Algebra

A triangle symbol stands for a mystery number. The equation 67 − ▲ = 32 + 17 looks tricky because of the symbol, but once you see it as a plain subtraction equation the answer comes in two steps: add the right side first, then find what was subtracted from 67.

Symbol-for-number equations (▲, ★, □) appear regularly in the very easy to easy band of OC MR. They are designed to check whether students can treat an unfamiliar symbol exactly like a variable. The most common mistake is trying to guess from the answer options instead of solving the right side first.

The examiner is checking whether students can simplify the right-hand side before solving, rather than trying all five options. Distractor D (49) is the value of the right side — students who forget to subtract it from 67 pick D instead of B.

A symbol (triangle, star, square) represents an unknown number. A simple one-step equation is given with arithmetic on one or both sides. Students must evaluate both sides and isolate the symbol. The distractors include the right-side total, one of the numbers in the equation, and values that result from common arithmetic errors.

Jack and Sarah share 12 strawberries; Sarah eats twice as many as Jack. How many did Sarah eat? This is the simplest form of ratio/algebra question: one variable, one multiplier, one total. The classic trap is option C (6) — just halving 12 — which ignores the 2:1 ratio entirely.

Sharing-in-a-ratio questions are among the most frequent easy algebra items in OC MR. They appear in the form 'X has twice / three times as many as Y; together they have Z — find one share.' Students who write a single equation (1 part + 2 parts = total) solve these in seconds.

The examiner is testing whether students can translate a ratio relationship into an equation and identify whose share is being asked. Option B (4) traps students who find Jack's share and stop. Option C (6) traps students who simply halve the total and ignore the ratio. Option E (24) traps students who double the total.

Two people (or groups) share a total. One has a multiple of the other. Students must set up parts (e.g. 1 part + 2 parts = 3 parts total), find the value of one part, then multiply to find the required share.

Age problems are one of the most satisfying question types in OC MR algebra: a single variable for the current age unlocks every time-shifted condition in the question. Students who try to guess using the answer options lose time; students who introduce one letter and write the equation solve in under 90 seconds.

Age problems appear in virtually every OC MR paper, spread across easy and medium difficulty. They are a reliable scoring opportunity — the setup is always the same, the algebra is short, and a quick check at the end confirms the answer.

The examiner is checking whether you can anchor on the present age as your variable, translate "in N years" and "N years ago" into expressions involving that variable, and write the multiplier or difference relationship as a single equation before solving.

Two time points are described — usually a future age and a past age — linked by a ratio (e.g. "in 12 years will be 3 times the age 2 years ago"). You must find the current age. The numbers are chosen so that trial-and-error with the answer options is slower than one-variable algebra.

A player starts with 1000 game points, buys 5 buildings at 106 each and 1 farm, and has 140 left. What did the farm cost? This is a classic working-backwards equation: start − buildings − farm = leftover. The trap is option C (470) — students who calculate 1000 − 530 = 470 correctly but forget to subtract the 140 points remaining.

Multi-step equation-solving word problems (start with a total, subtract known amounts, solve for the unknown) appear frequently at the easy-to-medium tier of OC MR. They reward students who set up the full equation before calculating rather than working forwards step by step.

The examiner tests whether students can identify all the known quantities (buildings cost, leftover) and use them to isolate the unknown (farm cost). Option C (470) is a deliberate trap for students who stop after computing 1000 − 530 without subtracting the leftover. Option E (754) traps students who multiply only 1 building.

A total is given. Several items at a known per-unit price are purchased, plus one item at an unknown price. A remainder is stated. Students must: (1) calculate the known total spend, (2) subtract it and the remainder from the starting total to find the unknown price.

Brian and Yarram together have 250 cards. Yarram has 18 more than Brian. How many does Yarram have? This is the classic 'sum and difference' problem — one of the most frequently tested easy OC MR equation types. The fastest method: halve the total (125), then split the difference (±9).

Sum-and-difference problems (total given + one person has X more than the other → find each person's value) appear in almost every OC MR paper as an easy or very-easy question. They are reliably solved by either algebra (let B = x, then B+18+B = 250) or the halving-and-splitting shortcut.

The examiner is testing whether students can combine two constraints simultaneously. Option B (116) is the most common trap — students correctly find Brian's value but forget the question asks for Yarram's. Option D (143) traps students who add 18 to 125 instead of splitting the difference symmetrically.

Two people/quantities share a known total. One has a fixed amount more than the other. Students must find one (or both) of the quantities. Standard form: 'Together they have T. Person A has D more than Person B. How many does Person A have?'

Now for a classic 'think of a number' equation — where two completely different sequences of operations are applied to the same unknown number and produce identical results. The key is translating each verbal sequence into algebra before doing any arithmetic.

"Think of a number" and verbal equation-setting problems appear consistently in the medium difficulty band of OC MR. They look harder than they are: the algebra is never more than two steps once both sides are written down correctly.

The examiner is testing whether you can translate two separate verbal operation sequences into two algebraic expressions and recognise that setting them equal gives you the equation to solve — rather than trying to work backwards from the options.

An unknown number undergoes two different sequences of operations (e.g. "multiply by 3 then add 8" vs. "multiply by 5 then subtract 4"). Both sequences produce the same result. You must find the original number.

12 × 35 = 4 × ▲ × 7. What is ▲? Two elegant approaches: (1) compute 12 × 35 = 420, simplify RHS to 28▲, divide to get 15. (2) Factor — 12 = 4 × 3 and 35 = 5 × 7, so LHS = 4 × 3 × 5 × 7; cancel the 4 and 7 that appear on both sides, leaving ▲ = 3 × 5 = 15. Method 2 is the elegant OC-level approach.

Missing-factor equations where both sides are products appear regularly at the medium level of OC MR. Students who factor both sides and cancel matching factors can solve them mentally. Method 1 (multiply LHS, divide) is reliable but slower; Method 2 (factorisation and cancellation) is the examiner's intended elegant path.

The examiner is testing whether students can recognise that 4 and 7 appear on both sides of the equation, so the missing number must account for the remaining factors of 12 × 35. Students who reach for a calculator approach (12 × 35 ÷ 28) get the right answer, but factorisation builds deeper number sense.

An equation of the form A × B = C × ▲ × D is given, where A, B, C, D are given numbers and ▲ is unknown. Students find ▲ either by computing both sides or by factoring A and B and cancelling factors that appear on both sides.

Same whole number in every box. Two constraints: □+4 > 8 → n > 4; □+□ < 14 → 2n < 14 → n < 7. Intersection: 5 ≤ n ≤ 6 → exactly 2 whole numbers work. n=4 fails the first (4+4=8 is NOT bigger than 8); n=7 fails the second (7+7=14 is NOT less than 14).

Simultaneous inequality puzzles (find all whole numbers satisfying two constraints) appear regularly at the medium level of OC MR. Students must convert both sentences into inequalities, find the intersection range, and then count the whole numbers in that range.

The examiner tests whether students can solve two inequalities simultaneously. The boundary values (n=4 and n=7) are deliberate traps: n=4 gives 8+0 not strictly greater than 8, and n=7 gives 14 not strictly less than 14. Students who use ≥ instead of > will incorrectly include a boundary value.

A puzzle presents two 'fill the same number in each box' sentences, one using 'bigger than' (strict >) and one using 'less than' (strict <). Students translate each into an inequality, find the intersection, and count valid whole numbers.

Four people, three age clues. Express everyone in terms of Lisa: Sam = Lisa+5; Dave = Sam−7 = Lisa−2; Fred = Dave+13 = Lisa+11. Fred − Lisa = 11. The chain shortcut: net difference = +5 − 7 + 13 = +11. Option E (15 = 7+5+13) traps students who just add all numbers.

Age-chain problems with 4 people connected by 3 relationships appear regularly at the medium level of OC MR. The key is to express everyone in terms of one person (usually the reference asked about) and combine the differences algebraically.

The examiner tests whether students can chain three age-gap clues correctly, tracking the signs (+/-). Option E (15) traps those who simply add 7+5+13. Option C (10) catches those who subtract the Sam-Lisa gap instead of adding it.

Several people are related by 'X is N years older/younger than Y' statements. Students express all ages relative to one person and compute the requested gap, taking care with the direction (older = add, younger = subtract).

Six cards (8, ◆, 4, 5, ⬡, 2) sum to 25. Known sum = 19, so ◆ + ⬡ = 6. Both > 0. Possible pairs: (1,5), (2,4), (3,3), (4,2), (5,1). Max difference = 4. Option E (6) traps students who use (6,0) — forgetting both must be greater than zero.

Two-unknown sum with a maximisation constraint is a medium OC MR pattern. Students find the fixed sum of the two unknowns from the total, then enumerate pairs satisfying the constraint (both > 0), and identify which pair gives the largest/smallest difference.

The examiner tests whether students can find the pair (5,1) or (1,5) — and why (6,0) is ruled out by the 'greater than zero' constraint. Option E (6) is the classic trap for students who forget the lower bound.

A set of cards with known values and two unknown symbols is given. The total is stated. Both unknowns are constrained (e.g. whole numbers > 0). Students find the unknowns' sum, enumerate valid pairs, and maximise or minimise a quantity.

Cho's money box is empty on Monday morning. By Wednesday she has $30, by Friday she has $80. She saves the same amount on Tuesday, Wednesday and Friday. The twist: Thursday's deposit is an unusual large number ($41) — not one of the equal amounts. Students must first find the equal daily amount from the Wed total, then use the Fri total to isolate Thursday.

Two-phase equation problems (find a repeated value from one constraint, then use that value plus a new constraint to find an unknown) appear regularly in OC MR at medium difficulty. They require careful reading to identify which days share an equal amount and which day is different.

The examiner tests whether students can (1) correctly use the Wednesday total to find the equal amount ($9), and (2) correctly set up the Friday equation using that value to isolate Thursday's deposit. Option C ($32) traps students who compute 80 − 30 − 9 = 41 but subtract incorrectly, and Option D ($39) comes from forgetting to subtract the Friday deposit.

A running total is given at several checkpoints over a week. Some days have a shared equal amount; one day has a different (unknown) amount. Students use the checkpoint totals to first find the equal amount, then solve for the unknown day's deposit.

Each letter = a different whole number > 1. NAN = N²A = 36; NUN = N²U = 18. Try N=3: U=2, A=4. UNA = 2×3×4 = 24. The elegant insight: dividing the two equations gives A/U = 2, then substituting back fixes all values uniquely.

Letter-product code questions appear at the difficult level of OC MR. Students set up two product equations sharing a squared variable, then use trial-and-error on the shared base (N) constrained to whole numbers > 1. Only one value of N yields whole-number results for both U and A.

The examiner tests systematic trial for the shared unknown N, then back-substitution to find U and A. Option D (36) traps students who confuse NAN's value with UNA's. The constraint 'different whole numbers > 1' rules out N=1 and eliminates repeated-digit solutions.

A code where words = product of letter values. Two 3-letter words share two letters and form a palindrome (e.g. NAN, NUN), giving equations with N². Students solve for N by trial, find the other letters, then compute a rearranged word.