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Selected-topic Practice Paper Showcase

TASC TASC General Mathematics Level 3 Free Online Pack 0 — Selected-topic Practice Paper Showcase

Read Selected-topic Practice Paper Showcase online for free, including every question, worked solution, marking note and diagnostic action. No public PDF download or checkout is provided.

TASC Selected-topic Practice Paper 2026 Edition - Pack 0 v1.1
Selected-topic Practice Paper Showcase is free to read in your browser. There is no public checkout or PDF download.

Exam-pack paper structure

This full-length showcase paper is available to read online.

Selected-topic Practice Paper Showcase

10 questions

100 marks

Estimated duration: Approximately 100 minutes suggested working time

Reading: Optional 15 minutes · Writing: Approximately 100 minutes suggested working time

Read Selected-topic Practice Paper Showcase online

Skill Align

Skill Align TASC General Mathematics Level 3 Selected-topic Practice Paper Pack 0 - 2026 Edition

Current for 2026 | Course code MTG315123. This condensed 100-mark paper samples selected content and criteria from the current course. It is not a full external-examination simulation and does not reproduce the official five-section, 180-mark architecture. Samples bivariate numerical and categorical data, sequences, loans and annuities, trigonometry, Earth geometry, shortest paths, critical paths and adjacency matrices; time-series seasonality, maximum flow and assignment are omitted.

Paper
Selected-topic Practice Paper Showcase
Preparation time
Optional 15 minutes
Suggested working time
Approximately 100 minutes suggested working time
Assessment
100 marks

Skill Align practice conditions: use a TASC-approved calculator and the current MTG315123 General Mathematics Information Sheet, which must be supplied or opened from the official resource link before the session. Spreadsheet software is permitted for questions that explicitly request it. Internet access beyond the supplied information sheet and all external communication are not permitted.

Selected-topic practice questions - 100 marks

Answer all questions. The optional 15-minute preparation period may be used to read, plan and annotate this booklet; do not begin writing responses until the suggested working period starts. Show method, working, units, interpretation, assumptions and reasonableness when requested. Network and trigonometry/Earth-geometry questions are both included for revision; they are not presented as the official Section E alternative choice.

Question 1

10 marks
Course alignment: Section B alignment; Module 2 Topic 1: Statistical analysis; Bivariate numerical data; Criterion 5. A training study records sessions x=2,4,6,8,10 and scores y=23,28,31,38,41. Calculator output gives hat y=18.4+2.3x. The observed x-range is 2 to 10 sessions.
(a) 3 marks
Interpret the slope in context.
(b) 3 marks
Calculate the residual for x=6.
(c) 4 marks
A coach wants to predict the score after 30 sessions. Explain whether the model should be used and identify one assumption.

Question 2

10 marks
Course alignment: Section B alignment; Module 2 Topic 1: Statistical analysis - bivariate categorical data; Bivariate categorical data; Criterion 5. A survey records travel mode: urban students 42 bus and 18 car; rural students 12 bus and 28 car.
(a) 3 marks
Complete the urban and rural row totals.
(b) 3 marks
Calculate the bus percentage within each location group.
(c) 4 marks
State whether the variables appear associated. Explain why this does not establish causation.

Question 3

10 marks
Course alignment: Section C alignment; Module 2 Topic 2: Growth and decay in sequences; Growth and decay in sequences; Criterion 6. A vehicle is valued at $28 000 and depreciates by 16% each year. Let V_n be its value after n years.
(a) 3 marks
Write an explicit model for V_n.
(b) 3 marks
Find the value after 3 years, to the nearest dollar.
(c) 4 marks
Find the percentage value lost after 3 years and comment on whether a constant-rate model is reasonable.

Question 4

10 marks
Course alignment: Section D alignment; Module 3 Topic 1: Investment, loans and annuities; Loans and reducing balances; Criterion 7. A monthly loan balance follows B_(n+1)=1.004B_n-850, with B_0=24000.
(a) 3 marks
State the monthly interest rate and repayment.
(b) 3 marks
Calculate B_1.
(c) 4 marks
Calculate B_2 to the nearest cent and explain when the final repayment would differ from $850.

Question 5

10 marks
Course alignment: Section E Trigonometry alignment; Module 3 Topic 2a: Trigonometry and Earth geometry; Non-right-angled triangles; Criterion 8. In triangle ABC, AB=7.2 km, AC=9.5 km and angle BAC=58^circ.
(a) 3 marks
Calculate BC to two decimal places.
(b) 3 marks
Calculate angle ABC to the nearest degree.
(c) 4 marks
Find the area and state an appropriate unit.

Question 6

10 marks
Course alignment: Section E Trigonometry alignment; Module 3 Topic 2a: Trigonometry and Earth geometry; Earth geometry; Criterion 8. Use Earth radius 6371 km. Two locations lie on latitude 42^circ S, with longitudes 147^circ E and 173^circ E.
(a) 3 marks
Find the longitude difference in radians.
(b) 3 marks
Find the radius of the parallel at 42^circ S.
(c) 4 marks
Estimate the east-west distance along the parallel.

Question 7

10 marks
Course alignment: Section E Networks alignment; Module 3 Topic 2b: Graphs, networks and decision mathematics; Shortest paths; Criterion 8. A road network has weighted edges AB=7, AC=4, BC=2, BD=5, CD=8, CE=7, DE=3. All weights are kilometres.
(a) 3 marks
List the current shortest labels from A to B and C after considering A.
(b) 3 marks
Find the shortest distance from A to D.
(c) 4 marks
Find the shortest distance and route from A to E, and justify that no shorter unsettled label remains.

Question 8

10 marks
Course alignment: Section E Networks alignment; Module 3 Topic 2b: Graphs, networks and decision mathematics; Project networks; Criterion 8. A project has activities A=3 starting at Start; B=5 and C=4 after A; D=2 and E=3 after both B and C; F=4 after both D and E; then Finish. Durations are days.
(a) 3 marks
Calculate the earliest start and finish times for C and E.
(b) 3 marks
Find the minimum project duration.
(c) 4 marks
Identify the critical path and state the float of B.

Question 9

10 marks
Course alignment: Section E Networks alignment; Module 3 Topic 2b: Graphs, networks and decision mathematics; Graphs and adjacency matrices; Criterion 8. An undirected graph has vertices A,B,C,D and edges AB, AC, BC, CD. Vertex order is A,B,C,D.
(a) 3 marks
Write the adjacency matrix.
(b) 3 marks
State the degree of each vertex.
(c) 4 marks
Determine whether an Euler trail exists and give one if it does.

Question 10

10 marks
Course alignment: Section D alignment; Module 3 Topic 1: Investment, loans and annuities; Annuities; Criterion 7. A savings account starts at A_0=0. At the end of each month it earns 0.4% interest and then receives a $250 deposit, so A_(n+1)=1.004A_n+250.
(a) 3 marks
Calculate the balances after the first two deposits.
(b) 3 marks
Use A_n=250((1.004^n-1) / (0.004)) to find A_(12).
(c) 4 marks
State two assumptions and evaluate whether this model is suitable for a one-year target.

TASC courses, assessment and certification are administered by the Tasmanian Assessment, Standards and Certification office. Skill Align is an independent publisher and is not affiliated with, authorised by, sponsored by, approved by or endorsed by TASC or the Tasmanian Government. This is original Skill Align selected-topic practice material, not an official TASC assessment.

Copyright (c) 2026 Skill Align. Free for personal, non-commercial online viewing at https://skillalign.au. You may share the Skill Align page link. Except as permitted by law or with Skill Align's prior written permission, the pack itself must not be resold, copied, redistributed, republished, automatically extracted, or uploaded to a question bank.

Worked Solutions And Marking Guide

Question 1

(a) Each extra session is associated with a predicted score increase of 2.3 points.

The slope is 2.3 predicted score points per additional session.

(b) -1.2 points.

The fitted value is 18.4+2.3(6)=32.2, so residual =31-32.2=-1.2.

(c) The prediction is an unsupported extrapolation; it assumes the linear pattern continues well beyond the observed range.

Thirty is outside 2 to 10. A numerical substitution alone does not make the prediction reliable.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Identifies the slope as the predicted change in score for one additional training session.
  2. Uses the numerical slope 2.3 with score-point units.
  3. States that each additional session is associated with a predicted increase of 2.3 score points.

Do not credit by itself: Omitting or contradicting this required achievement: Identifies the slope as the predicted change in score for one additional training session. Stating a result without establishing: States that each additional session is associated with a predicted increase of 2.3 score points.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Substitutes x=6 into 18.4+2.3x.
  2. Obtains the fitted score 32.2.
  3. Calculates observed minus fitted, 31-32.2=-1.2 points.

Do not credit by itself: Omitting or contradicting this required achievement: Substitutes x=6 into 18.4+2.3x. Stating a result without establishing: Calculates observed minus fitted, 31-32.2=-1.2 points.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Notes that 30 sessions is outside the observed range 2 to 10.
  2. Classifies use at x=30 as substantial extrapolation.
  3. States the assumption that the fitted linear pattern continues beyond the observed range.
  4. Concludes that the prediction is unsupported or unreliable despite being calculable.

Do not credit by itself: Omitting or contradicting this required achievement: Notes that 30 sessions is outside the observed range 2 to 10. Stating a result without establishing: Concludes that the prediction is unsupported or unreliable despite being calculable.

Question 2

(a) Urban 60; rural 40.

Add the two modes in each row.

(b) Urban 70%; rural 30%.

Use bus count divided by its row total.

(c) They appear associated because the conditional percentages differ by 40 percentage points; an observational table cannot rule out confounding or establish cause.

Compare like row percentages and distinguish association from causation.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Adds the urban counts 42+18.
  2. Adds the rural counts 12+28.
  3. States the row totals as urban 60 and rural 40.

Do not credit by itself: Omitting or contradicting this required achievement: Adds the urban counts 42+18. Stating a result without establishing: States the row totals as urban 60 and rural 40.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Calculates the urban conditional percentage 42 / 60 × 100.
  2. Calculates the rural conditional percentage 12 / 40 × 100.
  3. States urban 70% and rural 30%.

Do not credit by itself: Omitting or contradicting this required achievement: Calculates the urban conditional percentage 42 / 60 × 100. Stating a result without establishing: States urban 70% and rural 30%.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Compares the conditional bus-use percentages using the 40-percentage-point difference.
  2. Concludes that location and travel mode appear associated.
  3. Recognises that the survey is observational rather than a controlled experiment.
  4. Explains that confounding or other group differences prevent a causal conclusion.

Do not credit by itself: Omitting or contradicting this required achievement: Compares the conditional bus-use percentages using the 40-percentage-point difference. Stating a result without establishing: Explains that confounding or other group differences prevent a causal conclusion.

Question 3

(a) V_n=28000(0.84)^n.

A 16% decrease leaves a multiplier of 0.84.

(b) $16,596.

28000(0.84)³=16595.712.

(c) Approximately 40.73% lost; the model is reasonable only while the percentage depreciation is approximately constant.

Compare the loss with the original value and state the modelling assumption.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Identifies the initial value V_0=28000.
  2. Converts 16% depreciation to the multiplier 1-0.16=0.84.
  3. Writes V_n=28000(0.84)^n.

Do not credit by itself: Omitting or contradicting this required achievement: Identifies the initial value V_0=28000. Stating a result without establishing: Writes V_n=28000(0.84)^n.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Substitutes n=3 into 28000(0.84)^n.
  2. Evaluates the unrounded value 16595.712.
  3. Rounds to $16,596.

Do not credit by itself: Omitting or contradicting this required achievement: Substitutes n=3 into 28000(0.84)^n. Stating a result without establishing: Rounds to $16,596.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Calculates the three-year loss 28000-16595.712=11404.288.
  2. Expresses the loss as 11404.288 / 28000 × 100approx40.73%.
  3. Identifies the constant annual percentage-depreciation assumption.
  4. Limits the model's reasonableness to a period in which the 16% rate remains plausible.

Do not credit by itself: Omitting or contradicting this required achievement: Calculates the three-year loss 28000-16595.712=11404.288. Stating a result without establishing: Limits the model's reasonableness to a period in which the 16% rate remains plausible.

Question 4

(a) 0.4% per month and $850 per month.

Read the multiplier and subtraction from the recurrence.

(b) $23,246.

1.004(24000)-850=23246.

(c) B_2=$22,488.98; the last payment is adjusted when the balance plus interest is less than $850.

Apply the recurrence again and interpret the terminal balance.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Interprets 1.004 as a monthly growth factor.
  2. Converts the growth factor to 0.4% monthly interest.
  3. Interprets the subtraction as an $850 monthly repayment after interest.

Do not credit by itself: Omitting or contradicting this required achievement: Interprets 1.004 as a monthly growth factor. Stating a result without establishing: Interprets the subtraction as an $850 monthly repayment after interest.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Calculates first-month interest 0.004(24000)=96.
  2. Uses the operation order 24000+96-850.
  3. Obtains B_1=$23,246.

Do not credit by itself: Omitting or contradicting this required achievement: Calculates first-month interest 0.004(24000)=96. Stating a result without establishing: Obtains B_1=$23,246.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Substitutes B_1=23246 into the recurrence.
  2. Calculates 1.004(23246)=23338.984.
  3. Obtains B_2=$22,488.98 to the nearest cent.
  4. Explains that the last repayment is reduced when the interest-added balance is below $850.

Do not credit by itself: Omitting or contradicting this required achievement: Substitutes B_1=23246 into the recurrence. Stating a result without establishing: Explains that the last repayment is reduced when the interest-added balance is below $850.

Question 5

(a) BCapprox8.34 km.

Use BC²=7.2²+9.5²-2(7.2)(9.5)cos58^circ.

(b) angle ABCapprox75^circ.

Using BCapprox8.34248 km, apply the cosine rule: cos B=(7.2²+8.34248²-9.5²) / (2(7.2)(8.34248)), so Bapprox74.953^circapprox75^circ.

(c) Approximately 29.0 km^2.

Use frac12(7.2)(9.5)sin58^circ.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Selects the cosine rule because two sides and the included angle are known.
  2. Substitutes BC²=7.2²+9.5²-2(7.2)(9.5)cos58^circ.
  3. Obtains BCapprox8.34 km to two decimal places.

Do not credit by itself: Omitting or contradicting this required achievement: Selects the cosine rule because two sides and the included angle are known. Stating a result without establishing: Obtains BCapprox8.34 km to two decimal places.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Selects the cosine rule using AB=7.2, AC=9.5 and the consequential value BCapprox8.34248.
  2. Substitutes cos B=(7.2²+8.34248²-9.5²) / (2(7.2)(8.34248)).
  3. Obtains angle ABCapprox74.953^circapprox75^circ.

Do not credit by itself: Omitting or contradicting this required achievement: Selects the cosine rule using AB=7.2, AC=9.5 and the consequential value BCapprox8.34248. Stating a result without establishing: Obtains angle ABCapprox74.953^circapprox75^circ.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Selects A=frac12absin C using sides 7.2, 9.5 and included angle 58^circ.
  2. Substitutes frac12(7.2)(9.5)sin58^circ.
  3. Evaluates the area as approximately 29.0.
  4. States the area unit km^2.

Do not credit by itself: Omitting or contradicting this required achievement: Selects A=frac12absin C using sides 7.2, 9.5 and included angle 58^circ. Stating a result without establishing: States the area unit km^2.

Question 6

(a) 26^circ=((13π) / (90))approx0.4538 rad.

Subtract longitudes and convert degrees to radians.

(b) 6371cos42^circapprox4735 km.

A parallel has radius Rcos(latitude).

(c) Approximately 2149 km.

Use arc length s=rtheta with the small-circle radius.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Calculates the longitude difference 173^circ-147^circ=26^circ.
  2. Converts using 26π / 180=13π / 90.
  3. States the decimal angle 0.4538 radians.

Do not credit by itself: Omitting or contradicting this required achievement: Calculates the longitude difference 173^circ-147^circ=26^circ. Stating a result without establishing: States the decimal angle 0.4538 radians.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Uses the parallel-radius relationship r=Rcos(latitude).
  2. Substitutes r=6371cos42^circ.
  3. Obtains rapprox4735 km.

Do not credit by itself: Omitting or contradicting this required achievement: Uses the parallel-radius relationship r=Rcos(latitude). Stating a result without establishing: Obtains rapprox4735 km.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Selects the arc-length model s=rtheta.
  2. Substitutes the small-circle radius and radian angle 6371cos42^circ(13π / 90).
  3. Evaluates the distance as approximately 2149.
  4. States the distance in kilometres along the parallel.

Do not credit by itself: Omitting or contradicting this required achievement: Selects the arc-length model s=rtheta. Stating a result without establishing: States the distance in kilometres along the parallel.

Question 7

(a) B=7, C=4.

Use the two edges incident with A.

(b) 11 km via A-C-B-D.

Compare candidate routes; 4+2+5=11.

(c) 11 km via A-C-E.

After C is settled, E receives label 4+7=11; no unsettled vertex has a label below 11.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Assigns B the direct tentative label 7.
  2. Assigns C the direct tentative label 4.
  3. States the two labels as B=7 and C=4 after considering A.

Do not credit by itself: Omitting or contradicting this required achievement: Assigns B the direct tentative label 7. Stating a result without establishing: States the two labels as B=7 and C=4 after considering A.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Settles C at distance 4 and updates B to 4+2=6.
  2. Compares the routes to D, obtaining 6+5=11 via B and 4+8=12 directly via C.
  3. States the shortest A-to-D route A-C-B-D with distance 11 km.

Do not credit by itself: Omitting or contradicting this required achievement: Settles C at distance 4 and updates B to 4+2=6. Stating a result without establishing: States the shortest A-to-D route A-C-B-D with distance 11 km.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Updates E from C to label 4+7=11.
  2. Shows that the competing route through D is at least 11+3=14.
  3. States route A-C-E with distance 11 km.
  4. Justifies finality because no unsettled label below 11 can improve E.

Do not credit by itself: Omitting or contradicting this required achievement: Updates E from C to label 4+7=11. Stating a result without establishing: Justifies finality because no unsettled label below 11 can improve E.

Question 8

(a) C: ES=3, EF=7; E: ES=8, EF=11.

A finishes at 3, so C runs from 3 to 7. E must wait for both B and C; B finishes at 8, so E runs from 8 to 11.

(b) 15 days.

F must wait for D and E. E controls its start at time 11, so F finishes at time 15.

(c) Critical path A-B-E-F; B has 0 days float.

The backward pass gives zero float to A, B, E and F; C and D each have one day of float.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Uses A's finish at 3 to give C: ES=3, EF=7.
  2. Uses B's finish at 8 and C's finish at 7 to set E: ES=max(8,7)=8.
  3. States E: EF=8+3=11.

Do not credit by itself: Omitting or contradicting this required achievement: Uses A's finish at 3 to give C: ES=3, EF=7. Stating a result without establishing: States E: EF=8+3=11.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Finds D: ES=8, EF=10 from its two predecessors.
  2. Sets F: ES=max(10,11)=11.
  3. States the minimum project duration EF_F=15 days.

Do not credit by itself: Omitting or contradicting this required achievement: Finds D: ES=8, EF=10 from its two predecessors. Stating a result without establishing: States the minimum project duration EF_F=15 days.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Completes the backward pass to identify zero float for A, B, E and F.
  2. Identifies A-B-E-F as the critical path.
  3. States that B has 0 days float.
  4. Checks that C and D each have 1 day float, consistent with the critical-path result.

Do not credit by itself: Omitting or contradicting this required achievement: Completes the backward pass to identify zero float for A, B, E and F. Stating a result without establishing: Checks that C and D each have 1 day float, consistent with the critical-path result.

Question 9

(a) begin(bmatrix)0&1&1&01&0&1&01&1&0&10&0&1&0end(bmatrix).

Place 1 for each listed undirected adjacency.

(b) A=2, B=2, C=3, D=1.

Row sums give degrees.

(c) Yes; for example C-A-B-C-D.

Exactly C and D have odd degree, so a trail starts at one and ends at the other.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Uses vertex order A, B, C, D with zeros on the diagonal.
  2. Places symmetric ones for edges AB,AC,BC,CD and zeros for absent edges.
  3. Writes the complete matrix begin(bmatrix)0&1&1&01&0&1&01&1&0&10&0&1&0end(bmatrix).

Do not credit by itself: Omitting or contradicting this required achievement: Uses vertex order A, B, C, D with zeros on the diagonal. Stating a result without establishing: Writes the complete matrix begin(bmatrix)0&1&1&01&0&1&01&1&0&10&0&1&0end(bmatrix).

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Counts degrees A=2 and B=2.
  2. Counts degrees C=3 and D=1.
  3. States the complete ordered degree list 2,2,3,1.

Do not credit by itself: Omitting or contradicting this required achievement: Counts degrees A=2 and B=2. Stating a result without establishing: States the complete ordered degree list 2,2,3,1.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. Identifies C and D as the only odd-degree vertices.
  2. Concludes that an Euler trail exists but an Euler circuit does not.
  3. Gives a trail such as C-A-B-C-D that uses every edge exactly once.
  4. Uses C and D as the two endpoints.

Do not credit by itself: Omitting or contradicting this required achievement: Identifies C and D as the only odd-degree vertices. Stating a result without establishing: Uses C and D as the two endpoints.

Question 10

(a) A_1=$250.00, A_2=$501.00.

Apply the recurrence twice in the stated operation order.

(b) A_(12)approx$3,066.89.

Substitute n=12 and retain unrounded values.

(c) It assumes a constant monthly rate and deposits made on time with no fees or withdrawals; it is suitable for an estimate if those conditions are realistic.

Identify model assumptions and connect them to the one-year use.

Detailed marking criteria

Part a (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Uses A_0=0 to calculate A_1=1.004(0)+250=250.
  2. Applies interest before the second deposit.
  3. Calculates A_2=1.004(250)+250=$501.00.

Do not credit by itself: Omitting or contradicting this required achievement: Uses A_0=0 to calculate A_1=1.004(0)+250=250. Stating a result without establishing: Calculates A_2=1.004(250)+250=$501.00.

Part b (3 marks)

Use the 3 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement. Apply consequential marking only to a later checkpoint that correctly uses the candidate's earlier value; do not restore the mark lost for the original error.

Mark-by-mark checkpoints:

  1. Substitutes n=12 into 250(1.004^n-1) / 0.004.
  2. Evaluates the geometric accumulation factor without premature rounding.
  3. Obtains A_(12)approx$3,066.89.

Do not credit by itself: Omitting or contradicting this required achievement: Substitutes n=12 into 250(1.004^n-1) / 0.004. Stating a result without establishing: Obtains A_(12)approx$3,066.89.

Part c (4 marks)

Use the 4 numbered checkpoints below, awarding one mark only for each independently demonstrated achievement.

Mark-by-mark checkpoints:

  1. States the assumption of a constant 0.4% monthly rate.
  2. States that every $250 deposit is made on time with no fees or withdrawals.
  3. Links suitability to the one-year horizon rather than indefinite extrapolation.
  4. Concludes that the estimate is suitable only if the stated conditions are realistic.

Do not credit by itself: Omitting or contradicting this required achievement: States the assumption of a constant 0.4% monthly rate. Stating a result without establishing: Concludes that the estimate is suitable only if the stated conditions are realistic.

Diagnostic Checklist

TopicQuestionsMarksMarks LostAction
Section B alignment | Module 2 Topic 1: Statistical analysis | Bivariate numerical data | Criterion 5 | interpret a least-squares model and its limitations - part a Q1(a) 3 ___ Q1(a): reproduce these exact achievements without reading the solution: Identifies the slope as the predicted change in score for one additional training session; then Uses the numerical slope 2.3 with score-point units; then States that each additional session is associated with a predicted increase of 2.3 score points.
Section B alignment | Module 2 Topic 1: Statistical analysis | Bivariate numerical data | Criterion 5 | interpret a least-squares model and its limitations - part b Q1(b) 3 ___ Q1(b): reproduce these exact achievements without reading the solution: Substitutes x=6 into 18.4+2.3x; then Obtains the fitted score 32.2; then Calculates observed minus fitted, 31-32.2=-1.2 points.
Section B alignment | Module 2 Topic 1: Statistical analysis | Bivariate numerical data | Criterion 5 | interpret a least-squares model and its limitations - part c Q1(c) 4 ___ Q1(c): reproduce these exact achievements without reading the solution: Notes that 30 sessions is outside the observed range 2 to 10; then Classifies use at x=30 as substantial extrapolation; then States the assumption that the fitted linear pattern continues beyond the observed range; then Concludes that the prediction is unsupported or unreliable despite being calculable.
Section B alignment | Module 2 Topic 1: Statistical analysis - bivariate categorical data | Bivariate categorical data | Criterion 5 | use row percentages to assess association - part a Q2(a) 3 ___ Q2(a): reproduce these exact achievements without reading the solution: Adds the urban counts 42+18; then Adds the rural counts 12+28; then States the row totals as urban 60 and rural 40.
Section B alignment | Module 2 Topic 1: Statistical analysis - bivariate categorical data | Bivariate categorical data | Criterion 5 | use row percentages to assess association - part b Q2(b) 3 ___ Q2(b): reproduce these exact achievements without reading the solution: Calculates the urban conditional percentage 42 / 60 × 100; then Calculates the rural conditional percentage 12 / 40 × 100; then States urban 70% and rural 30%.
Section B alignment | Module 2 Topic 1: Statistical analysis - bivariate categorical data | Bivariate categorical data | Criterion 5 | use row percentages to assess association - part c Q2(c) 4 ___ Q2(c): reproduce these exact achievements without reading the solution: Compares the conditional bus-use percentages using the 40-percentage-point difference; then Concludes that location and travel mode appear associated; then Recognises that the survey is observational rather than a controlled experiment; then Explains that confounding or other group differences prevent a causal conclusion.
Section C alignment | Module 2 Topic 2: Growth and decay in sequences | Growth and decay in sequences | Criterion 6 | construct and evaluate a geometric depreciation model - part a Q3(a) 3 ___ Q3(a): reproduce these exact achievements without reading the solution: Identifies the initial value V_0=28000; then Converts 16% depreciation to the multiplier 1-0.16=0.84; then Writes V_n=28000(0.84)^n.
Section C alignment | Module 2 Topic 2: Growth and decay in sequences | Growth and decay in sequences | Criterion 6 | construct and evaluate a geometric depreciation model - part b Q3(b) 3 ___ Q3(b): reproduce these exact achievements without reading the solution: Substitutes n=3 into 28000(0.84)^n; then Evaluates the unrounded value 16595.712; then Rounds to $16,596.
Section C alignment | Module 2 Topic 2: Growth and decay in sequences | Growth and decay in sequences | Criterion 6 | construct and evaluate a geometric depreciation model - part c Q3(c) 4 ___ Q3(c): reproduce these exact achievements without reading the solution: Calculates the three-year loss 28000-16595.712=11404.288; then Expresses the loss as 11404.288 / 28000 × 100approx40.73%; then Identifies the constant annual percentage-depreciation assumption; then Limits the model's reasonableness to a period in which the 16% rate remains plausible.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Loans and reducing balances | Criterion 7 | interpret a loan recurrence - part a Q4(a) 3 ___ Q4(a): reproduce these exact achievements without reading the solution: Interprets 1.004 as a monthly growth factor; then Converts the growth factor to 0.4% monthly interest; then Interprets the subtraction as an $850 monthly repayment after interest.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Loans and reducing balances | Criterion 7 | interpret a loan recurrence - part b Q4(b) 3 ___ Q4(b): reproduce these exact achievements without reading the solution: Calculates first-month interest 0.004(24000)=96; then Uses the operation order 24000+96-850; then Obtains B_1=$23,246.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Loans and reducing balances | Criterion 7 | interpret a loan recurrence - part c Q4(c) 4 ___ Q4(c): reproduce these exact achievements without reading the solution: Substitutes B_1=23246 into the recurrence; then Calculates 1.004(23246)=23338.984; then Obtains B_2=$22,488.98 to the nearest cent; then Explains that the last repayment is reduced when the interest-added balance is below $850.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Non-right-angled triangles | Criterion 8 | apply the cosine rule and included-angle area formula - part a Q5(a) 3 ___ Q5(a): reproduce these exact achievements without reading the solution: Selects the cosine rule because two sides and the included angle are known; then Substitutes BC²=7.2²+9.5²-2(7.2)(9.5)cos58^circ; then Obtains BCapprox8.34 km to two decimal places.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Non-right-angled triangles | Criterion 8 | apply the cosine rule and included-angle area formula - part b Q5(b) 3 ___ Q5(b): reproduce these exact achievements without reading the solution: Selects the cosine rule using AB=7.2, AC=9.5 and the consequential value BCapprox8.34248; then Substitutes cos B=(7.2²+8.34248²-9.5²) / (2(7.2)(8.34248)); then Obtains angle ABCapprox74.953^circapprox75^circ.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Non-right-angled triangles | Criterion 8 | apply the cosine rule and included-angle area formula - part c Q5(c) 4 ___ Q5(c): reproduce these exact achievements without reading the solution: Selects A=frac12absin C using sides 7.2, 9.5 and included angle 58^circ; then Substitutes frac12(7.2)(9.5)sin58^circ; then Evaluates the area as approximately 29.0; then States the area unit km^2.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Earth geometry | Criterion 8 | calculate distance along a parallel - part a Q6(a) 3 ___ Q6(a): reproduce these exact achievements without reading the solution: Calculates the longitude difference 173^circ-147^circ=26^circ; then Converts using 26π / 180=13π / 90; then States the decimal angle 0.4538 radians.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Earth geometry | Criterion 8 | calculate distance along a parallel - part b Q6(b) 3 ___ Q6(b): reproduce these exact achievements without reading the solution: Uses the parallel-radius relationship r=Rcos(latitude); then Substitutes r=6371cos42^circ; then Obtains rapprox4735 km.
Section E Trigonometry alignment | Module 3 Topic 2a: Trigonometry and Earth geometry | Earth geometry | Criterion 8 | calculate distance along a parallel - part c Q6(c) 4 ___ Q6(c): reproduce these exact achievements without reading the solution: Selects the arc-length model s=rtheta; then Substitutes the small-circle radius and radian angle 6371cos42^circ(13π / 90); then Evaluates the distance as approximately 2149; then States the distance in kilometres along the parallel.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Shortest paths | Criterion 8 | apply Dijkstra-style shortest-path reasoning - part a Q7(a) 3 ___ Q7(a): reproduce these exact achievements without reading the solution: Assigns B the direct tentative label 7; then Assigns C the direct tentative label 4; then States the two labels as B=7 and C=4 after considering A.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Shortest paths | Criterion 8 | apply Dijkstra-style shortest-path reasoning - part b Q7(b) 3 ___ Q7(b): reproduce these exact achievements without reading the solution: Settles C at distance 4 and updates B to 4+2=6; then Compares the routes to D, obtaining 6+5=11 via B and 4+8=12 directly via C; then States the shortest A-to-D route A-C-B-D with distance 11 km.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Shortest paths | Criterion 8 | apply Dijkstra-style shortest-path reasoning - part c Q7(c) 4 ___ Q7(c): reproduce these exact achievements without reading the solution: Updates E from C to label 4+7=11; then Shows that the competing route through D is at least 11+3=14; then States route A-C-E with distance 11 km; then Justifies finality because no unsettled label below 11 can improve E.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Project networks | Criterion 8 | complete forward/backward passes and identify critical activities - part a Q8(a) 3 ___ Q8(a): reproduce these exact achievements without reading the solution: Uses A's finish at 3 to give C: ES=3, EF=7; then Uses B's finish at 8 and C's finish at 7 to set E: ES=max(8,7)=8; then States E: EF=8+3=11.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Project networks | Criterion 8 | complete forward/backward passes and identify critical activities - part b Q8(b) 3 ___ Q8(b): reproduce these exact achievements without reading the solution: Finds D: ES=8, EF=10 from its two predecessors; then Sets F: ES=max(10,11)=11; then States the minimum project duration EF_F=15 days.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Project networks | Criterion 8 | complete forward/backward passes and identify critical activities - part c Q8(c) 4 ___ Q8(c): reproduce these exact achievements without reading the solution: Completes the backward pass to identify zero float for A, B, E and F; then Identifies A-B-E-F as the critical path; then States that B has 0 days float; then Checks that C and D each have 1 day float, consistent with the critical-path result.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Graphs and adjacency matrices | Criterion 8 | interpret adjacency and Euler trails - part a Q9(a) 3 ___ Q9(a): reproduce these exact achievements without reading the solution: Uses vertex order A, B, C, D with zeros on the diagonal; then Places symmetric ones for edges AB,AC,BC,CD and zeros for absent edges; then Writes the complete matrix begin(bmatrix)0&1&1&01&0&1&01&1&0&10&0&1&0end(bmatrix).
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Graphs and adjacency matrices | Criterion 8 | interpret adjacency and Euler trails - part b Q9(b) 3 ___ Q9(b): reproduce these exact achievements without reading the solution: Counts degrees A=2 and B=2; then Counts degrees C=3 and D=1; then States the complete ordered degree list 2,2,3,1.
Section E Networks alignment | Module 3 Topic 2b: Graphs, networks and decision mathematics | Graphs and adjacency matrices | Criterion 8 | interpret adjacency and Euler trails - part c Q9(c) 4 ___ Q9(c): reproduce these exact achievements without reading the solution: Identifies C and D as the only odd-degree vertices; then Concludes that an Euler trail exists but an Euler circuit does not; then Gives a trail such as C-A-B-C-D that uses every edge exactly once; then Uses C and D as the two endpoints.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Annuities | Criterion 7 | model and evaluate regular investments - part a Q10(a) 3 ___ Q10(a): reproduce these exact achievements without reading the solution: Uses A_0=0 to calculate A_1=1.004(0)+250=250; then Applies interest before the second deposit; then Calculates A_2=1.004(250)+250=$501.00.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Annuities | Criterion 7 | model and evaluate regular investments - part b Q10(b) 3 ___ Q10(b): reproduce these exact achievements without reading the solution: Substitutes n=12 into 250(1.004^n-1) / 0.004; then Evaluates the geometric accumulation factor without premature rounding; then Obtains A_(12)approx$3,066.89.
Section D alignment | Module 3 Topic 1: Investment, loans and annuities | Annuities | Criterion 7 | model and evaluate regular investments - part c Q10(c) 4 ___ Q10(c): reproduce these exact achievements without reading the solution: States the assumption of a constant 0.4% monthly rate; then States that every $250 deposit is made on time with no fees or withdrawals; then Links suitability to the one-year horizon rather than indefinite extrapolation; then Concludes that the estimate is suitable only if the stated conditions are realistic.

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