Unit 1 Review — General Maths

This review covers all three topics of Unit 1: Consumer Arithmetic (percentages, income, simple and compound interest, comparing financial options), Shape and Measurement (units, perimeter, area, surface area, volume, similarity), and Linear Equations and Graphs (solving equations, graphing, models, simultaneous equations). Allow approximately 90–120 minutes for this review.

Review Questions

Q1 — GST and discount
Fluency
A laptop has a pre-GST price of $1299.

(a) Find the total price including 10% GST.

(b) The laptop is then discounted by 15%. Find the sale price before GST is applied.

(a) Price including 10% GST:

Total = $1299 × 1.10 = $1428.90

(b) Sale price after 15% discount (pre-GST):

Discount = $1299 × 0.15 = $194.85

Sale price = $1299 − $194.85 = $1104.15

(Or: $1299 × 0.85 = $1104.15)
Q2 — Simple interest
Fluency
$5500 is invested at 4.2% p.a. simple interest for 3 years.

(a) Find the interest earned.

(b) Find the total amount at the end of the investment.

P = $5500, r = 0.042, t = 3 years

(a) Interest earned:

I = Prt = 5500 × 0.042 × 3 = $693

(b) Total amount:

A = P + I = $5500 + $693 = $6193
Q3 — Compound interest (quarterly)
Fluency
$9000 is invested at 5% p.a. compounding quarterly for 2 years. Find the final amount.

P = $9000, r = 0.05, n = 4 (quarterly), t = 2 years

A = P(1 + r/n)^nt

A = 9000 × (1 + 0.05/4)^4×2

A = 9000 × (1.0125)⁸

(1.0125)⁸ = 1.10449 (approx)

A = 9000 × 1.10449 = $9940.41
Q4 — Net income
Understanding
Sam earns $78 000 per year. He pays $17 172 in income tax and 10% superannuation is withheld.

(a) Find Sam's net (take-home) income.

(b) Express net income as a percentage of gross income.

(a) Net income:

Superannuation = 10% × $78 000 = $7800

Total deductions = $17 172 + $7800 = $24 972

Net income = $78 000 − $24 972 = $53 028

(b) Net as % of gross:

(53 028 ÷ 78 000) × 100 = 68.0%
Q5 — Comparing investments
Understanding
$12 000 is invested for 5 years. Two options are available:
- Option A: 5.5% p.a. simple interest
- Option B: 5.2% p.a. compounding monthly
(a) Find the final amount for Option A.

(b) Find the final amount for Option B.

(c) Find the Effective Annual Rate (EAR) for Option B.

(a) Option A — Simple interest:

I = 12000 × 0.055 × 5 = $3300

Final amount = $12 000 + $3300 = $15 300

(b) Option B — Compound interest monthly:

A = 12000 × (1 + 0.052/12)^12×5

= 12000 × (1.004333...)⁶⁰

(1.004333)⁶⁰ = 1.29596 (approx)

A = 12000 × 1.29596 = $15 551.52

(c) EAR for Option B:

EAR = (1 + 0.052/12)¹² − 1

= (1.004333)¹² − 1 = 1.05328 − 1 = 0.05328

EAR = 5.33% (compared to nominal 5.2%)

Option B gives a higher final amount ($15 551.52 vs $15 300).

Q6 — Doubling investment: compound growth

Problem Solving

$8000 is invested at 4% p.a. compounding annually. Find the first year in which the investment exceeds $12 000. Use a table of values to show your working.

A = 8000 × (1.04)^t

We need A > 12 000, so (1.04)^t > 1.5

Year (t)	Amount ($)
5	8000 × (1.04)⁵ = 8000 × 1.2167 = $9733
8	8000 × (1.04)⁸ = 8000 × 1.3686 = $10 949
10	8000 × (1.04)¹⁰ = 8000 × 1.4802 = $11 842
11	8000 × (1.04)¹¹ = 8000 × 1.5394 = $12 315

At t = 10: $11 842 < $12 000

At t = 11: $12 315 > $12 000

The investment first exceeds $12 000 in Year 11.

Q7 — Unit conversions
Fluency
Convert the following:

(a) 3.2 km to metres

(b) 850 mm to centimetres

(c) 4500 cm² to m²

(d) 4.7 m² to cm²

(a) 3.2 km to m:

3.2 × 1000 = 3200 m

(b) 850 mm to cm:

850 ÷ 10 = 85 cm

(c) 4500 cm² to m²:

4500 ÷ 10 000 = 0.45 m²

(1 m = 100 cm, so 1 m² = 10 000 cm²)

(d) 4.7 m² to cm²:

4.7 × 10 000 = 47 000 cm²
Q8 — Area of a trapezium
Fluency
Find the area of a trapezium with parallel sides of 7 cm and 13 cm, and a perpendicular height of 8 cm.

Area = ½ × (a + b) × h

Area = ½ × (7 + 13) × 8

Area = ½ × 20 × 8

Area = 80 cm²
Q9 — Cylinder surface area and volume
Understanding
A cylinder has radius 4.5 cm and height 12 cm. Find:

(a) The total surface area.

(b) The volume.

r = 4.5 cm, h = 12 cm

(a) Total surface area:

SA = 2πr² + 2πrh

= 2π(4.5)² + 2π(4.5)(12)

= 2π(20.25) + 2π(54)

= 40.5π + 108π

= 148.5π

= 466.6 cm²

(b) Volume:

V = πr²h = π × (4.5)² × 12

= π × 20.25 × 12

= 243π

= 763.4 cm³
Q10 — Composite volume
Understanding
A composite solid consists of a cylinder (radius 3 m, height 8 m) with a hemisphere (radius 3 m) sitting on top. Find the total volume of the solid.

Volume of cylinder:

V_cyl = πr²h = π × 9 × 8 = 72π m³

Volume of hemisphere:

V_hemi = ½ × &frac43;πr³ = ⅔πr³ = ⅔π × 27 = 18π m³

Total volume:

V = 72π + 18π = 90π = 282.7 m³
Q11 — Similar prisms and scale factors
Understanding
Two similar rectangular prisms are related by a scale factor of 2.5. The smaller prism has dimensions 4 cm × 3 cm × 6 cm.

(a) Find the dimensions of the larger prism.

(b) State the volume scale factor.

(c) Find the volume of the larger prism.

(a) Dimensions of larger prism:

Each dimension is multiplied by 2.5:

4 × 2.5 = 10 cm; 3 × 2.5 = 7.5 cm; 6 × 2.5 = 15 cm

Larger prism: 10 cm × 7.5 cm × 15 cm

(b) Volume scale factor:

Volume scale factor = (linear scale factor)³ = 2.5³ = 15.625

(c) Volume of larger prism:

V_small = 4 × 3 × 6 = 72 cm³

V_large = 72 × 15.625 = 1125 cm³

(Check: 10 × 7.5 × 15 = 1125 cm³ ✓)
Q12 — Cone and cylinder composite surface area
Problem Solving
A cone (radius 5 m, height 12 m) sits on top of a cylinder (radius 5 m, height 8 m). Find:

(a) The total external surface area (lateral surfaces of both shapes plus the base of the cylinder only — the flat circle where they meet is internal).

(b) The cost to paint the external surface at $4.50 per m².

(a) Total external surface area:

Lateral surface of cylinder:

= 2πrh = 2π × 5 × 8 = 80π m²

Base of cylinder (circular bottom):

= πr² = π × 25 = 25π m²

Lateral surface of cone (slant height first):

Slant height l = √(r² + h²) = √(25 + 144) = √169 = 13 m

Lateral SA of cone = πrl = π × 5 × 13 = 65π m²

Total:

SA = 80π + 25π + 65π = 170π = 534.1 m²

(b) Painting cost:

Cost = 534.1 × $4.50 = $2403.45
Q13 — Garden measurement problem
Problem Solving
A rectangular garden is 12 m × 8 m. A circular pond of radius 2.5 m sits in the centre.

(a) Find the garden area excluding the pond.

(b) Find the length of fencing needed for the entire rectangular perimeter.

(c) If garden turf costs $18 per m², find the total cost to turf the garden area excluding the pond.

(a) Garden area excluding pond:

Rectangle area = 12 × 8 = 96 m²

Pond area = πr² = π × (2.5)² = 6.25π = 19.63 m²

Turfed area = 96 − 19.63 = 76.37 m²

(b) Rectangular perimeter:

P = 2(12 + 8) = 2 × 20 = 40 m

(c) Turf cost:

Cost = 76.37 × $18 = $1374.66
Q14 — Solving linear equations
Fluency
(a) Solve: 5x − 3 = 3x + 11

(b) Solve: (2x + 1)/3 = (x − 2)/2

(a) 5x − 3 = 3x + 11:

5x − 3x = 11 + 3

2x = 14

x = 7

(b) (2x + 1)/3 = (x − 2)/2:

Multiply both sides by 6 (LCM of 3 and 2):

2(2x + 1) = 3(x − 2)

4x + 2 = 3x − 6

4x − 3x = −6 − 2

x = −8
Q15 — Properties of a linear function
Fluency
For the function y = −3x + 9, find:

(a) The gradient m

(b) The y-intercept c

(c) The x-intercept

(d) The y-value when x = −2

(a) Gradient: m = −3

(b) y-intercept: c = 9 (when x = 0, y = 9)

(c) x-intercept (set y = 0):

0 = −3x + 9 → 3x = 9 → x = 3

(d) y when x = −2:

y = −3(−2) + 9 = 6 + 9 = 15
Q16 — Perpendicular line equation
Understanding
Find the equation of the line through (2, −5) that is perpendicular to y = 4x + 1.

The gradient of y = 4x + 1 is m = 4.

Perpendicular gradient = −1/4 (negative reciprocal).

Using point-gradient form with (2, −5):

y − (−5) = −¼(x − 2)

y + 5 = −¼x + ½

y = −¼x + ½ − 5

y = −¼x − &frac92;

Or equivalently: y = −0.25x − 4.5
Q17 — Linear model for pool draining
Understanding
A swimming pool holds 9600 litres at the start. After 5 hours of draining, it holds 2800 litres. Assuming the volume decreases linearly:

(a) Write a linear model for volume V (litres) at time t (hours).

(b) Find when the pool reaches 500 litres.

(a) Linear model:

Points: (0, 9600) and (5, 2800)

Gradient = (2800 − 9600) ÷ (5 − 0) = −6800 ÷ 5 = −1360

V = 9600 − 1360t

(b) When V = 500 litres:

9600 − 1360t = 500

1360t = 9100

t = 9100 ÷ 1360 = 6.69 hours (approx. 6 hours 41 minutes)
Q18 — Simultaneous equations: ticket sales
Understanding
Adult tickets cost $28 and child tickets cost $14. A total of 35 tickets were sold and the total revenue was $700. Find the number of adult and child tickets sold.

Let a = number of adults, c = number of children.

Equation 1 (total tickets): a + c = 35 ...(1)

Equation 2 (total revenue): 28a + 14c = 700 ...(2)

From (1): c = 35 − a

Substitute into (2): 28a + 14(35 − a) = 700

28a + 490 − 14a = 700

14a = 210

a = 15

c = 35 − 15 = 20

15 adult tickets and 20 child tickets were sold.

Check: 28 × 15 + 14 × 20 = 420 + 280 = $700 ✓
Q19 — Comparing two training plans
Problem Solving
Two gym training plans are available:
- Plan A: $120 joining fee + $15 per week
- Plan B: $40 joining fee + $25 per week
(a) Find the number of weeks at which both plans cost the same.

(b) Which plan is cheaper for a full year (52 weeks)?

(a) Cost equations:

Plan A: C_A = 15w + 120

Plan B: C_B = 25w + 40

Set equal: 15w + 120 = 25w + 40

80 = 10w → w = 8 weeks

(b) Cost at 52 weeks:

Plan A: 15 × 52 + 120 = 780 + 120 = $900

Plan B: 25 × 52 + 40 = 1300 + 40 = $1340

Plan A is cheaper at 52 weeks (saving $440).

Plan B is cheaper for fewer than 8 weeks; Plan A is cheaper for more than 8 weeks.
Q20 — Finding k from gradient condition
Problem Solving
A line passes through the points (k, 2) and (3, −k) and has a gradient of 3. Find k and write the equation of the line.

Finding k:

Gradient = (y₂ − y₁) ÷ (x₂ − x₁) = 3

(−k − 2) ÷ (3 − k) = 3

−k − 2 = 3(3 − k)

−k − 2 = 9 − 3k

−k + 3k = 9 + 2

2k = 11

k = 5.5

So k = 5.5 and the points are (5.5, 2) and (3, −5.5).

Equation of the line:

Using point (3, −5.5) and m = 3:

y − (−5.5) = 3(x − 3)

y + 5.5 = 3x − 9

y = 3x − 14.5
Q21 — Ratio dimensions: SA and volume
Problem Solving
A rectangular block has dimensions in ratio 1 : 2 : 3, with the smallest dimension being 4 cm.

(a) State the dimensions of the block.

(b) Find the surface area.

(c) Find the volume.

(d) If the block is scaled by a factor of 1.5, find the new volume.

(a) Dimensions:

Smallest = 4 cm (ratio 1), so dimensions are 4 × 8 × 12 cm.

(b) Surface area:

SA = 2(lw + lh + wh) = 2(4×8 + 4×12 + 8×12)

= 2(32 + 48 + 96) = 2 × 176 = 352 cm²

(c) Volume:

V = 4 × 8 × 12 = 384 cm³

(d) New volume after scale factor 1.5:

Volume scale factor = 1.5³ = 3.375

New volume = 384 × 3.375 = 1296 cm³
Q22 — Linear depreciation model
Problem Solving
A car is purchased for $32 000 and depreciates at a constant $3200 per year.

(a) Write a linear model for the car’s value V(t) in dollars after t years.

(b) Find the value after 6 years.

(c) When does the car’s value reach $8000?

(d) Briefly comment on whether linear depreciation is a realistic model.

(a) Linear model:

V(t) = 32 000 − 3200t

(b) Value after 6 years:

V(6) = 32 000 − 3200 × 6 = 32 000 − 19 200 = $12 800

(c) When V = $8000:

32 000 − 3200t = 8000

3200t = 24 000

t = 7.5 years

(d) Realism of linear depreciation:

Linear depreciation assumes the car loses the same dollar amount each year, which does not reflect reality. Cars typically depreciate faster in their early years (e.g., reducing by a percentage of their current value — exponential/percentage depreciation). Linear depreciation is a simplification useful for accounting purposes but less accurate as a predictive model.
Q23 — Compound interest: finding principal for a target
Problem Solving
A person wants to have $50 000 in 10 years. Two investment options are available:
- Option A: 5% p.a. compounding annually
- Option B: 4.8% p.a. compounding monthly
(a) Find the principal required for each option.

(b) Which option requires less initial investment?

(a) Finding principal for each option:

A = P(1 + r/n)^nt → P = A ÷ (1 + r/n)^nt

Option A (5%, annual):

P = 50 000 ÷ (1.05)¹⁰

(1.05)¹⁰ = 1.62889

P = 50 000 ÷ 1.62889 = $30 696.79

Option B (4.8%, monthly):

P = 50 000 ÷ (1 + 0.048/12)¹²⁰

= 50 000 ÷ (1.004)¹²⁰

(1.004)¹²⁰ = 1.61414

P = 50 000 ÷ 1.61414 = $30 974.34

(b) Option A requires less investment ($30 697 vs $30 974 — a difference of $277.55).

Despite the higher rate, Option A (annual compounding) requires a smaller principal because the nominal 5% annual rate is effectively higher than the monthly-compounded 4.8%.
Q24 — Triangular distance using trigonometry
Problem Solving
Three towns form a triangle. Town A to Town B is 85 km. Town A to Town C is 110 km. The angle at A (between AB and AC) is 72°.

(a) Find the distance from B to C.

(b) Find the area of the triangle formed by the three towns.

(c) If B is due north of A, find the bearing of C from B.

(a) Distance BC (cosine rule):

a = AB = 85 km, b = AC = 110 km, angle A = 72°

BC² = AB² + AC² − 2 × AB × AC × cos A

BC² = 85² + 110² − 2 × 85 × 110 × cos(72°)

BC² = 7225 + 12 100 − 18 700 × 0.3090

BC² = 19 325 − 5778.3 = 13 546.7

BC = √13 546.7 = 116.4 km

(b) Area of triangle:

Area = ½ × AB × AC × sin A

= ½ × 85 × 110 × sin(72°)

= 4675 × 0.9511 = 4446 km²

(c) Bearing of C from B:

B is due north of A (A to B bearing = 000°).

The angle at A between north (direction to B) and direction to C is 72°.

So C lies on a bearing of 072° from A.

Find angle ABC using sine rule:

sin(angle B)/AC = sin A/BC

sin B = 110 × sin(72°) ÷ 116.4 = 110 × 0.9511 ÷ 116.4 = 0.8985

Angle B = sin⁻¹(0.8985) = 64.0°

From B, looking south (towards A) is bearing 180°. C is at angle B = 64.0° to the right (east) of the line BA from B’s perspective.

Direction from B to A = 180°. Bearing of C from B = 180° − 64.0° = 116.0°
Q25 — Combined financial and measurement problem
Problem Solving
A developer buys a 120 m × 80 m rectangular block of land for $1 200 000. They plan to build on 60% of the land and leave the rest as open space.

(a) Find the total land area.

(b) Find the price per m² paid for the land.

(c) The developer finances the purchase with a loan at 6% p.a. compounding monthly for 3 years. Find the total amount owing at the end of 3 years (interest only — no repayments).

(d) If the development sells for $2 400 000 after 3 years, find the gross profit (sale price minus total amount owing on loan).

(a) Total land area:

Area = 120 × 80 = 9600 m²

(b) Price per m²:

Price/m² = $1 200 000 ÷ 9600 = $125/m²

(c) Loan amount owing after 3 years:

P = $1 200 000, r = 0.06, n = 12, t = 3 years

A = 1 200 000 × (1 + 0.06/12)³⁶

= 1 200 000 × (1.005)³⁶

(1.005)³⁶ = 1.19668

A = 1 200 000 × 1.19668 = $1 436 016

(d) Gross profit:

Gross profit = Sale price − loan amount

= $2 400 000 − $1 436 016 = $963 984

Unit 1 Review — General Maths

Review Questions

Q1 — GST and discount

Q2 — Simple interest

Q3 — Compound interest (quarterly)

Q4 — Net income

Q5 — Comparing investments

Q6 — Doubling investment: compound growth

Q7 — Unit conversions

Q8 — Area of a trapezium

Q9 — Cylinder surface area and volume

Q10 — Composite volume

Q11 — Similar prisms and scale factors

Q12 — Cone and cylinder composite surface area

Q13 — Garden measurement problem

Q14 — Solving linear equations

Q15 — Properties of a linear function

Q16 — Perpendicular line equation

Q17 — Linear model for pool draining

Q18 — Simultaneous equations: ticket sales

Q19 — Comparing two training plans

Q20 — Finding k from gradient condition

Q21 — Ratio dimensions: SA and volume

Q22 — Linear depreciation model

Q23 — Compound interest: finding principal for a target

Q24 — Triangular distance using trigonometry

Q25 — Combined financial and measurement problem