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 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.
- 2.A rectangular garden has a length of 12.4 m and a width of 7.5 m, each measured correct to 1 decimal place. Calculate the upper bound for the area of the garden.
- 3.A photography studio offers 4 backdrops and 3 outfits for a portrait session. The plain grey backdrop, one of the 4 backdrops, cannot be used with the formal suit outfit, one of the 3 outfits. Work out how many different backdrop-and-outfit combinations are possible.
- 4.A gym increases its membership price by 1/5. The original price is £60. Work out the new price.
- 5.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.
- 6.A café offers sandwiches with one filling from 5 choices and one type of bread from 4 choices. Cheese and mustard, which is one of the 5 fillings, is not available on gluten-free bread, which is one of the 4 breads. Work out how many different sandwiches are possible.
- 7.A tap fills a 20 litre bucket in 2 minutes 30 seconds. Work out the rate of flow, in litres per minute.
- 8.A florist has 5 different flowers. Freya picks 3 of them to make a small bunch, and the order in which she picks them does not matter. Work out how many different bunches she could make.
- 9.The error interval for the mass of a suitcase, m kg, is given as 22.5 ≤ m < 23.5. Write down the mass of the suitcase, correct to the nearest whole number.
- 10.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
- 11.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.
- 12.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.
- 13.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.
- 14.Write these decimals in order, starting with the smallest: 0.6, 0.55, 0.601, 0.5, 0.56
- 15.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.
Answer key
- (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.
- (b) 93.9975 — Each measurement was rounded to 1 decimal place, so the error is half of 0.1: length is 12.35 ≤ L < 12.45, and width is 7.45 ≤ W < 7.55. The upper bound for the area comes from multiplying the upper bounds of both dimensions: 12.45 × 7.55 = 93.9975 m². Using the lower bound of both dimensions instead, 12.35 × 7.45 = 92.0075 m², gives the lower bound of the area rather than the upper one. Multiplying the two given rounded values directly, 12.4 × 7.5 = 93, forgets that a rounded measurement is not exact and needs its own error interval. Bounding only the length and leaving the width at its given value, 12.45 × 7.5 = 93.375, misses that the width also has an upper bound of its own.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (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.
- (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.
- (c) 19 — Without any restriction there would be 5 × 4 = 20 different sandwiches. The restriction removes exactly one combination, cheese and mustard on gluten-free bread, so subtract 1: 20 − 1 = 19. 20 comes from ignoring the restriction completely. 15 comes from removing the gluten-free bread altogether, as if none of the fillings were available on it, 5 × 3 = 15. 16 comes from removing the cheese and mustard filling completely, as if it were not available on any bread, 4 × 4 = 16.
- (a) 8 litres per minute — Method: write the time as a decimal number of minutes, then divide the volume by the time. Working: 30 seconds = 30/60 minute = 0.5 minute, so 2 minutes 30 seconds = 2.5 minutes. Rate = 20 ÷ 2.5 = 8 litres per minute. Answer: 8 litres per minute. (10 litres per minute comes from ignoring the extra 30 seconds and dividing by 2 minutes only. 8.7 litres per minute comes from misreading 2 minutes 30 seconds as 2.3 minutes instead of 2.5 minutes. 0.125 litres per minute comes from dividing the time by the volume instead of the volume by the time.)
- (d) 10 — Method: picking 3 flowers from 5 leaves 2 flowers behind, so counting the different pairs that could be left out counts the bunches, and those pairs can be listed systematically. Working: number the flowers 1 to 5; the first flower can be left out alongside any of the 4 flowers after it, the second alongside any of the 3 after it, the third alongside any of the 2 after it and the fourth alongside the last one, so the number of pairs left out is 4 + 3 + 2 + 1 = 10. Answer: 10. The distractors: 60 comes from working out 5 × 4 × 3 and treating the three picks as an ordered selection when the order does not matter; 30 comes from dividing that product by 2 instead of by the 6 orders in which three chosen flowers could have been picked; 15 comes from multiplying the 5 flowers by the 3 flowers picked instead of counting the selections.
- (b) 23 kg — Method: the rounded value sits exactly in the middle of the error interval. Working: the interval 22.5 ≤ m < 23.5 stretches 0.5 either side of the rounded value, so the rounded value is 23. Answer: 23 kg. (22 kg comes from rounding the lower bound down instead of finding the middle of the interval. 22.5 kg comes from giving the lower bound itself rather than the rounded value. 23.5 kg comes from giving the upper bound itself rather than the rounded value.)
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (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.
- (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.
- (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) 0.5, 0.55, 0.56, 0.6, 0.601 — Compare the decimals by giving them all the same number of decimal places first: 0.600, 0.550, 0.601, 0.500, 0.560. Ordering these from smallest to largest gives 0.500, 0.550, 0.560, 0.600, 0.601, which is 0.5, 0.55, 0.56, 0.6, 0.601. Comparing the digits as though they were whole numbers, reading 0.601 as "601" and 0.5 as "5", without padding to the same number of decimal places, gives the wrong order 0.5, 0.6, 0.55, 0.56, 0.601, because it ignores the place value of each digit. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 0.601, 0.6, 0.56, 0.55, 0.5. Misreading the close values 0.55 and 0.56 and swapping them gives 0.5, 0.56, 0.55, 0.6, 0.601. So the correct order, smallest to largest, is 0.5, 0.55, 0.56, 0.6, 0.601.
- (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.
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