Part A: Mathematics
Algebra and relationships
Key takeaways
- An inequality is solved like an equation, except that multiplying or dividing both sides by a negative number reverses the inequality sign.
- Simultaneous equations are solved for the one set of values that satisfies all of them at once: by elimination, by substitution, or graphically where the lines intersect.
- A quadratic ax² + bx + c = 0 is solved by factorisation where possible, otherwise by the quadratic formula x = (-b ± √(b² - 4ac)) / 2a.
- The 2×2 matrix [[a, b], [c, d]] has determinant ad - bc, and it has an inverse only when that determinant is not zero.
- A logarithm finds an unknown power, which is how the time for a sum to grow is found: KES 100,000 at 8% compound interest doubles in n = ln(2) / ln(1.08), about 9 years.
Business mathematics builds every technique in this paper on algebra: expressing a relationship between costs, revenue or quantities as an equation, then solving it. This chapter is the toolkit the rest of the paper draws on.
Equations and inequalities
Linear equations and inequalities
A linear equation, one where the variable appears only to the first power, is solved by isolating the variable on one side. A linear inequality is solved the same way, with one rule to watch: multiplying or dividing both sides by a negative number reverses the inequality sign. Graphically, a linear equation in two variables (y = mx + c) plots as a straight line; an inequality plots as a region on one side of that line.
Solving a linear inequality
3x - 7 < 11
3x < 18
x < 6 (dividing by a positive number: sign unchanged)
Simultaneous equations
Simultaneous equations are two or more equations sharing the same variables, solved for the one set of values that satisfies all of them at once, algebraically by elimination or substitution, or graphically by the point where the lines intersect.
Solving simultaneous equations by elimination
2x + y = 11
x - y = 1
Adding the two equations eliminates y: 3x = 12, so x = 4
Substituting back: y = 11 - 2(4) = 3
Quadratic equations
A quadratic equation, ax² + bx + c = 0, is solved by factorisation where possible, or otherwise by the quadratic formula. Its graph is a parabola, opening upward when a is positive, downward when a is negative, a shape that recurs whenever this paper models a cost or profit function with a squared term.
The quadratic formula
x = (-b ± √(b² - 4ac)) / 2a
For x² - 5x + 6 = 0: a=1, b=-5, c=6
x = (5 ± √(25-24)) / 2 = (5 ± 1) / 2, so x = 3 or x = 2
The exponential function
The exponential function y = e^x, where e is Euler's number (approximately 2.71828), grows at a rate proportional to its own current value at every point, which is what makes it the natural model for continuous growth processes such as continuous compound interest or population growth. Its graph never touches the x-axis and rises ever more steeply as x increases.
Absolute and relative errors
Every measurement or rounded figure used in a calculation carries some error. Absolute error is the plain difference between the true value and the measured or rounded value; relative error expresses that same difference as a proportion of the true value, which is what makes it meaningful to compare the accuracy of two very differently sized figures.
Absolute and relative error
True value: 250; measured value: 245
Absolute error = |250 - 245| = 5
Relative error = 5 / 250 = 0.02, or 2%
Algebraic expressions
Factorisation and BODMAS
Factorise: x² - 9 = (x - 3)(x + 3)
Simplify using BODMAS: 4 + 2 × (6 - 3)² = 4 + 2 × 9 = 4 + 18 = 22
Matrices
Order and types of matrices
A matrix is a rectangular array of numbers arranged in rows and columns; its order is stated as rows × columns (a 2×3 matrix has 2 rows and 3 columns). Common types include the square matrix (equal rows and columns), the identity matrix (1s on the leading diagonal, 0s elsewhere, the matrix equivalent of the number 1), and the zero matrix (every entry 0).
Matrix operations
Matrices are added or subtracted entry by entry, and must be the same order to do so. Matrix multiplication is row-by-column: each entry of the product is the sum of the products of corresponding entries from a row of the first matrix and a column of the second, which requires the number of columns in the first matrix to equal the number of rows in the second.
Determinants of 2x2 matrices
For a 2×2 matrix, the determinant is a single number computed directly from its four entries.
Determinant of a 2x2 matrix
For matrix [[a, b], [c, d]], determinant = ad - bc
For [[4, 3], [2, 5]]: determinant = (4×5) - (3×2) = 20 - 6 = 14
Inverses of 2x2 matrices
A 2×2 matrix has an inverse only if its determinant is non-zero. The inverse is found by swapping the two entries on the leading diagonal, negating the other two entries, and dividing every entry by the determinant.
Inverse of a 2x2 matrix
For [[4, 3], [2, 5]], determinant = 14 (from above)
Inverse = (1/14) × [[5, -3], [-2, 4]]
Matrices in business problems
A system of simultaneous equations can be written as a matrix equation AX = B, and solved as X = A⁻¹B once A's inverse is known, a method that scales far better than substitution once a business problem involves three or more unknowns (for example, allocating three products across shared production constraints).
Logarithms
Logarithms and anti-logarithms
A logarithm answers the question 'to what power must the base be raised to give this number'. If b^y = x, then log_b(x) = y. An antilogarithm reverses the operation, recovering x from a known logarithm and base.
Natural logarithm
The natural logarithm, written ln(x), is the logarithm with base e (see The exponential function, above). It appears throughout continuous-growth business calculations, most directly in finding the time needed for a continuously compounded sum to reach a target value.
Logarithms in business
Using logarithms to find time to double an investment
How many years for KES 100,000 to double at 8% compound interest?
200,000 = 100,000 (1.08)^n, so 2 = (1.08)^n
n = ln(2) / ln(1.08) ≈ 9.0 years
Exam tip
This chapter is rarely examined as a standalone question; it is the toolkit later chapters lean on. A candidate weak on factorisation or BODMAS will lose marks throughout the paper, not just here, so treat fluency in this chapter as a prerequisite, not an isolated topic.
Self-check
(a) Solve the simultaneous equations 3x + 2y = 16 and x - y = 2. (4 marks)
(b) Solve the quadratic equation x^2 - 7x + 10 = 0 by factorisation. (4 marks)
(c) For the matrix [[3, 2], [5, 4]], compute the determinant and the inverse. (5 marks)
(d) A sum of KES 50,000 is invested at 6% compound interest per annum. Using logarithms, compute how many years it takes to triple in value. (4 marks)
(e) A quantity's true value is 480 units, but it is measured as 470 units. Compute the absolute error and the relative error, expressed as a percentage. (3 marks)
Show worked answer
(a) From x - y = 2, x = y + 2. Substituting: 3(y + 2) + 2y = 16, so 3y + 6 + 2y = 16, giving 5y = 10, y = 2. Then x = 2 + 2 = 4. (4 marks)
(b) x^2 - 7x + 10 = (x - 5)(x - 2) = 0, so x = 5 or x = 2. (4 marks)
(c) Determinant = (3 x 4) - (2 x 5) = 12 - 10 = 2. Inverse = (1/2) x [[4, -2], [-5, 3]] = [[2, -1], [-2.5, 1.5]]. (5 marks)
(d) 3 = (1.06)^n, so n = ln(3) / ln(1.06) ≈ 18.9 years. (4 marks)
(e) Absolute error = |480 - 470| = 10 units. Relative error = 10 / 480 x 100 ≈ 2.08%. (3 marks)