Algebra

When two ages add to a total and one person is a fixed number of years older, let the younger age be x — the elder is x + 3 and their sum gives you one quick equation.

Sum-and-difference age items appear in the very-easy band of Selective MR — solid introduction to one-variable algebra before timeline age puzzles.

The examiner checks whether you can define two unknowns with one letter using a stated age gap, then use the combined total to solve.

Two siblings’ ages sum to a given number this year, with a fixed year gap between them. You find each current age.

9 × □ = 108 → □ = 12. Then △ + 12 = 36 → △ = 24. Two steps, each a single equation.

Two-step symbol substitution questions are among the most common easy questions in Selective MR. The first equation gives you one symbol’s value; you plug it into the second equation to find the other.

The examiner checks whether students (1) solve the first equation correctly by division, (2) substitute the result into the second equation, and (3) subtract rather than confusing the two symbols.

Two equations use two different symbols. The first equation can be solved directly; its result is substituted into the second to find the unknown symbol.

Division with remainder means: total = (number of groups) × (group size) + remainder. So 117 = 14 × 8 + 5.

Translating a division-with-remainder word problem into a number sentence appears in the very-easy band of Selective MR. The main confusion is between which number is the quotient and which is the remainder.

The examiner tests whether students understand the structure of division with remainder and can write it as a multiplication-plus-remainder equation rather than a subtraction or an addition-first expression.

A sharing scenario with a leftover is described in words. Students must identify the correct algebraic expression from five options that all use the same four numbers.

Monique spent 4/5 of her money and had $80 left. What you SEE ($80) is the part that WASN'T spent — that's 1/5 of the original. So $80 = 1/5, which means the whole = $80 × 5. Working backwards from a fraction that remains is the key move.

Fraction inverse problems ('find the whole given a remaining fraction') appear regularly in NSW Selective MR. Students who jump to $80 × 4/5 or $80 + 4/5 make the classic trap error. The correct approach: identify the remaining fraction (1 − 4/5 = 1/5), then divide $80 by that fraction.

The examiner tests whether students can identify that 'spent 4/5' means '1/5 is left', set up $80 = (1/5) × original, and solve by multiplying $80 × 5 = $400. The distractor $320 (= 80 × 4) comes from multiplying by the spent fraction instead of the remaining fraction.

A person starts with an unknown amount and spends a stated fraction of it. The remaining amount is given. Students must find the original total by recognising the remaining amount equals the leftover fraction of the original.

Outfits = T-shirts × Skirts. Nidhi’s total tells you the skirt count (24 ÷ 3 = 8). Chris has the same skirts but fewer T-shirts (2 instead of 3), so her total is 2 × 8.

Counting principle (multiply choices) questions with a reverse-engineer step appear in the easy-to-medium band of Selective MR. The key algebraic move is dividing the known total by one factor to find the hidden factor.

The examiner tests whether students (1) recognise that outfits = T-shirts × skirts, (2) use Nidhi’s data to find the shared skirt count (24 ÷ 3 = 8), and (3) apply that skirt count to Chris’s different T-shirt number (2 × 8 = 16). Option D (23) traps students who subtract 1 from Nidhi’s total instead of recalculating.

Two people share one quantity (e.g. skirts) but differ in another (e.g. T-shirts). One person’s outfit count is given. Students find the shared quantity and use it to compute the other person’s outfit count.

Equal numbers of $1 and $2 coins totalling $54 — form one equation with one unknown, solve for n, then calculate weight separately for each coin type.

Two-variable word problems where both quantities are equal (so really one variable) appear in the easy-to-medium band of Selective MR. Students who set up the equation correctly find this a quick calculation question.

The examiner tests whether students (1) correctly form the equation n + 2n = 54, (2) solve to get n = 18, (3) apply two different weights for each coin type, and (4) add both weights — not confuse weight with value.

Two items are held in equal quantities. Their combined value is given. Each item has a separate weight. Students find the total weight.

Large = 36. Two large = three medium, so medium = 24. Two medium = three small, so small = 16.

Chain-of-equivalences questions (packet A = packet B = packet C) appear regularly in Selective MR and require careful step-by-step scaling rather than guessing.

The examiner checks whether students work through the chain correctly: Large → Medium → Small, without skipping a step or mixing up which packet is larger.

A starting quantity is given for the largest unit, and two equivalence rules connect large–medium and medium–small. Students calculate the smallest unit.

Reading-plan questions are hidden equations: two different daily rates must cover the same book — write pages = rate × days for each plan and set them equal.

Two-scenario equation items appear in the medium band of Selective MR — the algebra is short once both sides are written as expressions in the unknown page count.

The examiner tests whether you can translate two verbal schedules into algebra and solve for the total (or daily rate) without guessing from options.

A student reads a fixed number of pages per day under one plan and leaves pages unread; under a faster daily rate they finish with a smaller remainder on the last day. Find the book length.

Three friends share marbles. One gives some to another. You know totals and a final equal-share condition. Work backwards: use the equal condition to find one person's yesterday count, then subtract to find the rest.

Multi-person redistribution puzzles appear in the medium band of Selective MR. The trick is recognising that giving marbles away doesn't change the total — the group total stays the same. Use the 'equal after giving' condition to pin down the giver's original count, then chain through.

The examiner checks whether students can: (1) isolate one person's count using a group-minus-subgroup trick, (2) use the 'after transfer they are equal' condition to find the giver's original count, (3) find Lan using a two-person total, and (4) correctly answer about Lan + Niki NOW (after the transfer), not yesterday.

Three people start with a known combined total. A sub-group total is also given. One person transfers a fixed amount to another, making them equal. Students must find a combined total for two people after the transfer.

Kira earns 8 points for the first coin and 3 points per coin after that. She needs at least 120. Write the total as an expression in n (number of coins), set it ≥ 120, and solve — rounding UP since you can't collect a fraction of a coin.

'At least' inequality questions appear regularly in Selective MR. The trap is forgetting to round UP to the next whole number when the inequality gives a decimal — the question asks for the SMALLEST number of coins, so you must always round the ceiling.

The examiner tests whether students can (1) model a two-rate scoring system as 8 + 3(n−1), (2) set up and solve an inequality for n, (3) correctly round up a non-integer result (37.33… → n−1 = 38, n = 39), and (4) verify by checking n=39 gives ≥ 120 and n=38 gives < 120.

A player earns a fixed bonus for the first item and a lower fixed amount for each subsequent item. A minimum total is required. Students find the smallest integer number of items needed to reach or exceed that total.

Hal takes a fraction from some packets, leaving a remainder. Irene and Khaled split that remainder with Khaled getting double. Two-step: find what's left, then use a ratio equation to split it.

Multi-person sharing questions with a ratio condition appear in the medium band of Selective MR. The key trap is computing Hal’s share incorrectly (confusing which packets he took from, or computing the fraction per packet vs total).

The examiner tests whether students (1) compute total stickers correctly, (2) apply 4/6 to 3 packets only (not all 4), (3) subtract to find what Irene + Khaled share, and (4) set up x + 2x = remainder to find Irene's share.

Several packets of items are shared. One person takes a fractional amount from a subset of packets. The rest is divided between two people in a given ratio. Students find one person’s final share.

Alex → spends some → gives half to Charlie → Charlie spends 1/4 of that on $9 lunch. Work backwards from the $9: un-quarter it, un-halve it, then subtract from $76.

Reverse-chain word problems (undo fractions in reverse order) appear in the medium band of Selective MR. Students who work forwards and try all options waste time. The reliable method is always to start from the known end result and undo each step.

The examiner tests whether students (1) recognise the problem requires working backwards, (2) correctly un-quarter $9 to get $36, (3) correctly un-halve $36 to get $72, and (4) subtract from $76 to get $4. The large distractors ($36, $40) trap students who stop partway through.

Money passes through two or more people via fractions. A final amount is known. Students work backwards through the chain to find an earlier amount.

She plants half, then three-quarters of what’s left, and 6 remain. Work backwards: 6 is one-quarter of the second section’s starting amount — multiply up twice to reach the original total.

Sequential fraction-of-remaining problems appear in the medium-to-difficult band of Selective MR. Students who try forward guessing waste time; working backwards is always faster when the final amount is given.

The examiner tests whether students (1) correctly interpret ‘three-quarters of the remaining’ (not three-quarters of the original), (2) work backwards using inverse operations, and (3) avoid the trap of stopping at 24 (what was left after section 1) rather than doubling again to reach 48.

A quantity is reduced by a fraction, then a further fraction of the remainder is removed. A final leftover amount is given. Students work backwards to find the original total.

Orange + lemon = 33. Plug each statement in as an equation — if the result isn’t a whole number, that statement is impossible.

Multi-statement ‘cannot be correct’ problems appear in the medium band of Selective MR. The key skill is setting up a simple equation for each statement and checking whether it gives a whole-number solution.

The examiner tests whether students (1) set up orange + lemon = 33 correctly, (2) substitute each statement as a second equation, (3) solve and check for a whole-number result, and (4) identify that Statement 1 gives a non-integer answer (11.5) making it impossible.

A total of two groups is given. Three statements describe possible relationships between the groups. Students find which statement(s) cannot be true.

Greatest difference = (top price as large as possible) − (bottom price as small as possible). Set □=9 and △=0.

Missing-digit maximisation/minimisation questions appear in the medium band of Selective MR. Students who set both unknowns to the same extreme (e.g. both to 9) miss the key insight: you need to push the two prices in opposite directions.

The examiner tests whether students (1) recognise they must maximise the first price and minimise the second, (2) substitute □=9 and △=0, and (3) subtract correctly with decimal alignment.

Two prices each have one missing digit. Students find the greatest (or least) possible difference between them.

Let Gabriella = 1. Then Finn = 8 and Hassan = 1/2. Finn ÷ Hassan = 8 ÷ (1/2) = 8 × 2 = 16.

Three-person ratio chain questions appear in the medium band of Selective MR. The key skill is choosing a common reference (Gabriella) and expressing all three quantities in terms of it, then dividing.

The examiner tests whether students (1) correctly set up Finn = 8G and Hassan = G/2, (2) divide Finn by Hassan (not multiply), and (3) simplify 8G ÷ (G/2) = 16 rather than guessing 8 − (1/2) = 7.5 or 8 + (1/2) = 8.5.

Three people’s quantities are linked through a common middle person. Students find how many times the largest is compared to the smallest.

Let Monday = M. Tuesday = M/2, Wednesday = M/4. Then M + M/2 + M/4 + 80 = 500, so 7M/4 = 420, giving M = 240.

Halving-sequence total questions appear regularly in Selective MR. They look wordy but reduce to a single equation once you express every day in terms of Monday.

The examiner checks whether students (1) correctly express Tuesday and Wednesday as fractions of Monday, (2) set up the total equation, and (3) combine the fractions before solving — rather than guessing or trying each option.

A total is given across several days/items. Each quantity is half the previous one. Students must find the first (largest) quantity.

3n < 30 means n ≤ 9. 6n > 40 means n ≥ 7. The whole numbers in both ranges are 7, 8, and 9. Their sum is 24.

Double-inequality ‘find all valid numbers and sum them’ questions appear regularly in Selective MR. Students who find only one boundary (instead of both) typically get a partial answer that matches a wrong option.

The examiner checks whether students (1) correctly set up and solve both inequalities, (2) identify ALL whole numbers in the intersection range (not just the boundary values), and (3) sum them rather than listing or counting them.

A whole number must satisfy two multiplication inequalities simultaneously. Students find the range of valid numbers and compute their sum.

Lenny got $1.60. Double it to get Jason’s amount ($3.20). Double again to get what Alan had left ($6.40). Subtract from $10 to get Kiana’s amount: $10 − $6.40 = $3.60.

Work-backwards chain questions (halving forward = doubling backward) appear regularly in the medium band of Selective MR. Students who try to work forwards by guessing Kiana’s amount typically run out of time testing each option.

The examiner checks whether students (1) recognise that the chain must be undone in reverse order, (2) correctly double at each backward step, and (3) subtract the remaining amount from the starting total to find the first gift.

A sequence of halvings connects four people. The last amount is given. Students must work backwards to find the first transfer.

Row 1 gives magic total = 51. The anti-diagonal (top-right to bottom-left: 16, middle-centre, 18) gives middle-centre = 17. The main diagonal (20, 17, ♦) gives ♦ = 14.

Magic square questions appear in the medium band of Selective MR. Students who try to find every missing cell waste time; the fastest path is always through the diagonal that shares the most known values.

The examiner checks whether students (1) calculate the magic total from the complete row, (2) use the anti-diagonal to pin down the centre cell, and (3) use the main diagonal to reach the target cell — without needing to fill in every blank.

A 3×3 magic square has some cells filled. Students must find the value of one specific missing cell using the equal-sums property of rows, columns, and diagonals.

Set up two equations: 3s + 5h = 27 and 5s + 7h = 41. Eliminate s by scaling both equations, subtract to find h = 3, substitute back to find s = 4. Then s + h = 7.

Two-variable simultaneous equations disguised as game-scoring or fruit-price problems appear in the medium band of Selective MR. Students who try trial-and-error with the first equation alone miss the second constraint and pick wrong values.

The examiner checks whether students can translate two word-problem conditions into simultaneous equations and solve them systematically — not just guess values that satisfy one equation.

Two scoring or pricing conditions are given (each involving two unknowns). Students must find the value of a specific combination of the two unknowns.

Each row in the table is a hidden equation: water level = starting water + volume of objects submerged. Subtract pairs of rows to eliminate unknowns one at a time until you find each object's volume.

Displacement table simultaneous-equations questions appear in the difficult band of Selective MR. The algebra looks hard but there are only three unknowns (W, b, s), and clever subtraction of rows eliminates variables quickly.

The examiner tests whether students can (1) translate each table row into W + nb + ms = level, (2) subtract rows to find that b = 3s (from rows 1−2), (3) subtract rows to get 2b = 12, b = 6, and (4) find W = 42, then correctly compute W + b + s = 50.

A measuring jug contains water. Objects (blocks, spheres) are submerged and the water level recorded. A table of three known (objects, level) pairs and one unknown pair is given. Students find the unknown water level using simultaneous equations.

Let the tens digit = A and units digit = B. Reversing gives 10B + A. The difference is 9(B − A). Set that equal to 72, solve for B − A = 8, then find the only valid digit pair.

Digit-reversal algebra problems appear in the difficult band of Selective MR. Timothy’s example in the stem (24 → 42, difference 18 = 9×2) gives the key pattern: the difference between a number and its reverse is always 9 × (difference of digits).

The examiner tests whether students (1) write the two-digit number algebraically as 10A + B, (2) set up the equation 9(B−A) = 72, (3) solve B−A = 8, and (4) find the only valid digit pair (A=1, B=9) since both must be single digits and A ≥ 1.

Two-digit number, its reverse, and the difference between them are described. Students find a specific digit of the original number.

Two scenarios give you two equations: C + S/2 = 120 and C + S/4 = 65. Subtract to get S/4 = 55, so full sweetcorn = 220 g and the can = 10 g. Three-quarters full = 10 + 165 = 175 g.

Two-equation ‘find the can and contents separately’ questions appear in the difficult band of Selective MR. Students who try to guess or use only one equation always end up choosing a wrong option.

The examiner checks whether students (1) set up two equations with two unknowns, (2) eliminate the can mass by subtracting, (3) scale the sweetcorn correctly for three-quarters, and (4) add the can mass back at the end.

A container’s total mass is given at two fill levels. Students find the empty-can mass and the full-contents mass, then calculate the total at a third fill level.