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.Work out (6 × 10⁷) + (3 × 10⁶). Give your answer in standard form.
- 2.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.
- 3.A padlock code is formed from 3 different digits chosen from 1, 2, 3, 4, 5 and 6 (no digit may be used twice in the same code). Work out how many different codes can be made.
- 4.A cheetah runs at a steady speed of 25 metres per second. Work out this speed in kilometres per hour.
- 5.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.
- 6.Work out 10% of 30% of £200.
- 7.A charity raises 8 × 10⁶ pounds. This total is shared equally among 2 × 10² local projects. Work out how much each project receives, in standard form.
- 8.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.
- 9.A market trader buys 8 melons at £1.35 each. He sells all of them for a total of £15.60. Work out his profit.
- 10.A jug holds 1.8 litres of juice. Priya pours out 3/4 of a litre to fill a glass. Work out how much juice is left in the jug, in millilitres.
- 11.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.
- 12.A flight departs at 21:35 and lands at 06:10 the following morning. Work out the total flight time.
- 13.Write these fractions in order, starting with the smallest: 2/3, 3/5, 5/6, 1/2
- 14.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.
- 15.Write these three numbers in order, starting with the smallest: 7/20, 0.3, 32%
Answer key
- (c) 6.3 × 10⁷ — To add numbers in standard form, first write them with the same power of 10. 6 × 10⁷ = 60 × 10⁶, so the sum is 60 × 10⁶ + 3 × 10⁶ = 63 × 10⁶ = 6.3 × 10⁷. A candidate who added the A values without adjusting for the different powers worked out 6 + 3 = 9 and kept the larger power, writing 9 × 10⁷. A candidate who added the powers of 10 as if multiplying wrote 9 × 10¹³. A candidate who added the A values but used the smaller power wrote 9 × 10⁶.
- (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) 120 — There are 6 choices for the first digit. The second digit must be different from the first, leaving 5 choices, and the third digit must differ from both of the first two, leaving 4 choices. By the product rule, the number of codes is 6 × 5 × 4 = 120. Allowing every digit to repeat, ignoring the 'no digit twice' rule entirely, gives 6 × 6 × 6 = 216. Adding the number of choices at each position instead of multiplying them, 6 + 5 + 4, gives 15. Treating the three chosen digits as one unordered set, rather than as digits in a fixed order on the padlock, divides by the 3! = 6 ways of arranging them: 120 ÷ 6 = 20.
- (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) 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.
- (d) £6 — First find 30% of £200, which is £60, then find 10% of that: £60 × 0.1 = £6. Adding the two percentages together instead of applying them one after the other, 10% + 30% = 40%, and finding 40% of £200 gives £80. Finding 30% of £200 = £60 correctly but stopping before applying the second percentage leaves £60 as the final answer. Finding only 10% of the original £200, ignoring the 30% entirely, gives £20.
- (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) 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.
- (a) £4.80 — The total cost of the melons is 8 × £1.35 = £10.80. Profit is the selling total minus the cost, so £15.60 − £10.80 = £4.80. Subtracting the price of just one melon from the selling total instead of the total cost of all eight, £15.60 − £1.35, gives £14.25. Miscalculating the cost of the melons as 8 × £1.30 = £10.40, dropping the 5p from each price, gives a profit of £15.60 − £10.40 = £5.20. Adding the selling total and the cost together instead of subtracting them, £15.60 + £10.80, gives £26.40.
- (d) 1050 ml — Convert both amounts to millilitres: 1.8 litres = 1800 ml and 3/4 litre = 750 ml. Subtracting gives 1800 − 750 = 1050 ml. Confusing 3/4 with 75% and converting it as 75 ml instead of 750 ml gives 1800 − 75 = 1725 ml. Adding the amount poured out instead of subtracting it gives 1800 + 750 = 2550 ml. Misreading 1.8 litres as 0.8 litres, losing the whole litre, gives 800 − 750 = 50 ml.
- (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) 8 hours 35 minutes — From 21:35 to midnight is 2 hours 25 minutes, and from midnight to 06:10 is a further 6 hours 10 minutes, giving a total of 8 hours 35 minutes. Misreading the departure time as 22:35 instead of 21:35 loses an hour from the calculation and gives 7 hours 35 minutes. Misreading the departure time as 20:35 instead of 21:35 gains an hour and gives 9 hours 35 minutes. Subtracting the times as if both fell on the same day, without crossing midnight, gives 15 hours 25 minutes.
- (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.
- (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.
- (b) 0.3, 32%, 7/20 — Method: convert every number to a decimal so they can be compared on the same scale. Working: 7/20 = 0.35, 0.3 stays as 0.3, and 32% = 0.32. Comparing 0.3, 0.32 and 0.35 in size gives the order 0.3, then 0.32, then 0.35. Answer: 0.3, 32%, 7/20. 7/20, 32%, 0.3 lists the numbers from largest to smallest instead of smallest to largest. 0.3, 7/20, 32% swaps 32% and 7/20, treating the fraction 7/20 as smaller even though 7/20 = 0.35 is bigger than 32% = 0.32. 32%, 0.3, 7/20 comes from moving the digits one place too far when converting the percentage, giving 0.032 instead of 0.32, which makes 32% look far smaller than it really is.
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