Solutions

(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)

Cost = 68 × 220 = M14 960

Amount remaining = 15 000 − 14 960 = M40

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) Draw pattern 5 following the sequence in the diagram.

(b) Complete the table.

Black dots sequence: 1, 1, 7, 7, 17, … (each value repeats twice)

White dots sequence: 0, 4, 4, 12, 12, … (each value repeats twice)

Pattern 4: 7 black + 12 white = 19 total

Pattern 5: 17 black + 12 white = 29 total

(c) (i) Pattern 6:

Black dots = 17 (same as pattern 5, since each value repeats twice)

White dots = 24 (next value after 12, 12 in the sequence)

(c) (ii) The nth term for the total number of dots.

Total sequence: 1, 5, 11, 19, 29, …

First differences: 4, 6, 8, 10

Second differences: 2, 2, 2 → quadratic of the form an² + bn + c

2a = 2 ⟹ a = 1

Subtracting n² from the sequence: 0, 1, 2, 3, 4 → linear term n − 1

nth term = n² + n − 1

(d) Total dots = 239.

n² + n − 1 = 239

n² + n − 240 = 0

(n + 16)(n − 15) = 0

n = 15 (since n must be positive)

(a) (i) Factorise x² − x − 20.

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

x² − x − 20 = (x − 5)(x + 4)

(a) (ii) Simplify ^{x2 − 25}⁄_{x² − x − 20}.

Numerator: x² − 25 = (x − 5)(x + 5)

Denominator: x² − x − 20 = (x − 5)(x + 4)

Cancel (x − 5):

= ^{x + 5}⁄_{x + 4}

(b) (i) 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.)

(b) (ii) Express g in terms of π, T and l.

T = 2π√(^l⁄_g)

^T⁄_2π = √(^l⁄_g)

^T2⁄_4π² = ^l⁄_g

g = ^4π2l⁄_T²

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

LCM of 4 and 5 is 20.

= ^{5(a + 2)}⁄₂₀ + ^{4(a + 3)}⁄₂₀

= ^{5a + 10 + 4a + 12}⁄₂₀

= ^{9a + 22}⁄₂₀

(a) (i) 1 180 pages × 50 lisente = 59 000 lisente = M590

Discount = 7.5% of 590

= 0.075 × 590 = M44.25

(a) (ii) Cost in 2021 was 50 lisente, an increase of 10% from 2020.

2021 cost = 110% of 2020 cost

50 = 1.10 × (2020 cost)

2020 cost = ⁵⁰⁄_1.10 = 45.4545… lisente

≈ M0.45 (to 2 s.f.)

(b) Pule gains M3 600 interest from M20 000 at 6% simple interest.

I = ^{P × R × T}⁄₁₀₀

3 600 = ^{20 000 × 6 × T}⁄₁₀₀

3 600 = 1 200T

T = 3 years

(a) P(8, −5) translated by ( 3 ; 2 ):

Image = (8 + 3, −5 + 2) = (11, −3)

(b) Q(2, 7) reflected in the line y = −1.

Vertical distance from y = 7 to y = −1 is 8 units.

Reflected point: y = −1 − 8 = −9

Image = (2, −9)

(c) Describe the transformation represented by ( 2   0 ; 0   1 ).

The x-coordinate is multiplied by 2; the y-coordinate is unchanged.

Stretch parallel to the x-axis with scale factor 2, y-axis invariant.

(d) N + ( 0   2 ; 1   0 ) = ( 1   4 ; 0   −1 )

N = ( 1   4 ; 0   −1 ) − ( 0   2 ; 1   0 )

= ( 1   2 ; −1   −1 )

(e) M = ( 2   −1 ; 3   5 ). Find M⁻¹.

Determinant = (2)(5) − (−1)(3) = 10 + 3 = 13

M⁻¹ = ¹⁄₁₃ ( 5   1 ; −3   2 )

Total marbles = 150. Red = 60. Green = ⁷⁄₃₀ × 150 = 35.

Blue = 150 − 60 − 35 = 55

(a) (i) P(red) = ⁶⁰⁄₁₅₀ = ²⁄₅

(a) (ii) P(blue) = ⁵⁵⁄₁₅₀ = ¹¹⁄₃₀

(b) Without replacement: P(one red, one green)

= P(R then G) + P(G then R)

= (⁶⁰⁄₁₅₀ × ³⁵⁄₁₄₉) + (³⁵⁄₁₅₀ × ⁶⁰⁄₁₄₉)

= ^{2 100}⁄_{22 350} + ^{2 100}⁄_{22 350}

= ^{4 200}⁄_{22 350}

= ²⁸⁄₁₄₉

(a) Complete the Venn diagram.

n(D ∩ P) = 8. The triple intersection (D ∩ S ∩ P) = 5, so D ∩ P only = 8 − 5 = 3.

n(S) = 15: S∩D only = 4, triple = 5, S only = 1. So S ∩ P only = 15 − 4 − 5 − 1 = 5.

n(P) = 17: D∩P only = 3, triple = 5, S∩P only = 5. So P only = 17 − 3 − 5 − 5 = 4.

(b) (i) n(P′) = everything outside P = 9 + 4 + 1 = 14

(b) (ii) n(P ∪ (D ∩ S)):

D ∩ S = 4 (only) + 5 (triple) = 9

P ∪ (D ∩ S) = P (17) + (D ∩ S not in P) (4) = 21

(c) Total elements = 20.

(10 − n) + n + (16 − n) + ¹⁄₃n = 20

26 − n + ¹⁄₃n = 20

26 − ²⁄₃n = 20

6 = ²⁄₃n

n = 9

(a) Given a + b = 9 and a² + b² = 71.

(a + b)² = a² + 2ab + b²

9² = 71 + 2ab

81 − 71 = 2ab ⟹ ab = 5

(ab)⁵ = 5⁵ = 3 125

(b) Solve (1 + 2a)² = 81.

1 + 2a = ±9

1 + 2a = 9 ⟹ a = 4

1 + 2a = −9 ⟹ a = −5

(c) (i) f(x) = x − ³⁄_x = −2.

x − ³⁄_x = −2

x² − 3 = −2x

x² + 2x − 3 = 0

(x + 3)(x − 1) = 0

x = −3 or x = 1

(c) (ii) Solve √(2b − 3) − ³⁄_{√(2b − 3)} = −2.

Let y = √(2b − 3). Then y − ³⁄_y = −2.

This matches part (i) with y in place of x.

So y = −3 or y = 1.

Since y = √(2b − 3) ≥ 0, we take y = 1.

√(2b − 3) = 1

2b − 3 = 1

b = 2

(a) State the modal class.

The class with the highest frequency is 20 < s ≤ 30 with frequency 21.

(b) Calculate the estimated mean score.

Midpoints: 5, 15, 25, 35, 45

Σfx = (3 × 5) + (18 × 15) + (21 × 25) + (13 × 35) + (5 × 45)

= 15 + 270 + 525 + 455 + 225 = 1 490

Mean = ^{1 490}⁄₆₀ = 24.833… ≈ 24.83

(c) (i) Cumulative frequency:

s ≤ 10: 3

s ≤ 20: 3 + 18 = 21

s ≤ 30: 21 + 21 = 42

s ≤ 40: 42 + 13 = 55

s ≤ 50: 55 + 5 = 60

(c) (ii) Draw the cumulative frequency curve using the values above.

(d) (i) Read from your graph the number of students who scored above 35 marks.

Find 35 on the x-axis, read up to the curve, then across to find the cumulative frequency (approximately 52).

Number above 35 = 60 − 52 = 8 students

(d) (ii) Interquartile range.

Lower quartile (Q1) at 15 on the cumulative axis → read across for the score.

Upper quartile (Q3) at 45 on the cumulative axis → read across for the score.

IQR = Q3 − Q1 ≈ 33 − 18 ≈ 15

(a) Find the height of the cylinder.

Total height = 1.4 m. Hemisphere radius = 0.3 m, so hemisphere height = 0.3 m.

Cylinder height = 1.4 − 0.3 = 1.1 m

(b) Total volume of the container.

Cylinder: V = πr²h = π(0.3)²(1.1) = 0.099π ≈ 0.311 m³

Hemisphere: V = ²⁄₃πr³ = ²⁄₃π(0.3)³ = 0.018π ≈ 0.057 m³

Total = 0.311 + 0.057 = 0.368 m³

(c) Water fills ³⁄₄ of the total volume.

Volume of water = ³⁄₄ × 0.368 = 0.276 m³

Since 0.276 < cylinder capacity (0.311), the water only fills part of the cylinder.

π(0.3)² × h = 0.276

0.2827h = 0.276

h = 0.975 m

(a) Find the area of square ABCD.

Rectangle area = 15 cm², width = 5 cm, so height = 3 cm.

Let square side = x. Then (x − 5)² + (x − 3)² = 10².

x² − 10x + 25 + x² − 6x + 9 = 100

2x² − 16x − 66 = 0

x² − 8x − 33 = 0

(x − 11)(x + 3) = 0 ⟹ x = 11 cm

Area = 11 × 11 = 121 cm²

(b) Similar triangles PQR and TSR.

QS = 7 cm, SR = 8 cm, so QR = 15 cm.

PR = 18 cm, RT = a cm.

^SR⁄_QR = ^TR⁄_PR

⁸⁄₁₅ = ^a⁄₁₈

a = ^{8 × 18}⁄₁₅ = 9.6 cm

PT = PR − RT = 18 − 9.6 = 8.4 cm

(a) Area of trapezium ABCD.

Area = ¹⁄₂ × (AB + CD) × BC

= ¹⁄₂ × (10.2 + 6) × 7.1

= ¹⁄₂ × 16.2 × 7.1

= 57.51 cm²

(b) Perpendicular from E to AB.

AE = 6.4 cm, ∠EAB = 59.5°.

Perpendicular = AE × sin(59.5°) = 6.4 × 0.8616 = 5.514 cm ≈ 5.51 cm

(c) Length BE.

Horizontal distance from E: AE × cos(59.5°) = 6.4 × 0.5075 = 3.248 cm

Horizontal distance from E to B = 10.2 − 3.248 = 6.952 cm

BE = √(6.952² + 5.514²)

= √(48.33 + 30.40)

= √78.73 ≈ 8.87 cm

(a) Complete the table for f(x) = 2^ax.

Reading from the graph: f(0) = 1, f(1) = 2, f(2) = 4.

Values: 1, 2, 4

(b) Find a.

f(x) = 2^ax and f(1) = 2, so 2^a = 2¹ ⟹ a = 1.

(c) Find p and q for y = ²⁄_x² − x.

p at x = −2: ²⁄_(−2)² − (−2) = ²⁄₄ + 2 = 0.5 + 2 = 2.5

q at x = 1: ²⁄_1² − 1 = 2 − 1 = 1

(d) Draw the graph of y = ²⁄_x² − x on the grid.

(e) (i) Solve ²⁄_x² − x = 3 from the graphs.

Find the x-coordinates where the two curves intersect at y = 3.

x ≈ −1.6 or x ≈ 0.7

(e) (ii) Solve 2^x − ²⁄_x² + x = 0.

Rearrange: 2^x = ²⁄_x² − x

Find the x-coordinates where the two curves intersect.

x ≈ 1 or x ≈ −1.5

(f) ²⁄_x² − x = 3 − 2x → 0 = ax³ + bx² − 2.

Multiply both sides by x²:

2 − x³ = 3x² − 2x³

Bring all to one side:

2 − x³ − 3x² + 2x³ = 0

x³ − 3x² + 2 = 0

Multiply by −1: −x³ + 3x² − 2 = 0

So 0 = −x³ + 3x² − 2, giving a = −1 and b = 3.