Grade 11 — Term 1

Real Numbers

What you must know

• Standard form bound for A: In A × 10ⁿ, A MUST satisfy 1 ≤ A < 10; writing 0.35 × 10⁴ or 35 × 10² is NOT valid standard form (must be 3.5 × 10³).
• Negative exponents in standard form: Numbers less than 1 have negative powers of 10 (e.g., 0.0048 = 4.8 × 10⁻³).
• Rational vs Irrational identification: √16 = 4 is rational; √8 = 2√2 is irrational; repeating decimals like 0.333... = 1/3 are rational.

Try one question

Which of the following numbers is an irrational number?

1. A. 22/7
2. B. √7
3. C. √16
4. D. 0.375

Show the answer

Answer: B. √7. √7 cannot be expressed as a fraction a/b where a and b are integers; its decimal expansion is non-terminating and non-recurring. 0.375 = 3/8 (rational), √16 = 4 (rational), and 22/7 is a ratio of integers (rational).

A model question written by Idasara on the exam pattern. It is not from a past paper.

Indices and Logarithms - I

What you must know

• Common base reduction: In equations like 4^(x+1) = 8^(x-1), convert both bases to 2: (2²)^(x+1) = (2³)^(x-1) ⟹ 2x + 2 = 3x - 3 ⟹ x = 5.
• Negative fractional exponent handling: 16^(-3/4) = 1 / 16^(3/4) = 1 / (⁴√16)³ = 1 / 2³ = 1/8.
• Addition of indices trap: aᵐ + aⁿ NEVER equals aᵐ⁺ⁿ; index laws apply to MULTIPLICATION and DIVISION, never to addition or subtraction.

Try one question

Simplify and find the value of: (27)^(2/3) × (16)^(1/2).

1. A. 48
2. B. 36
3. C. 24
4. D. 18

Show the answer

Answer: B. 36. 27^(2/3) = (3³)^(2/3) = 3² = 9. 16^(1/2) = √16 = 4. Product = 9 × 4 = 36.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Indices and Logarithms - II

What you must know

• Bar characteristic arithmetic: In 2̄.4560, the characteristic -2 is negative but the mantissa +0.4560 is POSITIVE: 2̄.4560 = -2 + 0.4560 = -1.5440.
• Dividing a bar characteristic: To divide 2̄.4560 by 3: Rewrite characteristic as a multiple of 3: 2̄.4560 = (3̄ + 1.4560); then (3̄ + 1.4560) / 3 = 1̄.4853.
• Anti-logarithm conversion: Look up the mantissa in the Antilogarithm table to obtain digits, then use the characteristic c to place the decimal point (multiply by 10^c).

Try one question

Solve for x in the logarithmic equation: log₂(x + 5) - log₂(x - 2) = 3.

1. A. 5
2. B. 3
3. C. 2
4. D. 4

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Answer: B. 3. Combine logs using quotient rule: log₂((x + 5)/(x - 2)) = 3 → (x + 5)/(x - 2) = 2³ = 8 → x + 5 = 8(x - 2) = 8x - 16 → 7x = 21 → x = 3.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Surface Area of Solids

What you must know

• Slant height vs Vertical height: In cone surface area formula πrl, l MUST be the slant height l = √(r² + h²), NEVER the vertical height h.
• Open vs Closed cylinder: Read question carefully: Open pipe has curved surface area ONLY (2πrh); closed one end has 2πrh + πr²; closed both ends has 2πrh + 2πr².
• Solid Hemisphere 3πr²: Remember that a SOLID hemisphere has a flat circular base (area πr²) in addition to its curved surface (2πr²), giving total area 3πr².

Try one question

A closed cylinder has a radius of 7 cm and a height of 10 cm. Using π = 22/7, calculate the total surface area of the cylinder.

1. A. 880 cm²
2. B. 594 cm²
3. C. 748 cm²
4. D. 440 cm²

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Answer: C. 748 cm². Total surface area = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 44 × 17 = 748 cm².

A model question written by Idasara on the exam pattern. It is not from a past paper.

Volume of Solids

What you must know

• Melting and Recasting Conservation: When a solid is melted and recast into a different shape (e.g., metal sphere melted into cylinder or small spheres), TOTAL VOLUME REMAINS STRICTLY CONSERVED: V_initial = V_final.
• Volume of Hollow Shells: Volume of material in hollow cylinder = π(R² - r²)h; Volume of material in hollow sphere = (4/3)π(R³ - r³).
• Logarithmic calculation in volume: Paper 2 Part B volume questions frequently require solving volume formulas using four-figure logarithm tables; maintain characteristics correctly.

Try one question

A solid right circular cone has a base radius of 6 cm and a vertical height of 14 cm. Using π = 22/7, find the volume of the cone.

1. A. 528 cm³
2. B. 264 cm³
3. C. 792 cm³
4. D. 1584 cm³

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Answer: A. 528 cm³. Volume of a cone V = 1/3 × πr²h = 1/3 × (22/7) × 6² × 14 = 1/3 × (22/7) × 36 × 14 = (1/3) × 22 × 36 × 2 = 22 × 12 × 2 = 528 cm³.

A model question written by Idasara on the exam pattern. It is not from a past paper.

Binomial Expressions

Algebraic Fractions

Areas of Plane Figures between Parallel Lines

What you must know

• Median of a Triangle divides Area in Half: A median of a triangle connects a vertex to the midpoint of the opposite side, dividing the triangle into two sub-triangles of EQUAL AREA (same base length and same perpendicular height).
• Trap of equal perimeter vs equal area: Figures between parallel lines having equal areas DO NOT necessarily have equal perimeters.
• Geometric construction proofs: Adding a construction line parallel to a base is the standard technique to apply this theorem in riders.

Try one question

Parallelogram ABCD and triangle ABE lie on the same base AB and between the same parallel lines AB and DC. If the area of triangle ABE is 35 cm², what is the area of parallelogram ABCD?

1. A. 70 cm²
2. B. 35 cm²
3. C. 17.5 cm²
4. D. 105 cm²

Show the answer

Answer: A. 70 cm². By theorem, the area of a triangle on the same base and between the same parallels as a parallelogram is half the area of the parallelogram. Area of parallelogram = 2 × Area of triangle = 2 × 35 = 70 cm².

A model question written by Idasara on the exam pattern. It is not from a past paper.

Percentages