Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Number worksheet — GCSE Higher
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- 1.A car is hired from 10:00 on Monday until 16:00 on Thursday. Work out the total length of the hire, in hours.
- 2.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 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.A cheetah runs at a steady speed of 25 metres per second. Work out this speed in kilometres per hour.
- 5.The density of a type of plastic is 0.9 g/cm³. Work out this density in kg/m³.
- 6.A cube-shaped storage tank has a volume of 15,625 cubic centimetres. Work out the length of one edge of the tank.
- 7.Work out the value of .
- 8.Maya buys 4 plants at £3.20 each and a bag of compost for £6.75. She pays with a £20 note. Work out her change.
- 9.A cube-shaped storage box has edges of length 7 cm. A shelf can hold a total volume of 2,000 cm³. Work out the greatest number of these boxes that will fit in that volume.
- 10.The rainfall in a town during April is recorded as 62.4 mm, correct to 1 decimal place. Write down the error interval for the actual rainfall, r mm.
- 11.A company's turnover is 5.6 × 10⁷ pounds, spread evenly across 3.5 × 10² shops. Work out the average turnover per shop, in standard form.
- 12.The diameter of an artificial silk fibre is 4 × 10⁻⁶ metres. One nanometre is 10⁻⁹ metres. Work out the diameter of the fibre in nanometres.
- 13.A driving instructor is working out how many current-style UK number plates are possible. The format is 2 letters, then 2 digits, then a space, then 3 letters (for example AB12 CDE). The letters I, O and Q are never used in any letter position, leaving 23 allowed letters, but every letter and every digit may repeat anywhere on the plate. Work out how many different number plates are possible, giving your answer in standard form to 2 significant figures.
- 14.A school council must choose a committee of 3 pupils from 8 volunteers. The three places on the committee are all the same, so only which pupils are chosen matters. Work out how many different committees could be formed.
- 15.The radius of a circular pond is given as 3.2 m, correct to 1 decimal place. Calculate the upper bound for the area of the pond, giving your answer correct to 3 significant figures.
Answer key
- (b) 78 hours — From 10:00 on Monday to 10:00 on Thursday is exactly 3 complete days, which is 3 × 24 = 72 hours. From 10:00 to 16:00 on the Thursday is a further 6 hours, giving a total of 72 + 6 = 78 hours. Counting Monday to Thursday as 4 full calendar days instead of 3 complete 24-hour periods gives 4 × 24 = 96 hours. Undercounting the number of complete days as 2 instead of 3 gives 2 × 24 + 6 = 54 hours. Subtracting the extra 6 hours instead of adding them to the 3 complete days gives 72 − 6 = 66 hours.
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (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.
- (b) 90 km/h — To convert metres per second to kilometres per hour, multiply by 3.6 (there are 3600 seconds in an hour and 1000 metres in a kilometre, and 3600 ÷ 1000 = 3.6): 25 × 3.6 = 90 km/h. Dividing by 3.6 instead of multiplying gives 25 ÷ 3.6 = 6.9 km/h (to 1 d.p.). Multiplying by 60 instead of 3.6, confusing the conversion from seconds to minutes with the conversion to hours, gives 25 × 60 = 1500 km/h. Multiplying by 3600 to convert seconds to hours but forgetting to convert metres to kilometres gives 25 × 3600 = 90000, which is a speed in metres per hour, not kilometres per hour.
- (a) 900 kg/m³ — Both parts of the compound unit change. There are 100 cm in a metre, so 1 m³ = 100³ = 1 000 000 cm³, and there are 1000 g in a kilogram. So 1 g/cm³ = 1 000 000 ÷ 1000 = 1000 kg/m³, and the plastic is 0.9 × 1000 = 900 kg/m³. 900 000 kg/m³ converts the volume but leaves the mass in grams, 0.0009 kg/m³ converts the mass but leaves the volume in cm³, and 0.09 kg/m³ uses the 100 cm in a metre without cubing it.
- (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.
- (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (c) £0.45 — Method: find the total cost, then subtract from £20. Working: 4 × £3.20 = £12.80. £12.80 + £6.75 = £19.55. Change = £20.00 − £19.55 = £0.45. Answer: £0.45. (£7.20 comes from forgetting to include the compost and subtracting only the plants' cost from £20. £1.45 comes from dropping the carry when adding the pence: 80p + 75p = £1.55, but only the 55p is written down, giving £18.55 instead of £19.55. £10.05 comes from buying only one plant instead of four, using £3.20 + £6.75.)
- (c) 5 — Method: work out the volume of one box, divide the total volume by it, then round down since a partial box cannot fit. Working: volume of one box = 7³ = 343 cm³. 2000 ÷ 343 = 5.83 (2 d.p.). Since only whole boxes fit, the greatest number is 5. Answer: 5. (6 comes from rounding 5.83 up to the nearest whole number instead of rounding down to the number of boxes that actually fit. 343 comes from giving the volume of one box instead of the number of boxes. 5.8 comes from leaving the division as a decimal instead of rounding down to a whole number of boxes.)
- (c) 62.35 ≤ r < 62.45 — Method: with a value rounded to 1 decimal place, the error interval reaches half of 0.1 either side. Working: half of 0.1 is 0.05, so the interval runs from 62.4 − 0.05 to 62.4 + 0.05. Answer: 62.35 ≤ r < 62.45. (62 ≤ r < 63 comes from rounding to the nearest whole number instead of 1 decimal place. 62.35 ≤ r ≤ 62.45 comes from including the upper bound with ≤ instead of excluding it with <. 62.3 ≤ r < 62.5 comes from using 0.1 either side instead of half of it.)
- (d) 1.6 × 10⁵ — 5.6 ÷ 3.5 = 1.6, and 7 − 2 = 5, so the average turnover per shop is 1.6 × 10⁵ pounds. Multiplying the exponents instead of subtracting them gives 7 × 2 = 14, so 1.6 × 10¹⁴. Adding the exponents instead of subtracting them gives 7 + 2 = 9, so 1.6 × 10⁹. Subtracting the coefficients instead of dividing them gives 5.6 − 3.5 = 2.1, so 2.1 × 10⁵.
- (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
- (b) 6.4 × 10⁸ — There are 5 letter positions, each with 23 choices, and 2 digit positions, each with 10 choices, and every position is independent because repeats are allowed. By the product rule, the total is 23⁵ × 10² = 6,436,343 × 100 = 643,634,300, which is 6.4 × 10⁸ to 2 significant figures. Using all 26 letters instead of the 23 that are actually allowed, ignoring the excluded letters entirely, gives 26⁵ × 10² = 1,188,137,600, which is 1.2 × 10⁹ to 2 significant figures. Adding the seven counts of choices instead of multiplying them, 23 + 23 + 10 + 10 + 23 + 23 + 23, gives 135, which is 1.4 × 10² to 2 significant figures — a total far too small for seven independent positions. Swapping which count of choices belongs to letters and which belongs to digits, working out 23² × 10⁵ instead of 23⁵ × 10², gives 52,900,000, which is 5.3 × 10⁷ to 2 significant figures.
- (a) 56 — Method: count the ordered selections with the product rule first, then divide by the number of different orders in which any one committee could have been picked. Working: there are 8 choices for a first pupil, 7 for a second and 6 for a third, giving 8 × 7 × 6 = 336 ordered selections; any particular three pupils could have been picked in 3 × 2 × 1 = 6 orders, so the number of different committees is 336 ÷ 6 = 56. Answer: 56. The distractors: 336 comes from stopping at 8 × 7 × 6 and treating the three places as distinct posts when they are identical; 168 comes from dividing that product by 2 rather than by the 6 orders in which three chosen pupils can be listed; 24 comes from multiplying the 8 volunteers by the 3 places instead of multiplying the choices at each stage.
- (a) 33.2 — The radius was rounded to 1 decimal place, so its error interval is 3.15 ≤ r < 3.25. The upper bound for the area uses the upper bound of the radius, squared: area = π × 3.25² ≈ 33.183, which rounds to 33.2 m² (3 s.f.). Using the given value of the radius directly instead of its upper bound, π × 3.2² ≈ 32.2, ignores that the radius itself has a range of possible values. Bounding the radius correctly but forgetting to square it, using area = π × 3.25 ≈ 10.2 instead of π × 3.25², drops the whole squaring step from the area formula. Using the LOWER bound of the radius instead of the upper one, π × 3.15² ≈ 31.2, finds the lower bound of the area, not the upper one.
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