UNIT 01
Number & Algebra
Quick Memory Point
LAWS OF INDICES → MADS:
Multiply → add exponents | Divide → subtract | Scientific: \(a \times 10^n\), \(1 \le a < 10\)
Multiply → add exponents | Divide → subtract | Scientific: \(a \times 10^n\), \(1 \le a < 10\)
Q 01
Number
Simplify: \(\dfrac{x^5 \cdot x^{-2}}{x^{3}}\)
⚠ Watch out — many students forget that dividing subtracts the exponent.
💡 Explanation
Numerator: \(x^5 \cdot x^{-2} = x^{5+(-2)} = x^3\). Then \(\dfrac{x^3}{x^3} = x^{3-3} = x^0 = \mathbf{1}\).
Key rule: Any non-zero base to the power 0 equals 1.
Key rule: Any non-zero base to the power 0 equals 1.
Q 02
Number
Write \(0.000347\) in standard form (scientific notation).
⚠ The decimal moves — count the jumps carefully. Negative exponent = small number.
💡 Explanation
Move the decimal 4 places right to get \(3.47\). Since the original number is < 1, the exponent is negative: \(3.47 \times 10^{-4}\).
Options A & D have wrong sign or wrong coefficient range (must satisfy \(1 \le a < 10\)).
Options A & D have wrong sign or wrong coefficient range (must satisfy \(1 \le a < 10\)).
Q 03
Algebra
Expand and simplify: \((2x - 3)^2\)
⚠ Most common mistake: forgetting the middle term. Use the identity, not just "square each term".
💡 Explanation
\((a-b)^2 = a^2 - 2ab + b^2\). Here \(a=2x,\ b=3\):
\((2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9\).
B is the classic trap — squaring each term and ignoring \(2ab\).
\((2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9\).
B is the classic trap — squaring each term and ignoring \(2ab\).
Q 04
Algebra
Solve for \(x\): \(\quad 3(2x - 4) = 2(x + 5)\)
⚠ Expand brackets first — don't try to divide before expanding.
💡 Explanation
Expand: \(6x - 12 = 2x + 10\) → \(4x = 22\) → \(x = \dfrac{22}{4} = \dfrac{11}{2}\).
Always expand first, then collect like terms on each side.
Always expand first, then collect like terms on each side.
UNIT 02
Functions & Graphs
Quick Memory Point
SLOPE-INTERCEPT: \(y = mx + c\)
\(m\) = RISE over RUN | DOMAIN = allowed \(x\) inputs | RANGE = possible \(y\) outputs
\(m\) = RISE over RUN | DOMAIN = allowed \(x\) inputs | RANGE = possible \(y\) outputs
Q 05
Functions
A linear function passes through \((0,\ 4)\) and \((3,\ -2)\). What is its equation?
⚠ Find slope first: \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). The y-intercept is given for free!
💡 Explanation
\(m = \dfrac{-2-4}{3-0} = \dfrac{-6}{3} = -2\). The \(y\)-intercept \(c = 4\) (given by point \((0,4)\)).
Equation: \(y = -2x + 4\).
Equation: \(y = -2x + 4\).
Q 06
Functions
Given \(f(x) = x^2 - 4x + 3\), what are the x-intercepts?
⚠ Set \(f(x) = 0\) and factorise. Don't confuse x-intercepts with the vertex.
💡 Explanation
\(x^2 - 4x + 3 = 0\) → \((x-1)(x-3) = 0\) → \(x = 1\) or \(x = 3\).
The vertex is at \(x = 2\) (axis of symmetry between the roots), not an x-intercept!
The vertex is at \(x = 2\) (axis of symmetry between the roots), not an x-intercept!
Q 07
Functions
Which transformation maps \(y = x^2\) to \(y = (x+3)^2 - 5\)?
⚠ Inside the bracket → horizontal shift (OPPOSITE direction!). Outside → vertical shift.
💡 Explanation
\((x+3)\) shifts the graph left by 3 (counter-intuitive — adding inside moves left!).
\(-5\) outside shifts the graph down by 5. Vertex moves from \((0,0)\) to \((-3,-5)\).
\(-5\) outside shifts the graph down by 5. Vertex moves from \((0,0)\) to \((-3,-5)\).
UNIT 03
Geometry & Measurement
Quick Memory Point
PYTHAGORAS: \(a^2 + b^2 = c^2\) (c = hypotenuse)
SIMILAR triangles → sides in ratio, angles equal
AREA SCALE FACTOR = (linear scale factor)2
SIMILAR triangles → sides in ratio, angles equal
AREA SCALE FACTOR = (linear scale factor)2
Q 08
Geometry
A right triangle has legs of length 5 cm and 12 cm. What is the length of the hypotenuse?
⚠ Use \(c = \sqrt{a^2 + b^2}\). This is a classic Pythagorean triple — recognise it!
💡 Explanation
\(c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) cm.
The triple 5-12-13 is one of the most tested. Also memorise: 3-4-5, 8-15-17.
The triple 5-12-13 is one of the most tested. Also memorise: 3-4-5, 8-15-17.
Q 09
Geometry
Two similar rectangles have sides in the ratio 3 : 5. What is the ratio of their areas?
⚠ Do NOT use the same ratio for area. Area scales by the square of the linear factor.
💡 Explanation
Area scale factor = (linear scale factor)² = \((3)^2 : (5)^2 = 9 : 25\).
Option D uses cubing — that's for volume, not area.
Option D uses cubing — that's for volume, not area.
Q 10
Geometry
A cone has radius 6 cm and slant height 10 cm. What is its total surface area? (Use \(\pi \approx 3.14\))
⚠ Total surface area of cone = base circle + lateral face = \(\pi r^2 + \pi r l\)
💡 Explanation
Base: \(\pi r^2 = 3.14 \times 36 = 113.04\) cm².
Lateral: \(\pi r l = 3.14 \times 6 \times 10 = 188.4\) cm².
Total: \(113.04 + 188.4 = \mathbf{301.44}\) cm². Option B gives only lateral; D gives only base.
Lateral: \(\pi r l = 3.14 \times 6 \times 10 = 188.4\) cm².
Total: \(113.04 + 188.4 = \mathbf{301.44}\) cm². Option B gives only lateral; D gives only base.
UNIT 04
Trigonometry
Quick Memory Point
SOH·CAH·TOA
Sin = Opp/Hyp · Cos = Adj/Hyp · Tan = Opp/Adj
Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) | Cosine rule: \(c^2 = a^2 + b^2 - 2ab\cos C\)
Sin = Opp/Hyp · Cos = Adj/Hyp · Tan = Opp/Adj
Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) | Cosine rule: \(c^2 = a^2 + b^2 - 2ab\cos C\)
Q 11
Trigonometry
In a right triangle, the angle is 30° and the hypotenuse is 20 cm. Find the side opposite to 30°.
⚠ Opposite / Hypotenuse → which ratio? Careful not to use cos accidentally.
💡 Explanation
\(\sin 30° = \dfrac{\text{opp}}{\text{hyp}}\) → \(\text{opp} = 20 \times \sin 30° = 20 \times 0.5 = 10\) cm.
Option A uses \(\cos 30°\) — that gives the adjacent side, not opposite.
Option A uses \(\cos 30°\) — that gives the adjacent side, not opposite.
Q 12
Trigonometry
In triangle ABC: \(a = 8,\ b = 5,\ C = 60°\). Find side \(c\) using the cosine rule.
⚠ The angle in cosine rule must be the angle BETWEEN the two known sides.
💡 Explanation
\(c^2 = a^2 + b^2 - 2ab\cos C = 64 + 25 - 2(8)(5)\cos 60°\)
\(= 89 - 80 \times 0.5 = 89 - 40 = 49\) → \(c = 7\).
\(= 89 - 80 \times 0.5 = 89 - 40 = 49\) → \(c = 7\).
Q 13
Trigonometry
Find the area of triangle PQR if \(PQ = 10\), \(QR = 7\), and \(\angle Q = 45°\).
⚠ Area = \(\dfrac{1}{2}ab\sin C\). Use the angle between the two known sides.
💡 Explanation
Area \(= \dfrac{1}{2} \times 10 \times 7 \times \sin 45° = 35 \times \dfrac{\sqrt{2}}{2} = \dfrac{35\sqrt{2}}{2} \approx 24.75\) cm².
Option A ignores the sin factor; B doubles the answer.
Option A ignores the sin factor; B doubles the answer.
UNIT 05
Statistics & Probability
Quick Memory Point
MEAN = sum ÷ count |
MEDIAN = middle value |
MODE = most frequent
P(A or B) = P(A) + P(B) − P(A and B)
INDEPENDENT: P(A and B) = P(A) × P(B)
P(A or B) = P(A) + P(B) − P(A and B)
INDEPENDENT: P(A and B) = P(A) × P(B)
Q 14
Statistics
Data set: \(\{3,\ 7,\ 7,\ 8,\ 10,\ 12,\ 15\}\). Find the median and mode.
⚠ The median is the MIDDLE value when ordered — not the average of all. Count carefully.
💡 Explanation
7 values → median = 4th value = 8. Mode = value appearing most = 7 (appears twice).
C gives the mean, not the median. Never confuse these three measures!
C gives the mean, not the median. Never confuse these three measures!
Q 15
Statistics
A bag has 4 red and 6 blue marbles. Two are drawn without replacement. What is the probability both are red?
⚠ "Without replacement" = the total changes after first draw. Do NOT use the same fraction twice.
💡 Explanation
P(1st red) = \(\dfrac{4}{10}\). After one red removed: P(2nd red) = \(\dfrac{3}{9}\).
P(both red) = \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\).
Option D uses \(\dfrac{4}{10} \times \dfrac{4}{10}\) — the with-replacement answer (wrong!).
P(both red) = \(\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}\).
Option D uses \(\dfrac{4}{10} \times \dfrac{4}{10}\) — the with-replacement answer (wrong!).
Q 16
Statistics
In a class of 30, 18 play football and 12 play tennis, with 5 playing both. How many play neither?
⚠ Use the inclusion-exclusion principle. Don't forget to subtract from the total!
💡 Explanation
\(|F \cup T| = 18 + 12 - 5 = 25\). Neither = \(30 - 25 = \mathbf{5}\).
Inclusion-exclusion: always subtract the overlap to avoid counting twice!
Inclusion-exclusion: always subtract the overlap to avoid counting twice!
BONUS
Mixed Challenge — High Stakes
Challenge Zone
These questions mix units. READ SLOWLY, identify what's being asked, pick the right formula, then calculate step by step. Don't rush.
Q 17
Algebra
Solve the simultaneous equations: \(2x + 3y = 12\) and \(x - y = 1\).
⚠ Substitution or elimination both work. Pick one method and stay consistent. Check your answer!
💡 Explanation
From eq. 2: \(x = y + 1\). Sub into eq. 1: \(2(y+1) + 3y = 12\) → \(5y = 10\) → \(y = 2\), \(x = 3\).
Check: \(2(3)+3(2)=12\) ✓ and \(3-2=1\) ✓
Check: \(2(3)+3(2)=12\) ✓ and \(3-2=1\) ✓
Q 18
Functions
The quadratic \(y = -2(x-1)^2 + 8\) has vertex at \((1, 8)\). What is the maximum value, and where does it cross the x-axis?
⚠ Negative leading coefficient → opens downward → vertex is a MAX. Set \(y=0\) to find x-intercepts.
💡 Explanation
\(-2 < 0\) → opens down → max at vertex = 8.
Set \(y = 0\): \(-2(x-1)^2 + 8 = 0\) → \((x-1)^2 = 4\) → \(x-1 = \pm 2\) → \(x = 3\) or \(x = -1\).
Set \(y = 0\): \(-2(x-1)^2 + 8 = 0\) → \((x-1)^2 = 4\) → \(x-1 = \pm 2\) → \(x = 3\) or \(x = -1\).
Q 19
Geometry
A ladder 10 m long leans against a wall. The base is 6 m from the wall. At what height does the ladder touch the wall, and what angle does it make with the ground?
⚠ Two parts: Pythagoras for height, then trig (sin or cos) for angle.
💡 Explanation
Height: \(\sqrt{10^2 - 6^2} = \sqrt{64} = 8\) m.
Angle with ground: \(\cos\theta = \dfrac{6}{10} = 0.6\) → \(\theta = \cos^{-1}(0.6) \approx 53.1°\).
Option B uses the wrong angle: \(36.9°\) is the angle at the top of the wall.
Angle with ground: \(\cos\theta = \dfrac{6}{10} = 0.6\) → \(\theta = \cos^{-1}(0.6) \approx 53.1°\).
Option B uses the wrong angle: \(36.9°\) is the angle at the top of the wall.
Q 20
Statistics
A fair die is rolled twice. What is the probability of getting a sum of exactly 7?
⚠ List the sample space for sum = 7 systematically. Total outcomes = 6 × 6 = 36.
💡 Explanation
Favourable pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs.
P = \(\dfrac{6}{36} = \dfrac{1}{6}\). Note: A and C are the same — watch for reformulated correct answers as distractors!
P = \(\dfrac{6}{36} = \dfrac{1}{6}\). Note: A and C are the same — watch for reformulated correct answers as distractors!
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