Solutions

(a) Complete the table for pattern 4.

Dots form the sequence 1, 4, 9, 16, … which is n².

Crosses form the sequence 5, 8, 11, 14, … which is 3n + 2.

Pattern 4: dots = 4² = 16; crosses = 3(4) + 2 = 14.

Sum = 16 + 14 = 30

(b) (i) Number of crosses needed for pattern 45:

Crosses = 3n + 2 = 3(45) + 2 = 135 + 2 = 137

(b) (ii) The nth term for the sum of crosses and dots:

Sum = n² + (3n + 2)

= n² + 3n + 2

(a) Balance before deposit = −M180. New balance = M1 630.

Deposit = New balance − Old balance = 1 630 − (−180) = 1 630 + 180 = M1 810

(b) (i) M1 100 paid at M220 per stack.

Number of stacks = ^{1 100}⁄₂₂₀ = 5

(b) (ii) Reserved M15 000.

Maximum whole stacks = 68 (since 68 × 220 = 14 960 ≤ 15 000 and 69 × 220 = 15 180 > 15 000)

Cost = 68 × 220 = M14 960

Amount remaining = 15 000 − 14 960 = M40

(c) Accumulated amount = M40 000. Company gives 28.95% and keeps the balance.

Employee's share = 0.2895 × 40 000 = M11 580

Balance kept by company = 40 000 − 11 580 = M28 420

Monthly payment = M980

Number of months = ^{28 420}⁄₉₈₀ = 29 months

Given A(−4, 6), B(8, −2), C(2, −2).

(a) Midpoint of AB:

Midpoint = (^{−4 + 8}⁄₂, ^{6 + (−2)}⁄₂) = (⁴⁄₂, ⁴⁄₂) = (2, 2)

(b) Equation of line AB in the form ax + by + c = 0:

Gradient = ^{−2 − 6}⁄_{8 − (−4)} = ⁻⁸⁄₁₂ = −²⁄₃

Using point A(−4, 6):

y − 6 = −²⁄₃(x + 4)

3(y − 6) = −2(x + 4)

3y − 18 = −2x − 8

2x + 3y − 10 = 0

(c) Area of triangle ABC:

Using the coordinate formula:

Area = ¹⁄₂ | x_A(y_B − y_C) + x_B(y_C − y_A) + x_C(y_A − y_B) |

= ¹⁄₂ | (−4)(−2 − (−2)) + 8((−2) − 6) + 2(6 − (−2)) |

= ¹⁄₂ | 0 + 8(−8) + 2(8) |

= ¹⁄₂ | −64 + 16 |

= ¹⁄₂ × 48 = 24

(a) Given a + b = 9 and a² + b² = 71.

(a + b)² = a² + 2ab + b²

9² = 71 + 2ab

81 − 71 = 2ab ⟹ ab = 5

(ab)³ = 5³ = 125

(b) Factorise x² − x − 20.

Two numbers that multiply to −20 and add to −1: −5 and +4.

x² − x − 20 = (x − 5)(x + 4)

(c) T = 2π√(^l⁄_g) with l = 22.8, g = 9.8.

T = 2π√(^22.8⁄_9.8)

= 2π × 1.5253…

= 9.5838…

≈ 9.58 (to 3 s.f.)

(d) Express ^{a + 2}⁄₄ + ^{a + 3}⁄₅ as a single fraction.

LCM of 4 and 5 is 20.

= ^{5(a + 2)}⁄₂₀ + ^{4(a + 3)}⁄₂₀

= ^{5a + 10 + 4a + 12}⁄₂₀

= ^{9a + 22}⁄₂₀

(e) (i) Table: days 1, 2, 3, 4, 5 with frequencies 5, 4, 6, 2, a. Total = 17 + a.

Cumulative frequencies: after day 1 = 5, after day 2 = 9, after day 3 = 15, after day 4 = 17, after day 5 = 17 + a.

For the median to equal 3, the middle value must fall in day 3's range (positions 10 to 15).

Middle position = ^{17 + a + 1}⁄₂ = ^{18 + a}⁄₂

For this to fall between 10 and 15:

10 ≤ ^{18 + a}⁄₂ ≤ 15

20 ≤ 18 + a ≤ 30

2 ≤ a ≤ 12

Greatest value of a = 12

(e) (ii) Expression for the mean in terms of a.

Sum of (day × frequency) = 1(5) + 2(4) + 3(6) + 4(2) + 5(a)

= 5 + 8 + 18 + 8 + 5a = 39 + 5a

Total frequency = 17 + a

Mean = ^{39 + 5a}⁄_{17 + a}

(a) State the modal traditional food.

The largest single sector on the pie chart is nyekoe at 96°.

(b) Total = 30 students.

Sector angles from the pie chart: qolosi = 36°, lipabi = 36°, sekelenyekoe = 60°, nyekoe = 96°, qhubu = 60°, lehala = 72°.

Check total: 36 + 36 + 60 + 96 + 60 + 72 = 360° ✓

nyekoe = ⁹⁶⁄₃₆₀ × 30 = 8 students

lehala = ⁷²⁄₃₆₀ × 30 = 6 students

(c) (i) P(lipabi):

³⁶⁄₃₆₀ = ¹⁄₁₀

(c) (ii) P(qhubu or qolosi):

^{60 + 36}⁄₃₆₀ = ⁹⁶⁄₃₆₀ = ⁴⁄₁₅

(a) Day 1 = x, Day 2 = x − 3, Day 3 = 3(x − 3) = 3x − 9.

Total = x + (x − 3) + (3x − 9) = 5x − 12

(b) Given total = 33:

5x − 12 = 33

5x = 45

x = 9

(c) Day 1: cost = M900, charged Mk per phone, repaired x phones.

Revenue = kx

Profit = kx − 900

Percentage profit = ^{kx − 900}⁄₉₀₀ × 100%

(a) Find the length of the side of square ABCD.

The rectangle has area 15 cm² and horizontal width 5 cm.

Height of rectangle = ¹⁵⁄₅ = 3 cm

Let the square's side = x.

Horizontal distance from the rectangle's right edge to vertex B = x − 5.

Vertical distance from the rectangle's top edge to vertex B = x − 3.

Applying Pythagoras to the right-angled triangle with hypotenuse 10 cm:

(x − 5)² + (x − 3)² = 10²

x² − 10x + 25 + x² − 6x + 9 = 100

2x² − 16x + 34 = 100

2x² − 16x − 66 = 0

x² − 8x − 33 = 0

(x − 11)(x + 3) = 0

x = 11 (since x > 0)

(b) (i) Area of the container in contact with water.

Cylinder diameter = 10 cm, so radius r = 5 cm.

Water depth = 4 cm.

Area in contact = base area + curved surface up to water level

= πr² + 2πrh

= π(25) + 2π(5)(4)

= 25π + 40π = 65π

= 65 × 3.142 = 204.23 cm²

≈ 204.2 cm²

(b) (ii) New water level after immersing a 6 cm cube.

Volume of cube = 6³ = 216 cm³

Base area of cylinder = π(5²) = 25π = 78.55 cm²

Rise in water level = ²¹⁶⁄_78.55 = 2.75 cm

New water level = 4 + 2.75 = 6.75 cm

(a) Simplify the ratio 0.35 : ²⁄₅ : ³⁄₈

Convert to decimals: 0.35 : 0.4 : 0.375

Multiply all by 1000: 350 : 400 : 375

Divide by 25: 14 : 16 : 15

(b) Population correct to 2 s.f. is 37 000.

Lower bound = 36 500

Upper bound = 37 500

36 500 ≤ 37 000 < 37 500

(c) (i) Bank A: M50 000 at 7.5% simple interest for 3 years.

Interest = ^{50 000 × 7.5 × 3}⁄₁₀₀

= 50 000 × 0.075 × 3 = M11 250

Total = 50 000 + 11 250 = M61 250

(c) (ii) Bank B: M40 000 at 3% compound interest for 3 years.

Amount = 40 000 × (1.03)³

(1.03)³ = 1.092727

Amount = 40 000 × 1.092727 = 43 709.08

Interest = 43 709.08 − 40 000 = M3 709.08

(a) (i) Measure the reflex angle AED.

Use a protractor on the printed diagram. The reflex angle is the angle outside the interior of the pentagon at vertex E.

(a) (ii) Construct loci:

Equidistant from AE and DE: the angle bisector of ∠AED through E.

Equidistant from A and B: the perpendicular bisector of line AB.

Shade the region inside the field that lies on the side of the perpendicular bisector closer to A, and on the side of the angle bisector closer to DE.

(b) Distance = 50 km, speed = 15 km/h.

Time = ⁵⁰⁄₁₅ = ¹⁰⁄₃ hours

= 3 hours 20 minutes

Given: BX = 10 m, ∠OBX = 50°, ∠XCO = 60°, ∠BXA = 20°.

(a) Find the size of angle PXC.

In right-angled triangle XOC: ∠XCO = 60°, ∠XOC = 90°.

∠OXC = 180° − 90° − 60° = 30°

P, X, O are collinear with XO and XP forming a straight line.

∠PXC = 180° − ∠OXC = 180° − 30° = 150°

(b) Length BO in right triangle BXO.

BX = 10 m is the hypotenuse, BO is adjacent to ∠OBX = 50°.

cos(50°) = ^BO⁄_BX

BO = 10 × cos(50°)

= 10 × 0.6428 = 6.43 m

(c) Length AX.

In triangle ABX: ∠ABX = 50° (since A, B, O are collinear), ∠BXA = 20°.

∠BAX = 180° − 50° − 20° = 110°

By Sine Rule:

^AX⁄_sin(50°) = ^BX⁄_sin(110°)

AX = ^{10 × sin(50°)}⁄_sin(110°)

= ^{10 × 0.7660}⁄_0.9397

= 8.15 m

40 people total. n(M) = 10, n(L) = 20, neither = 12.

n(M ∪ L) = 40 − 12 = 28

n(M ∩ L) = n(M) + n(L) − n(M ∪ L) = 10 + 20 − 28 = 2

So y = 2

n(M only) = 10 − 2 = 8 ⟹ x = 8

n(L only) = 20 − 2 = 18 ⟹ z = 18

(b) (i) n(M ∪ L)′:

Number of people not in M or L = 12

(b) (ii) n(M′ ∪ L):

M′ = people not in M = 18 + 12 = 30

The only people NOT in M′ ∪ L are those in M only = 8

n(M′ ∪ L) = 40 − 8 = 32

(c) Landline only = L ∩ M′

(a) (i) Describe a single transformation that maps A onto B.

Triangle A vertices: (1, 3), (3, 3), (3, 6).

Triangle B vertices: (−5, 1), (−3, 1), (−3, 4).

Each vertex of B = corresponding vertex of A + (−6, −2).

Check: (1, 3) + (−6, −2) = (−5, 1) ✓

So the transformation is a translation by the vector (−6, −2).

(a) (ii) Describe a single transformation that maps A onto C.

Triangle C vertices: (4, 0), (7, 0), (4, 2).

Using the 90° clockwise rotation rule about (a, b): (x, y) → (a + (y − b), b − (x − a))

We need (3, 3) → (4, 0):

a + 3 − b = 4 ⟹ a − b = 1

b − 3 + a = 0 ⟹ a + b = 3

Solving: a = 2, b = 1

Check other vertices:

(3, 6) → (2 + 5, 1 − 1) = (7, 0) ✓

(1, 3) → (2 + 2, 1 + 1) = (4, 2) ✓

So the transformation is a 90° clockwise rotation about the point (2, 1).

(b) Enlarge B with centre (1, 1) and scale factor ¹⁄₂.

For each vertex: new position = centre + k(vertex − centre)

(−5, 1) → (1 + 0.5(−6), 1 + 0.5(0)) = (−2, 1)

(−3, 1) → (1 + 0.5(−4), 1 + 0.5(0)) = (−1, 1)

(−3, 4) → (1 + 0.5(−4), 1 + 0.5(3)) = (−1, 2.5)

Draw the triangle with these new vertices on the grid.