Printable · GCSE Foundation · ages 14-16
Number worksheet — GCSE Foundation
Fifteen questions across the number statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Non-calculator
Number worksheet — GCSE Foundation
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- 1.Write down the decimal that is equal to 3/5.
- 2.A machine fills bags of sugar and shows the mass of each bag to the nearest 10 g. A checker rejects any bag whose actual mass is less than 996 g. One bag shows a mass of 1,000 g on the machine. Decide whether this bag could be rejected, and give a reason for your answer.
- 3.By listing systematically, work out how many two-digit multiples of 5 can be made using the digits 0, 3 and 5, if each digit can be used at most once and the number cannot start with 0.
- 4.A student writes 0.08 as the fraction 8/10, reading the 8 as if it stood in the tenths column and ignoring the zero. Work out the correct fraction that 0.08 is equal to, giving your answer in its simplest form.
- 5.Grace buys three items costing £9.95, £19.90 and £4.99. Work out an estimate for the total cost, by rounding each price to the nearest pound.
- 6.Work out 5 + 3 × (9 − 6)
- 7.Work out the value of √49 + ∛27
- 8.A jug holds 1.8 litres of juice. Priya pours out 3/4 of a litre to fill a glass. Work out how much juice is left in the jug, in millilitres.
- 9.Work out 3 + 4 × 2²
- 10.Work out an estimate for 6.8 × 41, by rounding each number to 1 significant figure.
- 11.A spinner has 4 colours and a coin has 2 sides. Work out how many different outcomes are possible when the spinner is spun once and the coin is tossed once.
- 12.A plumber charges a call-out fee of £45 and then £28 for each hour she works. She is at a house from 09:15 until 12:45. Work out the total charge.
- 13.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 14.Work out 3.7 × 24.
- 15.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
Answer key
- (b) 0.6 — Method: convert the fraction to an equivalent fraction with denominator 10, then read off the decimal. Working: 3/5 = 6/10 (multiplying numerator and denominator by 2) = 0.6. Answer: 0.6. 0.35 comes from combining the digits 3 and 5 directly after the decimal point instead of converting the fraction. 0.53 comes from writing the numerator and denominator digits in the wrong order. 1.67 comes from flipping the fraction to 5/3 before converting to a decimal.
- (b) Yes — the actual mass could be as low as 995 g — Method: a mass shown to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure on the display, so compare the smallest mass the bag can have with the checker's limit of 996 g. Working: 1,000 − 5 = 995, so the actual mass of the bag can be as low as 995 g, and 995 g is below the 996 g limit, so a bag showing 1,000 g on the machine can still be rejected. Answer: Yes — the actual mass could be as low as 995 g. The distractors: 990 g comes from going a whole 10 g below the display instead of half of it; 999.5 g comes from treating the display as being to the nearest gram, when it is to the nearest 10 g; the claim that the mass is exactly 1,000 g treats a rounded display as an exact measurement.
- (b) 3 — Method: list all valid two-digit numbers that can be made without starting with 0, then keep only the ones that are multiples of 5. Working: the two-digit numbers possible are 30, 35, 50 and 53. A number is a multiple of 5 only if it ends in 0 or 5: 30 ends in 0, 35 ends in 5, 50 ends in 0, but 53 ends in 3. So there are 3 multiples of 5. Answer: 3. 4 comes from including 53 as a multiple of 5 without checking that its last digit is not 0 or 5. 2 comes from leaving out 50, wrongly assuming 0 cannot be used as the second digit either. 6 comes from listing every two-digit arrangement of the three digits, including ones that start with 0, without applying either restriction.
- (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
- (b) £35 — Method: round each price to the nearest pound, then add the rounded prices. Working: £9.95 rounds to £10, £19.90 rounds to £20 and £4.99 rounds to £5, and £10 + £20 + £5 gives the estimate. Answer: £35. The distractors: £40 comes from rounding each price up to the nearest £10 rather than to the nearest pound, giving £10 + £20 + £10; £32 comes from cutting the pence off each price instead of rounding it, giving £9 + £19 + £4; £34.84 is the exact total, worked out in full when the question asks for an estimate.
- (a) 14 — Brackets first: 9 − 6 = 3. Then multiply: 3 × 3 = 9. Then add: 5 + 9 = 14. So the answer is 14. A candidate who worked out (5 + 3) × (9 − 6) = 8 × 3 = 24 added before multiplying, ignoring the priority of operations outside the bracket. A candidate who dropped the brackets and worked out 5 + 3 × 9 − 6 = 5 + 27 − 6 = 26 multiplied by the 9 itself instead of by the bracket's value of 3, losing the grouping the brackets give. A candidate who forgot to add the 5 and only worked out 3 × (9 − 6) = 3 × 3 = 9 dropped a term from the calculation.
- (a) 10 — √49 = 7 and ∛27 = 3, so √49 + ∛27 = 7 + 3 = 10. Treating the cube root as dividing by 3 instead of finding the cube root gives 27 ÷ 3 = 9, then 7 + 9 = 16. Multiplying the two roots instead of adding them gives 7 × 3 = 21. Ignoring the cube root symbol and using 27 as it stands gives 7 + 27 = 34.
- (d) 1050 ml — Convert both amounts to millilitres: 1.8 litres = 1800 ml and 3/4 litre = 750 ml. Subtracting gives 1800 − 750 = 1050 ml. Confusing 3/4 with 75% and converting it as 75 ml instead of 750 ml gives 1800 − 75 = 1725 ml. Adding the amount poured out instead of subtracting it gives 1800 + 750 = 2550 ml. Misreading 1.8 litres as 0.8 litres, losing the whole litre, gives 800 − 750 = 50 ml.
- (d) 19 — 2² = 4, then 4 × 4 = 16, then 3 + 16 = 19. Adding before multiplying gives 3 + 4 = 7, then 7 × 4 = 28 — multiplication comes before addition. Squaring the product instead of just the 2 gives 4 × 2 = 8, then 8² = 64, then 3 + 64 = 67. Working strictly left to right throughout gives 3 + 4 = 7, then 7 × 2 = 14, then 14² = 196.
- (c) 280 — Method: round each number to 1 significant figure, then multiply the rounded numbers. Working: 6.8 rounds to 7 because the next digit is 8, and 41 rounds to 40 because its next digit is 1, so the estimate is 7 × 40 = 280. Answer: 280. The distractors: 240 comes from cutting 6.8 down to 6 instead of rounding it up to 7, giving 6 × 40 = 240; 350 comes from rounding 41 up to 50 when the digit after its first significant figure is less than 5, giving 7 × 50 = 350; 28 comes from multiplying the leading digits only and losing the place value of the 40, which makes the estimate ten times too small.
- (c) 8 — Each of the 4 spinner outcomes can be paired with each of the 2 coin outcomes, so multiply: 4 × 2 = 8. 6 comes from adding the two numbers instead of multiplying them. 4 comes from using only the spinner outcomes and ignoring the coin. 2 comes from using only the coin outcomes and ignoring the spinner.
- (d) £143 — From 09:15 to 12:45 is 3 hours 30 minutes, which is 3.5 hours. The charge for the time is 3.5 × £28 = £98, and the call-out fee is added on once: £98 + £45 = £143. £98 leaves out the call-out fee altogether, £129 charges for 3 hours instead of 3.5 hours, and £255.50 adds the fee to the hourly rate before multiplying, which charges the £45 for every hour.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (b) 88.8 — Multiply as whole numbers first, ignoring the decimal point: 37 × 24. Split it as 37 × 20 = 740 and 37 × 4 = 148, so 37 × 24 = 740 + 148 = 888. 3.7 has 1 decimal place and 24 has none, so the answer needs 1 decimal place: 88.8. Counting the 2 digits in "3.7" as though that were the number of decimal places gives 8.88 instead of 1 decimal place. Leaving the decimal point out altogether gives 888. Misreading 37 × 4 as 138 rather than 148 gives a running total of 878, placed with 1 decimal place as 87.8. So 3.7 × 24 = 88.8.
- (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
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