KCSE Mathematics — Practice Paper 1

Structured practice across algebra, indices and surds, geometry, trigonometry, statistics and commercial arithmetic, worked line by line so you can see where each method mark falls.

How to use this: attempt every question before reading a solution, and write your working as if a marker will read it — because in this subject the working carries most of the marks.

Section A

1. Simplify: $\dfrac{3^{5} \times 3^{-2}}{3^{4}}$ (2 marks)

Numerator:  3⁵ × 3⁻²  =  3⁽⁵⁻²⁾  =  3³
Divide:     3³ ÷ 3⁴   =  3⁽³⁻⁴⁾  =  3⁻¹
                       =  1/3

Where the marks sit: 1 for correctly combining the indices, 1 for the final value.

Common mistake: treating $3^{-2}$ as $-9$. A negative index means a reciprocal, not a negative number.

2. Solve for $x$: $2x^{2} - 7x + 3 = 0$ (3 marks)

Factorise:  2x² - 7x + 3
Find two numbers multiplying to (2 × 3) = 6 and adding to -7:  -6 and -1

2x² - 6x - x + 3 = 0
2x(x - 3) - 1(x - 3) = 0
(2x - 1)(x - 3) = 0

x = 1/2   or   x = 3

Where the marks sit: 1 for a valid factorisation method, 1 for each root.

Check: substitute $x = 3$: $2(9) - 21 + 3 = 18 - 21 + 3 = 0$ ✓

3. Rationalise the denominator: $\dfrac{6}{\sqrt{3}}$ (2 marks)

 6      6     √3     6√3
---- = ---- × ---- = ----- = 2√3
 √3     √3    √3      3

Where the marks sit: 1 for multiplying by $\sqrt{3}/\sqrt{3}$, 1 for simplifying to $2\sqrt{3}$.

Common mistake: stopping at $\frac{6\sqrt3}{3}$. Simplify fully — the final mark is for the simplified surd.

4. The angles of a triangle are in the ratio $2 : 3 : 4$. Find the largest angle (3 marks)

Total parts = 2 + 3 + 4 = 9
Angle sum of a triangle = 180°
One part = 180 ÷ 9 = 20°

Largest angle = 4 × 20° = 80°

Where the marks sit: 1 for total parts, 1 for the value of one part, 1 for the answer.

Check: $40° + 60° + 80° = 180°$ ✓

5. A trader bought an item for KSh 4,500 and sold it for KSh 5,400. Calculate the percentage profit (3 marks)

Profit = 5400 - 4500 = KSh 900

                profit
% profit  =  ----------- × 100
             cost price

          =  (900 / 4500) × 100
          =  20%

Where the marks sit: 1 for the profit, 1 for dividing by cost price, 1 for the answer.

Common mistake: dividing by the selling price. Percentage profit is always measured against what you paid.

6. Solve the simultaneous equations (4 marks)

$$3x + 2y = 16 \qquad\qquad x - y = 3$$

From the second equation:  x = y + 3

Substitute into the first:
3(y + 3) + 2y = 16
3y + 9 + 2y   = 16
5y            = 7
y             = 1.4

x = 1.4 + 3 = 4.4

Where the marks sit: 1 for making a variable the subject, 1 for substituting, 1 for the first value, 1 for the second.

Check: $3(4.4) + 2(1.4) = 13.2 + 2.8 = 16$ ✓

7. A ladder 5 m long leans against a vertical wall with its foot 3 m from the base. Find the angle it makes with the ground (3 marks)

The ladder is the hypotenuse; 3 m is adjacent to the angle θ.

cos θ = adjacent / hypotenuse = 3/5 = 0.6
θ     = cos⁻¹(0.6)
      = 53.13°
      ≈ 53.1°

Where the marks sit: 1 for choosing cosine (correct ratio for adjacent and hypotenuse), 1 for $\cos\theta = 0.6$, 1 for the angle.

Common mistake: using tangent. Tangent needs the opposite side, which here you would have to find first — a longer route to the same answer, and a wrong one if you use 3 and 5 directly.

8. The marks of 8 students are: 12, 15, 18, 20, 22, 25, 28, 32. Find the mean and the median (3 marks)

Mean   = (12+15+18+20+22+25+28+32) ÷ 8
       = 172 ÷ 8
       = 21.5

Median = mean of the 4th and 5th values (8 values, so no single middle)
       = (20 + 22) ÷ 2
       = 21

Where the marks sit: 1 for the sum, 1 for the mean, 1 for the median.

Common mistake: taking the 4th value alone as the median. With an even count there is no middle term — average the two central values.

9. Find the area of a circle sector with radius 7 cm and angle 90° (3 marks)

(Take $\pi = 22/7$)

                θ
Area  =  ----- × π r²
          360

      =  (90/360) × (22/7) × 7²
      =  (1/4) × (22/7) × 49
      =  (1/4) × 154
      =  38.5 cm²

Where the marks sit: 1 for the sector formula, 1 for substituting, 1 for the answer with units.

10. A matrix $\mathbf{A} = \begin{pmatrix} 3 & 1 \ 2 & 4 \end{pmatrix}$. Find its determinant and hence its inverse (4 marks)

det A = (3 × 4) - (1 × 2) = 12 - 2 = 10

              1   ( d  -b )      1   (  4  -1 )
A⁻¹  =  --------- (       )  =  ---- (        )
          det A   (-c   a )      10  ( -2   3 )

          (  0.4  -0.1 )
     =    ( -0.2   0.3 )

Where the marks sit: 1 for the determinant, 1 for swapping $a$ and $d$, 1 for negating $b$ and $c$, 1 for dividing by the determinant.

Common mistake: forgetting to swap the leading diagonal. The rule is swap $a$ and $d$, negate $b$ and $c$ — in that order.

Where marks are actually lost in Paper 1

1. Answers with no working. A correct answer alone often scores 1 of 3. The method marks are separate and you cannot claim them retrospectively.
2. Missing units. Areas need cm², angles need degrees. It is one mark, and it is free.
3. Rounding too early. Carry full accuracy through and round only at the end, or the final accuracy mark can be lost even with perfect method.
4. Not checking. Substituting your answer back takes fifteen seconds and catches most arithmetic slips.

Where to go next

• KCSE Physics Practice Paper 1 — applies this algebra to real quantities
• KCSE Chemistry Practice Paper 1 — mole calculations use the same proportional reasoning

Ask StudyAI to generate more questions on any topic here, then submit your working — it marks line by line and tells you which step lost the mark.