20 questions with no calculator: the four operations, fractions, percentages, order of operations, factors and estimation.
✏️ Paper 1 non-calculator warm-up — Foundation
Paper 1 is worth exactly as much as each calculator paper, and it is the one most Foundation students practise least. This sheet is a warm-up for it: twenty questions drawn from the number and ratio statements that turn up on the non-calculator paper year after year — column arithmetic with decimals and negatives, adding and subtracting fractions, percentages of amounts, priority of operations, factors and multiples, and estimating by rounding to one significant figure. Do it with a pen and nothing else. Twenty-five minutes is about right, but do not stop early if you need longer; the point is to find out which of these have quietly gone rusty, not to score well. Mark it, then go back to the topic page for whichever one cost you the most.
- 1.Write the ratio 5 : 8 in the form 1 : n.
- 2.A textbook is reduced from £60 to £45. Work out the percentage reduction.
- 3.A plank of wood is 5 1/4 m long. Pieces of length 3/4 m are cut from it. Work out how many complete pieces of 3/4 m can be cut from the plank.
- 4.The width, the length and the height of a box are in the ratio 3:4:5. The length of the box is 16 cm. Work out the height of the box.
- 5.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 6.Write 1/5 as a percentage.
- 7.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 8.Write 350 ml : 1.4 l as a ratio in its simplest form.
- 9.Write the ratio 27 : 45 in its simplest form.
- 10.Which of these numbers is a common factor of 18 and 24?
- 11.A tank contains 120 litres of water. Water is drained out at a rate of 8 litres per minute for 6 minutes, and then a hose adds 15 litres. Work out how much water is left in the tank.
- 12.Grace drinks 1/3 of a bottle of water in the morning and another 1/3 of the same bottle in the afternoon. Work out what fraction of the bottle she has drunk altogether.
- 13.Write 12 as a product of its prime factors.
- 14.A gardener has 42 tulip bulbs and 56 daffodil bulbs. She plants them in rows, with every row containing the same number of tulip bulbs and the same number of daffodil bulbs, and no bulbs left over. Work out the greatest number of rows she can plant.
- 15.A rectangular patio measures 90 cm by 120 cm. Ben wants to cover it exactly with identical square tiles, as large as possible, with no tiles cut. Work out the side length of the largest square tile he can use.
- 16.Write 200 as a product of its prime factors, using index notation.
- 17.Work out 0.6 × 0.04.
- 18.Work out (−3) × 4 + 2 × (−5)
- 19.Write 0.36 as a fraction in its simplest form.
- 20.Simplify the ratio 45 : 30 : 75 to its simplest form.
Answer key
- (b) 1:1.6 — To write 5 : 8 in the form 1 : n, divide both parts by 5, the first number, so that it becomes 1: 5 ÷ 5 = 1 and 8 ÷ 5 = 1.6, giving 1 : 1.6. Dividing both parts by 8 instead gives 0.6 : 1 (5 ÷ 8 = 0.625, rounded to 0.6) — the first part is no longer 1, so this is not in the required form. Dividing 5 by 8 but writing the result after the 1 gives 1 : 0.6, which divides in the wrong direction: n must come from 8 ÷ 5, not 5 ÷ 8. A slip in the division 8 ÷ 5, rounding it to 1.5 instead of the correct 1.6, gives 1 : 1.5.
- (d) 25% — Method: percentage decrease = decrease ÷ original amount × 100. Working: the reduction is £60 − £45 = £15, and 15 ÷ 60 = 0.25, so 0.25 × 100 = 25%. Answer: 25%. The distractors: 15% comes from quoting the £15 reduction as though pounds and per cent were the same thing; 33% comes from dividing the £15 by the new price £45 instead of by the original £60, which gives 33% to the nearest per cent; 75% is the new price written as a percentage of the old one, which is what is still paid rather than what has been taken off.
- (a) 7 — Convert the mixed number to an improper fraction: 5 1/4 = 21/4. Dividing by 3/4 means multiplying by its reciprocal, 4/3: 21/4 × 4/3 gives 84/12, which simplifies to 7. So exactly 7 complete pieces of 3/4 m can be cut. Ignoring the 1/4 m and dividing only the whole number, 5 ÷ 3/4, gives 20/3, which is 6 complete pieces with some wood left over. Multiplying by 3/4 instead of its reciprocal, 21/4 × 3/4, gives 63/16, which is 3 complete pieces. Misreading 5 1/4 as the fraction 5/4, then dividing by 3/4, gives 5/3, which is only 1 complete piece. So 7 complete pieces can be cut from the plank.
- (a) 20 cm — Method: match the measurement you are given to its own part of the ratio, use it to find the value of one part, then multiply by the parts belonging to the measurement asked for. Working: the length is the second measurement listed, so it matches 4 parts and one part = 16 ÷ 4 = 4 cm; the height is 5 parts, so 5 × 4 = 20. Answer: 20 cm. The distractors: 12 cm is the width, which is the 3-part measurement; 4 cm is the value of one part only; 80 cm comes from multiplying the 16 cm by 5 without first dividing by the 4 parts the length is worth.
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (a) 20% — Method: a percentage is a number of parts per hundred, so divide the numerator by the denominator and multiply by 100. Working: 1 ÷ 5 = 0.2, and 0.2 × 100 = 20; the same result comes from scaling 1/5 to 20/100. Answer: 20%. The distractors: 5% comes from reading the denominator 5 straight off as the percentage; 25% comes from confusing 1/5 with 1/4, the fraction met most often in percentage work; 50% comes from a place value slip in the division, writing 1 ÷ 5 as 0.5 instead of 0.2.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (d) 1:4 — Convert 1.4 l to millilitres: 1.4 l = 1400 ml. The ratio is 350 : 1400. Divide both parts by 350: 350 ÷ 350 = 1 and 1400 ÷ 350 = 4, giving 1 : 4. Misreading 1.4 l as 14 (moving the decimal point) gives 350 : 14, which simplifies to 25 : 1 — a very different, implausible ratio. Dividing by 175 instead of 350 gives 2 : 8, which still shares a common factor of 2, so it is not fully simplified. Swapping the order gives 4 : 1, litres to millilitres the wrong way round.
- (b) 3 : 5 — The highest common factor of 27 and 45 is 9. Divide both parts by 9: 27 ÷ 9 = 3 and 45 ÷ 9 = 5, giving 3 : 5. Giving 9 : 15 divides by 3, a common factor but not the highest one, so it is not yet in simplest form. Giving 5 : 3 swaps the order of the two parts. Giving 27 : 45 has not been simplified at all.
- (a) 6 — Method: list the factors of each number and compare them. Working: the factors of 18 are 1, 2, 3, 6, 9, 18; the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. The only option that appears in both lists is 6. 8 is a factor of 24 but not of 18. 9 is a factor of 18 but not of 24. 12 is a factor of 24 but not of 18. Answer: 6.
- (c) 87 litres — Work out how much water is drained: 8 × 6 = 48 litres. Subtract this from the starting amount: 120 − 48 = 72 litres. Then add the 15 litres from the hose: 72 + 15 = 87 litres. Subtracting the 15 litres instead of adding it, as though the hose also removed water, gives 120 − 48 − 15 = 57 litres. Stopping after the drain step, without adding the hose water back in, leaves the working at 72 litres. Adding the rate and the time instead of multiplying them, 8 + 6 = 14 litres drained, and then working from there gives 120 − 14 + 15 = 121 litres. So 87 litres of water is left in the tank.
- (c) 2/3 — Method: fractions with the same denominator are added by adding the numerators and leaving the denominator alone, because the parts are already the same size. Working: 1/3 + 1/3 has numerators 1 + 1 = 2 and the denominator stays as 3, giving 2/3. Answer: 2/3. The distractors: 2/6 comes from adding the denominators as well as the numerators, 1 + 1 over 3 + 3; 2/9 comes from adding the numerators but multiplying the denominators, 1 + 1 over 3 × 3; 1/9 comes from multiplying throughout instead of adding, 1 × 1 over 3 × 3.
- (a) 2² × 3 — Method: divide repeatedly by the smallest prime that goes in, until 1 is reached, then write the primes used as a product with indices. Working: 12 ÷ 2 = 6, 6 ÷ 2 = 3 and 3 ÷ 3 = 1, so the primes used are 2, 2 and 3, which is written as 2² × 3. Answer: 2² × 3. The distractors: 2 × 6 comes from stopping at the first factor pair without splitting the 6, which is not prime; 2 × 3 comes from listing each prime once and losing the repeat, and it multiplies to 6 rather than 12; 2 × 3² puts the index on the wrong prime and multiplies to 18.
- (b) 14 — Method: the greatest number of identical rows is the highest common factor of the two bulb totals, found by taking every prime factor the two totals share. Working: 42 = 2 × 3 × 7 and 56 = 2 × 2 × 2 × 7, so the prime factors common to both are 2 and 7, giving a highest common factor of 2 × 7 = 14. 2 comes from taking only the common factor 2 and forgetting the common factor 7. 7 comes from taking only the common factor 7 and forgetting the common factor 2. 168 is the lowest common multiple of 42 and 56, not their highest common factor. Answer: 14.
- (a) 30 cm — The tile's side length must be a common factor of 90 and 120. The factors of 90 include 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90; the factors of 120 include 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. The highest number common to both lists is 30, so the largest square tile has a side length of 30 cm. Picking 15 cm, a common factor but not the largest, gives tiles that are smaller than necessary. Picking 10 cm, also a common factor but smaller still, wastes even more of the possible tile size. Working out the lowest common multiple instead of the highest common factor gives 360 cm, a length far bigger than either side of the patio. So the largest square tile Ben can use has a side length of 30 cm.
- (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (c) 0.024 — Multiply the digits ignoring the decimal points: 6 × 4 = 24. Count the total number of decimal places in the two numbers being multiplied: 0.6 has 1 decimal place and 0.04 has 2, giving 3 in total. Place the decimal point in 24 so that there are 3 digits after it: 0.024. Counting only 2 decimal places instead of 3 gives 0.24. Counting 4 decimal places instead of 3 gives 0.0024. Counting only 1 decimal place instead of 3 — in effect moving the point in just one of the two numbers, as if the calculation were 6 × 0.4 — gives 2.4. So 0.6 × 0.04 = 0.024.
- (a) −22 — Method: both multiplications are carried out before the addition, and a positive multiplied by a negative is negative. Working: (−3) × 4 = −12 and 2 × (−5) = −10, so the calculation becomes −12 + (−10) = −22. Answer: −22. The distractors: 22 comes from ignoring the minus signs and working out 3 × 4 + 2 × 5 = 22; 50 comes from working from left to right with no priority at all, giving −12 + 2 = −10 and then −10 × (−5) = 50; −2 comes from taking 2 × (−5) as +10, so that −12 + 10 = −2.
- (b) 9/25 — Method: write the decimal over the matching power of ten, then divide the numerator and the denominator by their highest common factor. Working: 0.36 has two digits after the point, so 0.36 = 36/100; the highest common factor of 36 and 100 is 4, and 36 ÷ 4 = 9 with 100 ÷ 4 = 25; since 9 and 25 share no factor greater than 1, the fraction is fully cancelled. Answer: 9/25. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 9/50 comes from dividing the numerator by 4 but the denominator by only 2; 36/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
What is on this worksheet?
The sheet holds 20 questions drawn from the MathsUK bank — the content areas covered: Number, Ratio, proportion and rates of change (statements N2, N3, N4, N10, N12, N14, R4, R5, R9). It is pitched at GCSE Foundation and takes about 25 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
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 20 questions before checking — about 25 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 20 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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