for Business

Paper 1 Conventional Questions (65%, 2h15m) | Paper 2 Multiple Choice (35%, 1h15m, 45 questions)

考試結構（2024 官方 Public Assessment）

卷別	題型	比重	時長	範圍
Paper 1（長答）	Section A 基礎題（8–11 題）+ Section B 挑戰題（4–7 題）	65%	2 小時 15 分	Compulsory Part
Paper 2（MC）	全選擇題（甲部基礎 + 乙部非基礎）	35%	1 小時 15 分	Compulsory Part
M1 微積分與統計（選修，2 選 1）	Section A 8–12 短題 + B 3–5 長題	單獨應考	2 小時 30 分	Calculus & Statistics
M2 代數與微積分（選修，2 選 1）	同上結構	單獨應考	2 小時 30 分	Algebra & Calculus

商科學生選 M1：M1 側重統計與基礎微積分，對經濟/商科/社科最有直接幫助；M2 偏工程向，除非學校要求否則商科生選 M1。Paper 1 絕不涉及冗長運算，重視解題步驟與精確度。

Compulsory Part 六大範疇（溫習地圖）

1. Number & Algebra（數與代數）
2. Measures, Shape & Space（度量、圖形與空間）
3. Data Handling（數據處理 / 統計）
4. Algebra & Calculus（M1/M2 延伸基礎）
5. 商科高關聯：Percentage / Ratio / Growth、Interest & Depreciation、Statistics & Probability

練習題庫 Math 卷見 練習題庫 Math，每章配對應題型。

Chapter 1 Number & Algebra ★★★

Must-memorise Identities

(a+b)² = a² + 2ab + b²        (a−b)² = a² − 2ab + b²
a² − b² = (a+b)(a−b)
a³ + b³ = (a+b)(a² − ab + b²)  a³ − b³ = (a−b)(a² + ab + b²)
Indices:  aᵐ·aⁿ = aᵐ⁺ⁿ   (aᵐ)ⁿ = aᵐⁿ   a⁻ⁿ = 1/aⁿ
Logs:  log(MN) = logM + logN   log(M/N) = logM − logN   log Mᵏ = k·logM

Estimation & Errors

• Maximum absolute error = ½ × scale interval
• Relative error = absolute error ÷ measured value
• Percentage error = relative error × 100%

Core Mechanism (How to use)

The identity toolkit lets you rewrite any algebraic expression into a factorable or expandable form so that equations become solvable. Logarithm laws turn products, quotients and powers into sums, differences and coefficients — this is exactly how we solve for an unknown hidden in an exponent (e.g. compound growth or half-life). Master the three factorisations (difference of squares, perfect squares, sum/difference of cubes) because they appear inside every harder algebra question. (白話：先把式子「整形」成能開方或能約掉的樣子，對數是用來把指數上的 x 拉下來的。)

Worked Example 1.1 — Indices (solve for x)

Worked Example 1.1 — Indices                                [Total 4 marks]
  Given: 2^(x+1) = 8^(x−1),  find x.
(1) Write 8 as a power of 2:  8 = 2³                          (1 mark)
(2) RHS = (2³)^(x−1) = 2^[3(x−1)] = 2^(3x−3)                  (1 mark)
(3) Equal bases ⇒ equal exponents:  x + 1 = 3x − 3            (1 mark)
(4) Rearrange:  1 + 3 = 3x − x  ⇒  4 = 2x  ⇒  x = 2          (1 mark)
    Check: LHS = 2³ = 8,  RHS = 8¹ = 8   ✓

Worked Example 1.2 — Logarithms (solve for x)

Worked Example 1.2 — Logarithms                            [Total 5 marks]
  Given: log x + log(x − 3) = 1,  log is base 10,  find x.
(1) Combine logs:  log[x(x − 3)] = 1                          (1 mark)
(2) Exponential form:  x(x − 3) = 10¹ = 10                    (1 mark)
(3) Expand:  x² − 3x − 10 = 0                                (1 mark)
(4) Factorise:  (x − 5)(x + 2) = 0  ⇒  x = 5 or x = −2      (1 mark)
(5) Reject x = −2 (log of a negative number is undefined)
    ∴ x = 5                                                    (1 mark)

Worked Example 1.3 — Estimation & maximum percentage error (4 marks)

Worked Example 1.3 — Estimation & error                  [Total 4 marks]
  A rectangular field is measured as 24.5 m by 12.3 m
  using a tape marked in 0.5 m intervals.
  (a) Find the maximum absolute error of EACH measurement.
  (b) Find the percentage error in the calculated area,
      correct to 3 significant figures.
  Measured length L* = 24.5,  width W* = 12.3
(1) Scale interval = 0.5 m
      max absolute error of each = ½ × 0.5 = 0.25 m        (1 mark)
(2) Bounds for the area:
      lower = (24.5 − 0.25)(12.3 − 0.25)
            = 24.25 × 12.05 ≈ 292.21
      upper = (24.5 + 0.25)(12.3 + 0.25)
            = 24.75 × 12.55 ≈ 310.36       (1 mark)
    Calculated area A* = 24.5 × 12.3 = 301.35
(3) Max absolute error in area ≈ 310.36 − 301.35 = 9.01
    (using the larger deviation from the centre value)
    Relative error ≈ 9.01 / 301.35 ≈ 0.03090               (1 mark)
(4) Percentage error ≈ 0.03090 × 100% ≈ 3.09%             (1 mark)
    ∴ (a) 0.25 m for each side;
      (b) percentage error in area ≈ 3.09%.

Worked Example 1.4 — Factorisation & solving a cubic (5 marks)

Worked Example 1.4 — Factorisation                       [Total 5 marks]
  Solve:  x³ − 3x² − 4x + 12 = 0.
(1) Group terms:  (x³ − 3x²) − (4x − 12) = 0              (1 mark)
(2) Take out common factors:
      = x²(x − 3) − 4(x − 3)
      = (x − 3)(x² − 4)                                    (1 mark)
(3) x² − 4 = (x − 2)(x + 2)   (difference of squares)     (1 mark)
(4) So (x − 3)(x − 2)(x + 2) = 0                          (1 mark)
(5) ∴ x = 3,  x = 2,  or  x = −2                         (1 mark)
    Check: x = 2 ⇒ 8 − 12 − 8 + 12 = 0 ✓
            x = −2 ⇒ −8 − 12 + 8 + 12 = 0 ✓

Worked Example 1.5 — Logarithms with change of base (6 marks)

Worked Example 1.5 — Logarithms (change of base)        [Total 6 marks]
  Solve:  log₃ x + log₉ x = 6.
  (log base 3 and log base 9.)
(1) Convert log₉ x to base 3:
      log₉ x = log₃ x / log₃ 9
      log₃ 9 = log₃(3²) = 2
      ⇒ log₉ x = (log₃ x) / 2                            (2 marks)
(2) Equation becomes:
      log₃ x + (log₃ x)/2 = 6                            (1 mark)
(3) Let y = log₃ x.  Then y + y/2 = 6
      (3/2)y = 6  ⇒  y = 4                               (1 mark)
(4) So log₃ x = 4  ⇒  x = 3⁴ = 81                        (1 mark)
(5) Check: x = 81 > 0, arguments valid.
      log₃ 81 = 4,  log₉ 81 = 2,  4 + 2 = 6 ✓            (1 mark)
    ∴ x = 81.

Common Trap (Chapter 1): When solving log equations you MUST reject roots that make any log argument ≤ 0, even if they solve the quadratic. Never split log(a + b) into log a + log b — that is an illegal move and loses all marks. For indices, always convert to the SAME base before equating exponents; mixing base 2 and base 8 without conversion fails completely. Extra traps: (i) Change-of-base must use the SAME new base in both numerator and denominator — log_b x = log_k x / log_k b, never mixing k. (ii) Maximum absolute error is HALF the scale interval (0.25 m, not 0.5 m); when a quantity is computed from two measured values (an area), find the error from the bounds, do not just add the two side-errors. (iii) Negative exponents: a^−n = 1/a^n, so 2^−3 = 1/8, never −8. (iv) Fractional indices: x^(1/2) = √x and x^(3/2) = x√x — keep root and power distinct, do not write x^(3/2) = 3√(x²). (v) When equating exponents, the bases must already be identical and positive; 2^(x+1) = 8^(x−1) is NOT solved by x+1 = x−1.

Chapter 2 Functions & Graphs ★★★

Quadratic Functions（每年必考）

For y = ax² + bx + c:

• Vertex form: y = a(x−h)² + k, vertex (h, k)
• Axis of symmetry: x = −b / 2a
• Discriminant Δ = b² − 4ac：Δ>0 two distinct real roots；Δ=0 one double root；Δ<0 no real roots
• Sum of roots = −b/a；Product of roots = c/a

Graph Transformations

Transformation	Effect on y = f(x)
y = f(x) + k	translate upward by k
y = f(x − h)	translate rightward by h
y = k·f(x)	vertical stretch (×k)
y = f(kx)	horizontal shrink to 1/k
y = −f(x) / f(−x)	reflect about x-axis / y-axis

Core Mechanism (How to use)

The discriminant Δ = b² − 4ac is the most reused single tool — it tells you how many x-intercepts a parabola has WITHOUT drawing it. For max/min problems, the vertex (turning point) gives the extreme value; because a parabola is symmetric, the optimum in business contexts (profit, cost, revenue) almost always sits at the vertex or at a boundary. Always state the axis of symmetry x = −b/2a first, then substitute to get the y-value. (白話：Δ 決定根有幾個；頂點決定最大/最小值。先寫對稱軸，再代回求 y。)

Worked Example 2.1 — Quadratic maximum (business context)

Worked Example 2.1 — Quadratic maximum                    [Total 5 marks]
  Daily profit P = −2x² + 80x − 300,  x = items sold.
  Find the maximum profit and the corresponding x.
  a = −2, b = 80, c = −300
(1) Axis of symmetry:  x = −b/(2a) = −80 / (2×−2) = 20      (1 mark)
(2) Substitute x = 20:  P = −2(20)² + 80(20) − 300          (1 mark)
(3) = −2(400) + 1600 − 300 = −800 + 1600 − 300 = 500       (2 marks)
(4) Since a = −2 < 0, parabola opens downward ⇒ maximum     (1 mark)
    ∴ Max profit = $500 when x = 20 items.

Worked Example 2.2 — Discriminant (nature of roots)

Worked Example 2.2 — Discriminant                           [Total 3 marks]
  Show that 2x² − 4x + 5 = 0 has no real roots.
  a = 2, b = −4, c = 5
(1) Δ = b² − 4ac = (−4)² − 4(2)(5)                          (1 mark)
(2) = 16 − 40 = −24                                          (1 mark)
(3) Since Δ < 0, the equation has no real roots.             (1 mark)

Worked Example 2.3 — Vertex form by completing the square (5 marks)

Worked Example 2.3 — Vertex form                        [Total 5 marks]
  Express y = 2x² − 12x + 7 in the form y = a(x − h)² + k,
  and hence state the coordinates of the vertex.
(1) Factor 2 from the x-terms:
      y = 2(x² − 6x) + 7                                 (1 mark)
(2) Complete the square inside the bracket:
      x² − 6x = (x − 3)² − 9                             (1 mark)
(3) Substitute back:
      y = 2[(x − 3)² − 9] + 7
      = 2(x − 3)² − 18 + 7
      = 2(x − 3)² − 11                                   (1 mark)
(4) Vertex form:  y = 2(x − 3)² − 11                    (1 mark)
(5) Vertex = (h, k) = (3, −11)                           (1 mark)
    Check: at x = 3,  y = 2(0) − 11 = −11 ✓
    Since a = 2 > 0, the vertex is a MINIMUM.

Worked Example 2.4 — Graph transformations (4 marks)

Worked Example 2.4 — Transformations of y = f(x)       [Total 4 marks]
  The graph of y = f(x) passes through (1, 4).
  State the new coordinates after each transformation.
  (a) y = f(x) + 3        (b) y = f(x − 2)
  (c) y = 2f(x)           (d) y = f(2x)
(1) (a) translate up 3 ⇒ (1, 4 + 3) = (1, 7)            (1 mark)
(2) (b) translate right 2 ⇒ (1 + 2, 4) = (3, 4)         (1 mark)
(3) (c) vertical stretch ×2 ⇒ (1, 4 × 2) = (1, 8)       (1 mark)
(4) (d) horizontal shrink to 1/2 ⇒ (1/2, 4) = (0.5, 4)  (1 mark)
    Note: y = f(kx) changes the x-coordinate to x/k.

Worked Example 2.5 — Range of k for real roots (5 marks)

Worked Example 2.5 — Discriminant & parameter k       [Total 5 marks]
  The equation kx² + 4x + 2 = 0 has real roots.
  Find the range of possible values of k (k ≠ 0).
  a = k,  b = 4,  c = 2
(1) For real roots we need Δ ≥ 0                         (1 mark)
(2) Δ = b² − 4ac = 4² − 4(k)(2)
      = 16 − 8k                                         (1 mark)
(3) 16 − 8k ≥ 0  ⇒  16 ≥ 8k  ⇒  k ≤ 2                   (1 mark)
(4) Also k ≠ 0 (otherwise it is not a quadratic)         (1 mark)
(5) For k < 0, Δ > 16 > 0 so real roots exist too.
    ∴ Range: k ≤ 2 and k ≠ 0.
      (i.e. k ∈ (−∞, 0) ∪ (0, 2])                        (1 mark)
    Check boundary k = 2: 2x² + 4x + 2 = 0 ⇒
      Δ = 16 − 16 = 0, one double root x = −1 ✓

Common Trap (Chapter 2): The axis formula is x = −b/(2a) — students often drop the negative sign and get the wrong vertex x (here −(−12)/(2·2) = 3, not −3). For "no real roots" conclusions you must state BOTH "Δ < 0" AND the sentence "no real roots"; writing only the number loses the conclusion mark. When a < 0 the vertex is a MAXIMUM, when a > 0 it is a MINIMUM — always say which one. Extra traps: (i) In completing the square, after factoring a from x-terms the constant must be ADDED inside and SUBTRACTED outside with the factor: 2[(x−3)² − 9] + 7, not 2(x−3)² + 7. (ii) y = f(x − h) shifts RIGHT by h (the sign flips); many write (1−2, 4) instead of (1+2, 4). (iii) y = f(kx) compresses the x-value to x/k, it does NOT multiply x by k. (iv) A "range of k" question needs the inequality AND any exclusion such as k ≠ 0; forgetting k ≠ 0 loses the final mark.

Chapter 3 Equations & Inequalities ★★

• Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a
• Quadratic inequality: sketch the parabola first, then read the solution range（先画图后写解）
• Linear programming: constraints → feasible region → test vertices for optimal value（商科应用高频：利润最大化）

Core Mechanism (How to use)

Linear programming works because the objective function z = ax + by is linear, so its largest/smallest value over a convex feasible region is always attained at a CORNER (vertex). That is why we only ever test vertices — never interior points. For quadratic inequalities, sketch the parabola first to see which region lies above or below the x-axis; this prevents the sign errors that pure algebra causes. (白話：利潤最大化只在「角點」出現，所以只驗頂點；不等式先畫圖再讀解。)

Worked Example 3.1 — Quadratic inequality

Worked Example 3.1 — Quadratic inequality                   [Total 4 marks]
  Solve x² − 5x + 6 < 0.
(1) Factorise:  x² − 5x + 6 = (x − 2)(x − 3)                (1 mark)
(2) Roots: x = 2, x = 3  (parabola opens upward, a = 1 > 0)  (1 mark)
(3) Sketch: parabola is below the x-axis BETWEEN the roots    (1 mark)
(4) ∴ Solution:  2 < x < 3                                   (1 mark)

Worked Example 3.2 — Linear programming (profit maximisation)

Worked Example 3.2 — Linear programming                   [Total 6 marks]
  A factory makes chairs (x) and tables (y).
  Profit $40/chair, $70/table.
  Constraints:  x + y ≤ 8,  2x + y ≤ 12,  x ≥ 0, y ≥ 0.
  Maximise P = 40x + 70y.
(1) Vertices of feasible region:  A(0,0) B(6,0) C(4,4) D(0,8)  (1 mark)
(2) Test each vertex in P = 40x + 70y:                       (3 marks)
      A: 40(0) + 70(0) = 0
      B: 40(6) + 70(0) = 240
      C: 40(4) + 70(4) = 160 + 280 = 440
      D: 40(0) + 70(8) = 560
(3) Largest value = 560 at D(0,8)                            (1 mark)
(4) Check D satisfies constraints:
      0 + 8 = 8 ≤ 8 ✓ ,  2(0) + 8 = 8 ≤ 12 ✓               (1 mark)
    ∴ Max profit = $560 by making 0 chairs and 8 tables.

Worked Example 3.3 — Quadratic formula (4 marks)

Worked Example 3.3 — Quadratic formula                  [Total 4 marks]
  Solve 2x² + 3x − 5 = 0, correct to 3 significant figures.
  a = 2, b = 3, c = −5
(1) Use x = [−b ± √(b² − 4ac)] / 2a                     (1 mark)
(2) Discriminant: b² − 4ac = 3² − 4(2)(−5)
      = 9 + 40 = 49                                       (1 mark)
(3) x = [−3 ± √49] / 4 = [−3 ± 7] / 4                    (1 mark)
(4) x = (−3 + 7)/4 = 1    or    x = (−3 − 7)/4 = −2.5    (1 mark)
    Check: x = 1 ⇒ 2 + 3 − 5 = 0 ✓
           x = −2.5 ⇒ 2(6.25) + 3(−2.5) − 5 = 0 ✓
    ∴ x = 1 or x = −2.5.

Worked Example 3.4 — Simultaneous (linear + quadratic) (5 marks)

Worked Example 3.4 — Simultaneous equations           [Total 5 marks]
  Solve:  y = x + 1        ... (1)
          x² + y² = 25      ... (2)
(1) Substitute (1) into (2):                               (1 mark)
      x² + (x + 1)² = 25
(2) Expand:  x² + x² + 2x + 1 = 25                         (1 mark)
      2x² + 2x − 24 = 0
(3) Divide by 2:  x² + x − 12 = 0                         (1 mark)
(4) Factorise:  (x + 4)(x − 3) = 0
      ⇒ x = −4  or  x = 3                                 (1 mark)
(5) When x = −4, y = −3;  when x = 3, y = 4               (1 mark)
    Check in (2): (−4)² + (−3)² = 16 + 9 = 25 ✓
                  3² + 4² = 9 + 16 = 25 ✓
    ∴ (x, y) = (−4, −3) or (3, 4).

Common Trap (Chapter 3): Always list ALL vertices including the intercepts such as (0,0), and test EVERY one — picking the vertex you "like" loses the method mark. For inequalities, the strict sign < gives an OPEN interval (no equals), while ≤ gives a closed one; mixing them flips the answer. Sketch the feasible polygon; do not trust mental algebra for the region. Extra traps: (i) Under the root in the quadratic formula write b² − 4ac EXACTLY, including the sign of c — with c = −5 the term becomes −4(2)(−5) = +40, not −40. (ii) The formula is [−b ± √(...)], the minus is on b only, not on the whole numerator. (iii) When solving simultaneous linear+quadratic, substitution then expanding (x+1)² = x²+2x+1 carefully; dropping the 2x term is a common slip. (iv) Always give BOTH coordinate pairs — a single pair earns only half the marks even if correct. (v) For linear programming, a constraint like x + y ≤ 8 also requires x ≥ 0, y ≥ 0 to bound the region; without them the feasible region is unbounded and no maximum exists.

Chapter 4 Sequences ★★★（复利/投资高频载体）

	Arithmetic (A.S.)	Geometric (G.S.)
nth term	a + (n−1)d	a·rⁿ⁻¹
Sum of n terms	(n/2)[2a + (n−1)d]	a(1 − rⁿ) / (1 − r)
Sum to infinity	—	a / (1 − r)，only when |r| < 1

Business application：分期存款、按揭还款、机器折旧均为 G.S. 模型。

Core Mechanism (How to use)

An arithmetic sequence has a CONSTANT difference d (linear growth — like fixed monthly additions to savings). A geometric sequence has a CONSTANT ratio r (exponential — like compound interest, population, depreciation). Choose the right model: "increases by $100 each year" → A.S.; "grows at 5% each year" → G.S. The sum-to-infinity only exists when |r| < 1. (白話：等差是「每次加固定數」，等比是「每次乘固定比率」；複息、折舊都用等比。)

Worked Example 4.1 — Arithmetic sum (savings plan)

Worked Example 4.1 — Arithmetic progression                   [Total 4 marks]
  A man saves $500 in month 1 and increases by $50 each month.
  Find the total saved in 2 years (24 months).
  a = 500, d = 50, n = 24
(1) Use S_n = (n/2)[2a + (n−1)d]                             (1 mark)
(2) S_24 = (24/2)[2(500) + 23(50)]                           (1 mark)
(3) = 12[1000 + 1150] = 12 × 2150 = 25800                   (1 mark)
(4) ∴ Total saved = $25,800                                  (1 mark)

Worked Example 4.2 — Geometric sum (future value of annuity)

Worked Example 4.2 — Geometric (investment)                [Total 5 marks]
  $1,000 is deposited at the END of each year for 10 years
  at 6% p.a. compounded annually. Find the future value.
  First deposit grows 10 yrs, last grows 1 yr ⇒ G.S.
  a = 1000(1.06)¹ ,  r = 1.06,  n = 10
(1) Future value S_10 = a(r¹⁰ − 1)/(r − 1)                   (1 mark)
(2) = 1000(1.06)[(1.06)¹⁰ − 1]/(1.06 − 1)                   (1 mark)
(3) (1.06)¹⁰ ≈ 1.7908477  (keep ≥ 4 sig. fig.)               (1 mark)
(4) = 1000(1.06)(0.7908477)/(0.06)
      ≈ 1060 × 13.18080 ≈ 13972                             (2 marks)
    ∴ Future value ≈ $13,972 (3 sig. fig.)

Worked Example 4.3 — Geometric depreciation of a machine (5 marks)

Worked Example 4.3 — Depreciation (G.S.)                [Total 5 marks]
  A machine is bought for $80,000 and depreciates at
  12% per annum. Find its value after 5 years, to the
  nearest dollar.
  Initial value P = 80000,  rate = 12% = 0.12
  Retained ratio r = 1 − 0.12 = 0.88
(1) Value after t years:  V = P·rᵗ                            (1 mark)
(2) V = 80000 × (0.88)⁵                                    (1 mark)
(3) (0.88)⁵ = 0.88² × 0.88² × 0.88
      = 0.7744 × 0.7744 × 0.88
      ≈ 0.599695 × 0.88 ≈ 0.52773                           (1 mark)
(4) V ≈ 80000 × 0.52773 ≈ 42218.4                          (1 mark)
(5) To nearest dollar:  V ≈ $42,218                        (1 mark)
    Check: 12% drop each year, after 5 yrs value should be
      well below half — 0.5277 of original is reasonable ✓
    ∴ Value after 5 years ≈ $42,218.

Worked Example 4.4 — Sum to infinity (4 marks)

Worked Example 4.4 — Sum to infinity (G.S.)            [Total 4 marks]
  Find the sum to infinity of  1 + 1/3 + 1/9 + 1/27 + ...
(1) This is a G.S. with a = 1,  r = (1/3) / 1 = 1/3      (1 mark)
(2) Check |r| = 1/3 < 1, so sum to infinity exists        (1 mark)
(3) S_∞ = a / (1 − r) = 1 / (1 − 1/3)                     (1 mark)
      = 1 / (2/3) = 3/2 = 1.5
(4) ∴ Sum to infinity = 1.5                               (1 mark)
    Note: each added term gets smaller, total approaches 1.5.

Common Trap (Chapter 4): Confusing A.S. and G.S. — read the wording carefully: "each year $100 more" = arithmetic; "each year 5% more" = geometric. For depreciation use r = 1 − rate, never add. Check |r| < 1 BEFORE using the infinity sum, or you get a nonsense infinite answer. Keep 4+ sig. fig. in intermediate powers; rounding (1.06)¹⁰ early to 1.79 loses the final mark. Extra traps: (i) In a "future value of annuity" the FIRST deposit earns interest for (n−1) years if paid at the END of the first year, so the series is a·r¹ + a·r² + ... + a·rⁿ, with first term a·r (not a). (ii) Depreciation is multiplicative decay r = 1 − rate; do not write V = P(1 + 0.12)⁵. (iii) Sum-to-infinity formula S_∞ = a/(1−r) only; if you mistakenly use the finite sum denominator (1−rⁿ) with n→∞ you must take the limit, which gives the same a/(1−r) — but writing the finite form without the limit check loses the method mark. (iv) For the nth term of a G.S., it is a·r^(n−1); the 10th term uses r⁹, not r¹⁰.

Chapter 5 Probability & Statistics ★★★（商科最高权重）

Probability

P(A∪B) = P(A) + P(B) − P(A∩B)
Independent:   P(A∩B) = P(A) × P(B)
Conditional:   P(A|B) = P(A∩B) / P(B)
Permutation:   nPr = n! / (n−r)!
Combination:   nCr = n! / [r!(n−r)!]

• 抽签不放回 → 用组合数或树状图，切勿当成独立事件

Statistics

Measure	Key points
Mean / Median / Mode	Mean 受极端值影响大；skewed data 用 median
Standard deviation σ	加常数 k → σ 不变；乘 k → σ 变 σ×|k|
Inter-quartile range	IQR = Q₃ − Q₁，box-and-whisker plot 必考
Standard score	z = (x − mean) / σ（跨组比较表现）
Normal distribution	68% within ±1σ，95% within ±2σ，99.7% within ±3σ

Sampling 陷阱：questionnaire at MTR station ≠ random sample（每年 MC 必考抽样偏差判断）。

Core Mechanism (How to use)

Probability is always a number between 0 and 1; if your answer exceeds 1 you made a sign or addition error. "Without replacement" means dependent events — use tree diagrams or combinations, never multiply independent probabilities. For statistics, the mean is pulled by outliers while the median is stable; that is why skewed business data (income, sales) is compared with median + IQR rather than mean alone. (白話：機率答案一定在 0–1；「不放回」就是依賴事件；偏態數據用中位數比較更穩。)

Worked Example 5.1 — Combination probability

Worked Example 5.1 — Combination probability                 [Total 5 marks]
  From 5 men and 4 women, select 3 at random.
  Find P(exactly 2 women).
  Total ways to choose 3 from 9:  ⁹C₃ = 84
(1) Favourable = choose 2 women from 4 AND 1 man from 5:      (1 mark)
      = ⁴C₂ × ⁵C₁ = 6 × 5 = 30
(2) P(exactly 2 women) = 30 / 84                             (1 mark)
(3) = 30/84 = 5/14 ≈ 0.3571                                  (1 mark)
(4) Check: 0 < 0.3571 < 1  ✓                                 (1 mark)
    ∴ P = 5/14 (or 0.357).

Worked Example 5.2 — Standard score comparison

Worked Example 5.2 — Standard score comparison             [Total 5 marks]
  A student scored 82 in Math (mean 70, σ 8)
  and 76 in Econ (mean 68, σ 5).
  Which subject is the better relative performance?
(1) Math z = (82 − 70)/8 = 12/8 = 1.5                        (2 marks)
(2) Econ  z = (76 − 68)/5 = 8/5 = 1.6                        (2 marks)
(3) Since 1.6 > 1.5, the relative performance is better
    in Economics.                                             (1 mark)
    ∴ Better relative performance in Economics.

Worked Example 5.3 — Tree diagram (without replacement) (5 marks)

Worked Example 5.3 — Tree diagram                      [Total 5 marks]
  A bag has 4 red and 6 blue balls. Two balls are drawn
  WITHOUT replacement. Find P(exactly one is red).
  First draw: P(R) = 4/10, P(B) = 6/10
(1) Branch 1: R then B:  (4/10) × (6/9) = 24/90          (2 marks)
      (after taking one red, 9 balls left, 6 blue)
(2) Branch 2: B then R:  (6/10) × (4/9) = 24/90          (1 mark)
      (after taking one blue, 9 balls left, 4 red)
(3) These are mutually exclusive, so add:
      P(exactly one red) = 24/90 + 24/90 = 48/90          (1 mark)
(4) = 48/90 = 8/15 ≈ 0.5333                              (1 mark)
    Check: 0 < 0.5333 < 1 ✓
    ∴ P(exactly one red) = 8/15 (or 0.533).

Worked Example 5.4 — Coded data, mean & SD (5 marks)

Worked Example 5.4 — Coded data                        [Total 5 marks]
  A data set has mean = 20 and standard deviation = 4.
  Each value is transformed by y = 3x − 5.
  Find the new mean and the new standard deviation.
(1) For y = ax + b:  new mean = a·(old mean) + b         (1 mark)
      mean_y = 3(20) − 5 = 60 − 5 = 55                    (1 mark)
(2) Standard deviation is unaffected by +b, scaled by |a|:
      new σ = |a| × old σ = 3 × 4 = 12                    (2 marks)
(3) Check logic: spread triples because values are
      stretched ×3; the −5 only shifts the centre.         (1 mark)
    ∴ New mean = 55,  new σ = 12.

Worked Example 5.5 — Normal distribution (4 marks)

Worked Example 5.5 — Normal distribution               [Total 4 marks]
  Let X ~ N(50, 8²). Find P(X < 66), using the
  empirical (68–95–99.7) rule.
  mean μ = 50,  σ = 8
(1) 66 = 50 + 16 = μ + 2σ                               (1 mark)
(2) By the rule, 95% of data lies within μ ± 2σ,
      i.e. between 34 and 66.                             (1 mark)
(3) Symmetry: half of the 5% outside lies above 66,
      so P(X > 66) = 2.5% = 0.025                        (1 mark)
(4) P(X < 66) = 1 − 0.025 = 0.975                       (1 mark)
    ∴ P(X < 66) = 0.975 (97.5%).

Common Trap (Chapter 5): Never treat "without replacement" as independent — P changes after each draw. Probability answers MUST lie in [0,1]; an answer like 1.3 means an error. For standard deviation, adding a constant to all data leaves σ UNCHANGED (only the mean shifts); multiplying scales σ by |k|. Always compare groups with BOTH a measure of centre (mean/median) and spread (IQR/σ) — one alone is insufficient for HKEAA conclusion marks. Extra traps: (i) On a tree diagram the denominators change after the first draw (10→9), not stay at 10. (ii) For "exactly one red" you must ADD the two distinct orders (R then B, and B then R); using only one order halves the answer. (iii) The empirical 68–95–99.7 rule gives TAIL probabilities: P(X > μ+2σ) = 2.5%, so P(X < μ+2σ) = 97.5%, not 95%. (iv) A standard score z = (x−mean)/σ is unit-free; comparing students across subjects uses z-scores, not raw marks. (v) "At least one" is usually faster via 1 − P(none).

Chapter 6 Coordinate Geometry & Trigonometry ★★

Distance = √[(x₁−x₂)² + (y₁−y₂)²]        Mid-point = ((x₁+x₂)/2, (y₁+y₂)/2)
Slope m = (y₁−y₂)/(x₁−x₂)                Perpendicular: m₁ × m₂ = −1
Circle x² + y² + Dx + Ey + F = 0:  centre (−D/2, −E/2), radius √(D²/4 + E²/4 − F)
Sine rule:  a/sinA = b/sinB = c/sinC
Cosine rule:  c² = a² + b² − 2ab·cosC
Area of triangle = ½·ab·sinC

Core Mechanism (How to use)

Coordinate geometry turns shapes into algebra: distance and slope tell you lengths and parallelism, and m₁×m₂ = −1 is the test for perpendicular lines (right angles). In trigonometry, the cosine rule handles non-right triangles when you know two sides and the included angle, while the sine rule works when you know a side and its opposite angle. Always draw and label the triangle with known sides/angles first. (白話：垂直看斜率乘積是否 −1；非直角三角形看已知什麼再用正弦/餘弦定律。)

Worked Example 6.1 — Circle equation

Worked Example 6.1 — Circle equation                        [Total 4 marks]
  Find the centre and radius of
  x² + y² − 6x + 4y − 12 = 0.
  General form: x² + y² + Dx + Ey + F = 0
  Here D = −6, E = 4, F = −12
(1) Centre = (−D/2, −E/2) = (3, −2)                          (1 mark)
(2) Radius² = D²/4 + E²/4 − F
             = 36/4 + 16/4 − (−12)                            (1 mark)
(3) = 9 + 4 + 12 = 25                                        (1 mark)
(4) Radius = √25 = 5                                          (1 mark)
    ∴ Centre (3,−2), radius 5.

Worked Example 6.2 — Cosine rule

Worked Example 6.2 — Cosine rule                          [Total 4 marks]
  Triangle ABC: AB = 7, AC = 5, ∠A = 60°.  Find BC.
  Let a = BC, b = AC = 5, c = AB = 7, angle A = 60°
(1) a² = b² + c² − 2bc·cosA                                  (1 mark)
(2) = 5² + 7² − 2(5)(7)cos60°                                (1 mark)
(3) = 25 + 49 − 70(0.5) = 74 − 35 = 39                       (1 mark)
(4) a = √39 ≈ 6.245  ⇒  BC ≈ 6.24 (3 sig. fig.)              (1 mark)

Worked Example 6.3 — Sine rule (4 marks)

Worked Example 6.3 — Sine rule                        [Total 4 marks]
  In △ABC,  a = 7,  b = 10,  ∠A = 30°.  Find ∠B.
(1) By the sine rule:  a / sinA = b / sinB              (1 mark)
(2) Substitute:  7 / sin30° = 10 / sinB                (1 mark)
(3) sinB = 10·sin30° / 7 = 10(0.5)/7 = 5/7 ≈ 0.7143   (1 mark)
(4) B = sin⁻¹(0.7143) ≈ 45.6°
      (acute solution; the obtuse 180° − 45.6° = 134.4°
       is impossible here since 30° + 134.4° > 180°)
    ∴ ∠B ≈ 45.6° (1 decimal place).                      (1 mark)

Worked Example 6.4 — Distance, midpoint & perpendicular bisector (5 marks)

Worked Example 6.4 — Coordinate geometry             [Total 5 marks]
  A(1, 2) and B(5, 8).
  (a) Find the distance AB.
  (b) Find the midpoint M of AB.
  (c) Find the equation of the perpendicular bisector of AB.
(1) (a) AB = √[(5−1)² + (8−2)²]
        = √[4² + 6²] = √[16 + 36] = √52 ≈ 7.21        (1 mark)
(2) (b) M = ((1+5)/2, (2+8)/2) = (3, 5)               (1 mark)
(3) (c) Slope of AB: m_AB = (8−2)/(5−1) = 6/4 = 1.5   (1 mark)
(4) Perpendicular slope m⊥ = −1 / 1.5 = −2/3          (1 mark)
(5) Bisector passes through M(3,5) with slope −2/3:
      y − 5 = (−2/3)(x − 3)                            (1 mark)
    ∴ Equation:  y = (−2/3)x + 7   (or 2x + 3y = 21).
    Check: midpoint (3,5) satisfies 2(3)+3(5)=21 ✓

Common Trap (Chapter 6): For perpendicular lines remember m₁×m₂ = −1, NOT m₁ = m₂ (that is for parallel). When squaring in the distance/circle formula, sign errors in the centre (−D/2) are common — keep the negative. In trig, the sine rule can give an ambiguous second angle (acute vs obtuse); always check which fits the triangle. Give angles to 1 decimal place and lengths to 3 sig. fig. Extra traps: (i) The circle centre is (−D/2, −E/2); with D = −6 you get −(−6)/2 = +3, do not drop the outer negative. (ii) Perpendicular slope is the NEGATIVE reciprocal −1/m, not just the reciprocal 1/m. (iii) For the ambiguous sine-rule case, always test whether the obtuse supplement keeps the angle sum under 180°; if ∠A + ∠B_obtuse > 180° the obtuse solution is invalid (as in 6.3). (iv) Midpoint averages coordinates; students sometimes subtract. (v) Cosine rule c² = a²+b²−2ab·cosC requires the angle C to be the one BETWEEN sides a and b.

Business-oriented Special Module（弥补无 M1/M2 短板）★★★

1. Compound Interest & Growth

Compound amount:   A = P(1 + r/n)^(nt)
Depreciation:      A = P(1 − r)^t
Real value = nominal value ÷ (1 + inflation rate)^t

Worked example：Deposit $50,000 at 4% p.a. compounded quarterly for 3 years: A = 50000 × (1 + 0.04/4)^12 = 50000 × (1.01)^12 ≈ $56,341

2. Percentage Change Chain（对接 Economics 弹性计算）

• Overall change = (1 + a%)(1 + b%) − 1，不可直接相加
• GDP growth rate、CPI 通胀率均为此模型

3. Statistical Graph Interpretation in English

Standard sentences for describing graphs:

• The bar chart shows that X accounted for the largest share (35%) of...
• There was a steady upward trend in ... from 2020 to 2024, rising by 12%.
• The median of Group A is higher than that of Group B, while Group A also shows a larger spread (IQR).

4. 歷屆真題演練（Compound Interest 商科高頻題）

題目（改編自卷二 MC / 卷一基礎題型）： A bank offers a savings plan: deposit $P at an annual interest rate of 3.6%, compounded monthly. After 5 years, the amount grows to $120,000. Find the initial deposit $P, correct to the nearest dollar.

評分步驟（每步 1 分）：

1. 列公式：A = P(1 + r/n)^(nt)
2. 代入：120000 = P(1 + 0.036/12)^(12×5) = P(1.003)^60
3. 計算因子：(1.003)^60 ≈ 1.1964
4. P = 120000 ÷ 1.1964 ≈ $100,301（最接近整數）
5. 檢查：答案須小於 120000（因含利息），若得出大於本金即符號錯。

常見失分：① r 與 n 單位不一致（年利率 3.6% 誤寫 0.36）；② 忘記對最終答案四捨五入到指定精確度；③ 用 simple interest 公式 A=P(1+rt) 處理「compounded」題。

5. Linear Programming — Profit Maximisation（商科應用長題）

題目（改編自卷一 Section B 利潤最大化題型）： A bakery makes cakes (x) and breads (y) daily. Each cake gives a profit of $30, each bread $20. The preparation time and oven time constraints (in hours) are:

• Preparation: 2x + y ≤ 40
• Oven: x + 3y ≤ 45
• Non-negativity: x ≥ 0, y ≥ 0

(a) Find the vertices of the feasible region. (2 marks) (b) Maximise the daily profit P = 30x + 20y, and state how many cakes and breads should be made. (4 marks)

評分步驟（完整給分）：

Question 5 — Linear programming (business)              [Total 6 marks]
(a) Find vertices:
(1) Intercepts and intersections of the two lines:        (2 marks)
      From 2x + y = 40 and x + 3y = 45:
      Solve: y = 40 − 2x, substitute ⇒ x + 3(40 − 2x) = 45
             x + 120 − 6x = 45  ⇒  −5x = −75  ⇒  x = 15
             y = 40 − 30 = 10
      Vertices: A(0,0), B(20,0), C(15,10), D(0,15)
(b) Maximise P = 30x + 20y:
(2) Test each vertex:                                     (3 marks)
      A: 30(0) + 20(0) = 0
      B: 30(20) + 20(0) = 600
      C: 30(15) + 20(10) = 450 + 200 = 650
      D: 30(0) + 20(15) = 300
(3) Largest value = 650 at C(15,10)                      (1 mark)
    ∴ Maximum daily profit = $650 by making 15 cakes
      and 10 breads.

Business note: In real DSE Section B, you may also be asked to "suggest a reason" for a changed constraint (e.g. oven breaks down) — just re-test the new vertices. The method mark is awarded for testing ALL vertices, not for the final number.

6. Statistics — Comparing Two Sales Groups（mean / median / IQR / standard score）

題目（改編自卷一統計長題）： Two shops report their daily sales (in $'000) over 7 days:

• Shop A: 12, 15, 14, 13, 16, 90, 15
• Shop B: 20, 22, 21, 19, 23, 20, 21

(a) Calculate the mean and the median for each shop. (4 marks) (b) Explain, using mean and median, which shop's data is skewed and why. (2 marks) (c) A day with sales $90 in Shop A is a promotion day. Find its standard score if Shop A's standard deviation is 27.1 (in $'000). (2 marks)

評分步驟（完整給分）：

Question 6 — Comparing two groups (statistics)         [Total 8 marks]
(a) Means and medians:
(1) Shop A sum = 12+15+14+13+16+90+15 = 175
    Mean_A = 175/7 = 25.0  ($'000)                          (1 mark)
(2) Shop A ordered: 12,13,14,15,15,16,90
    Median_A = 15 (4th value)                               (1 mark)
(3) Shop B sum = 20+22+21+19+23+20+21 = 146
    Mean_B = 146/7 ≈ 20.857                                (1 mark)
(4) Shop B ordered: 19,20,20,21,21,22,23
    Median_B = 21 (4th value)                              (1 mark)
(b) Skewness:
(5) For Shop A, mean (25.0) > median (15), so the data is
    positively (right) skewed by the outlier $90.           (2 marks)
    Shop B: mean ≈ 20.9 ≈ median 21, roughly symmetric.
(c) Standard score:
(6) z = (x − mean)/σ = (90 − 25.0)/27.1 = 65/27.1 ≈ 2.40   (2 marks)
    ∴ The promotion day is about 2.4 standard deviations
      above Shop A's mean.

Business note: HKEAA wants you to compare BOTH centre and spread. A lone mean is never enough — pair it with median (for skewness) or IQR/σ (for spread). The outlier $90 in Shop A inflates the mean but not the median; that gap is the key conclusion.

更多商科統計與概率實戰題見 練習題庫 Math。

7. Compound Interest — Advanced Variations（商科高頻進階）★★★

Beyond the basic A = P(1 + r/n)^(nt), business papers frequently test effective annual rate, comparison of plans, and nominal-vs-effective reasoning.

Worked Example 7.1 — Effective annual rate (EAR) (4 marks)

Worked Example 7.1 — Effective annual rate           [Total 4 marks]
  A bank advertises 6% p.a. compounded monthly.
  Find the effective annual rate, correct to 2 d.p.
(1) Monthly rate = 0.06 / 12 = 0.005                         (1 mark)
(2) Growth factor for one year = (1.005)¹²                   (1 mark)
(3) (1.005)¹² ≈ 1.0616778                                    (1 mark)
(4) EAR = 1.0616778 − 1 = 0.0616778
      ≈ 6.17% (2 d.p.)                                       (1 mark)
    ∴ Effective annual rate ≈ 6.17%.
    (Higher than the nominal 6% because of monthly compounding.)

Worked Example 7.2 — Comparing two investment plans (6 marks)

Worked Example 7.2 — Compare two plans              [Total 6 marks]
  Plan A: 5% p.a. compounded annually for 8 years.
  Plan B: 4.8% p.a. compounded monthly for 8 years.
  Which gives more on a $100,000 deposit?
(1) Plan A: A = 100000(1 + 0.05)⁸                           (1 mark)
      (1.05)⁸ ≈ 1.477455 ⇒ A ≈ $147,746                     (1 mark)
(2) Plan B: monthly rate = 0.048/12 = 0.004                 (1 mark)
      A = 100000(1.004)⁹⁶
      (1.004)⁹⁶ ≈ 1.467999 ⇒ A ≈ $146,800                   (1 mark)
(3) Compare: 147746 > 146800                                (1 mark)
(4) Difference ≈ 147746 − 146800 = $946                    (1 mark)
    ∴ Plan A gives more by about $946, even though its
      nominal rate is lower — because annual compounding
      avoids the smaller monthly compounding advantage here.
    Check: recompute (1.004)⁹⁶ with more sig. fig. to confirm.

Worked Example 7.3 — Finding the time to double (5 marks)

Worked Example 7.3 — Doubling time                       [Total 5 marks]
  Money is invested at 8% p.a. compounded annually.
  Estimate how many years to double, then refine exactly.
  Use A = P(1.08)ⁿ, want A = 2P ⇒ (1.08)ⁿ = 2.
(1) Try n = 9: (1.08)⁹ ≈ 1.9990  (just under 2)            (1 mark)
(2) Try n = 10: (1.08)¹⁰ ≈ 2.1589 (over 2)                 (1 mark)
(3) So doubling occurs between 9 and 10 years.              (1 mark)
(4) Refine with logs:  n·log(1.08) = log 2
      n = log2 / log1.08 ≈ 0.30103 / 0.03342 ≈ 9.006       (1 mark)
(5) ∴ About 9.01 years (≈ 9 years to the nearest year).
    Check: 1.08⁹·⁰⁰⁶ ≈ 2.000 ✓                              (1 mark)
    (Rule of 72 estimate: 72/8 = 9 years — a quick check.)

8. Percentage Change Chains & Reverse Percentage ★★★

Successive percentage changes MUST be multiplied as factors, never added. A 10% increase then a 10% decrease does NOT return to the start.

Worked Example 8.1 — Two successive changes (4 marks)

Worked Example 8.1 — Successive % changes             [Total 4 marks]
  A price rises 15% then falls 10%. Find the overall
  percentage change.
(1) Factor after +15%: 1 + 0.15 = 1.15                  (1 mark)
(2) Factor after −10%: 1 − 0.10 = 0.90                  (1 mark)
(3) Overall factor = 1.15 × 0.90 = 1.035               (1 mark)
(4) Overall change = 1.035 − 1 = 0.035 = +3.5%
    ∴ Net increase of 3.5% (NOT +5%).                    (1 mark)

Worked Example 8.2 — Reverse percentage (discount) (4 marks)

Worked Example 8.2 — Reverse percentage               [Total 4 marks]
  After a 20% discount, a laptop costs $6,400.
  Find the original price.
(1) Discounted price = 80% of original = 0.80 × P       (1 mark)
(2) 0.80P = 6400                                        (1 mark)
(3) P = 6400 / 0.80 = 8000                              (1 mark)
(4) ∴ Original price = $8,000.
    Check: 20% of 8000 = 1600; 8000 − 1600 = 6400 ✓      (1 mark)

Worked Example 8.3 — Markup then discount chain (5 marks)

Worked Example 8.3 — Markup + discount chain          [Total 5 marks]
  A shop buys goods for $C, marks up by 40%, then
  offers a 25% sale discount. Express the final selling
  price as a percentage of C, and state profit/loss.
(1) After 40% markup:  price = C × 1.40                (1 mark)
(2) After 25% discount: price = 1.40C × (1 − 0.25)
      = 1.40C × 0.75 = 1.05C                            (2 marks)
(3) Final price = 105% of C                             (1 mark)
(4) Profit = 1.05C − C = 0.05C = 5% profit             (1 mark)
    ∴ Despite a 25% discount, the shop still makes 5%
      profit because the 40% markup dominated the chain.
    Check: C = 100 ⇒ 140 ⇒ ×0.75 = 105 ✓

9. Inflation, Real Value & Depreciation Comparison ★★★

Nominal money loses purchasing power over time. The real value adjusts a future amount back to today's prices.

Worked Example 9.1 — Real value under inflation (5 marks)

Worked Example 9.1 — Real value                     [Total 5 marks]
  You will receive $50,000 in 4 years. Inflation runs
  at 3% p.a. Find the present (real) value of this sum.
(1) Real value = nominal ÷ (1 + i)ᵗ                        (1 mark)
      = 50000 ÷ (1.03)⁴                                   (1 mark)
(2) (1.03)⁴ = 1.03² × 1.03² = 1.0609 × 1.0609 ≈ 1.1255    (1 mark)
(3) Real value ≈ 50000 / 1.1255 ≈ 44424                   (1 mark)
(4) ∴ About $44,424 in today's purchasing power.
    The $50,000 nominal has lost ≈ $5,576 of real value.   (1 mark)
    Check: 44424 × (1.03)⁴ ≈ 44424 × 1.1255 ≈ 50000 ✓

Worked Example 9.2 — Depreciation vs inflation (6 marks)

Worked Example 9.2 — Asset vs money value           [Total 6 marks]
  A machine bought for $100,000 depreciates 10% p.a.
  Inflation is 4% p.a. Find, after 3 years:
  (a) the nominal resale value of the machine;
  (b) its real value in today's dollars.
(1) (a) V = 100000 × (1 − 0.10)³ = 100000 × 0.9³          (1 mark)
      0.9³ = 0.729 ⇒ V = $72,900                           (1 mark)
(2) (b) Real value = 72900 ÷ (1.04)³                      (1 mark)
      (1.04)³ ≈ 1.124864                                  (1 mark)
      Real ≈ 72900 / 1.124864 ≈ 64785                     (1 mark)
(3) So in today's purchasing power the machine is worth
      about $64,785, far below its $100,000 original cost. (1 mark)
    ∴ Nominal resale $72,900; real value ≈ $64,785.
    Check: depreciation and inflation BOTH reduce real worth.

Worked Example 9.3 — Simple vs compound interest (5 marks)

Worked Example 9.3 — Simple vs compound             [Total 5 marks]
  $10,000 at 5% p.a. for 3 years. Compare simple and
  compound (annual) interest totals.
(1) Simple:  A = P(1 + rt) = 10000(1 + 0.05×3)            (1 mark)
      = 10000 × 1.15 = $11,500                            (1 mark)
(2) Compound: A = 10000(1.05)³                            (1 mark)
      (1.05)³ = 1.157625 ⇒ A = $11,576.25                 (1 mark)
(3) Difference = 11576.25 − 11500 = $76.25                (1 mark)
    ∴ Compound yields $76.25 more; the gap widens with
      time and rate. Never use simple-interest formula
      when the word "compounded" appears.

Business note (Sections 7–9): The single most common business-module error is mixing up nominal and effective rates, or adding successive percentages. Always convert every percentage change to a MULTIPLICATIVE factor (1 ± rate), multiply the factors, then subtract 1 for the net change. For inflation/real-value questions, divide by (1+i)ᵗ; for depreciation, multiply by (1−rate)ᵗ. Keep a "Rule of 72" sanity check (doubling time ≈ 72/rate%) for compound-growth answers. 更多商科金融數學實戰題見 練習題庫 Math。

Marking Scheme Insights（評卷官視角）★★★

HKEAA Long-Question Marking Iron Rules

1. State a reason for every step — write "By the cosine rule...", "By sum of G.S....", "Since Δ < 0...". A correct number with no stated reason often earns 0 method marks.
2. Conclusion sentence is compulsory — end with "∴ The claim is agreed/disagreed because...". Missing the concluding sentence loses the final mark even if the number is right.
3. Intermediate values: keep ≥ 4 significant figures; round ONLY the final answer to 3 sig. fig. (unless a specific precision like "correct to the nearest dollar" is stated).
4. Units are mandatory in every final answer ($ , cm², km/h, years).
5. A "show" / "hence" sub-step cannot be skipped — even if you know the final answer, you must display the requested intermediate derivation or lose that method mark.

Comparison 1 — Discriminant (student vs full mark)

Question: Show that x² + 2x + 5 = 0 has no real roots.   [3 marks]

--- Student answer (earns 1/3) ---
x² + 2x + 5 = 0
x = −1 ± √(−4)          ← writes complex root, no statement
(no conclusion sentence)

--- Full-mark answer (earns 3/3) ---
a = 1, b = 2, c = 5
Δ = b² − 4ac = 2² − 4(1)(5) = 4 − 20 = −16      (2 marks: formula + value)
Since Δ < 0, the equation has no real roots.       (1 mark: conclusion)

Comparison 2 — Probability (student vs full mark)

Question: From 3 red and 2 blue balls, draw 2 without replacement.
Find P(both red).                                    [4 marks]

--- Student answer (earns 1/4) ---
P = 3/5 × 3/5 = 9/25        ← treated as WITH replacement (independent)
(no combination working, no check)

--- Full-mark answer (earns 4/4) ---
Total ways ⁵C₂ = 10                                  (1 mark)
Favourable ³C₂ = 3                                   (1 mark)
P = 3/10 = 0.3                                       (1 mark)
Check 0 < 0.3 < 1 ✓                                  (1 mark)
∴ P(both red) = 3/10.

Takeaway: The difference is never the final number alone — it is the DISPLAYED method (formula → substitution → value → check → conclusion). Train yourself to write all five parts every time; that is what separates Level 5 from Level 3.

Past-Paper Practice (MC + Structured, full solutions) ★★★

Each question below includes a COMPLETE solution with reasoning, not just the answer. Use these to rehearse the HKEAA "formula → substitution → value → check → conclusion" rhythm. For more, see 練習題庫 Math.

A. Multiple-Choice (Paper 2 style) — full reasoning

MC Q1. Simplify (2x³y²)² ÷ (4x²y).

• A. x⁴y³ / 2
• B. x⁴y / 2
• C. 4x⁴y³
• D. x⁶y⁴ / 2

Solution:
(2x³y²)² = 4x⁶y⁴
Divide by (4x²y):  (4x⁶y⁴) / (4x²y) = x⁶⁻² · y⁴⁻¹ = x⁴y³
So result = x⁴y³, but we must keep the coefficient: 4/4 = 1.
Wait — re-read: (2x³y²)² = 4x⁶y⁴; ÷(4x²y) ⇒ (4/4)·x⁴·y³ = x⁴y³.
None of A–D show coefficient 1 except... A is x⁴y³/2 (wrong coeff),
B is x⁴y/2, C is 4x⁴y³, D is x⁶y⁴/2.
Correct simplified form is x⁴y³, which equals (x⁴y³) — closest matching
the structure of A but with denominator 1. Recheck: actually the intended
answer is x⁴y³, i.e. option A misprinted; the proper choice is x⁴y³.
Answer: x⁴y³  (if forced among options, the factor form is x⁴y³).

Note: This illustrates why you must simplify fully before matching; many MC traps hide a coefficient of 1.

MC Q2. If f(x) = x² − 6x + 5, the minimum value of f(x) is

• A. −4
• B. 5
• C. −5
• D. 0

Solution:
Vertex at x = −b/(2a) = 6/2 = 3.
f(3) = 3² − 6(3) + 5 = 9 − 18 + 5 = −4.
Since a = 1 > 0, parabola opens up ⇒ minimum = −4.
Answer: A.

MC Q3. Solve 3^(2x) = 9^(x−1).

• A. x = 1
• B. x = 2
• C. x = −1
• D. no real solution

Solution:
9 = 3², so RHS = (3²)^(x−1) = 3^(2x−2).
Equal bases ⇒ 2x = 2x − 2 ⇒ 0 = −2, contradiction.
So no real solution.
Answer: D.

MC Q4. A die is rolled twice. P(total = 7) is

• A. 1/6
• B. 1/12
• C. 5/36
• D. 1/9

Solution:
Total outcomes = 6 × 6 = 36.
Ways to make 7: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) = 6 ways.
P = 6/36 = 1/6.
Answer: A.

MC Q5. In △ABC, a = 8, b = 6, ∠C = 60°. Area is

• A. 12√3
• B. 24
• C. 12
• D. 24√3

Solution:
Area = ½·a·b·sinC = ½·8·6·sin60° = 24·(√3/2) = 12√3.
Answer: A.

B. Structured Questions (Paper 1 style) — full solutions

Structured Q1 — Quadratic & discriminant (7 marks). The equation x² + (k − 2)x + (k + 1) = 0 has equal roots. (a) Find the value(s) of k. (4 marks) (b) Hence solve the equation. (3 marks)

Structured Q1 — full solution                      [Total 7 marks]
(a) Equal roots ⇒ Δ = 0.
    a = 1, b = k − 2, c = k + 1
(1) Δ = (k − 2)² − 4(1)(k + 1) = 0                     (1 mark)
(2) Expand: k² − 4k + 4 − 4k − 4 = k² − 8k = 0        (1 mark)
(3) k(k − 8) = 0  ⇒  k = 0  or  k = 8                 (1 mark)
(4) Both give Δ = 0, so two valid values.              (1 mark)
(b) Solve:
(5) If k = 0: x² − 2x + 1 = 0 ⇒ (x − 1)² = 0 ⇒ x = 1  (1.5 marks)
(6) If k = 8: x² + 6x + 9 = 0 ⇒ (x + 3)² = 0 ⇒ x = −3 (1.5 marks)
    ∴ k = 0 gives x = 1 (double root);
      k = 8 gives x = −3 (double root).

Structured Q2 — Statistics comparison (8 marks). The monthly profits (in $'000) of two branches are:

• Branch X: 30, 32, 31, 29, 33, 31, 30
• Branch Y: 20, 45, 22, 21, 23, 19, 50

(a) Find the mean and median of each branch. (4 marks) (b) State, with a reason, which branch is more consistent (stable). (2 marks) (c) The value 50 in Branch Y is an error; if removed, would the mean of Y increase or decrease? Explain. (2 marks)

Structured Q2 — full solution                      [Total 8 marks]
(a)
(1) X sum = 30+32+31+29+33+31+30 = 216
    Mean_X = 216/7 ≈ 30.857 ('000)                       (1 mark)
(2) X ordered: 29,30,30,31,31,32,33
    Median_X = 31 (4th)                                  (1 mark)
(3) Y sum = 20+45+22+21+23+19+50 = 200
    Mean_Y = 200/7 ≈ 28.571 ('000)                       (1 mark)
(4) Y ordered: 19,20,21,22,23,45,50
    Median_Y = 22 (4th)                                  (1 mark)
(b)
(5) Branch X is more consistent: its values (29–33) are
    tightly clustered, while Y has extreme values 45 and 50
    far from its centre. X's mean ≈ median (symmetric-ish),
    Y's mean > median (right-skewed).                    (2 marks)
(c)
(6) The value 50 is ABOVE Y's mean (28.57), so removing a
    value greater than the mean LOWERS the mean.          (1 mark)
(7) New Y sum = 200 − 50 = 150 over 6 values ⇒ 25,
    which is less than 28.57 ⇒ mean decreases.            (1 mark)
    ∴ Mean of Y decreases after removing 50.

Structured Q3 — Compound interest & graph description (8 marks). A company invests $200,000 at 5% p.a. compounded quarterly. (a) Find the amount after 4 years, to the nearest dollar. (4 marks) (b) Write one English sentence describing the growth of the investment over time. (2 marks) (c) If the inflation rate is 2% p.a., find the real value of the final amount in today's dollars. (2 marks)

Structured Q3 — full solution                      [Total 8 marks]
(a)
(1) A = P(1 + r/n)^(nt), P = 200000, r = 0.05,
    n = 4, t = 4                                          (1 mark)
(2) A = 200000(1 + 0.05/4)^(16) = 200000(1.0125)^16      (1 mark)
(3) (1.0125)^16 ≈ 1.2198895                              (1 mark)
(4) A ≈ 200000 × 1.2198895 ≈ 243977.9
    ≈ $243,978 (nearest dollar)                           (1 mark)
(b)
(5) "The investment grows exponentially, reaching about
    $243,978 after four years, which is an increase of
    roughly 22% above the initial $200,000."              (2 marks)
(c)
(6) Real value = 243978 / (1.02)⁴                        (1 mark)
    (1.02)⁴ ≈ 1.082432 ⇒ real ≈ 243978/1.082432 ≈ 225254
    ≈ $225,254 (today's purchasing power)                 (1 mark)
    ∴ Real value ≈ $225,254.

更多歷屆真題與模擬題見 練習題庫 Math。

Comparison 3 — Quadratic vertex (student vs full mark)

Question: The profit is P = −x² + 10x − 16. Find the max profit.  [4 marks]

--- Student answer (earns 1/4) ---
P max when x = 5, answer = 9.
(only the number, no axis or substitution shown)

--- Full-mark answer (earns 4/4) ---
a = −1, b = 10
Axis of symmetry: x = −b/(2a) = −10/(−2) = 5        (1 mark)
P(5) = −25 + 50 − 16 = 9                            (2 marks)
Since a = −1 < 0, parabola opens down ⇒ maximum      (1 mark)
∴ Maximum profit = 9 (units) when x = 5.

Comparison 4 — Compound interest (student vs full mark)

Question: $1,000 at 8% p.a. compounded half-yearly for 2 yrs.
Find the amount.                                            [4 marks]

--- Student answer (earns 1/4) ---
A = 1000(1.08)^2 = 1166.4      ← used annual rate, ignored half-yearly

--- Full-mark answer (earns 4/4) ---
Half-yearly rate = 0.08/2 = 0.04; periods = 2×2 = 4    (1 mark)
A = 1000(1.04)^4                                     (1 mark)
(1.04)^4 ≈ 1.1698586                                 (1 mark)
A ≈ $1,169.86 (nearest cent)                         (1 mark)
∴ Amount ≈ $1,169.86.

Takeaway: The difference is never the final number alone — it is the DISPLAYED method (formula → substitution → value → check → conclusion). Train yourself to write all five parts every time; that is what separates Level 5 from Level 3.

English Writing Standard for Long Questions

1. Every step must state the reason: By the cosine rule, ... / By similar triangles (AAA), ...
2. Conclusion sentence is compulsory: ∴ The claim is agreed/disagreed because...
3. Keep intermediate values in 4+ significant figures; round only the final answer to 3 sig. fig. unless specified
4. Write units in every final answer ($, cm², km/h)

Careless Error Checklist（考前默读）

• [ ] 负号：移项、展开括号后逐项检查
• [ ] 单位：cm ↔ m、% ↔ 小数换算
• [ ] MC 题：算完后代回验算 / 用排除法双保险
• [ ] Probability 答案是否在 0–1 之间；总和是否合理
• [ ] 图形题：答案与图的直观大小是否吻合

M1 商科核心題型精練（延伸單元選 M1 必看）★★★

商科生應選 M1（微積分與統計）而非 M2（代數與微積分）。以下為 M1 在商科應用中最常出、最易失分的兩類題型，務必練熟。

1. 假設檢定 P 值解讀（統計）

關鍵觀念：P 值是在「原假設成立」前提下，觀察到目前樣本（或更極端結果）的機率。

• 若 P = 0.03 < 0.05（顯著水平 α 一般取 0.05）：拒絕原假設——代表樣本提供足夠證據，顯示與原假設不符。
• ⚠️ 常見錯誤：寫「P < 0.05 證明原假設不成立」。正確說法是「有顯著證據拒絕原假設」；統計上從不「證明」某事成立，只是否定或不否定。
• 若 P = 0.20 > 0.05：未能拒絕原假設（樣本證據不足，而非「證明原假設成立」）。

Worked Example（4 分） 某品牌聲稱顧客滿意度 ≥ 80%。隨機抽查 100 人，滿意者 72 人。在 5% 顯著水平下檢定。

步驟：
H₀: p = 0.80（滿意度達標）          H₁: p < 0.80
樣本比例 p̂ = 72 / 100 = 0.72
檢定統計量 z = (0.72 − 0.80) / √(0.80 × 0.20 ÷ 100)
            = −0.08 / 0.04 = −2.0
P 值 = P(Z < −2.0) ≈ 0.0228 < 0.05
∴ 拒絕 H₀，樣本提供足夠證據顯示滿意度低於 80%。（結論須配 P 值與 α 的比較）

評分重點：結論必須同時寫「拒絕／未能拒絕 H₀」＋「因 P 值與 α 比較」＋「用題目語境解釋」，三者缺一不可。

2. 邊際分析（微積分商科應用）

觀念：利潤 π(q) = 收益 R(q) − 成本 C(q)。最優產量發生在邊際利潤 = 0，即 dπ/dq = 0，等價於 dR/dq = dC/dq（邊際收益 = 邊際成本）。

• 二階條件：d²π/dq² < 0 才為最大利潤（確認是極大而非極小）。

Worked Example（5 分） 設成本 C(q) = q³ − 12q² + 60q + 100，收益 R(q) = 70q（每單位售 $70）。

(1) 利潤 π(q) = R − C = 70q − (q³ − 12q² + 60q + 100)
             = −q³ + 12q² + 10q − 100              (1 分)
(2) dπ/dq = −3q² + 24q + 10                        (1 分)
(3) 令 dπ/dq = 0：−3q² + 24q + 10 = 0
    ⇒ 3q² − 24q − 10 = 0                           (1 分)
    q = [24 ± √(576 + 120)] / 6 = [24 ± √696] / 6 ≈ [24 ± 26.38] / 6
    正根 q ≈ 50.38 / 6 ≈ 8.40                       (1 分)
(4) d²π/dq² = −6q + 24；於 q ≈ 8.40，
    d²π/dq² ≈ −6(8.40) + 24 = −26.4 < 0 ⇒ 最大利潤  (1 分)
∴ 最優產量約 8.4 單位。

3. M2 商科通常不需深練的題型

以下題型多見於 M2（代數與微積分），商科申請一般不要求熟練，備考時間有限可優先取捨：

• 剛體轉動慣量（rotational inertia）、曲面面積／體積的參數積分
• 傅里葉級數（Fourier series）、複變函數進階應用

提醒：是否需讀 M2 取決於目標課程要求；商科主流（如 BBA、IBBA）明確傾向 M1。實際以該課程當年入學要求為準。

商科配套計算專題（無 M1/M2 必補）★★★

經濟計算題省略文字說明直接扣步驟分；每步必寫英文一句解釋。下列五個題型皆不屬 M1/M2 範圍，而是 Compulsory Part 數值運算 + 經濟詞彙的結合，商科生必須獨立掌握（無 M1/M2 補足）。通用寫作順序：列公式 → 代入數值 → 計算 → 標單位 → 英文總結句。更多實戰題見 練習題庫 Math。

1. 需求彈性 PED (Price Elasticity of Demand)

Formula（公式）

PED = %ΔQd ÷ %ΔP
判斷： |PED| > 1 → elastic（需求對價格敏感）
       |PED| < 1 → inelastic（需求對價格不敏感）
       |PED| = 1 → unitary elastic（單一彈性）

Worked Example 1 — PED（完整 DSE 英文步驟）

Worked Example — Price Elasticity of Demand         [Total 5 marks]
  Price rises from $10 to $12, and quantity demanded
  falls from 100 to 80. Find PED and state the type.
(1) % change in price = (12 − 10) / 10 × 100% = 20%  (1 mark)
(2) % change in quantity demanded = (80 − 100) / 100
      × 100% = −20%                                   (1 mark)
(3) PED = %ΔQd ÷ %ΔP = −20% ÷ 20% = −1.0             (1 mark)
(4) |PED| = |−1.0| = 1.0, so demand is unitary elastic(1 mark)
(5) Therefore the price elasticity of demand is −1.0
      (|PED| = 1.0), meaning demand is unitary elastic.
                                                    (1 mark)

扣分雷區：① 單位遺漏直接扣步驟分（%ΔP 必須寫 20% 而非 20）。② 符號書寫錯誤：%ΔQd = −20% 的負號不可漏，漏負號會被判斷 elastic 類型錯誤；「÷」建議用標準符號，勿以 / 取代。③ 結尾缺少英文總結句 "Therefore the price elasticity of demand is ..." 直接扣結論分。④ 彈性一律以 |PED| 大小判類型，但 PED 本身為負值，不可寫成 +1.0。

2. 通脹率 Inflation rate

Formula（公式）

Inflation rate = (CPIₜ − CPIₜ₋₁) ÷ CPIₜ₋₁ × 100%
（CPIₜ = 本年消費物價指數，CPIₜ₋₁ = 上一年）

Worked Example 2 — Inflation（完整 DSE 英文步驟）

Worked Example — Inflation rate from CPI            [Total 4 marks]
  CPI in year t−1 is 110, and in year t is 115.
  Find the inflation rate.
(1) Identify base and current CPI:
      CPIₜ₋₁ = 110,  CPIₜ = 115                     (1 mark)
(2) Change in CPI = 115 − 110 = 5                    (1 mark)
(3) Inflation rate = 5 ÷ 110 × 100% ≈ 4.545%        (1 mark)
(4) Therefore the inflation rate is approximately
      4.55%, correct to 3 significant figures.       (1 mark)

扣分雷區：① 單位遺漏直接扣步驟分（Inflation rate 必須標 %）。② 符號書寫錯誤：分母必須是「上一年」CPIₜ₋₁，寫成 CPIₜ 會計算錯誤；「÷」建議用標準符號。③ 結尾缺少英文總結句 "Therefore the inflation rate is ..." 扣結論分。④ 通脹率為約數百分比，勿寫成倍數（如「升了 1.045 倍」是錯讀）。

3. GDP 名義 / 實質

Formula（公式）

real GDP = nominal GDP ÷ GDP deflator × 100
real GDP growth rate = (real GDPₜ − real GDPₜ₋₁) ÷ real GDPₜ₋₁ × 100%
（base-year deflator = 100）

Worked Example 3 — Real GDP（完整 DSE 英文步驟）

Worked Example — Real GDP from deflator           [Total 5 marks]
  Nominal GDP = $500 billion, GDP deflator = 125
  (base year deflator = 100). Find real GDP and
  state its meaning.
(1) Use real GDP = nominal GDP ÷ deflator × 100    (1 mark)
(2) Real GDP = 500 ÷ 125 × 100 = 4 × 100 = 400     (2 marks)
(3) Real GDP = $400 billion in base-year prices     (1 mark)
(4) Therefore the real GDP is $400 billion, meaning
      output valued at base-year prices is below the
      nominal $500 billion (prices rose by 25%).     (1 mark)

扣分雷區：① 單位遺漏直接扣步驟分（real GDP 必須寫 $ billion）。② 符號書寫錯誤：real GDP = nominal ÷ deflator × 100，不可漏 ×100，否則數值小 100 倍；「÷」建議用標準符號。③ 結尾缺少英文總結句 "Therefore the real GDP is ..." 扣結論分。④ deflator > 100 表示物價上升，此時 real GDP < nominal GDP，切勿反向解讀為產出萎縮。

4. 匯率換算 Exchange rate

Formula（公式）

Linked Exchange Rate band: 1 USD = HKD 7.75 to 7.85
USD amount = HKD amount ÷ exchange rate
HKD amount = USD amount × exchange rate

Worked Example 4 — Exchange rate（完整 DSE 英文步驟）

Worked Example — HKD to USD conversion            [Total 4 marks]
  Convert HKD 7,850 to USD using the rate
  1 USD = HKD 7.85 (upper bound of the band).
(1) Identify the rate: 1 USD = HKD 7.85            (1 mark)
(2) USD = HKD amount ÷ rate
      = 7850 ÷ 7.85 = 1000                         (2 marks)
(3) Therefore HKD 7,850 is equal to USD 1,000 at
      the rate of 1 USD = HKD 7.85.                (1 mark)

扣分雷區：① 單位遺漏直接扣步驟分（必須寫 USD 1,000 而非 1000，且標明所用匯率）。② 符號書寫錯誤：HKD→USD 用「除」rate，USD→HKD 用「乘」rate，方向反了數值錯近 60 倍；「÷」建議用標準符號。③ 結尾缺少英文總結句 "Therefore HKD ... is equal to USD ..." 扣結論分。④ 港元聯繫匯率區間為 7.75–7.85，超出區間的 rate 在 DSE 經濟題中視為錯誤假設。

5. 複利 Compound interest (商科)

Formula（公式）

A = P(1 + r/n)^(nt)
P = principal（本金）, r = annual rate（年利率，小數）
n = compounding per year（每年複息次數）
t = years（年數）
註：複利 ≠ 單利，指數增長；
    單利公式為 A = P(1 + rt)。

Worked Example 5 — Compound interest（完整 DSE 英文步驟）

Worked Example — Compound interest (business)     [Total 5 marks]
  Deposit $20,000 at 6% p.a. compounded quarterly
  for 3 years. Find the final amount.
(1) Identify values: P = 20000, r = 0.06,
      n = 4, t = 3                                   (1 mark)
(2) Periods nt = 4 × 3 = 12; rate per period
      r/n = 0.06 / 4 = 0.015                        (1 mark)
(3) A = 20000(1 + 0.015)^12 = 20000(1.015)^12      (1 mark)
(4) (1.015)^12 ≈ 1.195618; A ≈ 20000 × 1.195618
      ≈ 23912.4                                      (1 mark)
(5) Therefore the final amount is approximately
      $23,912, correct to the nearest dollar.
      Note: compound interest grows exponentially,
      unlike simple interest A = P(1 + rt).         (1 mark)

扣分雷區：① 單位遺漏直接扣步驟分（final amount 必須寫 $）。② 符號書寫錯誤：r 必須轉為小數 0.06，不可寫 6；n 是每年複息次數，季度為 4；「÷」建議用標準符號。③ 結尾缺少英文總結句 "Therefore the final amount is ..." 扣結論分。④ 複利≠單利：若誤用單利 A = P(1 + rt) = 20000(1.18) = 23600，與複利結果不同；DSE 見 "compounded" 必用複利公式。

經濟計算題通用扣分守則（總則）

扣分雷區（總則）：經濟計算題省略文字說明直接扣步驟分；每步必寫英文一句解釋（如 "Percentage change in price = ..."）。單位遺漏直接扣步驟分；符號書寫錯誤（如把 ÷ 寫成 / 雖可接受但建議標準符號；− 與 – 需區分）會被判卷歧義；結尾必須有英文總結句 "Therefore the ... is ..."。公式代入順序固定為：先列公式 → 代入數值 → 計算 → 標單位 → 英文總結。更多商科計算實戰題見 練習題庫 Math。

商科計算統一書寫範式（五步滿分格式，所有題型嚴格複刻）★★★

以 PED 為例，需求價格彈性、收入彈性、通脹率、GDP 名義/實際換算、匯率換算、單利複利全部套用：

Question data: Price rises from $22 to $28, quantity demanded falls from 120 units to 66 units

Step 1  Write down the official formula
Price elasticity of demand = % change in quantity demanded / % change in price

Step 2  Calculate percentage change in quantity
%dQd = [(66 - 120) / 120] x 100% = -45%

Step 3  Calculate percentage change in price
%dP = [(28 - 22) / 22] x 100% = 27.27% (approx.)

Step 4  Calculate absolute value of PED
PED = | -45% / 27.27% | = 1.65 (approx.)

Step 5  Final written conclusion (mandatory for full marks)
Since PED > 1, the demand for this good is elastic. Price change brings a more
significant change in quantity demanded.

統一扣分紅線：只計算數值、省略公式、缺少英文總結句，整道題直接扣除 50% 步驟分——DSE 數學嚴格按分步採分。

每類計算題易錯坑點（寫在每個模板下方）★★★

PED 計算：

• ❌ 坑點 1：忘記取絕對值，PED 數值寫成負數失分；
• ❌ 坑點 2：百分比變化分母誤用期末數量，標準公式分母永遠是初始數值；
• ❌ 坑點 3：算出數值後不寫彈性判定語句，丟失步驟分。

通脹率：分母用基期物價指數，不是報告期；結論句必寫 "The inflation rate is x%, indicating a sustained increase in general price level."

匯率換算：分清直接標價/間接標價方向，換算後必須標貨幣單位。

複利：區分計息週期（annually / quarterly / monthly），n 為期數不是年數；最終金額保留題目要求位數。

數學整卷答題分步得分拆解表 ★★★

一道 6 分計算大題得分構成：

得分項	分值
公式書寫	2 分
代入數據計算	2 分
英文文字結論	2 分

絕大多數學生只計算數字，丟掉公式 + 結論 4 分——這是數學常年拿不到 5** 的核心原因。

全課本通用答題書寫規範 ★★★

1. 幾何圖像、坐標曲線必須標注英文符號說明；
2. 應用題計算結果必須配套一句英文文字解讀；
3. 嚴格遵照題目要求保留小數位數，私自四捨五入、保留位數錯誤一律扣分；
4. 代數推演不要跳步，閱卷按演算步驟逐個給分。

應用題英文描述萬能句式庫（L2 背誦）

1. The calculated value indicates that ...
2. The figure shows a positive / negative correlation between the two variables.
3. This numerical result matches the corresponding economic theoretical trend.

統計專項強化（商科數據分析核心考點）★★★

頻數分佈、標準差、折線統計圖文字描述句式，配套經濟數據解讀：

• 趨勢描述：The line graph shows a steady upward trend in ... from [year] to [year].
• 離散程度：A larger standard deviation indicates that the data values are more spread out from the mean, implying higher risk in business context.
• 對比結論：Dataset A has a higher mean but larger standard deviation than Dataset B, so A brings higher expected return with higher volatility.

Unit Test Prompt

You are an HKDSE Mathematics (Compulsory Part) marker. For the chapter I specify, design 6 questions: 3 MC + 2 short conventional + 1 long application question with business/economic context (compound interest, statistics or linear programming).
After I answer, check every calculation step, point out omitted steps that would lose marks under HKEAA marking, and identify any careless-error patterns.

Chapter: [e.g. Probability & Statistics]