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 children's slide is built as a right-angled triangle: the vertical ladder side is 2.4 m, the horizontal base along the ground is 3.2 m, and the sloping slide is the third side. Work out the exact length of the sloping slide.
- 2.Work out −4.5 + 2.8.
- 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.A sector of a circle has radius 15 cm and angle 216°. Using π = 3.14, work out the arc length of the sector.
- 5.At a bus station one bus leaves every 6 minutes and another leaves every 9 minutes. The two buses leave together at the start of the day. Work out how long it is until they next leave together.
- 6.Work out 1 − 1/2 − 1/4 − 1/8 − 1/16. Give your answer as a fraction.
- 7.A water tank is a cube with edges of length 2 m. Work out how many cubic centimetres the tank holds when it is full.
- 8.A tin of paint has a mass of 5.672 kg. Round this mass to 1 decimal place.
- 9.A water butt holds 200 litres and is being drained at a steady 8 litres per minute. Write down the function for the amount of water y, in litres, left after x minutes.
- 10.A wooden block is a cuboid measuring 2 cm by 10 cm by 15 cm. Work out the volume of the block.
- 11.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 12.Work out the value of 3a − 2b when a = 5 and b = 4
Answer key
- (d) 4 m — The sloping slide is the hypotenuse of a right-angled triangle with the other two sides 2.4 m and 3.2 m. By Pythagoras' Theorem, hypotenuse² = 2.4² + 3.2² = 5.76 + 10.24 = 16. Square root: √16 = 4 m. Adding the two sides instead of squaring them gives 2.4 + 3.2 = 5.6 m. Truncating each squared side to a whole number before adding (5.76 to 5 and 10.24 to 10) gives √15 ≈ 3.87 m. Forgetting to take the square root and leaving the sum of the squares as the answer gives 5.76 + 10.24 = 16, written as 16 m.
- (d) −1.7 — Since the numbers have different signs, find the difference between their sizes: 4.5 − 2.8 = 1.7, then keep the sign of the number further from zero. So −4.5 + 2.8 = −1.7. A candidate who drops the negative sign gets 1.7. A candidate who adds the magnitudes instead of finding the difference gets −(4.5 + 2.8) = −7.3. A candidate who takes the smaller digit from the larger in the tenths column, doing 8 − 5 = 3 instead of borrowing to make 15 − 8 = 7, gets 2.3 and so −2.3.
- (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.
- (b) 56.52 cm — Arc length = (216 ÷ 360) × 2 × 3.14 × 15 = 0.6 × 94.2 = 56.52 cm. (28.26 cm comes from leaving out the factor of 2, using πr instead of 2πr; 94.2 cm comes from finding the full circumference and forgetting the angle fraction; 113.04 cm comes from using the diameter, 30 cm, instead of the radius.)
- (a) 18 minutes — Method: the buses leave together again after a number of minutes that is a multiple of both intervals, and the first such time is the lowest common multiple. Working: the multiples of 6 are 6, 12, 18, 24 … and the multiples of 9 are 9, 18, 27 … The first value in both lists is 18, which is 6 × 3 and 9 × 2. Answer: 18 minutes. The distractors: 54 minutes comes from multiplying 6 by 9, which does give a common multiple but not the lowest one; 3 minutes is the highest common factor of 6 and 9 rather than their lowest common multiple; 15 minutes comes from adding the two intervals together.
- (a) 1/16 — Method: terms can only be subtracted once they share a denominator, so write every term over the largest denominator, 16, and then subtract the numerators in order from left to right. Working: 1 = 16/16, 1/2 = 8/16, 1/4 = 4/16 and 1/8 = 2/16, so the numerators give 16 − 8 − 4 − 2 − 1 = 1, over a denominator of 16. Answer: 1/16. The distractors: 1/8 comes from stopping one term early, after 16 − 8 − 4 − 2 = 2; 3/16 comes from a sign slip on the last term, adding it instead of subtracting it, which gives 2 + 1 = 3; 15/16 comes from working from the right-hand end as though the last four terms were bracketed together, so that only a single sixteenth is taken away from 1.
- (a) 8,000,000 cm³ — Method: change the edge length into centimetres first and then cube it, because 1 m = 100 cm and a volume needs that conversion applied to all three dimensions. Working: 2 m = 2 × 100 = 200 cm, so the volume is 200 × 200 × 200. 200 × 200 = 40,000 and 40,000 × 200 = 8,000,000. Answer: 8,000,000 cm³. The distractors: 8,000 cm³ comes from converting 2 m to 20 cm and cubing that; 80,000 cm³ comes from cubing in metres to get 8 m³ and then multiplying by 10,000, the conversion factor for an area rather than the 1,000,000 a volume needs; 8 cm³ comes from cubing the 2 without converting at all and simply writing cm³ because the question asked for that unit.
- (c) 5.7 kg — Method: to round to 1 decimal place, keep one digit after the decimal point and let the digit in the second decimal place decide whether that digit stays as it is or goes up. Working: 5.672 has 6 in the first decimal place and 7 in the second decimal place; 7 is 5 or more, so the 6 goes up to 7 and the digits beyond the first decimal place are dropped. Answer: 5.7 kg. The distractors: 5.6 kg comes from chopping the digits after the first decimal place off instead of rounding them, which is truncation rather than rounding; 6.0 kg comes from rounding to the nearest whole kilogram instead of to 1 decimal place; 5.0 kg comes from chopping everything after the decimal point off, so the mass is both truncated and given to the wrong degree of accuracy.
- (c) y = 200 − 8x — Method: in a linear model the amount present at the start is the constant term and the steady rate of change is the gradient, which is negative because the amount is falling. Working: at x = 0 minutes there are 200 litres, so the constant term is 200; 8 litres are lost every minute, so after x minutes 8x litres have gone and the amount left is y = 200 − 8x. Answer: y = 200 − 8x. The distractors: y = 8x + 200 treats the draining as filling, so the butt would gain 8 litres a minute; y = 200x − 8 swaps the two numbers over, using the starting 200 litres as the rate per minute and the 8 litres per minute as the starting amount; y = −8x − 200 makes the starting amount negative as well as the rate, so the butt would begin 200 litres in deficit.
- (d) 300 cm³ — Method: the volume of a cuboid is length × width × height. Working: 2 × 10 = 20, then 20 × 15 = 300. Answer: 300 cm³. The distractors: 27 cm³ comes from adding the three edges, 2 + 10 + 15, instead of multiplying them; 400 cm² comes from working out the surface area, 2 × (2 × 10 + 2 × 15 + 10 × 15) = 400, which answers a different question and carries a different unit; 150 cm³ comes from multiplying 10 × 15 and leaving the 2 cm edge out of the calculation altogether.
- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (d) 7 — Method: substitute both values, work out the two multiplications first, and only then subtract. Working: 3a = 3 × 5 = 15 and 2b = 2 × 4 = 8, so the expression becomes 15 − 8 = 7. Answer: 7. The distractors: 23 comes from adding the two products instead of subtracting, giving 15 + 8; −7 comes from subtracting the wrong way round and working out 8 − 15; 52 comes from working from left to right instead of following the order of operations, giving 3 × 5 = 15, then 15 − 2 = 13, then 13 × 4.
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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