Unit 11
Algebraic / Equation Reasoning
About this unit
Set up and solve real-world algebra problems without formal algebraic notation. You will work with simultaneous equations, age relationships, work rates, maximisation problems, and counting arrangements — all presented as practical word problems that require systematic equation thinking.
What types of questions will you face?
- 1Simultaneous equations from word problems (e.g. pens + notebooks at two prices)
- 2Maximisation: given limited quantities of multiple resources, find how many complete items can be made
- 3Work rate problems: if one worker takes X hours, another takes Y hours, how long do they take together?
- 4Age problems: relationships between ages now and in the past/future
- 5Counting arrangements: how many ways can items be arranged given restrictions?
- 6Shape equation puzzles: shapes represent numbers — find the value of each shape to make all equations true
Skills you will build
- Translating a word problem into one or two equations
- Solving simultaneous equations using substitution or elimination
- Using the "limiting resource" principle for maximisation problems
- Applying the combined work rate formula (1/A + 1/B = 1/total)
- Setting up and solving age equations with a single unknown
- Calculating permutations and combinations with restrictions
By the end of this unit, you will be able to
- Solve any two-variable simultaneous equation from a word problem
- Calculate how many complete items can be made from given resource quantities
- Solve work rate and shared task problems efficiently
- Answer age and relationship problems using clear algebraic thinking
Difficulty profile
Medium difficulty (avg 3.07). Shape equation puzzles are Easy; work rate and age problems are Medium; complex maximisation and counting problems with multiple restrictions are Difficult.
Exam tip: Algebraic / Equation Reasoning
For simultaneous equations: label your unknowns clearly (pen = p, notebook = n), write both equations, then subtract or substitute to eliminate one variable. For maximisation: divide your supply of each resource by what one item needs — the smallest result is your answer.
Sample Questions
A farmer, some goats, two possible outputs — this is a classic two-equation word problem in disguise. Name your unknowns, write the equations, subtract.
Two-variable word problems from real-life contexts (animals, products, items) appear regularly in the easy-to-medium band of official NSW papers — instantly recognisable once you spot the two unknowns.
The examiner checks whether you can translate "total items" and "total output" into two simultaneous equations, then solve by elimination rather than trial-and-error on the answer choices.
A total count of objects is split into two groups with different outputs (e.g. 1 kid vs 2 kids). You are given the total objects and the total output, and asked to find the size of one group.
Best approach: Step 1: find the active group size (subtract the "inactive" ones). Step 2: write x + y = total active and x + 2y = total output. Step 3: subtract equation 1 from equation 2 to get y instantly. Step 4: find x = total active − y. Step 5: verify by checking total output.
Question
A farmer has 245 female goats.
This year all but 9 of them have had kids.
None of them have had more than two kids.
357 kids have been born in total.
How many of the goats have had one kid only?
- A115
- B121
- C124
- D133
Decided on your answer? Check how you went below.
Algebraic reasoning on Selective often means turning words into a short chain of numbers — no formal algebra symbols required.
Combined-work-rate questions appear in the harder band of Selective TS but follow a fixed recipe: convert times to “fraction of job per hour,” add, then invert.
The examiner checks whether you can combine individual rates correctly (1/time each) rather than averaging the two times or adding hours naively.
Two people (or machines) complete the same job alone in different times. You find how long they take working together at constant rate.
Best approach: Rates: Zara = 1/4 job per hour, Ethan = 1/6. Together = 1/4 + 1/6. Time = 1 ÷ combined rate. Keep fractions tidy with a common denominator.
Question
Zara can paint a fence in 4 hours. Her brother Ethan can paint the same fence in 6 hours. How long will it take them to paint the fence together?
- A3 hours 20 minutes
- B3 hours
- C2 hours 30 minutes
- D2 hours 24 minutes
Decided on your answer? Check how you went below.
Letter-score puzzles are equations in disguise: each word is a sum, and repeated letters force unique whole-number solutions.
Hidden letter-value chains show up on Selective mocks — reliable marks when you start from the word with the fewest different letters.
You must solve for each letter’s value step by step (whole numbers ≥ 1), then sum the target word — without guessing random values.
Several words and total scores are given. A new word uses only letters you can crack from the chain (e.g. CAT, HAT, CHAT, BAT → BATCH).
Best approach: Pick the most constrained word first (repeated letters help). Substitute known values into the next equation. Build a small table before scoring the final word.
Question
In a word game, each letter has a whole-number score of at least 1. A word's score is the sum of its letters' scores.
- CAT scores 6.
- HAT scores 7.
- CHAT scores 9.
- BAT scores 5.
What is the score for BATCH?
- A8
- B9
- C10
- D11
Decided on your answer? Check how you went below.
15 beanies, 8 scarves, 12 gloves — 35 items, but far fewer than 35 children. All scarves also have gloves. Half of scarves also have beanies. You need to count each child exactly once by placing them into non-overlapping groups and adding up.
Three-set Venn diagram counting problems are a top-difficulty NSW Selective TS type. Students who naively add 15+8+12=35 get it badly wrong. The key is to use the subset/intersection clues to build non-overlapping groups, then sum those groups (plus the 'none' group).
The examiner tests whether students can (a) use 'all scarves also have gloves' to place the scarf group entirely inside gloves; (b) use 'half of scarves have beanies' to split scarves into 4 with beanie and 4 without; (c) use the direct clue '4 wear beanie+gloves but no scarf'; (d) back-calculate beanie-only as 15-8=7; and (e) add 6 non-overlapping groups including the 'no accessories' group of 2.
Total counts for three overlapping attributes are given, plus several intersection facts. One attribute is a subset of another. The question asks for the total number of people (or items), which requires summing non-overlapping regions.
Best approach: Step 1: Note which group is a subset of another (scarves ⊆ gloves). Step 2: Fill in the triple overlap and double overlaps using the given numbers. Step 3: Back-calculate 'only one attribute' groups by subtracting placed members from the totals. Step 4: Add all non-overlapping groups plus the 'none' group.
Question
When the children in Freezetown arrived for class yesterday, the teacher saw that 15 were wearing beanies, 8 were wearing scarves and 12 were wearing gloves.
On closer inspection, he could see that all of the children wearing scarves were also wearing gloves, and that half of the children with a scarf also had a beanie. There were 4 children wearing a beanie who also had gloves but no scarf. There were 2 children with no beanie, no scarf and no gloves.
How many children arrived for class yesterday?
- A18
- B19
- C20
- D21
Decided on your answer? Check how you went below.
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