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) 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.
- (d) £4.80 — Method: find the error interval, then check which value falls outside it. Working: half of 20p is 10p, so the actual cost, c, satisfies £4.50 ≤ c < £4.70. £4.80 is above £4.70, so it could not be the actual cost. Answer: £4.80. (£4.50 is a genuine possible cost — it sits at the included lower boundary. £4.65 is a genuine possible cost, below the £4.70 upper boundary. £4.55 is a genuine possible cost, well inside the interval.)
- (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.
- (d) 1.6 × 10⁵ — 5.6 ÷ 3.5 = 1.6, and 7 − 2 = 5, so the average turnover per shop is 1.6 × 10⁵ pounds. Multiplying the exponents instead of subtracting them gives 7 × 2 = 14, so 1.6 × 10¹⁴. Adding the exponents instead of subtracting them gives 7 + 2 = 9, so 1.6 × 10⁹. Subtracting the coefficients instead of dividing them gives 5.6 − 3.5 = 2.1, so 2.1 × 10⁵.
- (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.
- (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.
- (b) £45.90 — The cost is 18 × £3.45 = £62.10. Profit = £108 − £62.10 = £45.90. A candidate who does not borrow in the tenths column, doing 1 − 0 = 1 instead of borrowing to make 10 − 1 = 9 and so leaving the units as 8 − 2 = 6, gets £46.10. A candidate who adds the cost to the selling price instead of subtracting gets £108 + £62.10 = £170.10. A candidate who gives the cost instead of the profit gets £62.10.
- (d) 6,000,000 — Method: round each number to 1 significant figure, then subtract. Working: 8,340,000 rounds to 8,000,000 (1 s.f.); 1,950,000 rounds to 2,000,000 (1 s.f.); 8,000,000 − 2,000,000 = 6,000,000. Answer: 6,000,000. 6,390,000 is the exact difference, found without rounding the numbers first. 8,000,000 comes from rounding the population correctly but forgetting to subtract the capital's population at all. 6,300,000 comes from rounding 8,340,000 to the nearest hundred thousand, 8,300,000, instead of to 1 significant figure, then subtracting the correctly rounded 2,000,000.
- (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.
- (c) 62.35 ≤ r < 62.45 — Method: with a value rounded to 1 decimal place, the error interval reaches half of 0.1 either side. Working: half of 0.1 is 0.05, so the interval runs from 62.4 − 0.05 to 62.4 + 0.05. Answer: 62.35 ≤ r < 62.45. (62 ≤ r < 63 comes from rounding to the nearest whole number instead of 1 decimal place. 62.35 ≤ r ≤ 62.45 comes from including the upper bound with ≤ instead of excluding it with <. 62.3 ≤ r < 62.5 comes from using 0.1 either side instead of half of it.)
- (b) 1.45 ≤ m < 1.55 — The flour's mass is labelled 1.5 kg, correct to the nearest 0.1 kg, so half of 0.1 kg is added to and subtracted from 1.5 kg to find the interval: 1.5 − 0.05 = 1.45 and 1.5 + 0.05 = 1.55, giving 1.45 ≤ m < 1.55. '1.4 ≤ m < 1.6' comes from taking the whole 0.1 kg as the margin either side, instead of half of it. '1.45 < m ≤ 1.55' comes from writing the inequality signs the wrong way round — the lower bound should be included and the upper bound excluded, not the other way round. '1.45 ≤ m ≤ 1.55' comes from including the upper bound, when the convention is that the upper bound is never actually reached.
- (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) £96.90 — Method: multiply the number of units by the cost per unit, then convert the result from pence to pounds. Working: 340 × 28.5p = 9690p; converting to pounds, 9690p ÷ 100 = £96.90. Answer: £96.90. £969.00 comes from misplacing the decimal point when converting pence to pounds, dividing by 10 instead of 100. £9.69 comes from misplacing the decimal point the other way, dividing by 1000 instead of 100. £102.00 comes from rounding the rate to 30p per unit before multiplying.
- (c) 20p — Turn each price into the same rate before comparing. The small bag is 400 g = 0.4 kg, so it costs £1.12 ÷ 0.4 = £2.80 per kg. The large bag costs £3.90 ÷ 1.5 = £2.60 per kg. The saving is £2.80 − £2.60 = £0.20, which is 20p per kg. 2p compares the prices per 100 g rather than per kilogram, £2.78 subtracts one bag price from the other without turning either into a rate, and £2.60 is the large bag's price per kilogram rather than the saving.
- (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.
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