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.
Number worksheet — GCSE Higher
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- 1.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.
- 2.The decimal 0.272727... repeats the block 27 for ever. Write 0.27 recurring as a fraction in its simplest form.
- 3.A plank has length a = 12 cm and a second plank has length b = 7 cm, each correct to the nearest centimetre. Work out the upper bound of a + b.
- 4.A florist has 60 red roses and 84 white roses. She wants to make identical bunches using all the flowers, with the greatest possible number of bunches. Work out how many red roses will be in each bunch.
- 5.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.
- 6.Write these fractions in order, starting with the smallest: 2/3, 3/5, 5/6, 1/2
- 7.A shape is made by joining a rectangle measuring 10 cm by 6 cm to a semicircle of diameter 6 cm along one of the rectangle's shorter sides. Work out the exact area of the shape, in terms of π.
- 8.Priya has 24 red beads and 36 blue beads. She makes identical bracelets, using every bead and with none left over. Work out the greatest number of bracelets she can make.
- 9.Work out 3³ + 2⁴.
- 10.A number is multiplied by 4, then 8 is added, giving the result 40. Work out the number.
- 11.Without using a calculator, estimate the value of √70 × ∛65, giving your answer to 1 significant figure.
- 12.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.
- 13.A three-digit code is made using the digits 1, 2 and 3, and each digit may be used more than once. Work out how many different three-digit codes can be made.
- 14.A company sells 4 × 10³ items per day, each priced at £2.50. It operates for 3 × 10² days a year. Work out the company's total revenue for the year. Give your answer in standard form.
- 15.Petrol costs £1.48 per litre. Sam buys 35 litres and pays with three £20 notes. Work out his change.
Answer key
- (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.
- (c) 3/11 — Method: let a letter stand for the recurring decimal, multiply by the power of ten that shifts exactly one repeating block past the point, subtract the original equation so that the recurring tail cancels, then solve and cancel. Working: let x = 0.272727...; the repeating block is two digits long, so multiply by 100 to give 100x = 27.272727...; subtracting gives 99x = 27, so x = 27/99; the highest common factor of 27 and 99 is 9, and 27 ÷ 9 = 3 with 99 ÷ 9 = 11. Answer: 3/11. The distractors: 27/100 comes from writing the repeating block over 100 instead of over 99, forgetting that subtracting x leaves 99x rather than 100x; 3/10 comes from rounding the decimal to one place and converting 0.3; 2/9 comes from treating only the 2 as recurring and converting 0.222... instead.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (d) 5 — Method: the greatest number of identical bunches is the highest common factor of the two flower totals; then divide the red roses by that number of bunches. Working: 60 = 2² × 3 × 5 and 84 = 2² × 3 × 7, so their highest common factor is 2² × 3 = 12. That means 12 bunches, and 60 ÷ 12 = 5 red roses in each. 7 is the number of white roses in each bunch, since 84 ÷ 12 = 7, not red roses. 12 is the number of bunches itself, not the number of red roses in one bunch. 20 comes from working out 60 ÷ 3 = 20, dividing by only part of the highest common factor. Answer: 5.
- (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.
- (a) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (b) (60 + 4.5π) cm² — Method: find the area of the rectangle and the area of the semicircle separately, then add them. Working: the rectangle has area 10 × 6 = 60 cm². The semicircle has radius 3 cm, so its area is half of π × 3² = half of 9π = 4.5π cm². Total area = (60 + 4.5π) cm². Answer: (60 + 4.5π) cm². (60 + 18π) cm² comes from using the diameter (6 cm) as the radius in the semicircle area formula: half of π × 6² = 18π. (60 + 9π) cm² comes from forgetting to halve the full circle's area: π × 3² = 9π. (60 + 3π) cm² comes from finding the semicircle's arc length instead of its area: half of 2 × π × 3 = 3π.
- (d) 12 — Method: if the bracelets are identical and no beads are left over, the number of bracelets must divide exactly into both totals, so it is the highest common factor of 24 and 36. Working: 24 = 2³ × 3 and 36 = 2² × 3²; taking the lower index of each shared prime gives 2² × 3 = 4 × 3 = 12. Each bracelet then has 2 red beads and 3 blue beads. Answer: 12. The distractors: 6 comes from taking each shared prime once rather than at its lower index, giving 2 × 3, which is a common factor but not the highest; 72 is the lowest common multiple of 24 and 36, from taking the higher index of each prime instead of the lower; 60 comes from adding the two bead totals instead of looking for a common factor.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (a) 8 — To undo 'multiply by 4, then add 8', reverse the operations in reverse order: subtract 8 first, then divide by 4. 40 − 8 = 32, and 32 ÷ 4 = 8, so the number is 8. A candidate who added 8 again instead of subtracting worked out 40 + 8 = 48, then 48 ÷ 4 = 12. A candidate who divided before subtracting, doing the inverse operations in the wrong order, worked out 40 ÷ 4 = 10, then 10 − 8 = 2. A candidate who multiplied instead of dividing at the last step worked out (40 − 8) × 4 = 32 × 4 = 128.
- (b) 30 — 70 is close to the perfect square 64, so √70 ≈ 8. 65 is close to the perfect cube 64, so ∛65 ≈ 4. Multiplying these estimates: 8 × 4 = 32, which rounds to 30 to 1 significant figure. Estimating ∛65 as 5 instead of 4, perhaps by confusing it with the nearby cube 125 = 5³ rather than the much closer 64 = 4³, and then multiplying by 8, gives 8 × 5 = 40. Adding the two estimates instead of multiplying them, 8 + 4 = 12, rounds to 10 to 1 significant figure. Rounding both estimates up to the next whole number using the wrong nearby power for each, taking √70 as 9 and ∛65 as 5, gives 9 × 5 = 45, which rounds to 50 to 1 significant figure.
- (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.
- (c) 27 — Each of the 3 digits can be chosen independently for each of the 3 positions, so multiply: 3 × 3 × 3 = 27. 9 comes from multiplying only two of the three positions, 3 × 3, and forgetting the third. 6 comes from working out 3 × 2 × 1 = 6, which counts codes where digits do not repeat, but the question allows repeated digits. 3 comes from considering only one digit position.
- (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) £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.)
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