Year 12 Review — General Maths

This comprehensive review covers all seven topics across Units 3 and 4 of General Mathematics. Use it as exam preparation to identify any areas needing further practice. Allow approximately 2–3 hours for the full review.

Bivariate Data

Q1 — Correlation
Fluency
A researcher studies the relationship between daily study hours (x) and exam score (y) for 8 students and finds a correlation coefficient of r = 0.87.

(a) Describe the direction and strength of this correlation.

(b) Can the researcher conclude that studying more hours causes higher exam scores? Explain.

(a) r = 0.87 indicates a strong positive linear relationship. As study hours increase, exam scores tend to increase.

(b) No. Correlation does not imply causation. A third variable (for example, student motivation or ability) may drive both increased study time and higher scores. An experiment with random assignment would be needed to establish causation.
Q2 — Least squares regression equation
Fluency
For a dataset of 10 students the following summary statistics are given: x̄ = 4 hours, ȳ = 60 marks, s_x = 2, s_y = 8, r = 0.75.

(a) Calculate the gradient b of the least-squares regression line.

(b) Calculate the y-intercept a.

(c) Write the equation of the regression line.

(a) b = r × (s_y/s_x) = 0.75 × (8/2) = 0.75 × 4 = 3.0

(b) a = ȳ − b × x̄ = 60 − 3.0 × 4 = 60 − 12 = 48

(c) ŷ = 48 + 3x
Q3 — Prediction and residual analysis
Understanding
Use the regression line ŷ = 48 + 3x from Q2. The data was collected from students studying between 1 and 7 hours per day.

(a) A student studies 6 hours and scores 64 marks. Calculate the predicted score and the residual.

(b) Interpret the sign of the residual from part (a).

(c) A friend wants to predict the score for a student who studies 9 hours. Is this interpolation or extrapolation? Explain why caution is needed.

(a) Predicted: ŷ = 48 + 3 × 6 = 66. Residual = actual − predicted = 64 − 66 = −2

(b) A negative residual means the actual score (64) is below the regression line. The line over-predicted for this student.

(c) Extrapolation. x = 9 is outside the data range of 1–7 hours. The linear relationship may not hold beyond observed data, so predictions could be unreliable.

Time Series

Q4 — Moving average smoothing
Fluency
Monthly revenue ($‘000s) for the first six months of the year: Jan = 18, Feb = 24, Mar = 21, Apr = 27, May = 24, Jun = 30.

Calculate the 3-point centred moving average for March, April, and May.

March: (18 + 24 + 21) ÷ 3 = 63 ÷ 3 = 21.0

April: (24 + 21 + 27) ÷ 3 = 72 ÷ 3 = 24.0

May: (21 + 27 + 24) ÷ 3 = 72 ÷ 3 = 24.0

Note: The moving average has smoothed the data, removing month-to-month variability and revealing an upward trend.
Q5 — Seasonal indices and deseasonalising
Understanding
Quarterly sales ($‘000s) for a business over three years are shown below.

Year Q1 Q2 Q3 Q4

Year 1 40 60 80 60

Year 2 44 66 88 66

Year 3 48 72 96 72

(a) Calculate the seasonal index (SI) for each quarter.

(b) Deseasonalise the Year 3, Quarter 3 value.

(a) Quarterly averages: Q1 = (40+44+48)/3 = 44; Q2 = (60+66+72)/3 = 66; Q3 = (80+88+96)/3 = 88; Q4 = (60+66+72)/3 = 66.
Overall mean = (44+66+88+66)/4 = 264/4 = 66
SI(Q1) = 44/66 = 0.667 | SI(Q2) = 66/66 = 1.000 | SI(Q3) = 88/66 = 1.333 | SI(Q4) = 66/66 = 1.000
Check: 0.667 + 1.000 + 1.333 + 1.000 = 4.000 ✓

(b) Deseasonalised Y3Q3 = actual ÷ SI(Q3) = 96 ÷ 1.333 = 72.0 ($‘000s)
Q6 — Forecasting with trend and seasonal indices
Problem Solving
Using the seasonal indices from Q5, the deseasonalised trend line (where t = 1 is Year 1, Q1) is: D_t = 57 + 1.5t.

(a) Forecast sales for Year 4, Quarter 1 (t = 13).

(b) Forecast sales for Year 4, Quarter 3 (t = 15).

(c) Explain why these forecasts might be less reliable as you project further into the future.

(a) Year 4, Q1 (t = 13):
Trend: D₁₃ = 57 + 1.5 × 13 = 57 + 19.5 = 76.5
Forecast = D₁₃ × SI(Q1) = 76.5 × 0.667 = $51.0 thousand

(b) Year 4, Q3 (t = 15):
Trend: D₁₅ = 57 + 1.5 × 15 = 57 + 22.5 = 79.5
Forecast = D₁₅ × SI(Q3) = 79.5 × 1.333 = $106.0 thousand

(c) Forecasts assume the current linear trend and seasonal pattern continue unchanged. Unexpected events (economic downturns, new competitors, changes in consumer behaviour) can invalidate the model. The further ahead you forecast, the more uncertainty accumulates.

Year	Q1	Q2	Q3	Q4
Year 1	40	60	80	60
Year 2	44	66	88	66
Year 3	48	72	96	72

Growth and Decay in Sequences

Q7 — Arithmetic sequences: salary growth
Fluency
An employee starts on a salary of $45,000 and receives a fixed annual raise of $2,500.

(a) Write a recurrence relation for the salary in Year n.

(b) Calculate the salary in Year 5.

(c) In which year does the salary first exceed $55,000?

(a) V₁ = 45 000; V_n+1 = V_n + 2500 (or t_n = 45 000 + (n−1) × 2500)

(b) t₅ = 45 000 + 4 × 2500 = 45 000 + 10 000 = $55,000

(c) t_n > 55 000 ⇒ 45 000 + (n−1) × 2500 > 55 000 ⇒ (n−1) > 4 ⇒ n > 5.
First exceeds $55,000 in Year 6 (t₆ = $57,500).
Q8 — Geometric sequences: compound growth
Fluency
A bank account starts with $5,000 and earns 4% interest per annum, compounded annually.

(a) Write a recurrence relation for the account balance after n years.

(b) Calculate the balance after 3 years, correct to the nearest cent.

(c) How many years will it take for the balance to double? (Use trial and error or logarithms.)

(a) V₀ = 5000; V_n+1 = 1.04 × V_n

(b) V₃ = 5000 × (1.04)³ = 5000 × 1.124864 = $5624.32

(c) Solve 5000 × 1.04ⁿ ≥ 10 000 ⇒ 1.04ⁿ ≥ 2.
Using logarithms: n ≥ log(2)/log(1.04) = 0.6931/0.01703 ≈ 17.67.
The balance first doubles after 18 years.
Q9 — Annuity recurrence relation and future value
Understanding
Annette deposits $400 at the end of each month into an account paying 0.5% per month interest.

(a) Write a recurrence relation for the account balance after n months.

(b) Calculate the balance after the first 3 months by iterating the recurrence relation.

(c) Use the future value formula to find the balance after 12 months. [(1.005)¹² ≈ 1.06168]

(a) V₀ = 0; V_n+1 = 1.005 × V_n + 400

(b)
V₁ = 1.005 × 0 + 400 = $400.00
V₂ = 1.005 × 400 + 400 = 402 + 400 = $802.00
V₃ = 1.005 × 802 + 400 = 806.01 + 400 = $1,206.01

(c) FV = D × ((1+r)ⁿ − 1)/r = 400 × (1.06168 − 1)/0.005 = 400 × (0.06168/0.005) = 400 × 12.336 = $4,934.40

Q10 — Comparing salary growth models

Problem Solving

Two job offers have different salary structures:

Option A: Starting salary $50,000, with a fixed raise of $1,000 per year.
Option B: Starting salary $46,000, with a 4% annual increase.

(a) Calculate both salaries for Years 1 to 6.

(b) In which year does Option B first exceed Option A?

(a) Annual salaries:

Year	Option A	Option B
1	$50,000	$46,000
2	$51,000	$47,840
3	$52,000	$49,754
4	$53,000	$51,744
5	$54,000	$53,814
6	$55,000	$55,966

(b) Option B first exceeds Option A in Year 6 ($55,966 > $55,000).

(c) Option A total (Y1–Y10): 10 terms of arithmetic series with a=$50,000, d=$1,000:
S₁₀ = 10/2 × (2×50,000 + 9×1,000) = 5 × 109,000 = $545,000
Option B total: 46,000 × (1.04¹⁰−1)/0.04 = 46,000 × (1.48024−1)/0.04 = 46,000 × 12.006 = $552,276
Over 10 years, Option B gives approximately $7,276 more in total earnings.

Earth Geometry and Time Zones

Q11 — Time zones
Fluency
Sydney operates on UTC+11 (daylight saving) and Los Angeles operates on UTC−8.

(a) When it is 3:00 PM Wednesday in Sydney, what is the time and day in Los Angeles?

(b) A flight departs Sydney at 10:00 AM Thursday (local time) and arrives in Los Angeles at 6:00 AM Thursday (local time). What is the actual flight duration?

(a) Sydney is 19 hours ahead of LA (11 − (−8) = 19).
LA time = Sydney time − 19 hours = 3:00 PM Wed − 19 h = 8:00 PM Tuesday

(b) Convert both times to UTC:
Departure: 10:00 AM Thu (UTC+11) → 10 − 11 = 11:00 PM Wednesday (UTC)
Arrival: 6:00 AM Thu (UTC−8) → 6 + 8 = 2:00 PM Thursday (UTC)
Duration = from 11:00 PM Wed to 2:00 PM Thu = 15 hours
Q12 — Great circle distance between two cities
Understanding
City A is located at 23°N and City B is at 42°S. Both cities lie on the same meridian (same longitude). Use Earth’s radius R = 6400 km.

(a) Find the angular separation between the two cities.

(b) Calculate the great circle distance between them.

(a) Since the cities are on opposite sides of the equator, the angular difference = 23° + 42° = 65°

(b) d = (θ/360) × 2πR = (65/360) × 2π × 6400 = 0.18056 × 40 212 ≈ 7260 km
Q13 — Navigation on Earth’s surface
Problem Solving
A ship is located at 20°S, 150°E. Use 1° of latitude ≈ 111 km for all calculations.

(a) The ship sails due north for 2220 km. State its new position (latitude and longitude).

(b) From its new position, the ship sails due east for 1110 km. State the final position. (At the equator, 1° longitude ≈ 111 km.)

(a) Distance north = 2220 km ÷ 111 km/° = 20° north.
Starting at 20°S, moving 20° north reaches the equator.
New position: 0° (Equator), 150°E

(b) Now at the equator, 1° longitude ≈ 111 km.
Change in longitude = 1110 ÷ 111 = 10° east.
New longitude = 150°E + 10° = 160°E
Final position: 0°, 160°E

Loans, Investments and Annuities

Q14 — Compound interest
Fluency
$5,000 is invested at 6% per annum, compounded monthly, for 2 years. [(1.005)²⁴ ≈ 1.12716]

(a) Write a recurrence relation for the investment balance.

(b) Calculate the final value, correct to the nearest cent.

(c) Calculate the total interest earned.

(a) Monthly rate = 6%/12 = 0.5% = 0.005.
V₀ = 5000; V_n+1 = 1.005 × V_n

(b) V₂₄ = 5000 × (1.005)²⁴ = 5000 × 1.12716 = $5,635.80

(c) Interest = $5,635.80 − $5,000.00 = $635.80
Q15 — Reducing-balance loan
Fluency
A personal loan of $10,000 is taken at 1% per month interest. Monthly repayments are $400.

(a) Write a recurrence relation for the outstanding balance.

(b) Calculate the balance after 3 monthly repayments.

(c) Explain why the loan balance is reducing month by month.

(a) V₀ = 10 000; V_n+1 = 1.01 × V_n − 400

(b)
V₁ = 1.01 × 10 000 − 400 = 10 100 − 400 = $9,700.00
V₂ = 1.01 × 9700 − 400 = 9797 − 400 = $9,397.00
V₃ = 1.01 × 9397 − 400 = 9490.97 − 400 = $9,090.97

(c) Month 1 interest = $10,000 × 0.01 = $100. Since the $400 repayment exceeds the $100 interest, $300 reduces the principal each period. Each month the interest decreases as the principal falls, so the loan is progressively paid down.
Q16 — Future value of an annuity
Understanding
Sofia deposits $500 at the end of each month into a savings account that pays 0.5% interest per month. [(1.005)¹² ≈ 1.06168]

(a) Write the future value formula and identify each variable.

(b) Calculate the account balance after 12 months.

(c) How much of the final balance is from interest earned (not her own deposits)?

(a) FV = D × ((1+r)ⁿ − 1) / r
where D = $500 (regular deposit), r = 0.005 (monthly rate), n = 12 (number of periods).

(b) FV = 500 × (1.06168 − 1) / 0.005 = 500 × (0.06168 / 0.005) = 500 × 12.336 = $6,168.00

(c) Total deposited = 500 × 12 = $6,000. Interest earned = $6,168 − $6,000 = $168.00
Q17 — Comparing savings strategies
Problem Solving
David is planning a 5-year savings strategy and considers two options, both with 0.5% per month interest [(1.005)⁶⁰ ≈ 1.3489]:
- Option A: Invest a one-time lump sum of $25,000 today.
- Option B: Deposit $450 at the end of each month for 60 months.
(a) Calculate the final value of each option after 5 years (60 months).

(b) Which option results in a larger balance, and by how much?

(c) Option B requires smaller initial capital. State one advantage of Option B despite giving a lower final balance.

(a)
Option A: FV = 25 000 × (1.005)⁶⁰ = 25 000 × 1.3489 = $33,722.50
Option B: FV = 450 × ((1.005)⁶⁰ − 1) / 0.005 = 450 × (0.3489/0.005) = 450 × 69.78 = $31,401.00

(b) Option A gives a larger balance by $33,722.50 − $31,401.00 = $2,321.50

(c) Option B does not require $25,000 upfront; it spreads deposits over 5 years ($27,000 total contributed), making it more accessible for someone without a large lump sum. It also provides flexibility month-to-month.

Graphs and Networks

Q18 — Graph theory and Euler paths
Fluency
A graph has vertices A, B, C, D, E and the following edges: AB, AC, BC, CD, DE, EA.

(a) State the degree of each vertex.

(b) Does an Euler circuit exist? Give a reason.

(c) Does an Euler path exist? If so, state the two endpoints and write one valid Euler path.

(a) Vertex degrees:
A: edges AB, AC, EA → degree 3
B: edges AB, BC → degree 2
C: edges AC, BC, CD → degree 3
D: edges CD, DE → degree 2
E: edges DE, EA → degree 2

(b) No Euler circuit. An Euler circuit requires all vertices to have even degree. Vertices A and C both have odd degree (3).

(c) Yes, an Euler path exists. Exactly 2 vertices have odd degree (A and C), so the Euler path runs from A to C (or C to A).
One valid path: A → B → C → D → E → A → C (uses all 6 edges exactly once)
Q19 — Dijkstra’s shortest path algorithm
Fluency
A weighted graph has vertices A, B, C, D, E with the following edge weights: AB = 3, AC = 7, BC = 2, BD = 5, CE = 4, DE = 3. Find the shortest path from A to E and its total length.

Dijkstra’s algorithm:

Start: A = 0 (all others ∞)
Visit A: update B = 3, C = 7
Visit B (dist 3): update C = min(7, 3+2) = 5; D = 3+5 = 8
Visit C (dist 5): update E = 5+4 = 9
Visit D (dist 8): E = min(9, 8+3) = 9 (no improvement)
Visit E (dist 9)

Shortest path: A → B → C → E
Total length = 3 + 2 + 4 = 9
Q20 — Minimum spanning tree
Understanding
A graph has vertices A, B, C, D, E with edge weights: AB = 5, AC = 3, BC = 6, BD = 4, CD = 2, CE = 7, DE = 5.

Use Kruskal’s algorithm to find the minimum spanning tree (MST). State each edge added in order and the total weight.

Kruskal’s — add edges in ascending weight, no cycles:

Sorted edges: CD=2, AC=3, BD=4, AB=5, DE=5, BC=6, CE=7

1. Add CD = 2: connects C–D ✓
2. Add AC = 3: connects A to {C,D} ✓
3. Add BD = 4: connects B to {A,C,D} ✓
4. AB = 5: would create cycle A–B–C–A ✗
5. Add DE = 5: connects E to {A,B,C,D} — all 5 vertices now connected ✓

MST edges: CD, AC, BD, DE
Total weight = 2 + 3 + 4 + 5 = 14
Q21 — Minimum spanning tree application
Problem Solving
A utility company needs to connect five substations P, Q, R, S, T with the minimum total cable. Possible connection costs (km) are: PQ = 8, PR = 12, PS = 5, QR = 7, QS = 9, QT = 10, RS = 6, RT = 11, ST = 4.

(a) Use Kruskal’s algorithm to find the minimum spanning tree.

(b) What is the minimum total cable length required?

(c) The company later decides substation T must connect directly to Q (for reliability). By how much does this increase the total cable length compared to the MST?

(a) Kruskal’s algorithm:

Sorted edges: ST=4, PS=5, RS=6, QR=7, PQ=8, QS=9, QT=10, RT=11, PR=12

1. Add ST = 4: S–T ✓
2. Add PS = 5: P joins {S,T} ✓
3. Add RS = 6: R joins {P,S,T} ✓
4. Add QR = 7: Q joins — all 5 connected ✓
MST: ST, PS, RS, QR

(b) Minimum total cable = 4 + 5 + 6 + 7 = 22 km

(c) In the MST, Q connects via QR=7, and T connects via ST=4. Adding a direct QT=10 link while keeping the MST would create a cycle Q–R–S–T–Q. To add QT while maintaining a spanning tree, we remove the most expensive edge in that cycle (ST=4) and add QT=10.
New MST total = 22 − 4 + 10 = 28 km (+6 km more than the optimal MST).

Networks and Decision Mathematics

Q22 — Critical path analysis

Fluency

A project has the following activities (duration in days, prerequisites in brackets):

Activity	Duration (days)	Prerequisites
A	3	—
B	5	—
C	4	A
D	6	B
E	3	C, D
F	2	E

(a) Find the Earliest Start Time (EST) for each activity.

(b) State the minimum project duration and the critical path.

(a) Earliest Start Times (forward pass):
A: 0 | B: 0 | C: 0+3 = 3 | D: 0+5 = 5
E: max(3+4, 5+6) = max(7, 11) = 11
F: 11+3 = 14

(b) Minimum project duration = 14 + 2 = 16 days
Critical path: B → D → E → F (0+5+6+3+2 = 16 days)

(c) Float (backward pass):
F: LST = 16−2 = 14; E: LST = 14−3 = 11; D: LST = 11−6 = 5; B: LST = 5−5 = 0
C: LFT = 11 (must finish before E starts), LST = 11−4 = 7; Float(C) = 7 − 3 = 4 days
A: LFT = 7, LST = 7−3 = 4; Float(A) = 4 − 0 = 4 days

Q23 — Two-person zero-sum game
Understanding
Two players compete in a game with the following payoff matrix (values show Row player’s gain):

Column B1 Column B2

Row A1 4 2

Row A2 1 6

(a) Determine whether a saddle point exists.

(b) If no saddle point exists, use mixed strategies. Find the optimal probabilities and the value of the game.

(c) Which player does the game favour, and why?

(a) Row minima: A1 → min(4,2)=2; A2 → min(1,6)=1. Maximin = 2 (Row plays A1).
Column maxima: B1 → max(4,1)=4; B2 → max(2,6)=6. Minimax = 4 (Column plays B1).
Maximin (2) ≠ Minimax (4) → No saddle point. Mixed strategies required.

(b) For matrix [[a,b],[c,d]] = [[4,2],[1,6]]:
D = (a+d) − (b+c) = (4+6) − (2+1) = 10 − 3 = 7
p (prob Row plays A1) = (d−c)/D = (6−1)/7 = 5/7 ≈ 0.714
q (prob Col plays B1) = (d−b)/D = (6−2)/7 = 4/7 ≈ 0.571
V = (ad−bc)/D = (4×6 − 2×1)/7 = (24−2)/7 = 22/7 ≈ 3.14

(c) V ≈ 3.14 > 0, so the game favours the Row player. On average, Row wins approximately 3.14 units per game when both players use their optimal mixed strategies.
Q24 — The Hungarian assignment algorithm
Understanding
Three workers (W1, W2, W3) can each be assigned to one of three tasks (T1, T2, T3). The cost matrix (in hours) is:

T1 T2 T3

W1 6 9 5

W2 8 4 7

W3 3 6 8

Use the Hungarian algorithm to find the optimal assignment and minimum total cost.

Step 1 — Row reduction (subtract row minimum):
W1 (min=5): [1, 4, 0] W2 (min=4): [4, 0, 3] W3 (min=3): [0, 3, 5]

Step 2 — Column reduction (all column minima already 0; no change):
Col T1 min=0, Col T2 min=0, Col T3 min=0.

Step 3 — Assign zeros (one zero per row and column):
W1→T3 (zero), W2→T2 (zero), W3→T1 (zero) — unique assignment found immediately.

Step 4 — Check: 3 lines cover all zeros = n = 3 → Optimal!

Optimal assignment: W1→T3 (5 hrs), W2→T2 (4 hrs), W3→T1 (3 hrs)
Minimum total cost = 5 + 4 + 3 = 12 hours

	Column B1	Column B2
Row A1	4	2
Row A2	1	6

	T1	T2	T3
W1	6	9	5
W2	8	4	7
W3	3	6	8

Q25 — Integrated networks problem

Problem Solving

A construction company is managing a large project with multiple aspects to optimise.

Part (a) — Critical path
The project activities are:

Activity	Duration (days)	Prerequisites
P	4	—
Q	6	P
R	3	—
S	5	Q, R
T	7	S
U	2	T

Find the project duration, the critical path, and the float for activity R.

Part (b) — Network flow
Materials flow through a supply network from Source (S) to Terminal (T) with the following capacities:

S → A: 20, S → B: 15
A → C: 12, A → T: 9
B → C: 10, B → T: 8
C → T: 18

Find the maximum flow from S to T using augmenting paths. Verify using the minimum cut theorem.

Part (c) — Assignment
Four workers (W, X, Y, Z) are to be assigned to four posts (P, Q, R, S). The cost matrix ($‘000s) is:

	P	Q	R	S
W	7	5	8	9
X	3	6	8	7
Y	6	8	7	3
Z	5	7	3	6

Use the Hungarian algorithm to find the optimal assignment and minimum total cost.

Part (a) — Critical path:
EST: P=0, R=0, Q=0+4=4, S=max(4+6, 0+3)=max(10,3)=10, T=10+5=15, U=15+7=22.
Project duration = 22+2 = 24 days.
Critical path: P → Q → S → T → U (4+6+5+7+2 = 24 days).
R: LFT = LST(S) = 10; LST(R) = 10−3 = 7. Float(R) = 7 − 0 = 7 days.

Part (b) — Network flow (augmenting paths):

Path 1: S→A→T: min(20, 9) = 9. Remaining: S→A=11, A→T=0.
Path 2: S→A→C→T: min(11, 12, 18) = 11. Remaining: S→A=0, A→C=1, C→T=7.
Path 3: S→B→C→T: min(15, 10, 7) = 7. Remaining: S→B=8, B→C=3, C→T=0.
Path 4: S→B→T: min(8, 8) = 8. Remaining: S→B=0, B→T=0.
Maximum flow = 9 + 11 + 7 + 8 = 35 units
Verify: minimum cut {S→A, S→B} = 20 + 15 = 35 ✓

Part (c) — Hungarian algorithm (4×4):

Row reduction (subtract row min — W:5, X:3, Y:3, Z:3):
W: [2, 0, 3, 4] X: [0, 3, 5, 4] Y: [3, 5, 4, 0] Z: [2, 4, 0, 3]

Column reduction (all column minima = 0; no further reduction needed).

Reduced matrix zeros: (W,Q), (X,P), (Y,S), (Z,R) — one zero per row and column.
Cover with 4 lines = n = 4 → Optimal assignment found immediately.

Optimal assignment: W→Q ($5k), X→P ($3k), Y→S ($3k), Z→R ($3k)
Minimum total cost = 5 + 3 + 3 + 3 = $14,000

Year 12 Review — General Maths

Bivariate Data

Q1 — Correlation

Q2 — Least squares regression equation

Q3 — Prediction and residual analysis

Time Series

Q4 — Moving average smoothing

Q5 — Seasonal indices and deseasonalising

Q6 — Forecasting with trend and seasonal indices

Growth and Decay in Sequences

Q7 — Arithmetic sequences: salary growth

Q8 — Geometric sequences: compound growth

Q9 — Annuity recurrence relation and future value

Q10 — Comparing salary growth models

Earth Geometry and Time Zones

Q11 — Time zones

Q12 — Great circle distance between two cities

Q13 — Navigation on Earth’s surface

Loans, Investments and Annuities

Q14 — Compound interest

Q15 — Reducing-balance loan

Q16 — Future value of an annuity

Q17 — Comparing savings strategies

Graphs and Networks

Q18 — Graph theory and Euler paths

Q19 — Dijkstra’s shortest path algorithm

Q20 — Minimum spanning tree

Q21 — Minimum spanning tree application

Networks and Decision Mathematics

Q22 — Critical path analysis

Q23 — Two-person zero-sum game

Q24 — The Hungarian assignment algorithm

Q25 — Integrated networks problem