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.The attendance at a football match is 23,000, correct to the nearest thousand. Which of these attendances is not possible for the match?
- 2.A cyclist travels 18 km in 45 minutes. Work out the average speed, in km/h.
- 3.A model railway is built at a scale where the ratio of model length to real length is 1 : 76. Given that a real carriage is 20.6 metres long, work out the length of the model carriage in centimetres, to 1 decimal place.
- 4.In a board game Mia moves forward by the total of two ordinary fair dice. Work out the probability that her total is 7. Give your answer as a fraction in its simplest form.
- 5.A school has 1200 pupils. A teacher wants to take a random sample of 60 of them. Write down which of these methods gives a random sample.
- 6.A scatter graph shows the number of hours of sunshine, x, and the number of visitors, y, at an outdoor swimming pool in Torquay on each of 15 days. The line of best fit passes through the points (5, 150) and (15, 350). Work out the estimated number of visitors on a day with 8 hours of sunshine, using the line of best fit.
- 7.A trapezium has parallel sides of 5 cm and 11 cm, and a perpendicular height of 4 cm. Work out the area of the trapezium.
- 8.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 9.Point C is at (5, 2). It is reflected in the x-axis. Work out the coordinates of the image of point C.
- 10.A fair spinner has 10 equal sections, numbered 1 to 10. The spinner is spun 150 times. Work out how many times you would expect it to land on a number greater than 7.
- 11.Grace has t books. Noah has 1 book fewer than Grace. Write down an expression for the number of books Noah has.
- 12.A coach travels 210 miles at an average speed of 70 mph. Work out how long the journey takes.
Answer key
- (d) 23,600 — 23,000 correct to the nearest thousand means the true attendance, a, satisfies 22,500 ≤ a < 23,500. 23,600 is outside this range, so it is not possible. 22,500 is the lower bound of the range, so it is possible. 23,499 is just below the upper bound, so it is possible. 23,000 is the given value itself, so it is possible.
- (b) 24 km/h — First convert 45 minutes to hours: 45 ÷ 60 = 0.75 hours. Then divide the distance by the time: 18 ÷ 0.75 = 24 km/h. Reading 45 minutes as 0.45 hours (writing the minutes after the decimal point instead of dividing by 60) gives 18 ÷ 0.45 = 40 km/h. Working out 18 × 0.75 = 13.5 multiplies by the time instead of dividing. Working out 18 ÷ 45 = 0.4 divides by 45 without ever converting the minutes to hours. The cyclist's average speed is 24 km/h.
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (a) 1/6 — Method: write the results of the two dice as ordered pairs, count the pairs whose scores add to the total asked for, divide by the number of ordered pairs there are, then cancel the fraction down. Working: there are 6 × 6 = 36 equally likely ordered pairs. The pairs whose scores add to 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1), which is 6 pairs, so the probability is 6/36. Dividing the top and the bottom by 6 gives 1/6. Answer: the probability is 1/6. The distractors: 7/36 comes from taking the number of favourable pairs to be 7 because 7 is the total asked for, confusing the size of a total with the number of ways of making it; 1/7 comes from using the 21 different combinations of two scores as the equally likely results, finding the 3 combinations 1 and 6, 2 and 5, 3 and 4, and cancelling 3/21; 6/11 comes from counting the 6 favourable pairs correctly but dividing by the 11 possible totals from 2 to 12 rather than by the 36 pairs.
- (a) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (d) 210 — 350 − 150 = 200. 200 ÷ 10 = 20, so the gradient is 20. Using the point (5, 150): 20 × 5 = 100, so 150 − 100 = 50 is the intercept, giving the line y = 20x + 50. At x = 8: 20 × 8 = 160, and 160 + 50 = 210, so the estimated number of visitors is 210. Choosing 160 stops after 20 × 8 = 160 and forgets to add the intercept of 50. Choosing 250 comes from averaging the two given y-values: 150 + 350 = 500, and 500 ÷ 2 = 250, instead of using the line's equation. Choosing 240 assumes the visitors are directly proportional to the hours of sunshine using the first point, 150 × 8 ÷ 5 = 240, which ignores that the line does not pass through the origin.
- (a) 32 cm² — Area of a trapezium = 1/2 × (sum of parallel sides) × height = 1/2 × (5 + 11) × 4 = 1/2 × 16 × 4 = 32 cm². (64 cm² comes from forgetting to halve the product; 44 cm² comes from using only the longer parallel side, 11 × 4, instead of both parallel sides; 20 cm² comes from adding all three given lengths together, 5 + 11 + 4, instead of using the area formula.)
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (b) (5, −2) — Reflecting in the x-axis keeps the x-coordinate the same and changes the sign of the y-coordinate, so (5, 2) → (5, −2). ((−5, 2) comes from reflecting in the y-axis instead, changing the sign of the x-coordinate; (−5, −2) comes from changing the sign of both coordinates, as for a reflection in the origin; (2, 5) comes from swapping the coordinates, which is what happens for a reflection in the line y = x.)
- (c) 45 — Three of the ten numbers (8, 9 and 10) are greater than 7, so the probability is 3/10, and 150 × 3/10 = 45. Writing 105 is wrong because 150 × 7/10 = 105 uses the seven numbers that are NOT greater than 7 (1 to 7), the opposite of what is asked. Writing 60 is wrong because it counts 7, 8, 9 and 10 as four numbers greater than 7, wrongly including 7 itself: 150 × 4/10 = 60. Writing 50 is wrong because 150 ÷ 3 = 50 divides by the count of favourable numbers instead of multiplying by the correct fraction of the spinner. The expected number of spins landing on a number greater than 7 is 45.
- (b) t − 1 — Method: "fewer than" means a subtraction, and the amount taken away is taken away from the quantity being compared with. Working: Grace has t books and Noah has one book fewer, so Noah has t with 1 taken off, written t − 1. Answer: t − 1. The distractors: t + 1 comes from reading "fewer" as "more" and adding instead of subtracting; 1 − t comes from writing the subtraction the wrong way round, taking Grace's books from the 1 rather than the 1 from Grace's books; t comes from giving the number of books Grace has, which is not the person the question asks about.
- (c) 3 hours — Time = distance ÷ speed, so 210 ÷ 70 = 3 hours. Working out 210 − 70 = 140 subtracts the speed from the distance instead of dividing. Working out 210 + 70 = 280 adds the two values together, which does not give a time. Working out 210 × 70 = 14700 multiplies distance and speed together instead of dividing. The journey takes 3 hours.
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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