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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Answer key: Number worksheet — GCSE Higher
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- (c) 25 cm — Method: for a cube, the edge length is the cube root of the volume. Working: 25 × 25 × 25 = 15,625, so the edge length is 25 cm. 5 cm comes from cube-rooting 125 instead of 15,625, misreading the number of digits. 50 cm comes from working out 25 × 2 = 50, doubling the correct edge length. 125 cm comes from taking the square root of the volume instead of the cube root, since 125 × 125 = 15,625 — that would be the side of a SQUARE of area 15,625, not the edge of a cube of that volume. Answer: 25 cm.
- (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) 27/80 — Method: write the total as a fraction of a litre, then divide by the number of glasses. Working: 1.35 = 27/20, so each glass holds 27/20 ÷ 4 = 27/80 of a litre. Answer: 27/80. 27/20 comes from converting the total correctly to a fraction but forgetting to divide by the number of glasses. 27/5 comes from multiplying the total by 4 instead of dividing. 17/50 comes from rounding 1.35 ÷ 4 to 0.34 before converting to a fraction.
- (b) £37 — One box costs £4 + £3 = £7. Five boxes cost 5 × £7 = £35. Adding the single £2 delivery fee gives £35 + £2 = £37. A candidate who added the £2 delivery fee to each box instead of once for the whole order worked out 5 × (£7 + £2) = 5 × £9 = £45. A candidate who forgot the £3 markup and used the shop's buying price worked out 5 × £4 + £2 = £22. A candidate who added the £3 markup only once, after multiplying the buying price by 5, worked out 5 × £4 + £3 + £2 = £25.
- (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.
- (d) 9 — Method: split into two cases — the soups with no restriction, and the mushroom soup on its own — then add the totals. Working: the 2 soups other than mushroom can be paired with any of the 4 sandwiches: 2 × 4 = 8. The mushroom soup can only be paired with the cheese sandwich: 1 combination. Total = 8 + 1 = 9. Answer: 9. 12 comes from working out 3 × 4 = 12 without applying the restriction at all. 8 comes from correctly finding the 2 unrestricted soups' 8 combinations, but forgetting to add back the 1 allowed mushroom-and-cheese combination. 11 comes from taking the unrestricted total of 12 and removing only 1 mushroom combination instead of all 3 disallowed ones.
- (d) 50 km/h — The error intervals are 99.5 ≤ distance < 100.5 and 1.95 ≤ time < 2.05. The minimum speed is 99.5 ÷ 2.05 ≈ 48.54 km/h, and the maximum speed is 100.5 ÷ 1.95 ≈ 51.54 km/h. These two bounds round to different whole numbers, 49 and 52, so the speed cannot be guaranteed to the nearest whole number — but every value between them rounds to 50 at the nearest 10, so 50 km/h is the value that can safely be guaranteed. Quoting 49 km/h uses only the minimum bound's rounding, without checking that the maximum bound rounds to something different. Quoting 52 km/h makes the same mistake using only the maximum bound instead. Quoting 48.54 km/h states one bound to the full accuracy a calculator shows, as if the smallest possible speed were the answer, when the true speed could be anything up to 51.54 km/h.
- (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.
- (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) 150 — Since 90 students represent 3 of the 5 equal parts, one part is 90 ÷ 3 = 30, and the whole year group is five parts: 30 × 5 = 150. Applying the fraction forwards to 90 instead of reversing it, 90 × 3/5 = 54, treats the given number as the whole rather than as three fifths of it. Finding one part correctly as 30 but forgetting to scale up to the whole year group leaves 30 as the final answer. Treating 90 as the whole year group and adding on 2/5 of 90 for the students who do not walk, 90 + (90 × 2/5) = 126, applies the missing fraction to the wrong base amount.
- (a) £18 — Method: the percentage acts as an operator on the bill, and 15% can be built from 10% and 5%, where 5% is half of 10%. Working: 10% of £120 is £120 ÷ 10 = £12, and 5% is half of that, £6, so the charge is £12 + £6 = £18. Answer: £18. The distractors: £8 comes from reading 15% as one fifteenth and working out £120 ÷ 15 = £8; £12 comes from finding 10% of the bill and stopping there; £138 comes from finding the charge correctly and then giving the new total, £120 + £18, rather than the charge the question asks for.
- (c) 4.1 × 10³, 3.2 × 10⁴, 2.9 × 10⁵ — The exponent decides the size first: 10³ is smaller than 10⁴, which is smaller than 10⁵, so the order is 4.1 × 10³, then 3.2 × 10⁴, then 2.9 × 10⁵. Reversing the whole list gives largest to smallest instead of smallest to largest. Comparing 3.2 × 10⁴ and 4.1 × 10³ by their coefficients alone, 3.2 against 4.1, and swapping them ignores that 10⁴ is bigger than 10³ regardless of the coefficient. Comparing 2.9 × 10⁵ and 3.2 × 10⁴ by their coefficients alone and swapping them makes the same mistake at the top of the list.
- (c) 12.44 — The error intervals are 155.5 ≤ mass < 156.5 and 11.5 ≤ volume < 12.5. To make a quotient as small as possible, use the SMALLEST possible numerator together with the LARGEST possible denominator: 155.5 ÷ 12.5 = 12.44 g/cm³. Using the lower bound for both mass and volume, 155.5 ÷ 11.5 ≈ 13.52, forgets that dividing by a smaller number makes the result bigger, not smaller — that pairing does not give a minimum at all. Dividing the two given rounded values directly, 156 ÷ 12 = 13, ignores that both measurements have their own error interval. Using the upper bound of mass with the upper bound of volume, 156.5 ÷ 12.5 = 12.52, takes both bounds the same way round; it is neither the minimum nor the maximum, since the maximum needs the largest mass with the smallest volume, 156.5 ÷ 11.5 ≈ 13.61.
- (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.
- (a) 8,350 — Rounding to the nearest 100 means the true number can be up to half of 100, which is 50, below the recorded figure before it would round down to a lower hundred. The smallest possible number is therefore 8,400 − 50 = 8,350. Adding 50 instead of subtracting it gives 8,450, which is the upper end of the interval rather than the smallest value, and 8,450 is not itself possible because it would round up to 8,500. Subtracting a whole 100 instead of half of it gives 8,300, going too far below the recorded value. Subtracting 10 instead of half of the rounding unit gives 8,390, treating the rounding unit as 100 but the tolerance as only 10.
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