Solutions

(a) (i) Write the type of numbers given to the elements of (P ∪ Q)′.

Find the elements outside both circles P and Q: {2, 3, 7, 11, 19, 17, 5, 13}. All of these numbers have only two distinct factors: 1 and themselves.

Prime numbers.

(a) (ii) Describe P ∩ Q.

Look at the intersection region shared by circles P and Q. There are no elements inside it.

Empty set (or null set / ∅).

(a) (iii) Find n(Q)′.

Count all elements that are not inside circle Q.

Elements in P only: {6, 12, 18} (3 elements)

Elements outside both circles: {2, 3, 7, 11, 19, 17, 5, 13} (8 elements)

Total count = 3 + 8 = 11

(b) In a class of 102 students, 70 joined Mathematics club, 60 joined Debate club, and 15 joined neither. Find the number of students who joined Mathematics club only.

1. Let the number of students who joined both clubs be x.

2. Set up the equation for the total class size:

(Math only) + (Debate only) + (Both) + (Neither) = Total

(70 − x) + (60 − x) + x + 15 = 102

3. Simplify and solve for x:

145 − x = 102 ⟹ x = 145 − 102 = 43

4. Calculate the number of students in the Mathematics club only:

Math only = 70 − 43 = 27

(a) Geometric sequence: 2, …, 18, 54, 162, …

1. Find the common ratio (r) by dividing consecutive terms: ⁵⁴⁄₁₈ = 3.

2. Find the 2nd term: 1st term × r = 2 × 3 = 6.

3. Find the 6th term: 4th term (162) × r × r = 162 × 3 × 3 = 1458.

(b) Find the nth term of the sequence −1, −1, 1, 5, 11, …

1. Find the first differences between terms: 0, 2, 4, 6

2. Find the second differences: 2, 2, 2

3. Since the second difference is constant, it is a quadratic sequence of the form an² + bn + c.

2a = second difference ⟹ 2a = 2 ⟹ a = 1

4. Subtract an² = n² from the original sequence terms:

For n = 1: −1 − 1² = −2

For n = 2: −1 − 2² = −5

For n = 3: 1 − 3² = −8

5. Find the linear rule for the remaining sequence −2, −5, −8, …:

The difference is −3, so the term rule is −3n + 1.

6. Combine both parts: n² − 3n + 1.

(c) Arithmetic progression: 1st term is 4, 5th term is 16, T_n = a + d(n − 1).

(i) Here a = 4. For n = 5:

16 = 4 + d(5 − 1) ⟹ 12 = 4d ⟹ d = 3

(ii) Substitute a = 4 and d = 3 into the formula:

T_n = 4 + 3(n − 1) = 4 + 3n − 3 = 3n + 1

(a) Teboho and Sebolelo share M72 000 in the ratio 3 : 5.

(i) Total parts = 3 + 5 = 8.

Share = ³⁄₈ × 72 000 = 3 × 9 000 = 27 000

(ii) Sebolelo has 5 out of 8 parts.

Percentage = ⁵⁄₈ × 100% = 62.5%

(b) Lebeha invested part of M10 000 at 2.5% p.a. and the rest at 1.9% p.a. simple interest. Total interest after 3 years was M678. Calculate the amount invested at 2.5%.

1. Let the amount invested at 2.5% be x. The remaining amount is (10 000 − x).

2. Write the equation for the total interest earned over 3 years:

(^{x × 2.5 × 3}⁄₁₀₀) + (^{(10 000 − x) × 1.9 × 3}⁄₁₀₀) = 678

3. Multiply the entire equation by 100 to clear denominators:

7.5x + 5.7(10 000 − x) = 67 800

7.5x + 57 000 − 5.7x = 67 800

4. Simplify and solve for x:

1.8x = 10 800 ⟹ x = ^{10 800}⁄_1.8 = 6 000

(a) Machine A finishes a pile every 25 mins; Machine B every 30 mins. They start at 8:00 a.m.

(i) Find the Lowest Common Multiple (LCM) of 25 and 30:

Prime factors: 25 = 5² and 30 = 2 × 3 × 5

LCM = 2 × 3 × 5² = 150 minutes.

Calculate the piles completed by each machine within 150 minutes:

Machine A: 150 ÷ 25 = 6 piles

Machine B: 150 ÷ 30 = 5 piles

Total piles = 6 + 5 = 11

(ii) Convert 150 minutes into hours: 150 minutes = 2 hours and 30 minutes.

Add this duration to the starting time of 8:00 a.m. → 10:30 a.m.

(b) Mxolisi buys 30 pairs of shoes for M18 000. Sells 15 for M750 each, 10 for M700 each, and 5 for M650 each. Calculate percentage profit.

1. Calculate total sales revenue:

(15 × 750) + (10 × 700) + (5 × 650) = 11 250 + 7 000 + 3 250 = 21 500

2. Calculate absolute financial profit: 21 500 − 18 000 = 3 500

3. Calculate percentage profit over cost:

Percentage Profit = ^{3 500}⁄_{18 000} × 100% ≈ 19.44%

(c) Sempe has 3200 sheep and sells 19.5% of them. Find the remaining sheep.

Percentage of remaining sheep = 100% − 19.5% = 80.5%.

Remaining sheep = ^80.5⁄₁₀₀ × 3200 = 80.5 × 32 = 2576.

Note: Visual measuring questions depend on the physical paper dimensions. Measure directly on your question sheet.

(a) Measure and write distance BC:

Measure with your physical metric ruler on the question sheet. (Typically yields around 4.5 cm to 5.0 cm depending on the exact page scale.)

(b) Scale construction if actual distance AB = 30 km:

Measure physical line length AB on paper (e.g., if measured length AB = 6 cm):

Scale Factor = ^{30 km}⁄_{6 cm} = 5 km per cm

Answer: 1 cm is to 5 km (substitute your exact measured paper length here if different).

(c) Construction of bridge M equidistant from AB and AC:

"Equidistant from two intersecting lines" defines an angle bisector. Position your compass point at vertex A, draw an arc intersecting lines AB and AC, then draw intersecting arcs from those points to construct the bisecting line through angle ∠BAC. Mark the intersection near the river as M.

(d) Describe the locus represented by the perpendicular bisector of line AC:

The locus of all points that are equidistant from point A and point C.

(a) (i) Show that AD = 18 cm:

Drop a perpendicular line from D horizontally to the vertical side BC to construct a right-angled triangle.

Horizontal base segment = 26.8 − 16 = 10.8 cm

Vertical side segment = 14.4 cm

Apply Pythagoras' theorem to find AD:

AD = √(10.8² + 14.4²) = √(116.64 + 207.36) = √324 = 18 cm

(a) (ii) Calculate the perimeter of the shaded region:

1. Outer straight edges = AB + BC + AD = 26.8 + 14.4 + 18 = 59.2 cm.

2. Curved semicircle arc length = ¹⁄₂ × π × d = ¹⁄₂ × 3.142 × 16 = 25.136 cm.

3. Total Perimeter = 59.2 + 25.136 = 84.336 cm.

(b) 48 cubes of side a cm fit exactly into a cuboid of dimensions (5a + 8) by 2a by (5a − 8). Find a.

1. Volume of one cube = a³.

2. Total Volume of 48 cubes = 48a³.

3. Equate this to the algebraic volume of the container:

(5a + 8)(2a)(5a − 8) = 48a³

4. Factor out and simplify using the difference of two squares:

2a × (25a² − 64) = 48a³

5. Divide both sides by 2a (since a > 0):

25a² − 64 = 24a²

6. Solve for a:

25a² − 24a² = 64 ⟹ a² = 64 ⟹ a = 8 cm

(a) Given 2r − 17 = 2p + 3, find the value of r − p.

Rearrange terms to group variables on the left:

2r − 2p = 3 + 17 ⟹ 2(r − p) = 20 ⟹ r − p = 10

(b) Factorise fully: (a − t)² − 4

Express as A² − B² where A = (a − t) and B = 2:

[(a − t) − 2][(a − t) + 2]

(a − t − 2)(a − t + 2)

(c) 360 cubes of 1 cm form a cuboid. The largest base area is a perfect square. Find the smallest possible height.

Volume = Base Area × Height = 360. To minimize the height, we need to maximize the base area such that it is a perfect square factor of 360.

Perfect square factors of 360 are: 1, 4, 9, 36.

The largest perfect square factor is 36.

Smallest Height = ³⁶⁰⁄₃₆ = 10 cm.

(d) Make r² the subject of 4k + 3r² = 10:

Isolate the term containing r²:

3r² = 10 − 4k ⟹ r² = ^{10 − 4k}⁄₃

Given f(x) = ^{2x − 7}⁄₃ and g(x) = x − ²⁄₃:

(a) (i) Find f(¹⁄₂):

Substitute x = ¹⁄₂ into f(x):

f(¹⁄₂) = ^{2(1/2) − 7}⁄₃ = ^{1 − 7}⁄₃ = ⁻⁶⁄₃ = −2

(a) (ii) Find x when f(x) = g(x):

Equate the two functions:

^{2x − 7}⁄₃ = x − ²⁄₃

Multiply the entire equation by 3 to clear the denominators:

2x − 7 = 3x − 2 ⟹ −7 + 2 = 3x − 2x ⟹ x = −5

(b) Solve x² + x = 132:

Rearrange into standard form ax² + bx + c = 0:

x² + x − 132 = 0

Find two numbers that multiply to −132 and add to 1: these are +12 and −11.

(x + 12)(x − 11) = 0 ⟹ x = −12 or x = 11

Given $\overrightarrow{PQ}=\left(\begin{array}{c}14\\ 7\end{array}\right)$, $\overrightarrow{PR}=\left(\begin{array}{c}\mathrm{-6}\\ \mathrm{-3}\end{array}\right)$, $\overrightarrow{RS}=\left(\begin{array}{c}12\\ \mathrm{-5}\end{array}\right)$:

(a) (i) Find QR:

Using vector addition rules: QR = QP + PR = −PQ + PR.

QR = ( −14 ; −7 ) + ( −6 ; −3 ) = ( −20 ; −10 )

(a) (ii) Find |RS|:

Calculate the magnitude of the vector using Pythagoras' theorem:

|RS| = √(12² + (−5)²) = √(144 + 25) = √169 = 17

17 units

(b) Triangle with base 22 units and area 110 units². Find the simplest ratio of height to base.

1. Calculate height using the area formula: Area = ¹⁄₂ × base × height.

110 = ¹⁄₂ × 22 × height ⟹ 110 = 11 × height ⟹ height = 10

2. Write the ratio of height to base: 10 : 22.

3. Simplify by dividing by 2 → 5 : 11.

(c) Work out ( 1.2   0 ; −2.7   2.1 ) − ( −3.3   −0.5 ; 2.4   0 ):

Subtract corresponding elements:

( 1.2 − (−3.3)   0 − (−0.5) ; −2.7 − 2.4   2.1 − 0 ) = ( 4.5   0.5 ; −5.1   2.1 )

(a) Explain what M150 represents for Re-Connector:

Fixed standing connection charge (service fee before buying data).

(b) Write the linear cost equations for both companies:

Re-Connector: M = 5n + 150

Re-User: M = 5.5n + 120

(c) Calculate the cost for 45 data bundles from Re-User:

Substitute n = 45 into the Re-User cost formula:

M = 5.5(45) + 120 = 247.5 + 120 = 367.5

M367.50

(d) Which company is cheaper for 66 data bundles? Justify:

Calculate the total cost for both companies at n = 66:

Re-Connector: M = 5(66) + 150 = 330 + 150 = 480

Re-User: M = 5.5(66) + 120 = 363 + 120 = 483

Re-Connector, because its total cost (M480) is cheaper than Re-User's total cost (M483) for 66 bundles.

(a) State the modal salary:

Look for the salary value with the highest frequency count (6).

M16 000

(b) (i) Total number of football players:

Sum the frequencies: 5 + 4 + 6 + 4 + 2 + 3 = 24.

(b) (ii) Calculate the mean salary:

Mean = ^{Σ (Salary × Players)}⁄_{Total Players}

Σ = (8000×5) + (12000×4) + (16000×6) + (20500×4) + (22500×2) + (28500×3)

Σ = 40 000 + 48 000 + 96 000 + 82 000 + 45 000 + 85 500 = 396 500

Mean = ^{396 500}⁄₂₄ ≈ 16520.83

M16 520.83 (or M16 500 to 3 s.f.)

(c) (i) Probability salary is M8 000:

5 out of 24 players earn M8 000.

⁵⁄₂₄

(c) (ii) Probability salary is at least M20 000:

Count the players earning M20 500, M22 500, or M28 500: 4 + 2 + 3 = 9 players.

Simplify fraction: ⁹⁄₂₄ = ³⁄₈.