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.Jamal invests £600 in a savings account paying 3% simple interest per year. Work out the total amount in the account after 4 years.
- 2.Write 0.35 as a fraction in its lowest terms.
- 3.Find the missing number: ▢ ÷ 15 = 24
- 4.Work out (−3) × 4 + 2 × (−5)
- 5.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 6.Two lighthouses flash at the start of the same minute. The first lighthouse flashes every 8 minutes and the second flashes every 12 minutes. Work out how many minutes it will be until they next flash together.
- 7.The angles of a triangle are in the ratio 2:3:4. Work out the size of the smallest angle.
- 8.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 9.Work out −4.5 + 2.8.
- 10.A sponsored walk raised £350 for charity. 20% of the money raised is spent on equipment. Work out how much is spent on equipment.
- 11.Work out 250 ÷ 1000.
- 12.Work out (−6) + (−4) × 3
- 13.The number 24 can be written as 2³ × 3, and the number 60 can be written as 2² × 3 × 5. Work out the lowest common multiple of 24 and 60.
- 14.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.
- 15.Jack buys 3 books, each costing £4.25, and pays with a £20 note. Work out how much change he receives.
- 16.Work out √25 + 4² − 12 ÷ 3
- 17.For any two whole numbers, the product of the numbers is equal to the product of their highest common factor and their lowest common multiple. The highest common factor of 6 and 8 is 2, and 6 × 8 = 48. Work out the lowest common multiple of 6 and 8.
- 18.Work out 20 − 8 ÷ 2 + 1
- 19.Two investors put money into a business in the ratio 3:5. The first investor puts in £1,200. Work out the total amount invested by both investors.
- 20.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
Answer key
- (c) £672 — Simple interest per year = 3% of £600 = £18. Over 4 years the interest is 18 × 4 = £72. Total in the account = £600 + £72 = £672. A student who gives just the interest, without adding it to the principal, writes £72. A student who adds only one year's interest instead of four gets £600 + £18 = £618. A student who wrongly compounds the interest each year gets 600 × 1.03⁴ = £675.31.
- (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (a) 360 — The inverse of ÷ 15 is × 15, so the missing number is 24 × 15 = 360. Subtracting instead of multiplying gives 24 − 15 = 9. Dividing by 15 again instead of multiplying gives 24 ÷ 15 = 1.6. Adding instead of multiplying gives 24 + 15 = 39.
- (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) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (b) 24 — List multiples of 8 and of 12: multiples of 8 are 8, 16, 24, 32; multiples of 12 are 12, 24, 36. The lowest number in both lists is 24, so the lighthouses next flash together after 24 minutes. Multiplying the two numbers together, 8 × 12, gives 96, which double-counts the common factor of 4 shared by 8 and 12. Working out the highest common factor instead of the lowest common multiple gives 4, far too soon a time for both lighthouses to line up again. Adding the two numbers, 8 + 12, gives 20, which is not even a multiple of either 8 or 12. So the lighthouses next flash together after 24 minutes.
- (c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (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.
- (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.
- (d) 0.25 — Dividing by 1000 moves every digit three place-value columns, so 250 ÷ 1000 = 0.25. A candidate who divides by 100 instead of 1000 gets 2.5. A candidate who divides by 10,000 instead of 1000 gets 0.025. A candidate who divides by 10 instead of 1000 gets 25.
- (b) −18 — Method: the multiplication is carried out before the addition, and a negative multiplied by a positive is negative. Working: (−4) × 3 = −12, so the calculation becomes (−6) + (−12) = −18. Answer: −18. The distractors: −30 comes from adding first and multiplying afterwards, giving (−6 + −4) × 3 = −10 × 3 = −30; 6 comes from treating (−4) × 3 as +12 on the grounds that a minus sign makes a product positive, giving −6 + 12 = 6; 18 comes from ignoring both minus signs and working out 6 + 4 × 3 = 18.
- (a) 120 — For the lowest common multiple, take each prime that appears in either factorisation, raised to the higher power. In 2³ × 3 and 2² × 3 × 5, the prime 2 appears with power 3 in one and power 2 in the other — take the higher, 2³; the prime 3 appears with the same power in both, 3¹; and the prime 5 appears only in the second factorisation, so use 5¹. Multiplying these, 2³ × 3 × 5, gives 120. Taking the lower power of 2 instead of the higher, and leaving out 5 altogether, gives the highest common factor, 12, instead. Multiplying the two original numbers together, 24 × 60, gives 1440, which double-counts every shared prime factor. Assuming the lowest common multiple is simply the larger of the two numbers gives 60, but 60 is not a multiple of 24 — 60 ÷ 24 does not divide exactly. So the lowest common multiple of 24 and 60 is 120.
- (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.
- (c) £7.25 — Find the total cost of the books first: 3 × 4.25 = 12.75, so the books cost £12.75 in total. Subtract this from the £20 note: 20.00 − 12.75 = 7.25, so the change is £7.25. Stopping after finding the cost and not subtracting it from £20 gives £12.75, which is the amount spent, not the change. Borrowing correctly in the pence column but forgetting to reduce the pounds column by 1 gives £8.25 instead of £7.25. Multiplying 3 × 4.25 as 12.25 instead of 12.75, a multiplication slip, makes the change come out £0.50 too high, at £7.75. So Jack receives £7.25 change.
- (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
- (d) 17 — 8 ÷ 2 = 4, then 20 − 4 = 16, then 16 + 1 = 17. Stopping after the subtraction and forgetting to add the final 1 leaves 16. Adding the 4 and the 1 together before subtracting gives 4 + 1 = 5, then 20 − 5 = 15 — the subtraction should use the 4 from the division, not a combined total. Working strictly left to right without giving division priority gives 20 − 8 = 12, then 12 ÷ 2 = 6, then 6 + 1 = 7.
- (c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
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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