KCSE Mathematics Paper 2 (121/2) sits on Monday 9 November 2026 at 8.00 a.m. and lasts 2 hours 30 minutes. It has the same shape as Paper 1: Section I has 16 short questions worth 50 marks, all compulsory, and Section II has 8 longer questions of 10 marks each, from which you choose any 5. Paper 2 leans harder on statistics, probability, matrices and transformations, loci, vectors, compound interest and trigonometry beyond right-angled triangles. Below are original practice questions in that style, each with full working, so you can check your method and not just the final number.
These are original CBC Guides questions in the KCSE format, not an official KNEC paper. For Paper 1 practice, go to KCSE Mathematics Paper 1.
Section I: Short Questions With Worked Answers
1. Compound interest (3 marks)
Wanjiru deposits KSh 80,000 in a SACCO account that pays 9% per annum compounded half-yearly. Find the amount after 2 years, to the nearest shilling.
Working: Half-yearly means the rate per period is 9 ÷ 2 = 4.5% and the number of periods is 2 × 2 = 4. A = 80,000 × (1.045)4. (1.045)2 = 1.092025, and squaring again gives 1.192519. A = 80,000 × 1.192519 = KSh 95,402. Marks are lost most often for using 9% with 2 periods; always convert both the rate and the time.
2. Relative error (3 marks)
A rectangle measures 6.4 cm by 3.5 cm, each measured to the nearest millimetre. Find the percentage error in its area, to 2 decimal places.
Working: The absolute error in each length is 0.05 cm. Maximum area = 6.45 × 3.55 = 22.8975; minimum area = 6.35 × 3.45 = 21.9075. Absolute error in area = (22.8975 − 21.9075) ÷ 2 = 0.495. Actual area = 6.4 × 3.5 = 22.4. Percentage error = 0.495 ÷ 22.4 × 100 = 2.21%.
3. Matrices (3 marks)
Given A = (3 2 / 4 3), find A−1 and use it to solve 3x + 2y = 7 and 4x + 3y = 10.
Working: det A = 9 − 8 = 1, so A−1 = (3 −2 / −4 3). Then x = 3(7) − 2(10) = 1 and y = −4(7) + 3(10) = 2. So x = 1, y = 2. Check: 3 + 4 = 7 and 4 + 6 = 10.
4. Circle equation (3 marks)
Find the centre and radius of the circle x2 + y2 − 6x + 4y − 12 = 0.
Working: Complete the square: (x − 3)2 − 9 + (y + 2)2 − 4 − 12 = 0, so (x − 3)2 + (y + 2)2 = 25. Centre (3, −2), radius 5.
5. Trigonometric equation (3 marks)
Solve 2 sin2θ − sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.
Working: Let s = sin θ: 2s2 − s − 1 = 0 factorises to (2s + 1)(s − 1) = 0, so s = 1 or s = −½. sin θ = 1 gives θ = 90°. sin θ = −½ gives θ = 210° and 330° (third and fourth quadrants). θ = 90°, 210°, 330°.
6. Binomial expansion (3 marks)
Expand (1 + x/2)5 up to the term in x3, and use it to estimate 1.025.
Working: 1 + 5(x/2) + 10(x/2)2 + 10(x/2)3 = 1 + 2.5x + 2.5x2 + 1.25x3. For 1.02, x/2 = 0.02 so x = 0.04: 1 + 0.1 + 0.004 + 0.00008 = 1.10408 (the true value is 1.1040808).
Section II: Longer Questions
7. Grouped data statistics (10 marks)
The marks of 40 Form Four candidates in a mock were grouped as follows: 30–39: 4, 40–49: 7, 50–59: 12, 60–69: 9, 70–79: 6, 80–89: 2. (a) Using an assumed mean of 54.5, find the mean. (b) Find the median. (c) Find the standard deviation.
Working: Midpoints are 34.5, 44.5, 54.5, 64.5, 74.5, 84.5, so d = x − 54.5 gives −20, −10, 0, 10, 20, 30. Σfd = −80 − 70 + 0 + 90 + 120 + 60 = 120. (a) Mean = 54.5 + 120 ÷ 40 = 57.5.
(b) The median is the 20th value. Cumulative frequencies are 4, 11, 23, so it lies in 50–59 (lower boundary 49.5). Median = 49.5 + (20 − 11) ÷ 12 × 10 = 49.5 + 7.5 = 57.0.
(c) Σfd2 = 4(400) + 7(100) + 0 + 9(100) + 6(400) + 2(900) = 1,600 + 700 + 900 + 2,400 + 1,800 = 7,400. Variance = 7,400 ÷ 40 − (120 ÷ 40)2 = 185 − 9 = 176. Standard deviation = √176 ≈ 13.27.
8. Probability (10 marks)
A bag holds 5 red and 3 green counters. Two are drawn one after the other without replacement. Find the probability that (a) both are red, (b) they are of different colours, (c) at least one is green.
Working: Draw a tree diagram. (a) P(RR) = 5/8 × 4/7 = 20/56 = 5/14. (b) P(RG) + P(GR) = 5/8 × 3/7 + 3/8 × 5/7 = 15/56 + 15/56 = 30/56 = 15/28. (c) P(at least one green) = 1 − P(RR) = 1 − 5/14 = 9/14. Using the complement saves time and is fully credited.
9. Transformations (10 marks)
Triangle PQR has vertices P(1, 1), Q(3, 1) and R(1, 4). It is mapped by the matrix M = (2 0 / 0 2) followed by a reflection in the line y = x. Find the coordinates of the final image and describe the single matrix that does both.
Working: M doubles every coordinate: P′(2, 2), Q′(6, 2), R′(2, 8). The reflection in y = x has matrix (0 1 / 1 0) and swaps coordinates: P″(2, 2), Q″(2, 6), R″(8, 2). The single matrix is (0 1 / 1 0)(2 0 / 0 2) = (0 2 / 2 0). Its determinant is −4, which tells you the area scale factor is 4 and orientation is reversed. The original area is ½ × 2 × 3 = 3, so the final image has area 12.
How To Use The Last Two Weeks Before Paper 2
- Do Section II questions under time: about 20 minutes each. Candidates who spend 40 minutes on Section I usually run short at the end.
- Show every line. In KCSE marking, method marks (M) are separate from accuracy marks (A), so a wrong final answer with correct working still scores.
- Keep your calculator in degree mode for trigonometry and check it at the start of the paper.
- Learn the standard results that appear almost every year: compound interest, the sine and cosine rules, the equation of a circle, and the cumulative frequency median.
Next, check the KCSE 2026 timetable and guide, and when results come out, read how cluster points are worked out.