CEED – General Awareness & Aptitude
Basic Maths – Study Notes
Why it matters
The CEED exam tests not only subject knowledge but also your numerical reasoning. A solid foundation in “Basic Maths” lets you tackle quantitative questions quickly and accurately.
1. Simple Explanation
Basic maths for the CEED consists of:
| Area | What you’ll see on the test |
|---|---|
| Arithmetic | Whole numbers, decimals, fractions, percentages, ratios |
| Algebra | Simple equations, linear forms, pattern‑finding |
| Geometry | Perimeter, area, volume (triangles, circles, rectangles, cubes, cylinders) |
| Data Interpretation | Tables, bar‑charts, pie‑charts |
| Time‑related problems | Speed, distance, work, rates |
The common denominator: quick mental calculation + clear logic. Practice the same patterns over and over – you’ll see the time factor drop dramatically.
2. Key Points
2.1 Arithmetic
| Concept | Quick Tips | Example |
|---|---|---|
| Multiplication | Use commutative property (a × b = b × a) to pick easier numbers. |
27 × 4 → 4 × 27 = 108 |
| Division | If a is divisible by b, the quotient is an integer. |
45 ÷ 5 = 9 |
| Decimals | Shift decimal places to convert to integers, solve, then shift back. | 0.6 × 0.5 → 6 × 5 = 30 → shift 2 places → 0.30 |
| Fractions | Reduce before multiplying or adding. | ½ + ⅓ = (3+2)/6 = 5/6 |
| Percentages | x% of y = (x/100) × y. |
20% of 250 = 0.20 × 250 = 50 |
| Ratios | Keep the simplest form. | 3:4 → 3/4 = 0.75 |
2.2 Algebra
| Concept | Quick Tips | Example |
|---|---|---|
| Linear Equation | Isolate the variable on one side. | 2x + 5 = 13 → 2x = 8 → x = 4 |
| Patterns | Look for arithmetic or geometric progressions. | Sequence: 2, 4, 8, 16 → next = 32 |
| Substitution | Replace variables with given numbers. | If y = 2x, then y = 2(3) = 6 |
2.3 Geometry
| Shape | Formula (quick reference) | Quick Trick |
|---|---|---|
| Rectangle | P = 2(l + w) , A = l × w |
Remember “Perimeter = 2 × (length + width)” |
| Square | P = 4a , A = a² |
All sides equal → 4×side |
| Triangle | P = a + b + c , A = ½ × base × height |
For right triangles, use Pythagoras if needed |
| Circle | P = 2πr , A = πr² |
Use π ≈ 3.14 or 22/7 |
| Cube | Surface Area = 6a² , Volume = a³ |
6 faces, each a² |
| Cylinder | Surface Area = 2πr(r + h) , Volume = πr²h |
Think “top + bottom + side” |
2.4 Data Interpretation
| Data Type | What to look for | Quick Calculation |
|---|---|---|
| Tables | Highest/lowest values, averages | Add columns, divide by count |
| Bar Charts | Compare heights, find differences | Visual difference → numeric |
| Pie Charts | Percentages, share of whole | x% of total |
| Line Graphs | Trends, growth/decline rates | Slope = Δy/Δx |
2.5 Time‑Related Problems
| Scenario | Formula | Example |
|---|---|---|
| Speed/Distance | speed = distance / time |
60 km in 2 hrs → 30 km/h |
| Work | Work = Rate × Time |
A worker does 5 units/hr → 3 hrs → 15 units |
| Compound Interest | A = P(1 + r/n)^(nt) |
10,000 at 5% p.a. for 3 yrs (n=1) → 10,000(1.05)^3 |
3. Worked Example – “Quantitative Aptitude (CEED style)”
Problem
A factory produces two types of widgets: A and B.
- 60% of the widgets are type A.
- Each type A widget costs $12, and each type B widget costs $15.
- Total cost of all widgets produced in a month is $48,000.
Find the total number of widgets produced.
Step‑by‑Step Solution
-
Let the total number of widgets =
N.
Then,- Number of A widgets =
0.60N - Number of B widgets =
0.40N
- Number of A widgets =
-
Cost equation
12(0.60N) + 15(0.40N) = 48,000 -
Simplify
7.2N + 6N = 48,000 13.2N = 48,000 -
Solve for N
N = 48,000 / 13.2 = 3,636.36