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.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.
- 2.Simplify (5² × 2³)³ ÷ (5³ × 2²)²
- 3.The highest common factor of two numbers is 4 and their lowest common multiple is 60. One of the numbers is 20. Work out the other number.
- 4.A 3-character PIN starts with one letter chosen from A, B, C, D and E, followed by two different digits chosen from 1 to 9 (no digit may be used twice in the same PIN). Work out how many different PINs are possible.
- 5.Work out the exact value of √(2² + 3²)
- 6.Write ∛(x²) as a single power of x.
- 7.Work out ∛64.
- 8.The number of visitors to a museum on Saturday is given as 1,800, correct to the nearest 100. Which of these could not be the actual number of visitors?
- 9.A number is divided by 5, then 6 is subtracted, giving the result −1. Work out the number.
- 10.A pair of trainers is priced at £80 online. The shop takes 25% off the price and then adds £10 for next-day delivery. Work out the total cost.
- 11.A plane flies 2340 km at an average speed of 780 km/h. It departs at 08:20. Work out the arrival time, using the 24-hour clock.
- 12.By first working out 2.9², then squaring your result, estimate 2.9⁴ to 2 significant figures.
- 13.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.
- 14.A bag holds 5 different sweets. Noah takes 2 of the sweets out of the bag together, so the order in which he takes them does not matter. Work out how many different pairs of sweets he could take.
- 15.A sponsored walk is 36 km long. Aisha has completed 8/12 of the walk. Work out how far she has walked.
Answer key
- (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) 2⁵ — Method: a power outside brackets multiplies the index of every factor inside them, and dividing powers of the same base subtracts their indices. Working: (5² × 2³)³ = 5⁶ × 2⁹ and (5³ × 2²)² = 5⁶ × 2⁴. Dividing gives 5⁶⁻⁶ × 2⁹⁻⁴, and since 5⁰ = 1 the whole expression reduces to 2⁵. Answer: 2⁵. The distractors: 2² comes from adding the outside index to each inside index instead of multiplying, which gives 5⁵ × 2⁶ over 5⁵ × 2⁴; 2¹³ comes from adding the indices of 2 when dividing, 9 + 4, instead of subtracting them; 2 comes from applying the outside power to the first factor inside each bracket only, leaving 5⁶ × 2³ over 5⁶ × 2².
- (a) 12 — Method: for any two numbers, their highest common factor multiplied by their lowest common multiple equals the product of the two numbers. This holds because the HCF collects every prime factor the two numbers share, and the LCM collects every prime factor that appears in either number, so between them they use each prime factor of the two numbers exactly once — the same primes as the product. Working: 4 × 60 = 240, and 240 ÷ 20 = 12. 15 comes from working out 60 ÷ 4 = 15, dividing the wrong pair of numbers. 16 comes from working out 20 − 4 = 16, subtracting the highest common factor instead of using the product rule. 240 is 4 × 60, the product of the highest common factor and the lowest common multiple, left un-divided by 20. Answer: 12.
- (c) 360 — There are 5 choices for the letter. The first digit can be any of the 9 digits from 1 to 9, giving 9 choices, and the second digit must differ from the first, leaving 8 choices. By the product rule, the number of PINs is 5 × 9 × 8 = 360. Allowing the second digit to repeat the first, ignoring the 'no digit twice' rule, gives 5 × 9 × 9 = 405. Adding the numbers of choices instead of multiplying them, 5 + 9 + 8, gives 22. Treating the pair of digits as an unordered choice, rather than as a first digit followed by a second digit in a fixed order, halves the digit count: 5 × (9 × 8 ÷ 2) = 180.
- (d) √13 — Method: everything under a root sign is worked out first, because a square root cannot be taken term by term across an addition. Working: 2² = 4 and 3² = 9, so the expression under the root is 4 + 9 = 13. As 13 is not a square number, the exact value is left in root form as √13. Answer: √13. The distractors: 5 comes from rooting each square separately and adding, 2 + 3, which treats the root of a sum of squares as the sum of the numbers; 13 comes from working out the sum under the root correctly and then forgetting to take the root; √5 comes from subtracting the two squares, 9 − 4, instead of adding them.
- (c) x⁽²⁄³⁾ — Method: a root can be written as a fractional index, with the root's index as the denominator and the power inside the root as the numerator. Working: the cube root gives a denominator of 3 and the square inside gives a numerator of 2, so ∛(x²) = x⁽²⁄³⁾. Answer: x⁽²⁄³⁾. The distractors: x⁽³⁄²⁾ comes from writing the fraction upside down, with the root's index on top; x⁽¹⁄⁶⁾ comes from treating the square as a second root and multiplying 1/3 by 1/2; x⁶ comes from multiplying the root's index by the power, 3 × 2, and keeping the result as a whole-number index.
- (b) 4 — Method: the cube root of a number is the value that multiplies by itself three times to give that number. Working: 4 × 4 × 4 = 64, so ∛64 = 4. Answer: 4. (8 comes from finding the square root of 64 instead of the cube root. 192 comes from multiplying 64 by 3 instead of finding the number that cubes to 64. 21.3 comes from dividing 64 by 3 instead of finding its cube root.)
- (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) 25 — Reverse the operations in reverse order: undo the subtraction by adding 6, then undo the division by multiplying by 5. −1 + 6 = 5, so the number divided by 5 equals 5, and 5 × 5 = 25 — checking, 25 ÷ 5 − 6 = 5 − 6 = −1. A candidate who subtracted 6 again instead of adding worked out −1 − 6 = −7, then −7 × 5 = −35. A candidate who multiplied by 5 before undoing the subtraction, doing the inverse operations in the wrong order, worked out −1 × 5 = −5, then −5 + 6 = 1. A candidate who multiplied by 5 but forgot to undo the subtraction at all worked out −1 × 5 = −5 and stopped there.
- (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.
- (a) 11:20 — Method: find the flight time using time = distance ÷ speed, then add this to the departure time. Working: 2340 ÷ 780 = 3 hours; 08:20 + 3 hours = 11:20. Answer: 11:20. 08:40 comes from dividing speed by distance instead of distance by speed, giving a flight time of 1/3 hour (20 minutes) rather than 3 hours. 11:00 comes from adding the 3-hour flight time to the hour of the departure time only, 8 + 3 = 11, and losing the 20 minutes. 03:00 comes from finding the flight time correctly but giving it as a clock time on its own, forgetting to add it to the departure time.
- (a) 71 — 2.9⁴ = (2.9²)². First, 2.9² = 8.41. Then square that: 8.41² = 70.7281, since 841² = 707281 and the decimal point moves four places. To 2 significant figures this rounds to 71, because the figure after the first two significant figures (7 and 0) is a 7, which rounds the 0 up to 1. Rounding the working down instead of up — taking 70.7281 to 70 — ignores that the next figure is 5 or more. Rounding 2.9 up to 3 before doing any working at all, then computing 3⁴ = 81, uses a much cruder approximation and overshoots the true value. Rounding 8.41 all the way down to 8 before squaring, 8² = 64, rounds far too aggressively and loses the accuracy needed for 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.
- (d) 10 — Method: list the pairs systematically, taking each sweet in turn and pairing it only with the sweets that come after it, so that no pair is written down twice. Working: numbering the sweets 1 to 5, the first sweet pairs with 4 others, the second pairs with 3 sweets that come after it, the third with 2 and the fourth with 1, so the total is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 20 comes from working out 5 × 4 and never halving, which counts each pair twice, once in each order; 25 comes from working out 5 × 5, which allows the same sweet to be chosen twice; 9 comes from adding the 5 choices and the 4 remaining choices instead of combining them as a selection of two.
- (d) 24 km — Method: a fraction acts as an operator, so finding 8/12 of a distance means dividing by the denominator and multiplying by the numerator. Working: 36 ÷ 12 = 3, so one twelfth of the walk is 3 km, and eight twelfths is 3 × 8 = 24 km. Answer: 24 km. The distractors: 12 km comes from working out the part of the walk still left, the other four twelfths, instead of the part already completed; 288 km comes from multiplying by the numerator without dividing by the denominator, giving 36 × 8 = 288; 4.5 km comes from dividing by the numerator instead of multiplying by it, giving 36 ÷ 8 = 4.5.
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