Unit 2 Review — General Maths

This review covers all four topics of Unit 2: Applications of Linear Equations (linear models, piecewise functions, step graphs, break-even), Applications of Trigonometry (right-angle trig, elevation/depression, bearings, sine/cosine rules, area), Matrices (operations, multiplication, inverse, systems), and Univariate Data (summary statistics, distributions, comparisons). Allow approximately 90–120 minutes for this review.

Review Questions

Q1 — Break-even analysis
Fluency
A business has total cost TC = 8x + 1600 and revenue R = 24x.

(a) Find the break-even quantity.

(b) Find the profit when x = 150 units.

(a) Break-even (TC = R):

8x + 1600 = 24x

1600 = 16x

x = 100 units

(b) Profit at x = 150:

Profit = R − TC = 24x − (8x + 1600) = 16x − 1600

= 16 × 150 − 1600 = 2400 − 1600 = $800
Q2 — Piecewise phone plan
Fluency
A mobile phone plan charges $40 per month. This includes 100 minutes of calls. Any additional minutes cost 10c each.

(a) Find the monthly bill for 80 minutes of calls.

(b) Find the monthly bill for 160 minutes of calls.

(a) Bill for 80 minutes:

80 min ≤ 100 min (within the included allowance)

Bill = $40.00

(b) Bill for 160 minutes:

Extra minutes = 160 − 100 = 60 minutes at $0.10 each

Extra charge = 60 × $0.10 = $6.00

Bill = $40 + $6 = $46.00
Q3 — Step graph parking
Understanding
A car park uses stepped hourly pricing:
- Up to 1 hour: $5
- More than 1 hour up to 2 hours: $9
- More than 2 hours up to 3 hours: $12
- More than 3 hours up to 4 hours: $15
(a) Find the cost for 1 hour 30 minutes.

(b) Find the cost for exactly 3 hours.

(a) Cost for 1 hr 30 min:

1.5 hr > 1 hr, and 1.5 hr ≤ 2 hr → falls in second bracket

Cost = $9

(b) Cost for exactly 3 hours:

3 hr falls in the “more than 2 hours up to 3 hours” bracket

Cost = $12
Q4 — Break-even with multiple questions
Understanding
A business has fixed monthly costs of $2800, variable costs of $16 per unit and revenue of $40 per unit.

(a) Find the break-even quantity.

(b) Find the profit when 180 units are produced and sold.

(c) What minimum selling price per unit gives a break-even point at 100 units?

(a) Break-even:

TC = 16x + 2800, R = 40x

40x = 16x + 2800 → 24x = 2800 → x = 2800/24 = 116.7 → 117 units

(b) Profit at 180 units:

Profit = (40 − 16) × 180 − 2800 = 24 × 180 − 2800 = 4320 − 2800 = $1520

(c) Minimum price for break-even at 100 units:

Let p = selling price. At break-even: R = TC

100p = 16 × 100 + 2800 = 1600 + 2800 = 4400

p = 4400 ÷ 100 = $44 per unit
Q5 — Comparing two suppliers
Problem Solving
Two suppliers offer the following pricing:
- Supplier A: $200 delivery charge + $8.50 per unit
- Supplier B: $50 delivery charge + $11 per unit
(a) Write cost equations for each supplier.

(b) Find the quantity at which both suppliers cost the same (the crossover point).

(c) Which supplier is cheaper for 100 units? For 200 units?

(a) Cost equations:

Supplier A: C_A = 8.50x + 200

Supplier B: C_B = 11x + 50

(b) Crossover point (C_A = C_B):

8.50x + 200 = 11x + 50

150 = 2.5x

x = 60 units

(c) Comparing costs:

At 100 units:

C_A = 8.50 × 100 + 200 = $1050

C_B = 11 × 100 + 50 = $1150

Supplier A is cheaper at 100 units (saving $100)

At 200 units:

C_A = 8.50 × 200 + 200 = $1900

C_B = 11 × 200 + 50 = $2250

Supplier A is cheaper at 200 units (saving $350)

Supplier A is cheaper for any order above 60 units.
Q6 — Right-angle trigonometry
Fluency
In a right-angled triangle, the angle is 47° and the adjacent side is 14 cm. Find:

(a) The opposite side.

(b) The hypotenuse.

(a) Opposite side:

tan(47°) = opp ÷ adj

opp = 14 × tan(47°) = 14 × 1.0724 = 15.01 cm

(b) Hypotenuse:

cos(47°) = adj ÷ hyp

hyp = 14 ÷ cos(47°) = 14 ÷ 0.6820 = 20.53 cm
Q7 — Sine rule
Fluency
In triangle ABC: angle A = 62°, angle B = 55° and side a = 24 m. Find side b.

Using the sine rule:

b/sin B = a/sin A

b/sin(55°) = 24/sin(62°)

b = 24 × sin(55°) ÷ sin(62°)

b = 24 × 0.8192 ÷ 0.8829

b = 22.27 m
Q8 — Cosine rule
Fluency
In triangle ABC: a = 8 m, b = 13 m, and angle C = 70°. Find side c.

c² = a² + b² − 2ab cos C

c² = 8² + 13² − 2 × 8 × 13 × cos(70°)

c² = 64 + 169 − 208 × 0.3420

c² = 233 − 71.14 = 161.86

c = √161.86 = 12.72 m
Q9 — Angle of elevation from two distances
Understanding
(a) From a point 50 m from the base of a tree (on level ground), the angle of elevation to the top is 36°. Find the height of the tree.

(b) From the other side of the tree, 100 m away from the base, what would the angle of elevation to the top be?

(a) Height of tree:

tan(36°) = h ÷ 50

h = 50 × tan(36°) = 50 × 0.7265 = 36.33 m

(b) Angle of elevation from 100 m:

tan(θ) = 36.33 ÷ 100 = 0.3633

θ = tan⁻¹(0.3633) = 20.0°
Q10 — Bearing and return journey
Understanding
A walker travels 6 km due north, then 8 km on a bearing of 115°.

(a) Find the direct distance from the starting point to the final position.

(b) Find the bearing needed to walk directly back to the start.

Finding final position components:

Leg 1 (6 km north): North = +6 km, East = 0

Leg 2 (8 km on bearing 115°):

North component = 8 × cos(115°) = 8 × (−0.4226) = −3.381 km

East component = 8 × sin(115°) = 8 × 0.9063 = 7.250 km

Total North from start: 6 − 3.381 = 2.619 km

Total East from start: 0 + 7.250 = 7.250 km

(a) Direct distance:

d = √(2.619² + 7.250²) = √(6.859 + 52.56) = √59.42 = 7.71 km

(b) Bearing to return:

The return journey goes west and south from final position.

To get from final position back to start: move −7.250 km east (i.e. west) and −2.619 km north (i.e. south).

Reference angle (from south towards west) = tan⁻¹(7.250 ÷ 2.619) = tan⁻¹(2.769) = 70.1°

Bearing = 180° + 70.1° = 250.1° (i.e. S 70.1° W)
Q11 — Triangular block of land
Problem Solving
A triangular block of land has two sides of 180 m and 240 m with an included angle of 74°.

(a) Find the length of the third side.

(b) Find the area of the block.

(c) Find the cost to fence the entire block at $22 per metre.

(a) Third side (cosine rule):

c² = 180² + 240² − 2 × 180 × 240 × cos(74°)

= 32 400 + 57 600 − 86 400 × 0.2756

= 90 000 − 23 811 = 66 189

c = √66 189 = 257.3 m

(b) Area:

Area = ½ × 180 × 240 × sin(74°)

= 21 600 × 0.9613 = 20 764 m²

(c) Fencing cost:

Perimeter = 180 + 240 + 257.3 = 677.3 m

Cost = 677.3 × $22 = $14 900.60
Q12 — Matrix notation, determinant and inverse
Fluency
Let A = [3 -1 / 2 5] (a 2×2 matrix with first row [3, −1] and second row [2, 5]).

(a) State the order of A.

(b) Find det(A).

(c) Find A⁻¹.

(a) Order: A is a 2 × 2 matrix (2 rows, 2 columns).

(b) Determinant:

det(A) = (3)(5) − (−1)(2) = 15 + 2 = 17

(c) Inverse:

For a 2×2 matrix [a b / c d], A⁻¹ = (1/det) × [d -b / -c a]

A⁻¹ = (1/17) × [5 1 / −2 3]

A⁻¹ = [5/17 1/17 / −2/17 3/17]

Or approximately: [0.294 0.059 / −0.118 0.176]
Q13 — Matrix addition and scalar multiplication
Fluency
(a) Add: [2 -3 / 1 4] + [5 0 / -2 3]

(b) Find 2 × [4 -1 / 0 3]

(a) Matrix addition:

Add corresponding elements:

Row 1: [2+5, −3+0] = [7, −3]

Row 2: [1+(−2), 4+3] = [−1, 7]

Result: [7 -3 / -1 7]

(b) Scalar multiplication:

Multiply each element by 2:

Row 1: [2×4, 2×−1] = [8, −2]

Row 2: [2×0, 2×3] = [0, 6]

Result: [8 -2 / 0 6]
Q14 — Matrix multiplication
Understanding
Multiply: [2 1 / -1 3] × [4 -2 / 1 5]. Show all working.

Let A = [2 1 / −1 3] and B = [4 −2 / 1 5].

The result C = AB is a 2×2 matrix where C_ij = row i of A • column j of B.

C₁₁ (row 1 of A • col 1 of B): (2)(4) + (1)(1) = 8 + 1 = 9

C₁₂ (row 1 of A • col 2 of B): (2)(−2) + (1)(5) = −4 + 5 = 1

C₂₁ (row 2 of A • col 1 of B): (−1)(4) + (3)(1) = −4 + 3 = −1

C₂₂ (row 2 of A • col 2 of B): (−1)(−2) + (3)(5) = 2 + 15 = 17

Result: [9 1 / -1 17]
Q15 — Solving a system using matrices
Understanding
Solve the system using matrices: 3x + 2y = 16 and x − y = 2.

Set up the matrix equation AX = B, find A⁻¹, then solve for X = A⁻¹B.

Setting up AX = B:

A = [3 2 / 1 -1], X = [x / y], B = [16 / 2]

Finding A⁻¹:

det(A) = (3)(−1) − (2)(1) = −3 − 2 = −5

A⁻¹ = (1/−5) × [−1 −2 / −1 3]

A⁻¹ = [1/5 2/5 / 1/5 -3/5]

Solving X = A⁻¹B:

x = (1/5)(16) + (2/5)(2) = 16/5 + 4/5 = 20/5 = 4

y = (1/5)(16) + (−3/5)(2) = 16/5 − 6/5 = 10/5 = 2

Solution: x = 4, y = 2

Check: 3(4) + 2(2) = 12 + 4 = 16 ✓; 4 − 2 = 2 ✓
Q16 — Singular matrix condition
Understanding
Matrix A = [a 2 / 3 a]. Find the value(s) of a for which A is singular (non-invertible).

A matrix is singular when its determinant equals zero.

det(A) = (a)(a) − (2)(3) = a² − 6

Set det(A) = 0:

a² − 6 = 0

a² = 6

a = ±√6

A is singular when a = √6 or a = −√6 (approximately ±2.449).
Q17 — Matrix solution: factory production
Problem Solving
Two factories produce widgets (w) and gadgets (g). Their production satisfies:

4w + 3g = 34 and 2w + 5g = 28

Solve for the production quantities using matrices.

Matrix setup AX = B:

A = [4 3 / 2 5], X = [w / g], B = [34 / 28]

Finding det(A):

det(A) = (4)(5) − (3)(2) = 20 − 6 = 14

Finding A⁻¹:

A⁻¹ = (1/14) × [5 -3 / -2 4]

Solving X = A⁻¹B:

w = (1/14)[5 × 34 + (−3) × 28] = (1/14)[170 − 84] = 86/14 = 6.14...

Hmm — let us check for integer solutions:

w = (5 × 34 − 3 × 28)/14 = (170 − 84)/14 = 86/14 = 43/7

g = (−2 × 34 + 4 × 28)/14 = (−68 + 112)/14 = 44/14 = 22/7

Check: 4(43/7) + 3(22/7) = 172/7 + 66/7 = 238/7 = 34 ✓

2(43/7) + 5(22/7) = 86/7 + 110/7 = 196/7 = 28 ✓

w = 43/7 ≈ 6.14 widgets; g = 22/7 ≈ 3.14 gadgets

(In a real context, production values would typically be whole numbers — this system yields fractional solutions.)
Q18 — Summary statistics
Fluency
Data set: 12, 15, 11, 18, 14, 16, 13, 20, 11, 15

Find: (a) mean; (b) median; (c) mode; (d) range.

Ordered data: 11, 11, 12, 13, 14, 15, 15, 16, 18, 20

(a) Mean:

Sum = 11 + 11 + 12 + 13 + 14 + 15 + 15 + 16 + 18 + 20 = 145

Mean = 145 ÷ 10 = 14.5

(b) Median:

10 values → median = average of 5th and 6th values

5th value = 14, 6th value = 15

Median = (14 + 15) ÷ 2 = 14.5

(c) Mode:

Both 11 and 15 appear twice. Mode = 11 and 15 (bimodal)

(d) Range:

Range = 20 − 11 = 9
Q19 — Quartiles, IQR and outliers
Fluency
Ordered data set: 4, 6, 8, 10, 12, 14, 16, 18, 20.

(a) Find Q1, Q3 and IQR.

(b) Use the 1.5 × IQR rule to identify any outliers.

9 values. Median = 5th value = 12.

Lower half (values below median): 4, 6, 8, 10

Upper half (values above median): 14, 16, 18, 20

(a) Quartiles:

Q1 = median of lower half = (6 + 8)/2 = 7

Q3 = median of upper half = (16 + 18)/2 = 17

IQR = Q3 − Q1 = 17 − 7 = 10

(b) Outlier boundaries:

Lower fence = Q1 − 1.5 × IQR = 7 − 15 = −8

Upper fence = Q3 + 1.5 × IQR = 17 + 15 = 32

All data values (4 to 20) fall within [−8, 32].

There are no outliers in this dataset.
Q20 — Normal distribution and z-scores
Understanding
Exam scores in a class follow a normal distribution with mean 68 and standard deviation 12.

(a) Approximately what percentage of scores fall between 56 and 80?

(b) Approximately what percentage of scores are above 92?

(c) Find the z-score for a mark of 74.

(d) Find the mark corresponding to z = −1.5.

Mean = 68, SD = 12

(a) Percentage between 56 and 80:

56 = 68 − 12 = μ − 1σ

80 = 68 + 12 = μ + 1σ

By the 68-95-99.7 rule, approximately 68% of scores fall within one standard deviation of the mean.

(b) Percentage above 92:

92 = 68 + 24 = μ + 2σ

95% of data falls within 2 SD. So 5% falls outside, with 2.5% in each tail.

Approximately 2.5% of scores are above 92.

(c) z-score for mark 74:

z = (74 − 68) ÷ 12 = 6/12 = 0.5

(The score of 74 is 0.5 standard deviations above the mean.)

(d) Mark at z = −1.5:

x = μ + z × σ = 68 + (−1.5) × 12 = 68 − 18 = 50
Q21 — Comparing datasets using box plot data
Understanding
Two classes completed the same test. Their results are summarised below:
- Class A: Min = 42, Q1 = 56, Median = 65, Q3 = 74, Max = 88
- Class B: Min = 38, Q1 = 50, Median = 70, Q3 = 80, Max = 95
Compare the two classes in terms of shape, centre and spread.

Centre:

Class B has a higher median (70) than Class A (65), suggesting Class B students performed better on average.

Spread:

IQR Class A = 74 − 56 = 18

IQR Class B = 80 − 50 = 30

Range Class A = 88 − 42 = 46; Range Class B = 95 − 38 = 57

Class B has greater spread (IQR = 30 vs 18; range = 57 vs 46), meaning more variability in performance.

Shape:

Class A: The median (65) is roughly centred between Q1 and Q3, and the whiskers are roughly symmetric, suggesting an approximately symmetric distribution.

Class B: The lower whisker (38–50, length 12) is shorter than the upper whisker (80–95, length 15), suggesting a slight positive (right) skew.

Summary: Class B achieved higher scores on average but with more variability; Class A’s results were more consistent though generally lower.
Q22 — Comparing two groups: mean and IQR
Understanding
Two groups recorded their daily step counts (thousands of steps) for 10 days:
- Group 1: 7, 9, 6, 8, 10, 7, 9, 8, 6, 10
- Group 2: 4, 12, 5, 11, 8, 3, 14, 6, 9, 8
(a) Find the mean and IQR for each group.

(b) Write a comparison paragraph.

Group 1 ordered: 6, 6, 7, 7, 8, 8, 9, 9, 10, 10

Mean₁ = (6+6+7+7+8+8+9+9+10+10)/10 = 80/10 = 8.0

Q1 = (7+7)/2 = 7.0; Q3 = (9+9)/2 = 9.0; IQR₁ = 2.0

Group 2 ordered: 3, 4, 5, 6, 8, 8, 9, 11, 12, 14

Mean₂ = (3+4+5+6+8+8+9+11+12+14)/10 = 80/10 = 8.0

Q1 = (5+6)/2 = 5.5; Q3 = (9+11)/2 = 10.0; IQR₂ = 4.5

(b) Comparison paragraph:

Both groups recorded the same mean daily step count of 8000 steps, suggesting their average activity levels were identical. However, Group 1 showed far more consistency, with an IQR of only 2000 steps, compared to Group 2’s IQR of 4500 steps. Group 2’s data was much more spread out (ranging from 3000 to 14 000 steps), indicating greater day-to-day variation in activity level. Group 1 maintained a more regular routine, while Group 2 had some very active days and some very inactive days.
Q23 — Comparing performance using z-scores
Problem Solving
A student scores 72 in English (class mean = 65, SD = 8) and 61 in Maths (class mean = 55, SD = 6). Which result represents the better performance relative to the class? Explain using z-scores.

Z-score for English:

z = (72 − 65) ÷ 8 = 7/8 = 0.875

Z-score for Maths:

z = (61 − 55) ÷ 6 = 6/6 = 1.000

Conclusion:

The student’s Maths score (z = 1.00) is relatively better than the English score (z = 0.875). While the student scored higher in absolute terms in English (72 vs 61), relative to each class the Maths result placed the student further above average. In Maths, the student was exactly one standard deviation above the class mean, compared to 0.875 standard deviations above the mean in English.
Q24 — Maximum area of a triangle
Problem Solving
A triangular park has two fixed sides of 120 m and 160 m. The included angle between them can vary.

(a) Write an expression for the area of the triangle in terms of the included angle θ.

(b) Find the value of θ that gives the maximum area.

(c) Find the maximum possible area.

(a) Area in terms of θ:

Area = ½ × 120 × 160 × sin(θ) = 9600 sin(θ) m²

(b) Angle for maximum area:

sin(θ) is maximised when sin(θ) = 1, which occurs at θ = 90°.

The maximum area occurs when θ = 90° (the two given sides are perpendicular).

(c) Maximum area:

Area_max = 9600 × sin(90°) = 9600 × 1 = 9600 m²

(Note: At 90° the triangle is right-angled, and the area = ½ × base × height = ½ × 120 × 160 = 9600 m² ✓)
Q25 — Matrix system and profit calculation
Problem Solving
A company produces two products x and y. The daily production satisfies:

3x + 2y = 24 and x + 4y = 16

(a) Solve for x and y using matrices.

(b) If product x gives a profit of $80 each and product y gives $60 each, find the total daily profit.

(a) Matrix setup AX = B:

A = [3 2 / 1 4], X = [x / y], B = [24 / 16]

det(A):

det(A) = (3)(4) − (2)(1) = 12 − 2 = 10

A⁻¹:

A⁻¹ = (1/10) × [4 -2 / -1 3]

Solving X = A⁻¹B:

x = (1/10)[(4)(24) + (−2)(16)] = (1/10)[96 − 32] = 64/10 = 6.4

y = (1/10)[(−1)(24) + (3)(16)] = (1/10)[−24 + 48] = 24/10 = 2.4

Check: 3(6.4) + 2(2.4) = 19.2 + 4.8 = 24 ✓; 6.4 + 4(2.4) = 6.4 + 9.6 = 16 ✓

(b) Total daily profit:

Profit = 80x + 60y = 80(6.4) + 60(2.4) = 512 + 144 = $656 per day

Unit 2 Review — General Maths

Review Questions

Q1 — Break-even analysis

Q2 — Piecewise phone plan

Q3 — Step graph parking

Q4 — Break-even with multiple questions

Q5 — Comparing two suppliers

Q6 — Right-angle trigonometry

Q7 — Sine rule

Q8 — Cosine rule

Q9 — Angle of elevation from two distances

Q10 — Bearing and return journey

Q11 — Triangular block of land

Q12 — Matrix notation, determinant and inverse

Q13 — Matrix addition and scalar multiplication

Q14 — Matrix multiplication

Q15 — Solving a system using matrices

Q16 — Singular matrix condition

Q17 — Matrix solution: factory production

Q18 — Summary statistics

Q19 — Quartiles, IQR and outliers

Q20 — Normal distribution and z-scores

Q21 — Comparing datasets using box plot data

Q22 — Comparing two groups: mean and IQR

Q23 — Comparing performance using z-scores

Q24 — Maximum area of a triangle

Q25 — Matrix system and profit calculation