Unit 2 Topic 1 Review — Applications of Linear Equations — Solutions

This review covers all four lessons in Applications of Linear Equations: Developing Linear Models, Piecewise Linear Graphs, Step Graphs, and Break-even Analysis. Allow approximately 60–75 minutes for this review.

Review Questions

Q1 — Taxi fare model
Fluency
A taxi company charges according to the model C = 3.20 + 2.10d, where C is the cost in dollars and d is the distance in kilometres.

(a) Find the cost for a 12 km trip.

(b) Find the distance travelled if the fare is $25.40.

(a) Cost for 12 km:

C = 3.20 + 2.10 × 12

C = 3.20 + 25.20

C = $28.40

(b) Distance when C = $25.40:

25.40 = 3.20 + 2.10d

2.10d = 25.40 − 3.20 = 22.20

d = 22.20 ÷ 2.10

d = 10.57 km (to 2 d.p.)
Q2 — Piecewise electricity bill
Fluency
An electricity plan charges $0.22 per kWh for the first 500 kWh used each month, and $0.31 per kWh for any usage above 500 kWh.

(a) Find the monthly bill for 350 kWh.

(b) Find the monthly bill for 720 kWh.

(a) Bill for 350 kWh:

350 kWh ≤ 500 kWh, so use the first rate only.

Bill = 350 × $0.22 = $77.00

(b) Bill for 720 kWh:

First 500 kWh: 500 × $0.22 = $110.00

Remaining: 720 − 500 = 220 kWh at $0.31

220 × $0.31 = $68.20

Total bill = $110.00 + $68.20 = $178.20
Q3 — Step graph parking costs
Fluency
A car park uses the following stepped pricing:
- Up to 1 hour: $4
- More than 1 hour up to 2 hours: $7
- More than 2 hours up to 3 hours: $10
- More than 3 hours up to 4 hours: $13
Find the parking cost for: (a) 45 minutes; (b) exactly 2 hours; (c) 2 hours 10 minutes; (d) 4 hours.

(a) 45 minutes:

45 min = 0.75 hr ≤ 1 hr → $4

(b) Exactly 2 hours:

2 hr falls in the “more than 1 hour up to 2 hours” bracket → $7

(c) 2 hours 10 minutes:

2 hr 10 min = 2.167 hr, which is more than 2 hr and up to 3 hr → $10

(d) 4 hours:

4 hr falls in the “more than 3 hours up to 4 hours” bracket → $13
Q4 — Break-even analysis
Fluency
A business has total cost TC = 7x + 1400 and revenue R = 21x, where x is the number of units produced and sold.

(a) Find the break-even quantity.

(b) Find the profit when x = 180 units.

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

7x + 1400 = 21x

1400 = 21x − 7x = 14x

x = 1400 ÷ 14 = 100 units

(b) Profit at x = 180:

Profit = R − TC = 21x − (7x + 1400)

Profit = 14x − 1400

Profit = 14 × 180 − 1400 = 2520 − 1400 = $1120
Q5 — Building a linear model from a table
Understanding
A plumber charges based on time. The table shows costs for different job lengths:

Time (hr) 0 1 2 3

Cost ($) 80 140 200 260

(a) Confirm the relationship is linear.

(b) Write the linear model for cost C in terms of time t.

(c) Find the cost for a job taking 4.5 hours.

(d) Interpret the gradient and y-intercept in context.

(a) Confirming linearity:

Check differences in cost: 140−80 = 60, 200−140 = 60, 260−200 = 60

The cost increases by a constant $60 for each additional hour → the relationship is linear. ✓

(b) Linear model:

Gradient m = $60/hr (increase per hour)

y-intercept c = $80 (cost at t = 0)

C = 60t + 80

(c) Cost for 4.5 hours:

C = 60 × 4.5 + 80 = 270 + 80 = $350

(d) Interpretation:

Gradient ($60/hr): The plumber charges $60 for each hour of work.

y-intercept ($80): There is a fixed call-out fee of $80, charged even before any time is worked.
Q6 — Piecewise gym membership
Understanding
A gym charges $15 per visit for the first 8 visits per month, then $9 per visit for any additional visits beyond 8.

(a) Write a piecewise model for monthly cost C in terms of number of visits v.

(b) Find the cost for 5 visits in a month.

(c) Find the cost for 14 visits in a month.

(d) How many visits in a month results in a total cost of exactly $111?

(a) Piecewise model:

C = 15v for 0 ≤ v ≤ 8

C = 120 + 9(v − 8) for v > 8

Which simplifies to: C = 9v + 48 for v > 8

(b) Cost for 5 visits:

5 ≤ 8, so C = 15 × 5 = $75

(c) Cost for 14 visits:

14 > 8, so C = 120 + 9 × (14 − 8) = 120 + 9 × 6 = 120 + 54 = $174

(d) Finding visits for C = $111:

Check first segment: 15v = 111 → v = 7.4 visits — not possible (not a whole number, but check if it falls in range). Since v must be a whole number and 7.4 is not, try second segment.

Actually, check if $111 occurs in the v > 8 segment:

120 + 9(v − 8) = 111 → 9(v − 8) = −9 → v − 8 = −1 → v = 7

But v = 7 is in the first segment: C = 15 × 7 = $105 ≠ $111.

Check v = 8 (boundary): C = 15 × 8 = $120 ≠ $111.

There is no whole-number of visits that gives exactly $111 under this pricing. If fractional visits were allowed, v = 7.4 gives $111 in the first segment.

Note: $111 falls between $105 (7 visits) and $120 (8 visits), so it is not achievable with a whole number of visits at this gym.
Q7 — Step graph mobile data plan
Understanding
A mobile phone plan uses stepped data pricing:
- 0–10 GB: $35/month
- 10–30 GB: $55/month
- 30–60 GB: $75/month
- More than 60 GB: $95/month
(a) Find the monthly cost for: 8 GB, 25 GB, 55 GB, 72 GB.

(b) Which plan tier is best for someone who uses an average of 40 GB per month?

(a) Monthly costs:

8 GB: Falls in 0–10 GB tier → $35/month

25 GB: Falls in 10–30 GB tier → $55/month

55 GB: Falls in 30–60 GB tier → $75/month

72 GB: Exceeds 60 GB tier → $95/month

(b) Best plan for 40 GB/month average:

40 GB falls in the 30–60 GB tier at $75/month.

The 10–30 GB plan ($55/month) would not cover 40 GB of usage.

The 30–60 GB plan at $75/month is the appropriate choice — it covers their usage and upgrading to the $95 plan would be unnecessary.
Q8 — Canteen break-even analysis
Understanding
A school canteen has fixed daily costs of $180 and variable costs of $2.40 per item. Items are sold for $7.50 each.

(a) Write equations for Total Cost (TC) and Revenue (R) in terms of items sold x.

(b) Find the break-even number of items.

(c) Find the profit when 50 items are sold.

(d) How many items must be sold to achieve a profit of $200?

(a) Equations:

TC = 2.40x + 180

R = 7.50x

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

2.40x + 180 = 7.50x

180 = 7.50x − 2.40x = 5.10x

x = 180 ÷ 5.10 = 35.29...

Break-even point = 36 items (must sell at least 36 to cover costs)

(c) Profit at 50 items:

Profit = R − TC = 7.50(50) − [2.40(50) + 180]

= 375 − [120 + 180] = 375 − 300 = $75

(d) Items needed for $200 profit:

Profit = R − TC = 5.10x − 180 = 200

5.10x = 380

x = 380 ÷ 5.10 = 74.5...

Must sell at least 75 items to achieve $200 profit.
Q9 — Water tank drainage model
Understanding
At the start of draining, a water tank holds 8400 litres. Two hours later it holds 5600 litres.

(a) Write a linear model for the volume V (litres) in terms of time t (hours).

(b) What is the drain rate per hour?

(c) When does the tank reach 1000 litres?

(d) State the practical domain of your model.

(a) Linear model:

Two points: (0, 8400) and (2, 5600)

Gradient = (5600 − 8400) ÷ (2 − 0) = −2800 ÷ 2 = −1400

V = 8400 − 1400t

(b) Drain rate:

The gradient of −1400 means the tank drains at 1400 litres per hour.

(c) When V = 1000 litres:

8400 − 1400t = 1000

1400t = 7400

t = 7400 ÷ 1400 = 5.29 hours (approx. 5 hours 17 minutes)

(d) Practical domain:

The tank cannot have a negative volume. It empties when:

8400 − 1400t = 0 → t = 6 hours

Domain: 0 ≤ t ≤ 6 hours
Q10 — Comparing two delivery options
Understanding
Two courier companies quote for delivery:
- Option A: $60 base charge + $0.35 per km
- Option B: $30 base charge + $0.50 per km
(a) Write cost equations for each option in terms of distance d (km).

(b) Find the distance at which both options cost the same.

(c) Which option is cheaper for a 200 km trip?

(a) Cost equations:

Option A: C_A = 0.35d + 60

Option B: C_B = 0.50d + 30

(b) Break-even distance (C_A = C_B):

0.35d + 60 = 0.50d + 30

60 − 30 = 0.50d − 0.35d

30 = 0.15d

d = 30 ÷ 0.15 = 200 km

(c) Cost at 200 km:

Both options cost the same at exactly 200 km.

C_A = 0.35 × 200 + 60 = 70 + 60 = $130

C_B = 0.50 × 200 + 30 = 100 + 30 = $130 ✓

At 200 km, both options cost $130 — neither is cheaper.

Note: For trips shorter than 200 km, Option B is cheaper; for trips longer than 200 km, Option A is cheaper.
Q11 — Piecewise overtime pay
Understanding
An employee is paid:
- $26/hr for the first 38 hours (normal time)
- $39/hr (time-and-a-half) for the next 4 hours (hours 39–42)
- $52/hr (double time) for any hours beyond 42
Find the weekly pay for: (a) 35 hours; (b) 40 hours; (c) 46 hours.

(a) 35 hours (normal time only):

Pay = 35 × $26 = $910

(b) 40 hours (normal + overtime):

Normal: 38 × $26 = $988

Overtime (time-and-a-half): (40 − 38) × $39 = 2 × $39 = $78

Total = $988 + $78 = $1066

(c) 46 hours (all three rates):

Normal: 38 × $26 = $988

Time-and-a-half: 4 × $39 = $156

Double time: (46 − 42) × $52 = 4 × $52 = $208

Total = $988 + $156 + $208 = $1352
Q12 — Income tax as a step function
Problem Solving
Income tax is calculated using these brackets:
- $0 – $18 200: Nil
- $18 201 – $45 000: 19c per $1 over $18 200
- $45 001 – $120 000: $5092 + 32.5c per $1 over $45 000
(a) Calculate the tax payable on an income of $27 500.

(b) Calculate the tax payable on an income of $65 000.

(a) Tax on $27 500:

$27 500 falls in the $18 201–$45 000 bracket.

Taxable amount above $18 200 = $27 500 − $18 200 = $9300

Tax = $9300 × 0.19 = $1767

Effective tax rate = 1767 ÷ 27 500 × 100 = 6.4%

(b) Tax on $65 000:

$65 000 falls in the $45 001–$120 000 bracket.

Taxable amount above $45 000 = $65 000 − $45 000 = $20 000

Tax = $5092 + (0.325 × $20 000)

Tax = $5092 + $6500 = $11 592

Effective tax rate = 11 592 ÷ 65 000 × 100 = 17.8%
Q13 — Production decision and machine upgrade
Problem Solving
A business has fixed costs of $3600/month and variable costs of $14/unit. Each unit sells for $38.

(a) Find the break-even quantity.

(b) The business currently produces 200 units/month. Find the current monthly profit.

(c) A machine upgrade costs an extra $500/month in fixed costs but reduces variable cost to $10/unit. Is the upgrade worth it at 200 units/month?

(d) At what monthly production volume does the upgraded setup become more profitable than the current setup?

(a) Break-even (current setup):

TC = 14x + 3600, R = 38x

38x = 14x + 3600 → 24x = 3600 → x = 150 units

(b) Profit at 200 units (current):

Profit = (38 − 14) × 200 − 3600 = 24 × 200 − 3600 = 4800 − 3600 = $1200/month

(c) Upgraded setup at 200 units:

New TC = 10x + (3600 + 500) = 10x + 4100

Profit_upgraded = (38 − 10) × 200 − 4100 = 28 × 200 − 4100 = 5600 − 4100 = $1500/month

Upgraded profit ($1500) > current profit ($1200) → Yes, the upgrade is worth it at 200 units.

(d) Finding the crossover volume:

Current profit = 24x − 3600

Upgraded profit = 28x − 4100

Set equal: 24x − 3600 = 28x − 4100

500 = 4x → x = 125 units

For x < 125: current setup is more profitable (lower fixed costs dominate).

For x > 125: upgrade is more profitable.

The upgrade becomes worthwhile above 125 units/month.
Q14 — Piecewise car journey model
Problem Solving
A car makes a journey: it travels at 80 km/h for 2 hours, stops for 30 minutes (0.5 hr), then travels at 100 km/h for 1.5 hours.

(a) Write a piecewise model for distance D (km) from the start in terms of time t (hours).

(b) Find the total distance travelled.

(c) At what time(s) is the car exactly 160 km from the start?

(a) Piecewise distance model:

Segment 1 (0 ≤ t ≤ 2): Travelling at 80 km/h

    D = 80t

At t = 2: D = 160 km (end of segment 1)

Segment 2 (2 < t ≤ 2.5): Stopped (distance stays at 160 km)

    D = 160

Segment 3 (2.5 < t ≤ 4): Travelling at 100 km/h from 160 km mark

    D = 160 + 100(t − 2.5)

Summary:

D(t) = 80t                          for 0 ≤ t ≤ 2

D(t) = 160                          for 2 < t ≤ 2.5

D(t) = 160 + 100(t − 2.5)    for 2.5 < t ≤ 4

(b) Total distance:

D(4) = 160 + 100(4 − 2.5) = 160 + 100 × 1.5 = 160 + 150 = 310 km

(c) When is D = 160 km?

In Segment 1: 80t = 160 → t = 2 hr

In Segment 2: D = 160 throughout 2 < t ≤ 2.5 → the car is at 160 km for the entire stop.

The car is exactly 160 km from the start for t = 2 hr through t = 2.5 hr (the entire stop period).
Q15 — Business profitability and pricing decision
Problem Solving
A business currently sells 450 units/month at $32 each. Variable costs are $12/unit and fixed costs are $4500/month.

(a) Find the current monthly profit.

(b) A price cut to $28 is expected to increase sales to 600 units/month. Would the price cut increase profit?

(c) Find the break-even quantity at the new price of $28.

(d) With current production fixed at 450 units, what selling price would produce a monthly profit of exactly $5000?

(a) Current monthly profit (at $32, 450 units):

R = 32 × 450 = $14 400

TC = 12 × 450 + 4500 = $5400 + $4500 = $9900

Profit = $14 400 − $9900 = $4500/month

(b) Profit at new price $28, 600 units:

R = 28 × 600 = $16 800

TC = 12 × 600 + 4500 = $7200 + $4500 = $11 700

Profit = $16 800 − $11 700 = $5100/month

$5100 > $4500 → Yes, the price cut increases profit (by $600/month).

(c) Break-even at $28:

R = TC: 28x = 12x + 4500

16x = 4500

x = 4500 ÷ 16 = 281.25 → 282 units (round up)

(d) Selling price for $5000 profit at 450 units:

Let p = selling price.

Profit = R − TC = 450p − (12 × 450 + 4500) = 5000

450p − 5400 − 4500 = 5000

450p − 9900 = 5000

450p = 14 900

p = 14 900 ÷ 450 = $33.11 per unit (to 2 d.p.)

Time (hr)	0	1	2	3
Cost ($)	80	140	200	260

Unit 2 Topic 1 Review — Applications of Linear Equations — Solutions

Review Questions

Q1 — Taxi fare model

Q2 — Piecewise electricity bill

Q3 — Step graph parking costs

Q4 — Break-even analysis

Q5 — Building a linear model from a table

Q6 — Piecewise gym membership

Q7 — Step graph mobile data plan

Q8 — Canteen break-even analysis

Q9 — Water tank drainage model

Q10 — Comparing two delivery options

Q11 — Piecewise overtime pay

Q12 — Income tax as a step function

Q13 — Production decision and machine upgrade

Q14 — Piecewise car journey model

Q15 — Business profitability and pricing decision