Grade 11 — Term 2

Share Market

What you must know

• Dividend calculated on Nominal Value: Dividend percentage is ALWAYS calculated on the NOMINAL (FACE) VALUE of the share, NEVER on the market purchase price.
• Investment calculation: Number of shares purchased = Total money invested / Market price per share.
• Capital Gain / Loss on Selling: Capital Gain = (Selling Market Price - Buying Market Price) × Number of shares; Total Income from transaction = Total Dividends + Capital Gain.

Try one question

An investor purchases 2,000 shares of a company with a market price of Rs. 30 per share. If the company declares an annual dividend of Rs. 4.50 per share, calculate the annual dividend income and the dividend yield on investment.

1. A. Rs. 9,000 and 12%
2. B. Rs. 6,000 and 10%
3. C. Rs. 4,500 and 15%
4. D. Rs. 9,000 and 15%

Show the answer

Answer: D. Rs. 9,000 and 15%. Annual dividend = 2,000 shares × Rs. 4.50 = Rs. 9,000. Total investment = 2,000 × 30 = Rs. 60,000. Dividend yield = (4.50 / 30) × 100% = 15% (or 9,000 / 60,000 × 100% = 15%).

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

The Midpoint Theorem

What you must know

• Both halves of the theorem: Midpoint theorem gives TWO conclusions simultaneously: parallel (DE // BC) AND half-length (DE = 1/2 BC); cite both when needed in proofs.
• Distinguishing theorem from converse: Use the theorem when given TWO midpoints; use the converse when given ONE midpoint and ONE parallel line.
• Rider citation standard: In Paper 2 Part B geometric proofs, write 'Midpoint theorem' or 'Converse of midpoint theorem' as formal reasons.

Try one question

In triangle ABC, D is the midpoint of AB and E is the midpoint of AC. If the length of BC is 18 cm, what is the length of DE?

1. A. 9 cm
2. B. 6 cm
3. C. 12 cm
4. D. 18 cm

Show the answer

Answer: A. 9 cm. By the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of its length: DE = 1/2 × BC = 1/2 × 18 = 9 cm.

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

Graphs

What you must know

• Coordinates of turning point: Always write the turning point as an ordered pair (x, y) with both coordinates clearly stated.
• Smooth parabolic curve: Plot points accurately and connect with a continuous smooth curve; NEVER join plotted points with straight ruler lines.
• Interval of increase/decrease: State intervals with correct inequality signs (e.g., 'function increases for x > 1').

Try one question

What are the coordinates of the turning point (vertex) of the quadratic graph y = (x - 3)² - 4, and does it represent a minimum or maximum value?

1. A. (-3, -4), minimum
2. B. (3, 4), maximum
3. C. (-3, 4), maximum
4. D. (3, -4), minimum

Show the answer

Answer: D. (3, -4), minimum. The vertex form of a parabola is y = a(x - h)² + k with turning point at (h, k). Here, h = 3, k = -4. Since a = 1 > 0, the parabola opens upwards and has a minimum at (3, -4).

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

Equations

Equiangular Triangles

What you must know

• Correct pairing of corresponding sides: Corresponding sides are ALWAYS opposite EQUAL ANGLES; write angle equality first before forming side ratios.
• Two angles equality is sufficient (AA): If two angles of one triangle equal two angles of another, the third angles are automatically equal (sum = 180°), proving equiangularity.
• Cross-multiplication in side ratios: If AB / PQ = BC / QR, then AB × QR = BC × PQ; standard algebraic method to prove geometric product relations in Paper 2 geometry riders.

Try one question

Triangles ABC and PQR are equiangular (similar) with angle A = angle P, angle B = angle Q, and angle C = angle R. If AB = 8 cm, PQ = 12 cm, and BC = 10 cm, find the length of QR.

1. A. 16 cm
2. B. 13.5 cm
3. C. 14 cm
4. D. 15 cm

Show the answer

Answer: D. 15 cm. In equiangular triangles, the ratios of corresponding sides are equal: AB/PQ = BC/QR → 8/12 = 10/QR → 2/3 = 10/QR → QR = (10 × 3)/2 = 15 cm.

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

Data Representation and Interpretation

What you must know

• Upper boundary plotting rule: Always plot cumulative frequency against UPPER class boundaries, NEVER against class mid-values.
• Start at zero frequency: The ogive curve must start on the horizontal axis at the lower boundary of the very first class interval (cumulative frequency = 0).
• Reading percentiles accurately: Show dashed projection lines from the vertical axis to the curve and down to the horizontal axis to secure method marks.

Try one question

The marks of 9 students in a test are: 42, 55, 60, 65, 68, 72, 75, 80, 88. Find the median mark and the interquartile range (IQR).

1. A. Median = 68, IQR = 15
2. B. Median = 72, IQR = 20
3. C. Median = 68, IQR = 20
4. D. Median = 65, IQR = 18

Show the answer

Answer: C. Median = 68, IQR = 20. Since n = 9, the median is the 5th value = 68. Lower half values: 42, 55, 60, 65 → Q₁ = (55 + 60)/2 = 57.5. Upper half values: 72, 75, 80, 88 → Q₃ = (75 + 80)/2 = 77.5. IQR = Q₃ - Q₁ = 77.5 - 57.5 = 20.

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

Geometric Progressions

What you must know

• Indices manipulation: When solving T_n = a r^(n-1), express both sides with matching prime bases to equate exponents (e.g., 2^(n-1) = 64 = 2⁶ => n - 1 = 6 => n = 7).
• Negative common ratio: When r < 0, terms oscillate alternately in sign (+, -, +, -, ...); compute powers of negative numbers carefully (e.g., (-2)³ = -8, (-2)⁴ = +16).
• Geometric Mean: For three terms a, b, c in GP: b / a = c / b => b² = ac => b = √(ac).

Try one question

The first term of a geometric progression is 5 and the common ratio is 2. What is the 7th term of this progression?

1. A. 256
2. B. 160
3. C. 320
4. D. 640

Show the answer

Answer: C. 320. The nth term of a geometric progression is T_n = a·r^(n-1). Here, a = 5, r = 2, n = 7: T₇ = 5 × 2^(7-1) = 5 × 2⁶ = 5 × 64 = 320.

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

Pythagoras' Theorem

What you must know

• Hypotenuse identification: The hypotenuse is ALWAYS the side opposite the right angle; when calculating a shorter leg, SUBTRACT: a² = c² - b².
• Converse proof structure: To prove ∠C = 90°, evaluate c² independently, evaluate a² + b² independently, show they are equal, then conclude 'by converse of Pythagoras' theorem'.
• Surd and exact form: If an answer does not result in an exact integer, express it in simplified surd form (e.g., √50 = 5√2) or round to requested decimal places.

Try one question

A 13-metre ladder rests against a vertical wall such that its foot is 5 metres away from the base of the wall. How high up the wall does the top of the ladder reach?

1. A. 12.5 m
2. B. 10 m
3. C. 12 m
4. D. 11 m

Show the answer

Answer: C. 12 m. Form a right triangle: hypotenuse c = 13 m, base b = 5 m. By Pythagoras' theorem: a² + b² = c² → h² + 5² = 13² → h² + 25 = 169 → h² = 144 → h = 12 m.

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