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.
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Answer key: Number worksheet — GCSE Foundation
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- (c) 20p — Turn each price into the same rate before comparing. The small bag is 400 g = 0.4 kg, so it costs £1.12 ÷ 0.4 = £2.80 per kg. The large bag costs £3.90 ÷ 1.5 = £2.60 per kg. The saving is £2.80 − £2.60 = £0.20, which is 20p per kg. 2p compares the prices per 100 g rather than per kilogram, £2.78 subtracts one bag price from the other without turning either into a rate, and £2.60 is the large bag's price per kilogram rather than the saving.
- (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.
- (c) £8.20 — Method: find the total cost, then subtract from the amount paid. Working: 35 × £1.48 = £51.80. Amount paid = 3 × £20 = £60.00. Change = £60.00 − £51.80 = £8.20. Answer: £8.20. (£58.52 comes from forgetting to multiply the price by the 35 litres and subtracting only £1.48 from £60. £7.50 comes from rounding £1.48 up to £1.50 before multiplying, giving a total of £52.50 instead of £51.80. £9.20 comes from miscarrying in the pence column when subtracting £51.80 from £60.00.)
- (a) 28.8 km/h — A compound unit is converted one part at a time. There are 3600 seconds in an hour, so in one hour the cyclist travels 8 × 3600 = 28 800 metres. There are 1000 metres in a kilometre, so 28 800 m = 28 800 ÷ 1000 = 28.8 km/h. 28 800 km/h leaves the distance in metres, 0.48 km/h converts the seconds to minutes rather than to hours, and 2.22 km/h divides by 3.6 instead of multiplying.
- (a) 3.8 × 10⁻³ — 0.0038 is less than 1, so the power of 10 is negative. Moving the decimal point 3 places gives A = 3.8, so 0.0038 = 3.8 × 10⁻³. A candidate who wrote 3.8 × 10³ used a positive power, which is only correct for numbers of 10 or more. A candidate who wrote 38 × 10⁻⁴ used a value of A outside the required range. A candidate who wrote 3.8 × 10⁻⁴ counted one place too many when moving the decimal point.
- (a) 1,850 — Method: find the error interval, then check which value falls outside it. Working: half of 100 is 50, so the actual number of visitors, v, satisfies 1,750 ≤ v < 1,850. 1,850 sits exactly on the excluded upper boundary, since a value of 1,850 would round to 1,900, not 1,800. Answer: 1,850. (1,750 is a genuine possible value — it sits at the included lower boundary. 1,799 is a genuine possible value, just below the upper boundary. 1,760 is a genuine possible value, well inside the interval.)
- (a) 42.5 ≤ t < 47.5 — Rounding to the nearest 5 minutes means the actual time can be up to half of 5 minutes, 2.5 minutes, below or above 45 before it would round to a different multiple of 5. The lower bound is 45 − 2.5 = 42.5 and the upper bound is 45 + 2.5 = 47.5. A time of exactly 47.5 minutes would round up to 50, not 45, so 47.5 is excluded while 42.5 does still round to 45. Writing 42.5 ≤ t ≤ 47.5 wrongly includes 47.5. Writing 40 ≤ t < 50 uses a whole rounding unit, 5, either side instead of half of it. Writing 44.5 ≤ t < 45.5 treats the rounding unit as 1 minute instead of 5 minutes.
- (d) £70 — Method: the discount is a percentage of the original price only, so work it out, subtract it, and add the fixed delivery charge afterwards. Working: 25% of £80 is £80 ÷ 4 = £20, so the discounted price is £80 − £20 = £60, and the total is £60 + £10 = £70. Answer: £70. The distractors: £60 comes from working out the discounted price and stopping there, leaving the delivery charge out of the total; £65 comes from taking £25 off the price instead of 25% of it, giving £80 − £25 = £55 and then £55 + £10 = £65; £67.50 comes from adding the delivery charge before the discount and reducing the whole amount, giving 75% of £90 = £67.50.
- (b) £300 — The increased price is 110% of the original, so the original price = £330 ÷ 1.1 = £300. A candidate who finds 10% of £330 and subtracts it, wrongly treating £330 as the original, gets £330 − £33 = £297. A candidate who adds 10% of £330 again instead of reversing the increase gets £330 + £33 = £363. A candidate who divides by 0.1 instead of 1.1 gets £3,300.
- (c) 4 × 10⁴ — 8 ÷ 2 = 4, and 6 − 2 = 4, so each project receives 4 × 10⁴ pounds. Multiplying the exponents instead of subtracting them gives 6 × 2 = 12, so 4 × 10¹². Adding the exponents instead of subtracting them gives 6 + 2 = 8, so 4 × 10⁸. Subtracting the coefficients instead of dividing them gives 8 − 2 = 6, so 6 × 10⁴.
- (c) 3 × 10⁶ — Daily revenue = (4 × 10³) × 2.5 = 1 × 10⁴ (£10,000). Multiplying by the number of days, (3 × 10²), gives annual revenue = (1 × 10⁴) × (3 × 10²) = 3 × 10⁶ (£3,000,000). A candidate who forgot to multiply by the price and just multiplied the number of items by the number of days worked out (4 × 10³) × (3 × 10²) = 1.2 × 10⁶. A candidate who added the number of days to the daily revenue instead of multiplying worked out 1 × 10⁴ + 3 × 10² = 1.03 × 10⁴. A candidate who misread £2.50 as £25 worked out a daily revenue of (4 × 10³) × 25 = 1 × 10⁵, giving an annual total of (1 × 10⁵) × (3 × 10²) = 3 × 10⁷.
- (c) 2.6 × 10⁵ — In standard form, A must satisfy 1 ≤ A < 10. Moving the decimal point in 260,000 to just after the 2 gives A = 2.6, and the decimal point moved 5 places, so 260,000 = 2.6 × 10⁵. A candidate who wrote 26 × 10⁴ used a value of A outside the required range, even though it has the same overall value. A candidate who wrote 2.6 × 10⁶ counted one place too many when moving the decimal point. A candidate who wrote 0.26 × 10⁶ used a value of A below 1, again outside the required range.
- (c) £384.00 — Adding 20% VAT means multiplying the price by 1.2: £320 × 1.2 = £384.00. Treating the 20% as a flat £20 rather than a percentage of the price, £320 + £20, gives £340.00. Working out the VAT amount alone, £320 × 0.2 = £64.00, and stopping there without adding it back to the original price gives just the VAT, not the total price. Misplacing the decimal point and using 2% instead of 20%, £320 × 1.02, gives £326.40.
- (c) 25 cm — Method: for a cube, the edge length is the cube root of the volume. Working: 25 × 25 × 25 = 15,625, so the edge length is 25 cm. 5 cm comes from cube-rooting 125 instead of 15,625, misreading the number of digits. 50 cm comes from working out 25 × 2 = 50, doubling the correct edge length. 125 cm comes from taking the square root of the volume instead of the cube root, since 125 × 125 = 15,625 — that would be the side of a SQUARE of area 15,625, not the edge of a cube of that volume. Answer: 25 cm.
- (c) £47.00 — First apply the 20% reduction: £65 × 0.8 = £52.00. Then take off the further £5: £52.00 − £5 = £47.00. Treating the 20% as a flat £20 rather than a percentage of the price, £65 − £20 − £5, gives £40.00. Applying the 20% reduction correctly but forgetting to take off the extra £5 leaves £52.00. Taking off the £5 first and then applying the 20% reduction to the smaller amount, (£65 − £5) × 0.8, gives £48.00.
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