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 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.
- 2.The density of a type of plastic is 0.9 g/cm³. Work out this density in kg/m³.
- 3.Petrol costs £1.48 per litre. Sam buys 35 litres and pays with three £20 notes. Work out his change.
- 4.A jumper is reduced by 15% in a sale to a price of £42.50. Work out the original price.
- 5.Round 0.006482 to 2 significant figures.
- 6.A charity raises money from a raffle and a cake sale in the ratio 5 : 3. Altogether the charity raises £320. Work out how much money the cake sale raised.
- 7.Work out (3 × 10²) × (2 × 10⁵). Give your answer in standard form.
- 8.The thickness of a sheet of card is 0.02384 cm. Write this thickness correct to 2 significant figures.
- 9.A gym increases its membership price by 1/5. The original price is £60. Work out the new price.
- 10.Rice is sold in a 400 g bag for £1.12 and in a 1.5 kg bag for £3.90. Work out how much less the rice in the larger bag costs per kilogram.
- 11.A train journey takes 45 minutes, correct to the nearest 5 minutes. Using t for the actual time of the journey in minutes, write down the error interval for t.
- 12.A van has a mass of 2,000 kg, correct to 1 significant figure. Using m for the mass of the van in kilograms, write down the error interval for m.
- 13.A netball team has 8 players. Two of them are chosen to be captains, and the two captains have equal standing. Work out how many different pairs of captains could be chosen.
- 14.Write these numbers in order, starting with the smallest: −1.4, 5/4, −6/5, 1.3, 0
- 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
- (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) 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) £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.)
- (d) £50.00 — The sale price is 85% of the original, so the original price = £42.50 ÷ 0.85 = £50.00. 15% of £42.50 is £6.375. A candidate who finds 15% of £42.50 and subtracts it from the sale price gets £42.50 − £6.375 = £36.125, which is £36.13 to the nearest penny. A candidate who adds 15% of £42.50 instead of reversing the decrease gets £42.50 + £6.375 = £48.875, which is £48.88 to the nearest penny. A candidate who divides by 0.15 instead of 0.85 gets £283.33.
- (c) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
- (d) £120 — Total parts = 5 + 3 = 8, so one part is worth 320 ÷ 8 = 40 pounds. The cake sale is 3 parts, so it raised 3 × 40 = 120 pounds. £200 comes from working out the raffle's share, 5 × 40, instead of the cake sale's share. £40 comes from finding the value of one part but forgetting to multiply by 3. £192 comes from dividing the total by 5 instead of 8 to find the value of one part, 320 ÷ 5 = 64, then multiplying by 3, 3 × 64 = 192.
- (d) 6 × 10⁷ — 3 × 2 = 6, and 2 + 5 = 7, so (3 × 10²) × (2 × 10⁵) = 6 × 10⁷. Multiplying the exponents instead of adding them gives 2 × 5 = 10, so 6 × 10¹⁰. Adding the coefficients instead of multiplying them gives 3 + 2 = 5, so 5 × 10⁷. Subtracting the exponents instead of adding them gives 5 − 2 = 3, so 6 × 10³.
- (d) 0.024 cm — Method: significant figures are counted from the first non-zero digit; the zeros in front of it only fix the place value and are not significant. Working: in 0.02384 the first significant figure is 2 and the second is 3, so the rounding is decided by the next digit, 8. As 8 is 5 or more, the second significant figure goes up from 3 to 4, in the same place value. Answer: 0.024 cm. The distractors: 0.023 cm comes from chopping the 84 off instead of rounding it; 0.02 cm comes from counting the leading zeros as significant figures, so the 2 is taken as the second figure and the rounding stops there; 0.0238 cm is 0.02384 correct to 3 significant figures, one figure too many.
- (b) £72 — One fifth of £60 = £12. New price = £60 + £12 = £72. A candidate who gives the increase instead of the new price gets £12. A candidate who subtracts the increase instead of adding it gets £60 − £12 = £48. A candidate who uses 1/4 instead of 1/5 gets £60 + £15 = £75.
- (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.
- (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.
- (a) 1,500 ≤ m < 2,500 — Method: a four-digit figure written to 1 significant figure has been rounded to the nearest 1,000, so the mass lies within half of 1,000, that is 500, of the figure given. Working: 2,000 − 500 = 1,500 and 2,000 + 500 = 2,500. The lower limit is included, because 1,500 kg rounds up to 2,000 kg to 1 significant figure, while 2,500 kg rounds up to 3,000 kg, so the upper limit is not. Answer: 1,500 ≤ m < 2,500. The distractors: 1,950 ≤ m < 2,050 comes from rounding to the nearest 100 instead of to 1 significant figure; 1,000 ≤ m < 3,000 goes a whole 1,000 either side instead of half of it; 1,500 < m ≤ 2,500 has the two limits the wrong way round.
- (b) 28 — Method: count the ordered choices with the product rule and then correct for the double counting, because the two captains have equal standing and so a pair is the same pair whichever captain is named first. Working: there are 8 players who could be named first and 7 who could be named second, giving 8 × 7 = 56 ordered choices; each pair has been counted twice, once in each order, so the number of pairs is 56 ÷ 2 = 28. Answer: 28. The distractors: 56 comes from stopping at 8 × 7 and never halving, which counts each pair of captains twice; 64 comes from working out 8 × 8, which allows the same player to be chosen as both captains; 16 comes from multiplying the 8 players by the 2 captaincies instead of pairing the players with one another.
- (d) −1.4, −6/5, 0, 5/4, 1.3 — Method: convert the fractions 5/4 and −6/5 to decimals so every number is written the same way, then compare all five decimals. Working: 5/4 = 1.25 and −6/5 = −1.2. Comparing −1.4, −1.2, 0, 1.25 and 1.3 in size gives the order −1.4, −1.2, 0, 1.25, 1.3. Answer: −1.4, −6/5, 0, 5/4, 1.3. −6/5, −1.4, 0, 5/4, 1.3 swaps the two negative numbers, treating −6/5 as more negative than −1.4 even though −1.2 is closer to zero than −1.4. 1.3, 5/4, 0, −6/5, −1.4 lists the numbers from largest to smallest instead of smallest to largest. −1.4, −6/5, 0, 1.3, 5/4 swaps 5/4 and 1.3, comparing the numerator 5 directly with 1.3 instead of converting 5/4 to the decimal 1.25 first.
- (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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