in Context
Data Analysis
Critical Weakness
Lines & Gradients
Volume & Area Ratios
Solving & Graphing
Foundational Skill
Completing the Square
Similarity & Congruence
Topic | Hard Injury | Concrete Example | Type |
---|---|---|---|
Algebraic Manipulation Bedok View P2 Q1 |
Sign handling failure with negative brackets |
−(3x + 5) = −3x + 5
✓ −(3x + 5) = −3x − 5
|
P-Layer |
Algebraic Manipulation Northbrooks P2 Q8 |
Cannot convert non-standard equations to quadratic form |
Given: 2/x + 3/(x+1) = 5
Attempted to solve directly
✓ Multiply by LCD: x(x+1) first
|
C-Layer |
Quadratic Functions Kranji P1 Q24 |
Sign errors when completing square with negative coefficient |
Given: y = −x² + 6x − 5
−(x² + 6x) − 5
✓ −(x² − 6x) − 5
|
P-Layer |
Geometric Proofs Regent P1 Q3 |
Applying theorems without verifying conditions |
Used Pythagoras without checking for right angle
✓ First verify: Is there a 90° angle?
|
C-Layer |
Indices & Standard Form Northbrooks P2 Q4 |
Complete freeze with complex index expressions |
Given: (p³q⁻²)² ÷ p⁻¹q³
Gave up immediately
✓ Apply 4-step SOP systematically
|
P/C-Layer |
Coordinate Geometry Anglo-Chinese P2 Q3 |
Missing conceptual link: parallel → equal gradients |
Line L₁: y = 3x + 2. Find L₂ parallel, through (1, 5)
Didn't recognize m₁ = m₂
✓ Immediately write m₂ = 3
|
C-Layer |
Mensuration Northbrooks P1 Q21 |
Missing volume-length relationship |
Similar cylinders, height ratio 2:3
Volume ratio = 2:3
✓ Volume ratio = 2³:3³ = 8:27
|
C-Layer |
THE PROBLEM: Expressions like (p³q⁻²)² ÷ p⁻¹q³ cause instant cognitive overload → complete freeze
Step 1 (Brackets): (p³q⁻²)² = p⁶q⁻⁴
Now we have: p⁶q⁻⁴ ÷ p⁻¹q³
Step 2 (Combine bases using aᵐ ÷ aⁿ = aᵐ⁻ⁿ):
Result: p⁷q⁻⁷
Final Answer (Positive Index): p⁷/q⁷
Follow the SOP exactly. Get 3 consecutive successes!
THE PROBLEM: Dennis knows "parallel" and "equation of line" but cannot connect them
Given: Line L₁ has equation y = 3x + 2
Find: Line L₂ parallel to L₁, passing through (1, 5)
Step 1: Gradient of L₁ = 3
Step 2: Gradient of L₂ = 3 (parallel!)
Step 3: Substitute (1, 5): 5 = 3(1) + c → c = 2
Final Answer: y = 3x + 2
No deep thinking. Only memorized trigger-action pairs.
Install correct behaviors through physical repetition.
Build concept ↔ concept and concept ↔ procedure links.
Before entering the exam hall, verify you can recall each of these instantly:
THEN: Find corresponding/alternate angles AND set gradients equal (m₁ = m₂)
THEN: Consider using Area = ½ × base × height to find perpendicular distance
The Logic: The "shortest distance" from a point to a line is always the perpendicular height (h). If you can see the problem as a triangle, you can use its area to find this height.
Example: In ΔABC, if you know the Area and the base AB, you can find the height from C to AB using:
h = (2 × Area) / base.
THEN: Multiply both sides by LCD (Lowest Common Denominator) immediately
THEN: Move everything to one side, set = 0, then solve quadratic
THEN: Write ratios immediately: Length (k), Area (k²), Volume (k³)
Quantity | Formula | Rule | Example (2:4=1:2) |
---|---|---|---|
Length (1-D) | (L₁/L₂ = k) | Ratio in 1 power | 1 : 2 |
Area (2-D) | (A₁/A₂ = k²) | Ratio in square | 1²:2² = 1:4 |
Volume (3-D) | (V₁/V₂ = k³) | Ratio in cube | 1³:2³ = 1:8 |
THEN: Work backwards from the answer, but write forwards in your solution
"DENNIS,"
You've trained for this.
Trust your preparation.
Execute with confidence.