Junior algebra problems (ages 11 to 13)

Answer: D, 20

1. Mean × how many = total. The five numbers total 5 × 12 = 60.
2. The four that remain total 4 × 10 = 40.
3. The removed number is the difference: 60 − 40 = 20 (D).
4. Check: removing 20 lowers the mean because 20 is above the old mean of 12, as expected.

Why this works: Means are awkward to combine, but totals just add and subtract. Converting mean × count into a total is the standard move.

Where it leads: Removing a number changes the mean towards or away from it: means are sensitive to every value, unlike medians.

Answer: D, 50p

1. Let a pencil cost x pence and a pen y pence: 4x + 3y = 230 and 3x + 4y = 260.
2. Add the equations: 7x + 7y = 490, so x + y = 70.
3. Subtract the first from the second: −x + y = 30, so y = x + 30.
4. Substitute: x + (x + 30) = 70, so 2x = 40, x = 20 and y = 50.
5. Check: 4 × 20 + 3 × 50 = 230. ✓ A pen costs 50p (D).

Why this works: When two equations are symmetric, their sum and difference are much simpler than the originals. Look for that before reaching for substitution.

Where it leads: Adding and subtracting equations (elimination) generalises to solving any system of linear equations.

Answer: 1

1. Call the terms t₁, t₂, …. The rule is t_n+2 = t_n + t_n+1.
2. t₇ = t₅ + t₆, so t₆ = 53 − 20 = 33.
3. Run the rule backwards, t_n = t_n+2 − t_n+1: t₄ = t₆ − t₅ = 33 − 20 = 13.
4. t₃ = 20 − 13 = 7, t₂ = 13 − 7 = 6, t₁ = 7 − 6 = 1.
5. Check forwards: 1, 6, 7, 13, 20, 33, 53. ✓ The first term is 1.

Why this works: A rule that builds forwards can usually be run backwards. Here t_n = t_n+2 − t_n+1, so the sequence is fixed by any two neighbouring terms.

Where it leads: Working backwards along a recurrence is always possible when each term is a sum of the previous two; the sequence is then determined by any two consecutive terms.

Answer: 48

1. Let Sam be s years old now; then Jo is 3s.
2. In 12 years Sam is s + 12 and Jo is 3s + 12.
3. Then Jo is twice Sam’s age: 3s + 12 = 2(s + 12) = 2s + 24.
4. So s = 12, and Jo is 36.
5. Check: in 12 years they are 24 and 48, and 48 = 2 × 24. ✓ Sum now: 12 + 36 = 48.

Why this works: Age problems become easy once every age is written in terms of one letter at one time, then shifted by the same number of years.

Where it leads: The age difference never changes, so at the time Jo is twice Sam’s age, Sam’s age equals the difference: a quicker route.

Answer: 6

1. Start with 3 < 2x − 5 < 17.
2. Add 5 to all three parts: 8 < 2x < 22.
3. Divide all three parts by 2: 4 < x < 11.
4. The inequalities are strict, so 4 and 11 are excluded: x = 5, 6, 7, 8, 9, 10.
5. That is 6 whole numbers.

Why this works: A double inequality can be solved in one go: whatever you do to one part, do to all three. Take care with strict inequalities at the ends.

Where it leads: Double inequalities can be solved all at once, but take care when multiplying by negatives: the signs flip.

Answer: C, 72 litres

1. The 30 litres took the tank from 1/3 full to 3/4 full.
2. Fraction added: 3/4 − 1/3 = 9/12 − 4/12 = 5/12 of the tank.
3. So 5/12 of the tank is 30 litres, and 1/12 is 30 ÷ 5 = 6 litres.
4. The whole tank is 12 × 6 = 72 litres.
5. Check: 1/3 of 72 = 24, plus 30 = 54 = 3/4 of 72. ✓ Answer 72 litres (C).

Why this works: The change in amount matches the change in fraction. Finding one twelfth first is the ‘unitary method’.

Where it leads: Fractions of an unknown whole are solved by finding what one part is worth: the unitary method.

Answer: 6

1. Work from the inside: x ◆ 3 = 2x − 3.
2. Then (x ◆ 3) ◆ x = (2x − 3) ◆ x = 2(2x − 3) − x.
3. Simplify: 4x − 6 − x = 3x − 6.
4. Solve 3x − 6 = 12: 3x = 18, so x = 6.
5. Check: 6 ◆ 3 = 9, and 9 ◆ 6 = 18 − 6 = 12. ✓ Answer 6.

Why this works: A made-up operation is just a rule for substituting. Apply it carefully from the inside out and an ordinary equation appears.

Where it leads: Invented operations test whether you follow definitions exactly. Is ◆ commutative? Associative? Checking such properties is the start of abstract algebra.

Answer: 3

1. Start with n. Double it: 2n. Add 6: 2n + 6.
2. Halve: n + 3. Subtract the starting number: n + 3 − n = 3.
3. The answer is always 3.

Why this works: Algebra shows why the trick works for every starting number at once: the n cancels out.

Where it leads: Design your own: any sequence of steps that ends by subtracting the right multiple of n gives a fixed answer. Mind-reading tricks are algebra in disguise.

Answer: C, 2500

1. There are 50 odd numbers from 1 to 99.
2. Pair them: 1 + 99 = 100, 3 + 97 = 100, …: 25 pairs, each summing to 100.
3. Sum = 25 × 100 = 2500 (C).

Why this works: Pairing from both ends gives equal pair sums, so the total is (number of pairs) × (pair sum).

Where it leads: Notice 2500 = 50²: the sum of the first n odd numbers is always n². You can see it by building squares out of L-shapes.

Answer: £1.50

1. Add the two facts: 5 pens and 5 pencils cost £4.10 + £3.40 = £7.50.
2. So 1 pen and 1 pencil cost £7.50 ÷ 5 = £1.50.
3. (If you want them separately: subtracting gives pen − pencil = £0.70, so a pen is £1.10 and a pencil £0.40.)

Why this works: Looking at what the question actually asks (the sum) shows a shortcut: combine the equations to get that sum directly.

Where it leads: Adding and subtracting equations is the idea behind elimination, and behind solving large systems with matrices.

Answer: 58

1. (a + b)² = a² + 2ab + b².
2. So a² + b² = (a + b)² − 2ab = 100 − 42.
3. The answer is 58. (The numbers are 3 and 7: 9 + 49 = 58.)

Why this works: Expressions like a² + b² can be built from the sum and product without finding a and b.

Where it leads: Every symmetric expression in a and b can be written using a + b and ab. This is why Vieta’s formulas are so powerful for quadratics.

Answer: the 675th term

1. The terms go up by 3, and the 1st term is 2 = 3 × 1 − 1, so the nth term is 3n − 1.
2. 3n − 1 = 2024 gives 3n = 2025, so n = 675.
3. 2024 is the 675th term.

Why this works: A sequence that goes up by a fixed amount has nth term (step) × n + (constant), which turns ‘which term?’ into a one-step equation.

Where it leads: Is 2026 in the sequence? (No: every term leaves remainder 2 on division by 3.) Remainders decide membership of arithmetic sequences.

Answer: C, 22

1. The age difference never changes: it was 4 years five years ago too.
2. Then Ana was twice Bo’s age and 4 more, so Bo was 4 and Ana was 8.
3. Now Bo is 9 and Ana is 13. Sum: 22 (C).

Why this works: The difference in ages is an invariant. When one person is twice the other’s age, the younger person’s age equals the difference.

Where it leads: When will Ana be 1.5 times Bo’s age? (When Bo is 8, which is 1 year ago.) The ratio of ages always moves towards 1 as time passes.

Answer: 12

1. 5 bananas = 4 pears, so 10 bananas = 8 pears.
2. 2 pears = 3 apples, so 8 pears = 12 apples.
3. So 10 bananas weigh the same as 12 apples.

Why this works: Converting step by step through a common unit (pears) is the same as multiplying exchange rates.

Where it leads: Chains of conversions are how currency exchange works; if a loop of exchanges returns more than you started with, that is an arbitrage opportunity.

Answer: 26

1. The two numbers are close together, so each is near √168 ≈ 13.
2. 12 × 14 = 168. ✓ (Also −14 × −12 = 168, if negative numbers are allowed; their sum is −26.)
3. For positive numbers the sum is 12 + 14 = 26.

Why this works: Two numbers close together have a product near the square of their average, so a square root gives a good first guess.

Where it leads: n(n + 2) = (n + 1)² − 1: so the product of consecutive even (or odd) numbers is always one less than a square.

Answer: 16

1. Mean 8 means the total is 40. Order the numbers a < b < 10 < d < e (10 is the median).
2. To make e as big as possible, make a, b and d as small as possible: a = 1, b = 2, d = 11 (d must be more than 10).
3. Then e = 40 − (1 + 2 + 10 + 11) = 16.
4. The largest possible value is 16.

Why this works: To push one value to its extreme, push all the others to their opposite extremes.

Where it leads: This ‘make everything else as small as possible’ reasoning is the extremal principle, used in optimisation and in many harder contest problems.

Answer: C, 5

1. The three rows contain all nine numbers, total 45, so each line totals 15.
2. The middle row, middle column and two diagonals all pass through the centre c. Together they total 4 × 15 = 60.
3. Those four lines cover every square once, except the centre, which is covered four times: 45 + 3c = 60.
4. So c = 5 (C).

Why this works: Adding several lines and comparing with the total of all the numbers isolates the square that is counted extra times.

Where it leads: Up to rotations and reflections there is only one 3 × 3 magic square. There are 880 of size 4 × 4 and over 275 million of size 5 × 5.

Answer: 3 hours

1. When the car sets off, the lorry is 60 km ahead.
2. Each hour the car gains 80 − 60 = 20 km.
3. 60 ÷ 20 = 3 hours. Check: the car has gone 240 km, the lorry 4 × 60 = 240 km. ✓

Why this works: Thinking about the gap and the speed at which it closes (the relative speed) turns a chase into one division.

Where it leads: Relative speed also solves problems with trains passing each other and with clock hands: the minute hand gains 330° per hour on the hour hand.

Answer: A, 4 km/h

1. Take the path to be 6 km (any length gives the same answer).
2. Up: 6 ÷ 3 = 2 hours. Down: 6 ÷ 6 = 1 hour.
3. Total: 12 km in 3 hours, so the average speed is 4 km/h (A), not 4.5.

Why this works: He spends longer at the slow speed, so the average is pulled below the halfway value 4.5.

Where it leads: For equal distances the average speed is the harmonic mean 2ab/(a + b). The harmonic mean is never more than the ordinary mean.

Answer: B, 9/10

1. Each denominator is n(n + 1): 2 = 1 × 2, 6 = 2 × 3, …, 90 = 9 × 10.
2. 1/(n(n + 1)) = 1/n − 1/(n + 1). So the sum is (1 − 1/2) + (1/2 − 1/3) + … + (1/9 − 1/10).
3. Everything cancels except 1 − 1/10 = 9/10 (B).

Why this works: Splitting each fraction into a difference makes the sum ‘telescope’: neighbouring terms cancel and only the ends survive.

Where it leads: Telescoping sums are a key tool in series. The infinite version 1/2 + 1/6 + 1/12 + … adds up to exactly 1.

Answer: 2

1. Input x gives output 3x − 4.
2. We need 3x − 4 = x, so 2x = 4 and x = 2.
3. Check: 3 × 2 − 4 = 2. ✓

Why this works: A ‘fixed point’ of a machine is an input that comes out unchanged; setting output = input gives the equation.

Where it leads: If you feed outputs back in, 3x − 4 runs away from 2, but a machine like x/3 + 4 homes in on its fixed point 6. Fixed points underpin many numerical methods.

Answer: 12

1. Middle row: 7 + x and x + 9.
2. Top: (7 + x) + (x + 9) = 16 + 2x.
3. 16 + 2x = 40, so x = 12.

Why this works: The middle number of the bottom row is used twice on the way to the top, so it is counted twice in the top brick.

Where it leads: With a bottom row of n bricks, the top is a sum of the bottom bricks weighted by a row of Pascal’s triangle.

Answer: C, 33

1. The first square uses 4; each extra square adds 3 (one side is shared). So n squares use 3n + 1.
2. 3n + 1 = 100 gives n = 33.
3. 33 squares (C).

Why this works: Seeing each new square as ‘3 more sticks’ gives the formula without having to draw the pattern.

Where it leads: A 2 by n rectangle of squares uses 7n + 2 sticks. For an n by n grid you need 2n(n + 1): a quadratic, because sticks go along two directions.

Answer: 36

1. Giving 6 marbles makes them equal, so Ali had 2 × 6 = 12 more than Bo.
2. Ali has twice as many as Bo, so the difference equals Bo’s amount: Bo has 12, Ali has 24.
3. Altogether: 36.

Why this works: ‘Giving 6 makes them equal’ means the gap is 12 (the giver goes down 6 and the receiver up 6).

Where it leads: The total never changes when marbles pass between them: another invariant that often gives a quick check.

Answer: B, 100%

1. A 25% rise multiplies the price by 1.25; 20% off multiplies by 0.8.
2. 1.25 × 0.8 = 1.
3. The final price is 100% of the original (B): it is back where it started.

Why this works: Percentage changes combine by multiplying. 1.25 = 5/4 and 0.8 = 4/5 are reciprocals, so they cancel exactly.

Where it leads: To undo a rise of p%, you need a fall of 100p/(100 + p)%, which is always smaller than p%.

Answer: 55

1. Write the numbers as m − 2, m, m + 2.
2. Their sum is 3m = 159, so m = 53.
3. The largest is 55 (51 + 53 + 55 = 159).

Why this works: Naming the middle number makes the ±2 cancel, so the sum is simply three times the middle.

Where it leads: Can three consecutive odd numbers ever add to an even number? (No: three odd numbers always have an odd sum.)

Answer: 2

1. n² < 50: since 7² = 49 and 8² = 64, n is from −7 to 7.
2. 2ⁿ > 50: 2⁵ = 32 is too small and 2⁶ = 64 works, so n ≥ 6 (negative n give fractions).
3. Both: n = 6 or 7, so 2 numbers.

Why this works: Solving each condition separately and then taking the overlap is the standard way to handle two conditions at once.

Where it leads: For large n, 2ⁿ grows far faster than n²: 2ⁿ > n² for every n ≥ 5. Proving this is a classic first use of induction.

Answer: 13

1. If all 30 animals were chickens there would be 60 legs.
2. There are 86 − 60 = 26 extra legs, and each cow has 2 more legs than a chicken.
3. So there are 26 ÷ 2 = 13 cows (and 17 chickens: 52 + 34 = 86. ✓)

Why this works: Starting from an extreme case (all chickens) and then adjusting is the same as solving simultaneous equations, but quicker.

Where it leads: This method is thousands of years old: a version appears in the ancient Chinese text Sunzi Suanjing with pheasants and rabbits.