12 short questions across all six content areas, designed as a lesson starter or a daily warm-up.
⏱️ Ten-minute starters — Foundation
Twelve short questions, two from each of the six content areas: number, algebra, ratio, geometry, probability and statistics. Nothing here needs more than a minute, and nothing needs a calculator. It is built as a lesson starter — print it, hand it out, take ten minutes, mark it together — and it works just as well as a daily warm-up at home in the fortnight before a mock. The value is in the spread rather than the depth: a student who has spent a week on algebra can find out in ten minutes what a week away from statistics has cost. Generate a new sheet whenever you want a fresh twelve on the same pattern.
- 1.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 2.A gardener wants to plant a tree so that it is the same distance from two straight garden walls that meet at a corner, and also the same distance from two ornamental posts standing 4 m apart. Which construction locates this point?
- 3.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 4.y is inversely proportional to x. When x = 5, y = 8. Work out the value of y when x = 10.
- 5.A laptop costs £720. A carrying case for it costs 1/9 of the price of the laptop. Work out the cost of the carrying case.
- 6.Solve the inequality x + 5 < 12.
- 7.Write down the exact value of tan 0°.
- 8.A shop sells identical pens. The total cost, C pence, of n pens is given by C = 12n. Write the ratio n : C in its simplest form.
- 9.A car's windscreen wiper blade is 40 cm long and sweeps through an angle of 110° as it wipes the glass. Using π = 3.14, work out the area of glass it wipes, to the nearest cm².
- 10.The ratio of the amount Noah has saved to the amount Grace has saved is 4:5. Noah has saved £200. Work out how much they have saved altogether.
- 11.A bag of flour is labelled 1.5 kg, correct to the nearest 0.1 kg. The true mass of the flour is m kg. Which inequality gives all the possible values of m?
- 12.Work out 2/5 of 45.
Answer key
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (a) where the angle bisector meets the posts' perpendicular bisector — Being equidistant from the two walls means lying on the angle bisector of the corner; being equidistant from the two posts means lying on the perpendicular bisector of the 4 m segment joining them. A single point satisfying both conditions is wherever those two loci cross. "where the angle bisector meets the line joining the posts" uses the straight line between the posts instead of its perpendicular bisector — a point on that line is not generally equidistant from both posts. "the perpendicular bisector of the posts, alone" satisfies only the posts condition, ignoring the walls entirely. "the angle bisector of the corner, alone" satisfies only the walls condition, ignoring the posts entirely.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (c) £80 — Method: a unit fraction acts as an operator, so finding 1/9 of a price means dividing that price by 9. Working: £720 ÷ 9 = £80. Answer: £80. The distractors: £6480 comes from multiplying by the denominator instead of dividing by it, giving £720 × 9 = £6480; £640 comes from working out what is left of the £720 once the case is paid for, £720 − £80, instead of the cost of the case itself; £72 comes from dividing by 10 instead of 9, treating one ninth as one tenth.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (a) 0 — tan 0° = 0, because there is no opposite side to consider when the angle itself is 0° — one of the exact values you need to know. 1 is the exact value of tan 45°, not tan 0°. √3 is the exact value of tan 60°. 1/√3 is the exact value of tan 30°.
- (b) 1:12 — C = 12n means that for every 1 pen there are 12 pence of cost, so n : C = 1 : 12, and the highest common factor of 1 and 12 is 1, so this is already in its simplest form. Writing C : n instead of n : C gives 12 : 1, the ratio the wrong way round. Reading C = 12n as '12 more than n' instead of '12 times n', so n = 1 gives C = 13, gives 1 : 13, from adding instead of multiplying. Choosing n = 12, so C = 12 × 12 = 144, gives the correct pair of values n : C = 12 : 144, but the ratio the right way round left unsimplified — 12 and 144 share a common factor of 12, which has not been cancelled.
- (d) 1535 cm² — The wiper sweeps out a sector of radius 40 cm, the blade length, through an angle of 110°. Sector area is angle ÷ 360 × π × radius²: 110 ÷ 360 × 3.14 × 1600 = 1535.1 cm², which rounds to 1535 cm². Forgetting to square the radius, using radius instead of radius², gives 38 cm². Using 110 ÷ 180 instead of 110 ÷ 360 for the fraction gives 3070 cm². Treating the 40 cm blade length as a diameter, so using a radius of 20 cm, gives 384 cm².
- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (c) 18 — 45 ÷ 5 = 9, and 2 × 9 = 18. A candidate who stops after finding one fifth gets 9. A candidate who uses 3/5 instead of 2/5 gets 27. A candidate who uses 4/5 instead of 2/5 gets 36.
What is on this worksheet?
The sheet holds 12 questions drawn from the MathsUK bank — the content areas covered: Number, Algebra, Ratio, proportion and rates of change, Geometry and measures, Probability, Statistics. It is pitched at GCSE Foundation and takes about 10 minutes to work through in full. It works as a class handout, a homework, or a warm-up before a test.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 12 questions before checking — about 10 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 12 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
Similar worksheets worth a look
- ✏️ Paper 1 non-calculator warm-up — Foundation · 20 questions · ~25 min
- 🧮 Paper 1 non-calculator warm-up — Higher · 20 questions · ~25 min
- 🎯 Grade 4 pass booster · 25 questions · ~40 min
- ⚖️ Foundation to Higher crossover check · 20 questions · ~35 min