Solutions — Developing Linear Models from Context
-
Q1 — Taxi fare model
Fluency(a) Identify components:
Flag fall (y-intercept, c) = $2.80. Rate per km (gradient, m) = $2.10/km.
Independent variable: d (distance in km). Dependent variable: C (cost in $).
C = 2.10d + 2.80(b) Cost for 8 km:
C = 2.10(8) + 2.80 = 16.80 + 2.80 = $19.60(c) Distance for cost of $28.70:
28.70 = 2.10d + 2.80
2.10d = 28.70 − 2.80 = 25.90
d = 25.90 ÷ 2.10 = 12.33 km (to 2 d.p.)
Check: C = 2.10(25.90/2.10) + 2.80 = 25.90 + 2.80 = $28.70 ✓ -
Q2 — Candle burning model
Fluency(a) Burn rate (gradient):
Change in height = 16.5 − 24 = −7.5 cm over 3 hours.
Burn rate = −7.5 ÷ 3 = −2.5 cm/hr (negative because candle gets shorter)(b) Linear model:
At t = 0: h = 24 cm (initial height). m = −2.5.
h = −2.5t + 24(c) When does the candle burn out?
Candle burns out when h = 0:
0 = −2.5t + 24
2.5t = 24
t = 24 ÷ 2.5 = 9.6 hours
Check: h = −2.5(9.6) + 24 = −24 + 24 = 0 ✓ -
Q3 — Salary from table of values
Fluency(a) Gradient (hourly rate):
Using any two points, e.g. (0, 200) and (1, 240):
m = (240 − 200) ÷ (1 − 0) = 40 ÷ 1 = $40/hr
Check with (1, 240) and (3, 320): m = (320 − 240) ÷ (3 − 1) = 80 ÷ 2 = $40/hr ✓(b) Linear equation:
At h = 0: S = $200 (base pay, y-intercept). m = 40.
S = 40h + 200(c) Salary after 6.5 hours:
S = 40(6.5) + 200 = 260 + 200 = $460 -
Q4 — Car fuel consumption
Understanding(a) Fuel equation:
The car consumes 9 L per 100 km = 0.09 L/km. Fuel decreases as distance increases, so gradient is negative.
Initial fuel: 72 L. m = −0.09 L/km.
F = −0.09d + 72 where F = fuel remaining (L), d = distance (km)(b) Fuel after 350 km:
F = −0.09(350) + 72 = −31.5 + 72 = 40.5 L(c) Range (maximum distance):
Tank empties when F = 0:
0 = −0.09d + 72
0.09d = 72
d = 72 ÷ 0.09 = 800 km(d) Domain:
Distance can only be non-negative, and the car can travel at most 800 km before running out of fuel.
Domain: 0 ≤ d ≤ 800 (km) -
Q5 — Electricity bill model
Understanding(a) Cost equation:
Usage rate (m) = $0.28/kWh. Fixed quarterly supply charge (c) = $85.
C = 0.28u + 85 where C = quarterly cost ($), u = usage (kWh)(b) Bill for 650 kWh:
C = 0.28(650) + 85 = 182 + 85 = $267(c) Usage for $200 bill:
200 = 0.28u + 85
0.28u = 200 − 85 = 115
u = 115 ÷ 0.28 ≈ 410.7 kWh(d) Gradient in context:
The gradient of 0.28 represents the cost per kilowatt-hour of electricity used — for every additional kWh consumed, the bill increases by $0.28. It is the unit rate charged for electricity usage. -
Q6 — Phone plan comparison
Understanding(a) Equations:
Plan A: $45/month base + $0.10/min → A = 0.10m + 45
Plan B: $30/month base + $0.20/min → B = 0.20m + 30
(m = minutes used per month)(b) Crossover point (equal cost):
Set A = B:
0.10m + 45 = 0.20m + 30
45 − 30 = 0.20m − 0.10m
15 = 0.10m
m = 150 minutes
Cost at crossover: A = 0.10(150) + 45 = 15 + 45 = $60
Crossover point: 150 minutes, $60/month(c) Better plan at 200 min/month:
Plan A: A = 0.10(200) + 45 = 20 + 45 = $65
Plan B: B = 0.20(200) + 30 = 40 + 30 = $70
At 200 min/month, Plan A is cheaper by $5/month.
(Note: Plan B is cheaper for usage below 150 min; Plan A is cheaper above 150 min.) -
Q7 — Temperature cooling model
Understanding(a) Confirm linear (check constant gradient):
From 0 to 5 min: ΔT/Δt = (65 − 80) ÷ (5 − 0) = −15 ÷ 5 = −3
From 5 to 10 min: ΔT/Δt = (50 − 65) ÷ (10 − 5) = −15 ÷ 5 = −3
From 10 to 15 min: ΔT/Δt = (35 − 50) ÷ (15 − 10) = −15 ÷ 5 = −3
All gradients equal −3 °C/min. The data is linear.(b) Linear model:
m = −3 °C/min. At t = 0: T = 80 °C (y-intercept).
T = −3t + 80(c) Temperature at 25 min:
T = −3(25) + 80 = −75 + 80 = 5°C(d) When does it reach 20°C?
20 = −3t + 80
3t = 80 − 20 = 60
t = 60 ÷ 3 = 20 minutes -
Q8 — Spring extension model
Understanding(a) Check linearity:
Natural length (0 kg): 12 cm. With 2 kg: 15 cm. With 5 kg: 19.5 cm.
From 0 to 2 kg: ΔL/Δm = (15 − 12) ÷ (2 − 0) = 3 ÷ 2 = 1.5 cm/kg
From 2 to 5 kg: ΔL/Δm = (19.5 − 15) ÷ (5 − 2) = 4.5 ÷ 3 = 1.5 cm/kg
Gradient is constant. The data is linear (consistent with Hooke’s Law).(b) Linear model:
m = 1.5 cm/kg. At mass = 0: L = 12 cm (natural length, y-intercept).
L = 1.5m + 12 where L = length (cm), m = mass (kg)(c) Length at 8 kg:
L = 1.5(8) + 12 = 12 + 12 = 24 cm(d) Extrapolation to 50 kg:
L = 1.5(50) + 12 = 75 + 12 = 87 cm (predicted).
Not reasonable. Hooke’s Law only holds within the elastic limit of the spring. At 50 kg, the spring would almost certainly exceed its elastic limit, permanently deform or break. Extrapolation this far beyond the data range is unreliable. -
Q9 — Swimming pool filling
Problem Solving(a) Fill rate:
Using the two data points (30, 900) and (90, 2700):
m = (2700 − 900) ÷ (90 − 30) = 1800 ÷ 60 = 30 L/min(b) Was the pool empty at the start?
Find c: using point (30, 900) in V = 30t + c:
900 = 30(30) + c = 900 + c
c = 900 − 900 = 0
Model: V = 30t
At t = 0: V = 0. Yes, the pool was empty at the start.
(The y-intercept of 0 confirms it started with no water.)(c) When is the 15 000 L pool full?
15000 = 30t
t = 15000 ÷ 30 = 500 minutes
Convert: 500 ÷ 60 = 8 hours 20 minutes
The pool is full after 500 minutes (8 hours and 20 minutes). -
Q10 — Business break-even analysis
Problem Solving(a) Cost and Revenue functions:
Total Cost: fixed costs + variable costs × quantity
TC = 18n + 4200 where n = number of units
Total Revenue: selling price × quantity
TR = 45n(b) Break-even quantity:
At break-even: TC = TR
18n + 4200 = 45n
4200 = 45n − 18n = 27n
n = 4200 ÷ 27 ≈ 155.6
Since you can’t sell a fraction of a unit, break-even at 156 units (the business must sell at least 156 units to cover all costs).
Check at 156: TC = 18(156) + 4200 = 2808 + 4200 = $7008; TR = 45(156) = $7020. TR > TC ✓(c) Profit at 250 units:
Profit = TR − TC = 45(250) − [18(250) + 4200]
= 11250 − [4500 + 4200]
= 11250 − 8700
= $2 550 profit(d) Units needed for $5 000 profit:
Profit = TR − TC = 45n − (18n + 4200) = 27n − 4200
Set profit = 5000:
27n − 4200 = 5000
27n = 9200
n = 9200 ÷ 27 ≈ 340.7
341 units must be sold to achieve at least $5 000 profit.
Check: Profit = 27(341) − 4200 = 9207 − 4200 = $5 007 ✓