Basic Mathematics and Algebra
Mathematics
Basic Mathematics and Algebra
Syllabus tag: KASNEB CPA | Foundation Level | CA15 Quantitative Analysis | Topic 1 Basic Mathematics and Algebra
Lesson objectives
By the end of this topic, you will be able to:
- Apply the laws of indices and logarithms
- Solve linear and quadratic equations
- Work with arithmetic and geometric progressions
- Apply progressions to depreciation and growth problems
- Rearrange formulae correctly
Why this matters
Every later topic in this paper — and the whole of Financial Management — rests on these operations. A candidate who cannot rearrange a formula reliably will lose marks in papers that have nothing to do with algebra.
Laws of indices
| Law | Example |
|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | 2⁵ × 2³ = 2⁸ = 256 |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | 3⁴ ÷ 3² = 3² = 9 |
| (aᵐ)ⁿ = aᵐⁿ | (5²)³ = 5⁶ |
| a⁰ = 1 | 7⁰ = 1 |
| a⁻ⁿ = 1 / aⁿ | 4⁻² = 0.0625 |
| a^(m/n) = ⁿ√(aᵐ) | 81^0.75 = 27 |
The fractional index is the one candidates avoid. 81^0.75 is the fourth root of 81 cubed, or 3³ = 27 — and it appears constantly in compound interest, where the rate for part of a period is required.
Logarithms
A logarithm answers: to what power must the base be raised?
| Law | Form |
|---|---|
| Product | log(ab) = log a + log b |
| Quotient | log(a/b) = log a − log b |
| Power | log(aⁿ) = n log a |
The power law is the useful one, because it brings an unknown exponent down where it can be solved.
How long for money to double at 8%?
1.08ⁿ = 2 n log 1.08 = log 2 n = log 2 / log 1.08 = 9.0065 years
Any question of the form "how long until..." or "what rate would achieve..." is solved this way.
Quadratic equations
x = [−b ± √(b² − 4ac)] / 2a
For 2x² − 11x + 12 = 0:
Discriminant = 121 − 96 = 25 Roots: (11 + 5) / 4 = 4 and (11 − 5) / 4 = 1.5
The discriminant tells you what to expect before you compute:
| Discriminant | Roots |
|---|---|
| Positive | Two distinct real roots |
| Zero | One repeated root |
| Negative | No real roots |
A negative discriminant in a break-even calculation means the business never breaks even at any output — which is a finding, not an error.
Arithmetic progressions
A constant amount is added each period.
nth term = a + (n − 1)d Sum = n/2 [2a + (n − 1)d]
Sales start at KES 5,000 and grow by KES 750 a month:
12th month = 5,000 + 11(750) = KES 13,250 Total over 12 months = 6[10,000 + 8,250] = KES 109,500
Straight-line depreciation is an arithmetic progression — the same amount comes off each year.
Geometric progressions
A constant factor multiplies each period.
nth term = arⁿ⁻¹ Sum = a(rⁿ − 1) / (r − 1)
Sales start at KES 8,000 and grow 15% a month, so r = 1.15:
8th month = 8,000 × 1.15⁷ = KES 21,280.16 Total over 8 months = 8,000(1.15⁸ − 1) / 0.15 = KES 109,814.55
Reducing balance depreciation is a geometric progression with r below 1, and so is compound interest with r above 1. The same formula serves both.
Sum to infinity, where |r| < 1: S∞ = a / (1 − r)
This is the perpetuity formula from Financial Management. A dividend of 6,000 in perpetuity discounted at 12% is worth 6,000 / 0.12 = 50,000 — which is the geometric sum of an infinite series, arrived at from a different direction.
:::checkpoint A company's costs rise by KES 400 a month while its revenue grows 3% a month from a similar base. Explain which progression each follows and what happens over a long period. :::
Rearranging formulae
The rule: do the same operation to both sides, and unwind in reverse order.
To make r the subject of A = P(1 + r)ⁿ:
A / P = (1 + r)ⁿ (A / P)^(1/n) = 1 + r r = (A / P)^(1/n) − 1
That result is the compound annual growth rate, used throughout this paper and in every appraisal question that asks for an implied rate of return.
:::checkpoint An investment of KES 400,000 grows to KES 726,000 over 5 years. Using the formula above, set out the steps to find the annual rate, without computing the final figure. :::