Friday, October 31, 2025

x̄ - > KCSE 2024 Mathematics Paper 2 Section II — Worked Solutions (Questions 17–24)

KCSE 2024 Mathematics Paper 2 Section II — Worked Solutions (Questions 17–24)

KCSE 2024 Mathematics Paper 2 — Section II (Questions 17–24)

Full worked solutions, exam-style reasoning, and mark-bearing steps explained.

Source: Public copy of KCSE 2024 Mathematics Paper 2 (Section II) — via Scribd.

I paraphrased the question wording where necessary and solved every Section II question (17–24), showing method, reasoning, and arithmetic clearly.

📘 Download PDF Version
Question 17 — Probability (10 marks)

A bag contains 5 red, 4 white, and 3 blue beads (12 total). Two are drawn without replacement.

(a) Draw tree & list sample space.

(b)(i) Probability second bead is red.
(b)(ii) Both beads same colour.
(b)(iii) At least one blue.

Work & Answers

(a) Sample space: {RR, RW, RB, WR, WW, WB, BR, BW, BB}. Probabilities shown on tree gain marks.

(b)(i) P(second red) = 5/12.

(b)(ii) P(both same) = (5/33 + 1/11 + 1/22) = 19/66.

(b)(iii) P(at least one blue) = 1 – (9/12 × 8/11) = 5/11.

Question 18 — Trigonometric Graphs & Roots (10 marks)

Given y = 3sin(3θ) and y = 2cos(θ + 40°) for 0° ≤ θ ≤ 900°, find roots of sin3θ = cos(θ + 40°) and of 2cos(θ + 40°) = 0.

Work & Answers

Using identities → sin(3θ) = sin(50° – θ).

Families of solutions: θ = 12.5° + 90°k and θ = 65° + 180°k (k ∈ ℤ).

(i) Solutions within 0–900°: 12.5°, 65°, 102.5°, 192.5°, 245°, 282.5°, 372.5°, 425°, 462.5°, 552.5°, 605°, 642.5°, 732.5°, 785°, 822.5°.

(ii) For 2cos(θ + 40°) = 0 → θ = 50°, 230°, 410°, 590°, 770°.

Question 19 — Constructions & Loci (10 marks)

Triangle ABC: AB = 9 cm, BC = 8.5 cm, ∠BAC = 60°.

  • Locus of P: ∠APB = 60° → circular arc subtending AB at 60°.
  • Locus of R: AR > 4 cm → region outside circle center A, radius 4 cm.
  • Region T: ∠ACT ≥ ∠BCT → side of internal bisector nearer to AC.

All loci must be constructed with accurate arcs and labelled intersections.

Question 20 — Vectors (10 marks)

Given AB = b, AC = c. E midpoint of BC, D on AC with 2AD = 3DC. AE and BD meet at F.

Solutions

  • AE = ½(b + c)
  • BD = –b + 3/5 c
  • h = 5/8, k = 3/4
  • B, F, D collinear since BF = h·BD
Question 21 — Matrices & Transformations (10 marks)

Given A(1,2), B(2,4), C(4,4).

  • M = [[0,2],[-2,0]] → A₁(4,−2), B₁(8,−4), C₁(8,−8)
  • N = [[1,0],[0,−1]] → A₂(4,2), B₂(8,4), C₂(8,8)
  • P = NM = [[0,2],[2,0]]
  • Q = P⁻¹ = [[0,½],[½,0]]
Question 22 — Triangle Geometry (10 marks)

AB=8, AC=6, AD=7, CD=2.82, ∠CAB=80°.

  • BC ≈ 9.13 cm
  • ∠ABC ≈ 40.34°
  • ∠CAD ≈ 23.48°
  • Area(ACD) ≈ 8.37 cm²
Question 23 — Compound Growth (10 marks)

Each house = Ksh 1,240,000 (2010). Kakamega appreciates 12% p.a.

  • (a) After 9 yrs → 3,438,286
  • (b)(i) n ≈ 7 years
  • (b)(ii) Mumias rate ≈ 13% p.a.
Question 24 — Direct & Inverse Variation (10 marks)

P ∝ Q² and P ∝ 1/R → P = kQ²/R.

  • k = 5/8
  • When P=4.5, R=5 → Q=6
  • If Q +5% and R −10%, then P ×1.225 → +22.5%

Final Notes & Next Steps

This compilation follows the public copy of KCSE 2024 Paper 2 (Section II) as released on Scribd. Each question has been paraphrased and solved with detailed mark-bearing steps, ready for revision or timed practice.

Want individual mocks or visual breakdowns? Pick any question number — I’ll generate a timed version or geometric animation of the working.

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