(a) Construct a multiplication table for modulo 9 on the set {2, 4, 6, 8}
| × | 2 | 4 | 6 | 8 |
|---|---|---|---|---|
| 2 | 4 | 8 | 3 | 7 |
| 4 | 8 | 7 | 6 | 5 |
| 6 | 3 | 6 | 0 | 3 |
| 8 | 7 | 5 | 3 | 1 |
(b) Using the table:
(a) In the diagram: ∠RPQ = 37°, ∠NQP = 136°, ∠MNS = 5x° and NS//PR
Using the properties of parallel lines:
In triangle PQR:
Since NS//PR, ∠MNS = ∠NPR (corresponding angles) Therefore: 5x = 44 x = 8.8°
(b) If 13sin θ - 3 = 2, where 0° < θ < 90°, evaluate (2-tan θ)/(1+2cos θ)
First, solve for sin θ: 13sin θ - 3 = 2 13sin θ = 5 sin θ = 5/13
Since 0° < θ < 90°, we can find cos θ using Pythagorean identity: sin² θ + cos² θ = 1 (5/13)² + cos² θ = 1 25/169 + cos² θ = 1 cos² θ = 144/169 cos θ = 12/13
Therefore: tan θ = sin θ/cos θ = (5/13)/(12/13) = 5/12
Substituting into the expression: (2 - tan θ)/(1 + 2cos θ) = (2 - 5/12)/(1 + 2(12/13)) = (24/12 - 5/12)/(1 + 24/13) = (19/12)/(37/13) = (19/12) × (13/37) = 247/444
The ages in years of a family of 10 members have statistics Σx = 290 and Σx² = 8469. Calculate:
(i) Mean Mean = Σx/n = 290/10 = 29 years
(ii) Variance Variance = (Σx²/n) - (mean)² = (8469/10) - (29)² = 846.9 - 841 = 5.9
(b) The fifth term of an arithmetic progression is three times the first term. If a quarter of the fifth term is 9, find:
Let first term = a, common difference = d Fifth term = a + 4d = 3a Therefore: 4d = 2a, so d = a/2
Quarter of fifth term = 9 (a + 4d)/4 = 9 a + 4d = 36 a + 4(a/2) = 36 a + 2a = 36 3a = 36 a = 12
Therefore: d = a/2 = 6
(i) The first term and common difference First term = 12, Common difference = 6
(ii) The sum of the first eight terms Sum = n/2[2a + (n-1)d] = 8/2[2(12) + 7(6)] = 4[24 + 42] = 4 × 66 = 264
Circle with center O, radius 5 cm, point P on circle, PT tangent, PT = 12 cm, line OT cuts circle at Q
(a) Perimeter of shaded portion In right triangle OPT: OT² = OP² + PT² = 5² + 12² = 25 + 144 = 169 OT = 13 cm
Since Q is on the circle: OQ = 5 cm Therefore: QT = OT - OQ = 13 - 5 = 8 cm
For the arc PQ, we need angle POQ: cos(∠POT) = OP/OT = 5/13 ∠POQ = ∠POT (same angle)
Arc length PQ = (θ/360°) × 2πr cos⁻¹(5/13) ≈ 67.38° Arc PQ = (67.38°/360°) × 2π × 5 ≈ 5.87 cm
Perimeter = Arc PQ + QT + TP = 5.87 + 8 + 12 = 25.87 cm
(b) Area of shaded portion Area = Area of sector POQ + Area of triangle QTP
Area of sector = (67.38°/360°) × π × 5² ≈ 14.68 cm² Area of triangle QTP = ½ × QT × PT = ½ × 8 × 12 = 48 cm²
Total area ≈ 14.68 + 48 = 62.68 cm²
Given P(-4,8) and Q(6,-3), find coordinates after transformations:
(a) Translation by vector (-7,5) followed by half turn about origin After translation: P' = (-4-7, 8+5) = (-11, 13), Q' = (6-7, -3+5) = (-1, 2) After half turn: P'' = (11, -13), Q'' = (1, -2)
(b) Reflection in line y = -0 followed by enlargement from origin with scale factor ½ After reflection in y = 0: P' = (-4, -8), Q' = (6, 3) After enlargement: P'' = (-2, -4), Q'' = (3, 1.5)
(c) Equation of line through Q' parallel to 2x - 3y + 1 = 0 Parallel line has same slope: 2x - 3y + c = 0 Substituting Q'(3, 1.5): 2(3) - 3(1.5) + c = 0 6 - 4.5 + c = 0 c = -1.5
Equation: 2x - 3y - 1.5 = 0 or 4x - 6y - 3 = 0
Age distribution of males in a village:
| Ages | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 9 | 15 | 26 | 18 | 15 | 12 | 6 | 5 |
(i) Draw a histogram - This would be a bar chart with age ranges on x-axis and frequencies on y-axis.
(ii) Estimate the mode The modal class is 21-30 (highest frequency = 26) Mode ≈ 25.5 years (midpoint of modal class)
(b) Right circular cone: base radius = hemisphere radius, height = 15 cm
Let r = radius of base and hemisphere
(i) Total volume Volume of cone = ⅓πr²h = ⅓πr²(15) = 5πr² Volume of hemisphere = ⅔πr³ Total volume = 5πr² + ⅔πr³ = πr²(5 + ⅔r)
(ii) Total surface area (π = 22/7) Surface area of hemisphere = 2πr² = 2 × (22/7) × r² = (44r²)/7 Surface area of cone = πrl where l = √(r² + 15²) = √(r² + 225) = (22/7) × r × √(r² + 225) Total surface area = (44r²)/7 + (22r√(r² + 225))/7 = (22r/7)(2r + √(r² + 225))