Unit 1
Algebra — Expressions & Equations
Quick Memory Points
FOIL = First · Outer · Inner · Last |
DIST = Distribute before collecting
ZERO-PRODUCT: if \(ab=0\) → \(a=0\) OR \(b=0\)
TRAP: \((a+b)^2 \neq a^2+b^2\) — middle term \(2ab\) is always there!
ZERO-PRODUCT: if \(ab=0\) → \(a=0\) OR \(b=0\)
TRAP: \((a+b)^2 \neq a^2+b^2\) — middle term \(2ab\) is always there!
Q 01
Algebra
Expand and simplify: \((x + 3)^2 - (x - 2)(x + 2)\)
⚠️ Most students forget the middle term when squaring a binomial.
💡 Worked Pattern
\((a+b)^2 = a^2 + \mathbf{2ab} + b^2\)
e.g. \((x+1)^2 = x^2 + 2x + 1\), not \(x^2+1\)
e.g. \((x+1)^2 = x^2 + 2x + 1\), not \(x^2+1\)
Q 02
Algebra
Factorise completely: \(6x^2 - 13x + 5\)
⚠️ With a leading coefficient ≠ 1, use the AC-method (product–sum method).
💡 AC Method
\(ax^2+bx+c\): find two numbers that multiply to \(ac\) and add to \(b\).
Here: \(a \cdot c = 6 \times 5 = 30\), need sum \(-13\) → try \(-10\) and \(-3\).
Here: \(a \cdot c = 6 \times 5 = 30\), need sum \(-13\) → try \(-10\) and \(-3\).
Q 03
Algebra
Solve for \(x\): \(\dfrac{2x+1}{3} - \dfrac{x-2}{2} = 1\)
⚠️ Find the LCM of denominators FIRST — the most common error is forgetting to multiply every term.
💡 Steps
LCM(3, 2) = 6. Multiply all terms by 6:
\(2(2x+1) - 3(x-2) = 6\)
\(2(2x+1) - 3(x-2) = 6\)
Unit 2 — Functions
Quick Memory Points
VERTEX of \(y=a(x-h)^2+k\) → point \((h,k)\) — note the flip sign for \(h\)!
DISCRIMINANT \(\Delta = b^2-4ac\): >0 two roots · =0 one root · <0 no real roots
DOMAIN-RANGE: domain = allowed \(x\), range = possible \(y\) outputs
DISCRIMINANT \(\Delta = b^2-4ac\): >0 two roots · =0 one root · <0 no real roots
DOMAIN-RANGE: domain = allowed \(x\), range = possible \(y\) outputs
Q 04
Functions
The vertex of \(y = 2(x - 3)^2 + 5\) is at which point?
⚠️ The vertex form is \(y = a(x - h)^2 + k\). Many students read the sign of \(h\) incorrectly.
💡 Sign Trap
\(y = 2(x-3)^2+5\) → vertex \((3, 5)\), NOT \((-3, 5)\).
Think: "What \(x\) makes the bracket zero?" → \(x=3\).
Think: "What \(x\) makes the bracket zero?" → \(x=3\).
Q 05
Functions
Use the quadratic formula to solve \(2x^2 - 4x - 3 = 0\). Give answers to 2 decimal places.
Quadratic formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
💡 Identify a, b, c
\(a = 2,\ b = -4,\ c = -3\)
\(\Delta = (-4)^2 - 4(2)(-3) = 16 + 24 = 40\)
\(\Delta = (-4)^2 - 4(2)(-3) = 16 + 24 = 40\)
Q 06
Functions
If \(f(x) = 3x^2 - 2x + 1\), find \(f(-2)\).
⚠️ Substituting negative numbers — watch out when squaring: \((-2)^2 = +4\), not \(-4\).
Unit 3 — Geometry & Measurement
Quick Memory Points
PYTHAGORAS: \(c^2 = a^2 + b^2\) — \(c\) is always the hypotenuse (longest side)
CIRCLE: Area \(= \pi r^2\), Circumference \(= 2\pi r\) — never mix up \(r\) and \(d\)!
SIMILAR: if sides scale by \(k\), areas scale by \(k^2\), volumes by \(k^3\)
CIRCLE: Area \(= \pi r^2\), Circumference \(= 2\pi r\) — never mix up \(r\) and \(d\)!
SIMILAR: if sides scale by \(k\), areas scale by \(k^2\), volumes by \(k^3\)
Q 07
Geometry
Two similar triangles have corresponding sides in ratio \(3:5\). If the area of the smaller triangle is \(27\text{ cm}^2\), what is the area of the larger triangle?
⚠️ The most common mistake: using the side ratio directly for areas instead of squaring it.
💡 Key Rule
Area ratio = (side ratio)² → \(\left(\dfrac{5}{3}\right)^2 = \dfrac{25}{9}\)
Area of large = \(27 \times \dfrac{25}{9} = 75\text{ cm}^2\)
Area of large = \(27 \times \dfrac{25}{9} = 75\text{ cm}^2\)
Q 08
Geometry
A cylinder has radius \(4\text{ cm}\) and height \(10\text{ cm}\). Find its total surface area in terms of \(\pi\).
Total surface area = 2 circles + curved surface. Don't forget both circular ends!
💡 Formula Breakdown
TSA \(= 2\pi r^2 + 2\pi r h\)
\(= 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi\)
\(= 2\pi(16) + 2\pi(4)(10) = 32\pi + 80\pi = 112\pi\)
Q 09
Geometry
In a right triangle, the two shorter sides are \(5\text{ cm}\) and \(12\text{ cm}\). What is the length of the hypotenuse?
Unit 4 — Trigonometry
Quick Memory Points
SOH-CAH-TOA: Sin=Opp/Hyp · Cos=Adj/Hyp · Tan=Opp/Adj
ANGLE UP: to find angle, use inverse → \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\)
SPECIAL: \(\sin 30°=0.5\), \(\cos 60°=0.5\), \(\tan 45°=1\)
ANGLE UP: to find angle, use inverse → \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\)
SPECIAL: \(\sin 30°=0.5\), \(\cos 60°=0.5\), \(\tan 45°=1\)
Q 10
Trigonometry
In right triangle \(ABC\) with \(\angle C = 90°\), \(BC = 8\), \(AC = 15\), \(AB = 17\). What is \(\sin A\)?
⚠️ Students often confuse opposite and adjacent. Always label sides relative to the angle being used.
💡 Label From Angle A
From angle A: Opposite = BC = 8, Hypotenuse = AB = 17
\(\sin A = \dfrac{\text{Opp}}{\text{Hyp}} = \dfrac{8}{17}\)
\(\sin A = \dfrac{\text{Opp}}{\text{Hyp}} = \dfrac{8}{17}\)
Q 11
Trigonometry
A ladder \(10\text{ m}\) long leans against a wall. The base is \(6\text{ m}\) from the wall. What angle does the ladder make with the ground? (Give answer to 1 decimal place.)
Unit 5 — Statistics & Probability
Quick Memory Points
MMM: Mean = sum/count · Median = middle · Mode = most frequent
IQR = Q3 − Q1 (always positive) · Outlier if beyond \(Q1 - 1.5\times IQR\) or \(Q3 + 1.5\times IQR\)
P(event): favourable outcomes ÷ total outcomes · Always between 0 and 1
IQR = Q3 − Q1 (always positive) · Outlier if beyond \(Q1 - 1.5\times IQR\) or \(Q3 + 1.5\times IQR\)
P(event): favourable outcomes ÷ total outcomes · Always between 0 and 1
Q 12
Statistics
Data set: \(3,\ 7,\ 7,\ 8,\ 10,\ 12,\ 15\). Find the interquartile range (IQR).
⚠️ With 7 values: Q1 = median of bottom 3, Q3 = median of top 3. Don't include the overall median.
💡 Step-by-Step
Ordered: 3, 7, 7 | 8 | 10, 12, 15
Q1 = 7, Q3 = 12 → IQR = 12 − 7 = 5
Q1 = 7, Q3 = 12 → IQR = 12 − 7 = 5
Q 13
Statistics
A bag has 4 red, 3 blue, and 3 green marbles. Two marbles are drawn without replacement. What is the probability both are red?
⚠️ "Without replacement" means the total changes for the second draw!
💡 Multiplication Rule
P(both red) = \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\)
Unit 6 — Inequalities & Linear Systems
Quick Memory Points
FLIP: multiplying or dividing by a negative number FLIPS the inequality sign
ELIMINATION: make coefficients equal, then add/subtract equations
SUBSTITUTION: isolate one variable, plug into other equation
ELIMINATION: make coefficients equal, then add/subtract equations
SUBSTITUTION: isolate one variable, plug into other equation
Q 14
Inequalities
Solve and state the solution set: \(-3x + 2 > 8\)
⚠️ #1 most common error: forgetting to flip the inequality when dividing by a negative.
💡 The Flip Rule
\(-3x > 6\) → divide both sides by \(-3\) → flip ">" to "<"
\(x < -2\)
\(x < -2\)
Q 15
Inequalities
Solve the system: \(\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}\)
💡 Substitution Hint
From equation 2: \(x = y + 1\). Substitute into eq. 1:
\(2(y+1) + 3y = 12\) → solve for \(y\).
\(2(y+1) + 3y = 12\) → solve for \(y\).
Bonus Round — Mixed Topics
Quick Memory Points
EXPONENT LAWS: \(a^m \cdot a^n = a^{m+n}\) · \(\dfrac{a^m}{a^n}=a^{m-n}\) · \((a^m)^n = a^{mn}\)
ZERO/NEGATIVE: \(a^0 = 1\) · \(a^{-n} = \dfrac{1}{a^n}\) (never equals zero or negative!)
SURDS: \(\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}\) but \(\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}\)
ZERO/NEGATIVE: \(a^0 = 1\) · \(a^{-n} = \dfrac{1}{a^n}\) (never equals zero or negative!)
SURDS: \(\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}\) but \(\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}\)
Q 16
Exponents
Simplify: \(\dfrac{x^5 \cdot x^{-2}}{x^3}\)
⚠️ Students often make errors with negative exponents in division. Apply laws step by step.
Q 17
Surds
Simplify: \(\sqrt{75} - 2\sqrt{12}\)
⚠️ You must simplify each surd to like terms before adding or subtracting.
💡 Simplify First
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) → \(2\sqrt{12} = 4\sqrt{3}\)
Result: \(5\sqrt{3} - 4\sqrt{3} = \sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) → \(2\sqrt{12} = 4\sqrt{3}\)
Result: \(5\sqrt{3} - 4\sqrt{3} = \sqrt{3}\)
Q 18
Functions
The line passes through \((2, 5)\) and \((-1, -1)\). What is its equation in the form \(y = mx + c\)?
⚠️ Find gradient first: \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). Then substitute ONE point to find \(c\).
Q 19
Algebra
Solve: \(x^2 - 5x + 6 = 0\). Which answer is correct?
Find two numbers that multiply to \(+6\) AND add to \(-5\). Both must be negative!
Q 20
Statistics
The mean of 5 numbers is 12. Four of the numbers are 8, 10, 14, and 16. What is the fifth number?
⚠️ Mean × count = total sum. Work backwards: find total, then subtract known values.
💡 Reverse Mean
Total sum = mean × count = \(12 \times 5 = 60\)
Known sum = \(8+10+14+16 = 48\)
Missing = \(60 - 48 = 12\)
Known sum = \(8+10+14+16 = 48\)
Missing = \(60 - 48 = 12\)
Quiz Complete! 🎉
Here's how you did:
—
out of 20