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 factory finds that the probability a randomly chosen light bulb is defective is 0.035. In a batch of 4,000 bulbs, work out how many bulbs you would expect to work correctly.
- 2.Triangle ABC is translated by the column vector with top number 3 and bottom number −5 to form triangle A′B′C′. Triangle A′B′C′ is then translated by the column vector with top number −7 and bottom number 2 to form triangle A″B″C″. Work out the single column vector that translates triangle ABC directly to triangle A″B″C″.
- 3.A sponsored walk raised £350 for charity. 20% of the money raised is spent on equipment. Work out how much is spent on equipment.
- 4.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 5.A coach travels the 54 miles from London to Brighton in 1 hour 30 minutes. Work out the average speed of the coach, in mph.
- 6.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
- 7.Solve 4x² − 9 = 0.
- 8.A plan is drawn to a scale of 1 : 25. A real fence is 10 m long. Work out the length of the fence on the plan, in centimetres.
- 9.Triangle ABC has a right angle at B and hypotenuse AC = 17 cm, with AB = 8 cm. Triangle DEF has a right angle at E and hypotenuse DF = 17 cm. Which extra fact about DEF would complete the RHS condition for proving that ABC ≅ DEF, with the vertices matching in that order?
- 10.A point has coordinates (−6, 9). Work out the sum of the x-coordinate and the y-coordinate.
- 11.Work out the value of .
- 12.A recipe for one cake needs 2/3 of a cup of sugar. Priya has 3 1/2 cups of sugar. Work out how many complete cakes she can make.
Answer key
- (a) 3,860 — The probability a bulb works correctly is the complement of being defective: 1 − 0.035 = 0.965. Expected number working correctly = 0.965 × 4,000 = 3,860. Using the probability of being defective instead of its complement gives 4,000 × 0.035 = 140, the expected number of DEFECTIVE bulbs, not working ones. Shifting the decimal point in the complement, using 0.0965 instead of 0.965, gives 4,000 × 0.0965 = 386. Assuming every bulb works, ignoring the 0.035 probability altogether, gives the full batch of 4,000.
- (d) (−4, −3) — The combined translation is the sum of the two column vectors, added component by component: top numbers 3 + (−7) = −4, bottom numbers −5 + 2 = −3, giving (−4, −3). (10, −7) subtracts the second vector from the first instead of adding them. (−4, 3) gets the top number right but makes a sign error on the bottom, treating −5 + 2 as +3. (4, −3) gets the bottom number right but makes a sign error on the top, treating 3 + (−7) as +4.
- (c) £70 — Method: 20% of an amount is 20/100 of it; a reliable route is to find 10% by dividing by 10 and then double it. Working: 10% of £350 is £350 ÷ 10 = £35, and 20% is twice as much, £35 × 2 = £70. Answer: £70. The distractors: £35 comes from finding 10% and stopping there; £17.50 comes from reading 20% as one twentieth and working out £350 ÷ 20 = £17.50; £280 comes from taking 20% off the money raised rather than finding 20% of it, giving £350 ÷ 5 = £70 and then £350 − £70 = £280.
- (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (d) 36 mph — First convert 1 hour 30 minutes to hours: 30 minutes is half an hour, so the time is 1.5 hours. Then divide the distance by the time: 54 ÷ 1.5 = 36 mph. Reading 1 hour 30 minutes as 1.3 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 54 ÷ 1.3 ≈ 41.54 mph. Working out 54 ÷ 30 = 1.8 divides by the number of minutes only, ignoring the hour. Working out 54 × 1.5 = 81 multiplies by the time instead of dividing. The coach's average speed is 36 mph.
- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
- (a) x = 3/2 or x = −3/2 — Method: the equation has an x² term and a number but no x term, so make x² the subject and then take the square root of both sides, keeping the negative root as well as the positive one. Working: adding 9 to both sides of 4x² − 9 = 0 gives 4x² = 9, and dividing both sides by 4 gives x² = 9/4. Square-rooting the top and the bottom of 9/4 gives 3/2, so x = 3/2 or x = −3/2, and each value checks out because 4 × 9/4 − 9 = 0. Answer: x = 3/2 or x = −3/2. The distractors: x = 3 or x = −3 comes from square-rooting both sides of 4x² = 9 without first dividing by the 4, so the coefficient of x² is ignored; x = 9/4 or x = −9/4 comes from stopping at x² = 9/4 and writing that value down as x, leaving the square root undone; x = 3/2 only comes from taking the positive square root of 9/4 and losing the negative solution.
- (d) 40 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 10 m = 1000 cm. 1000 ÷ 25 = 40 cm. Wrong options: 0.4 cm comes from dividing 10 by 25 without converting metres to centimetres first; 25000 cm comes from multiplying instead of dividing (1000 × 25); 4 cm comes from converting 10 m to 100 cm (using ×10 instead of ×100) before dividing.
- (d) DE = 8 cm — Method: use the stated correspondence ABC ≅ DEF to work out which side in DEF matches the known side AB in ABC. Working: the correspondence sends A to D, B to E and C to F, so AB corresponds to DE; the right angles at B and E and the equal hypotenuses AC = DF = 17 cm are already given, so DE = 8 cm supplies the third ingredient — Right angle, Hypotenuse, Side. Options: EF = 8 cm matches AB to the wrong side, since EF corresponds to BC, and BC = √(17² − 8²) = 15 cm, not 8 cm; BC = 8 cm states something about triangle ABC rather than the missing fact about DEF, and it is false as well, since BC = 15 cm; angle D = angle A does follow once the triangles are congruent, but RHS is completed by a matching side, not by a matching angle. Answer: DE = 8 cm.
- (b) 3 — −6 + 9 = 3. A candidate who ignores the negative sign on the x-coordinate and adds the two positive values gets 6 + 9 = 15. A candidate who treats the y-coordinate as negative too gets −6 + (−9) = −15. A candidate who works out 9 − 6 correctly as 3 but then writes the answer with the wrong sign gets −3.
- (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (b) 5 — Method: divide the total amount of sugar by the amount needed for one cake, then round down because a part-used amount of sugar cannot make an extra whole cake. Working: 3 1/2 ÷ 2/3 = 7/2 × 3/2 = 21/4 = 5.25; only 5 complete cakes can be made, since the leftover 0.25 of a portion is not enough for a 6th cake. Answer: 5. 5.25 gives the exact result of the division without rounding down to a whole number of cakes. 7 comes from multiplying 3.5 by 2 and ignoring the need to also divide by 3 as part of dividing by the fraction 2/3. 6 comes from rounding 5.25 up to the nearest whole number instead of down, wrongly assuming a 6th cake could be made from the leftover sugar.
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.
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