Solutions

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

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

Teboho's share = ³⁄₈ × 72 000 = 3 × 9 000 = 27 000

(ii) Sebolelo has 5 parts out of 8 total parts.

Percentage = ⁵⁄₈ × 100% = 62.5%

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

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

2. Write down the total interest equation over 3 years (I = ^{P × R × T}⁄₁₀₀):

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

2.5 × 3 × x + 1.9 × 3 × (10 000 − x) = 678 × 100

7.5x + 5.7(10 000 − x) = 67 800

3. Expand brackets and group like terms:

7.5x + 57 000 − 5.7x = 67 800

1.8x = 67 800 − 57 000 ⟹ 1.8x = 10 800

4. Isolate x:

x = ^{10 800}⁄_1.8 = 6 000

(c) A car's value depreciates at 5% per year. The current value is M83 800.48. Calculate its value 7 years ago.

1. Use the depreciation compound interest formula: Current Value = Original Value × (1 − r)ⁿ

2. Substitute known numbers:

83 800.48 = P × (1 − 0.05)⁷

83 800.48 = P × (0.95)⁷

3. Using a calculator: (0.95)⁷ ≈ 0.698337

P = ^{83 800.48}⁄_0.698337 = 120 000

(a) Show that 3x + 5y > 45:

Total mass = 6x + 10y. Since it's more than 90 kg:

6x + 10y > 90

Divide the entire inequality by 2:

3x + 5y > 45 (Proven)

(b) Write two inequalities:

More large (y) than small (x): y > x

Not more than 8 large packets: y ≤ 8

(c) Determine the feasible region:

Plot boundary lines on the grid:

3x + 5y = 45 (passes through (0, 9) and (15, 0))

Line y = x, and horizontal line y = 8.

Shade out the unwanted parts to leave region R unshaded, bounded by these lines.

(d) Find the smallest total mass of beans the farmer can sell:

The feasible region is bounded by:

3x + 5y > 45 (strict), y > x, y ≤ 8, and x ≥ 0, y ≥ 0.

Since y ≤ 8, and we need y > x, the maximum y is 8 and the largest valid x is 7.

Test integer points near the boundary of the feasible region:

The lowest valid integer point where 3x + 5y > 45 and y > x is x = 2, y = 8:

3(2) + 5(8) = 6 + 40 = 46 > 45 ✓

Total mass = 6(2) + 10(8) = 12 + 80 = 92 kg

Check the next candidate: x = 1, y = 8 gives 3 + 40 = 43, which is not > 45. ✗

Check y = 9: not allowed since y ≤ 8. ✗

So the smallest total mass is 92 kg.

(a) Calculate the sector angle for the speed range 60 < s ≤ 70:

Total number of cars = 3 + 14 + 6 + 12 + 15 = 50. The frequency from 60 to 70 is the combined sum of columns 60 < s ≤ 64 and 64 < s ≤ 70: 6 + 12 = 18.

Sector Angle = ¹⁸⁄₅₀ × 360° = 129.6°

(b) Estimate the number of cars whose speed was less than 68 km/h:

Includes all columns up to 64: 3 + 14 + 6 = 23 cars. Add a proportional fraction of the 64 < s ≤ 70 class:

Cars = 23 + (^{68 − 64}⁄_{70 − 64} × 12) = 23 + (⁴⁄₆ × 12) = 23 + 8 = 31

(c) Two cars are chosen at random. Probability that one is > 70 km/h and the other is ≤ 60 km/h:

Total = 50. Number of cars > 70 = 15. Number of cars ≤ 60 = 3 + 14 = 17.

Order matters (either first car is slow and second is fast, or vice versa):

P = (¹⁷⁄₅₀ × ¹⁵⁄₄₉) + (¹⁵⁄₅₀ × ¹⁷⁄₄₉) = 2 × (²⁵⁵⁄₂₄₅₀) = ⁵¹⁰⁄₂₄₅₀ = ⁵¹⁄₂₄₅

(d) Calculate an estimate of the mean speed:

Find interval midpoints (x): 45, 55, 62, 67, 75.

Σf·x = (3 × 45) + (14 × 55) + (6 × 62) + (12 × 67) + (15 × 75)

Σf·x = 135 + 770 + 372 + 804 + 1125 = 3206

Estimated Mean = ³²⁰⁶⁄₅₀ = 64.12 km/h

(e) Complete the histogram:

Calculate frequency density (FD = ^Frequency⁄_Width) for each column block:

40–50: ³⁄₁₀ = 0.3 (given reference)

50–60: ¹⁴⁄₁₀ = 1.4

60–64: ⁶⁄₄ = 1.5

64–70: ¹²⁄₆ = 2.0

70–80: ¹⁵⁄₁₀ = 1.5

Draw rectangles matching these heights on your provided grid sheet.

(f) Draw a frequency polygon:

Join the midpoints of the top of each histogram bar with straight lines, starting and ending at the horizontal axis.

(a) (i) Area of the sector:

Area = ^θ⁄₃₆₀ × πr² = ¹²⁰⁄₃₆₀ × 3.142 × 14² = ¹⁄₃ × 3.142 × 196 = 205.28 cm²

205.3 cm²

(a) (ii) Show that arc length is 29.3 cm:

Arc Length = ¹²⁰⁄₃₆₀ × 2 × π × r = ¹⁄₃ × 2 × 3.142 × 14 = 29.325 cm → 29.3 cm (Proven)

(a) (iii) Calculate the radius of the cone:

The base circumference of the cone is exactly equal to the sector's arc length:

2πr_cone = 29.325 ⟹ r_cone = ^29.325⁄_{2 × 3.142} ≈ 4.67 cm

(b) Volume of water needed to fill up the cone:

1. Using similar triangles, volume scale factor is the cube of the linear scale factor:

k³ = (^r_top⁄_rwater)³ = (³⁄₂)³ = ²⁷⁄₈

2. Total Volume = Water Volume × ²⁷⁄₈ = 33.5 × ²⁷⁄₈ = 113.0625 cm³

3. Water needed to top up = Total Volume − Current Water = 113.0625 − 33.5 = 79.5625 cm³

79.6 cm³

(c) Volume of cylinder:

Volume = πr²h = 3.142 × 0.8² × 2.5 = 3.142 × 0.64 × 2.5 = 5.0272 m³

5.03 m³

(a) Flower is neither white nor yellow:

Combined probability of being white or yellow = 0.4 + 0.5 = 0.9. Remainder = 1 − 0.9 = 0.1.

(b) Plant produces flowers with white colors:

P(produces flowers) × P(white)

= 0.7 × 0.4

= 0.28

(c) Plant produces flowers which are either white or yellow:

P(produces flowers) × P(white or yellow)

= 0.7 × (0.4 + 0.5)

= 0.7 × 0.9

= 0.63

Given 4 sin(x − 3)° = 3 + 2y.

(a) Find y when x = 33°:

4 sin(33 − 3)° = 3 + 2y ⟹ 4 sin(30°) = 3 + 2y

Since sin(30°) = 0.5:

4(0.5) = 3 + 2y ⟹ 2 = 3 + 2y ⟹ −1 = 2y ⟹ y = −0.5

(b) Make x the subject of the formula:

4 sin(x − 3)° = 3 + 2y

sin(x − 3)° = ^{3 + 2y}⁄₄

(x − 3) = sin⁻¹(^{3 + 2y}⁄₄)

x = sin⁻¹(^{3 + 2y}⁄₄) + 3

(a) (i) Complete the Venn Diagram:

Degree and Masters = 9. The overlap intersection of D and M already contains 3, so the missing overlapping partition element = 9 − 3 = 6.

Total Degree = 18. The circle D currently contains 4 + 3 + 6 = 13. Left over pure degree value = 18 − 13 = 5.

Total exactly one certificate = 20. This means (pure D) + (pure H) + (pure M) = 20 ⟹ 5 + (pure H) + 7 = 20 ⟹ pure H = 8.

Fill in these values in their respective regions.

(a) (ii) Masters only count:

7

(a) (iii) Honours only set notation:

H ∩ D′ ∩ M′

(a) (iv) Find n(H′):

Count everything outside circle H: 5 + 6 + 7 + 8 = 26.

(b) Shade region (D′ ∪ H′) ∩ M:

This matches everything inside circle M except for the small intersection piece where all three sets meet (D ∩ H ∩ M).

Shade everything inside M except that central piece.

(a) Matrix operation:

( 1.2   0 ; −2.7   2.1 ) − ( −3.3   −0.5 ; 2.4   0 )

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

= ( 4.5   0.5 ; −5.1   2.1 )

(b) (i) Draw image translation: Move each vertex of triangle A 4 units left and 6 units down.

New coordinates: (1, 1) → (−3, −5), (2, 3) → (−2, −3), (1, 3) → (−3, −3).

(b) (ii) Find vector magnitude:

√((−4)² + (−6)²) = √(16 + 36) = √52 ≈ 7.21 units

(c) Describe mapping transformation from A onto B:

Triangle B is larger than A, so it's an enlargement. Ray lines through corresponding corners intersect at the origin (0, 0). Base length goes from 1 to 3 units, so scale factor is 3.

Enlargement, centre (0, 0), scale factor 3.

(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: 0, 2, 4, 6

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

3. Quadratic sequence of the form an² + bn + c. Since 2a = 2 ⟹ a = 1.

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

For n = 1: −1 − 1 = −2

For n = 2: −1 − 4 = −5

For n = 3: 1 − 9 = −8

5. Find the linear rule for −2, −5, −8, …: the difference is −3, so the 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) T_n = 4 + 3(n − 1) = 4 + 3n − 3 = 3n + 1.

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

(a) Measure and write the distance BC:

Measure with your metric ruler on the printed map. Record the value in cm.

(b) Find the scale in cm to km:

The actual distance from A to B is 30 km.

Measure the physical length of AB on the map (e.g., if AB = 6 cm):

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

1 cm is to 5 km (substitute your exact measured AB length).

(c) Bridge 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) Calculate the length of AB:

In right-angled triangle △ABD, use Pythagoras' theorem:

AB = √(AD² + BD²) = √(75² + 56²) = √(5625 + 3136) = √8761 ≈ 93.6 m

(a) (ii) Calculate the size of angle BCD:

In triangle △BCD, we know:

BD = 56 m (opposite angle C), BC = 62 m (opposite angle D), ∠BDC = 46°.

Use the Sine Rule:

^sin∠BCD⁄_BD = ^sin∠BDC⁄_BC

^sin∠BCD⁄₅₆ = ^sin(46°)⁄₆₂

sin∠BCD = ^{56 × sin(46°)}⁄₆₂ ≈ ^{56 × 0.71934}⁄₆₂ ≈ 0.6496

∠BCD = sin⁻¹(0.6496) ≈ 40.5°

(b) Calculate the shortest distance from D to AB:

Shortest distance is a perpendicular line dropped from D to line AB. First find angle ∠DAB:

tan∠DAB = ⁵⁶⁄₇₅ ⟹ ∠DAB ≈ 36.75°

Shortest Distance = AD × sin∠DAB = 75 × sin(36.75°) ≈ 75 × ⁵⁶⁄_93.6 ≈ 44.87 m

44.9 m

Given f(x) = 3x − 1, g(x) = x² − 1, h(x) = 3^x.

(a) (i) Find h(2):

h(2) = 3² = 9

(a) (ii) Find f⁻¹(x):

y = 3x − 1

y + 1 = 3x

x = ^{y + 1}⁄₃

f⁻¹(x) = ^{x + 1}⁄₃

(a) (iii) Find fg(x):

fg(x) = 3(x² − 1) − 1 = 3x² − 3 − 1 = 3x² − 4

(b) (i) Find x when h(x) = 81:

3^x = 81 ⟹ 3^x = 3⁴ ⟹ x = 4

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

x² − 1 = 3x − 1

x² − 3x = 0

x(x − 3) = 0

x = 0 or x = 3

(a) Rate expression for second field:

⁸⁄_{x + 20}

(b) Form an equation in x and show it simplifies to 3x² + 64x − 240 = 0:

⁶⁄_x − ⁸⁄_{x + 20} = 1.5

Multiply out by common denominator x(x + 20):

6(x + 20) − 8x = 1.5x(x + 20)

6x + 120 − 8x = 1.5x² + 30x

−2x + 120 = 1.5x² + 30x

1.5x² + 32x − 120 = 0

Multiply the entire equation by 2 to clear decimal coefficients:

3x² + 64x − 240 = 0 (Proven)

(c) Solve 3x² + 64x − 240 = 0:

Use the quadratic formula:

x = ^{−64 ± √(642 − 4(3)(−240))}⁄₂₍₃₎ = ^{−64 ± √(4096 + 2880)}⁄₆ = ^{−64 ± √6976}⁄₆

x = ^{−64 ± 83.522}⁄₆ ⟹ x ≈ 3.25 or x ≈ −24.59

x = 3.25 or x = −24.6

(d) Find rate on the first field:

Use valid positive option x = 3.2538:

Rate = ⁶⁄_3.2538 ≈ 1.84

1.84 litres per hectare

(a) Complete the table:

Plug missing inputs into y = x² − 2x + 3:

For x = −1.5: (−1.5)² − 2(−1.5) + 3 = 2.25 + 3 + 3 = 8.25

For x = 0: 3

For x = 2: 2² − 2(2) + 3 = 3

(b) Draw graph: Plot these points symmetrically onto your provided Cartesian axis plane.

(c) Find minimum value:

At vertex point (1, 2), the smallest y value is 2.

(d) Tangent gradient estimate at x = 1.5:

Draw a straight line touching the curve perfectly at x = 1.5.

Calculate gradient ^Δy⁄_Δx ≈ 1.

1 (values around 0.8 to 1.2 are accepted)

(e) Solve x² − 2x + 1 = ²⁄₃x + 6 by drawing a suitable line:

Rearrange the target equation to match the drawn curve y = x² − 2x + 3:

x² − 2x + 1 = ²⁄₃x + 6

Add 2 to both sides:

x² − 2x + 3 = ²⁄₃x + 8

So draw the straight line y = ²⁄₃x + 8 on the same axes.

Read the x-coordinates where this line intersects the curve:

x ≈ −1.1 or x ≈ 3.7