Printable · GCSE Higher · ages 14-16
Limits of accuracy and bounds worksheet — GCSE Higher
Fifteen questions on "limits of accuracy and bounds" — DfE statement N16. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Answer key: Limits of accuracy and bounds worksheet — GCSE Higher
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- (b) No, the true mass could be as high as 852.5 kg — Method: find the upper bound of the true mass and compare it with the weight limit. Working: the display is correct to the nearest 5 kg, so half of 5 kg is 2.5 kg, and the true mass, m kg, satisfies 847.5 ≤ m < 852.5. Part of that interval lies above 850 kg, so the parcels are not definitely within the limit. Answer: the true mass could be as high as 852.5 kg, which is above the limit. ("Yes, the display reads 850 kg, which is not above the limit" compares the limit with the displayed value instead of with the largest value the true mass could take. "Yes, the true mass is at least 847.5 kg and at most 850 kg" uses the correct half unit below but caps the interval at the limit instead of at 852.5 kg. "No, 850 kg on the display rounds up to 855 kg" wrongly treats the displayed value as if it rounds again.)
- (b) Yes, because 14.8 cm rounds to 15 cm to the nearest cm — Method: a recorded measurement is not an exact length; it stands for every length that rounds to it, so the two records agree if one rod can produce both. Working: Ben's record of 14.8 cm to the nearest 0.1 cm means the rod is between 14.75 cm and 14.85 cm, and 14.8 is nearer to 15 than to 14, so a rod of that length is recorded as 15 cm to the nearest centimetre. Both records can therefore come from the same rod. Answer: Yes, because 14.8 cm rounds to 15 cm to the nearest cm. The distractors: the claim that 14.8 cm rounds to 15.0 cm to 1 decimal place is false, since 14.8 cm is already written to 1 decimal place and stays 14.8 cm; the claim that it rounds to 14 cm is false, because 14.8 is 0.2 away from 15 and 0.8 away from 14; the claim that the two lengths are not the same treats each record as an exact length, when each is only a rounded record of one rod.
- (b) 14.5 ≤ l < 15.5 — A measurement given to the nearest metre could have been rounded from anywhere up to half a metre below or above it: 15 − 0.5 = 14.5 and 15 + 0.5 = 15.5. Every value from 14.5 up to (but not reaching) 15.5 rounds to 15, so the error interval is 14.5 ≤ l < 15.5, with the lower bound included and the upper bound excluded. Making both ends strict, 14.5 < l < 15.5, wrongly excludes 14.5 itself, even though 14.5 does round to 15. Making both ends inclusive, 14.5 ≤ l ≤ 15.5, wrongly includes 15.5, which actually rounds up to 16, not 15. Using a whole metre either side instead of half a metre, giving 14 ≤ l < 16, comes from forgetting that the error is only half the rounding unit.
- (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.)
- (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.
- (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.)
- (c) 23.375 — The error intervals are 7.5 ≤ base < 8.5 and 4.5 ≤ height < 5.5. The upper bound of the area uses the upper bound of both the base and the height, then halves the product: 8.5 × 5.5 ÷ 2 = 23.375 cm². Using the lower bound of both dimensions instead, 7.5 × 4.5 ÷ 2 = 16.875, gives the lower bound of the area rather than the upper one. Multiplying the two upper bounds together but forgetting to halve for the triangle formula, 8.5 × 5.5 = 46.750, treats the triangle as if it were a rectangle. Using the given values directly without applying any bound at all, 8 × 5 ÷ 2 = 20.000, ignores that each rounded measurement has its own range of possible values.
- (b) 20 — Each length has its own error interval: 11.5 ≤ a < 12.5 and 6.5 ≤ b < 7.5. The upper bound of a sum is found by adding the upper bounds of both quantities: 12.5 + 7.5 = 20. Bounding only one of the two lengths and adding the other quantity's given value unbounded, 12.5 + 7 = 19.5, misses that both measurements carry their own uncertainty. Adding the lower bounds instead of the upper bounds, 11.5 + 6.5 = 18, gives the lower bound of the sum, not the upper one. Using a whole centimetre of error either side instead of half a centimetre, (12 + 1) + (7 + 1) = 21, comes from forgetting the error is half the rounding unit.
- (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.)
- (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) 18.5 ≤ T < 18.7 — Method: the error interval reaches half the rounding unit either side of the recorded value. Working: half of 0.2 is 0.1, so the interval runs from 18.6 − 0.1 to 18.6 + 0.1. Answer: 18.5 ≤ T < 18.7. (18.4 ≤ T < 18.8 comes from using the full rounding unit, 0.2, either side instead of half of it. 18.5 ≤ T ≤ 18.7 comes from including the upper bound with ≤ instead of excluding it with <. 18.6 ≤ T < 18.8 comes from treating the recorded value as the start of the interval and adding the whole rounding unit, 0.2, above it.)
- (b) No — their possible jump lengths do not overlap — Method: each recorded jump stands for the lengths within half of 0.1 m, that is 0.05 m, of the figure recorded, and Priya is right only if the two ranges overlap. Working: Priya's jump is at least 3.8 − 0.05 = 3.75 m and below 3.85 m, because a jump of 3.85 m would have been recorded as 3.9 m; Nadia's jump is at least 3.85 m and below 3.9 + 0.05 = 3.95 m. Every length Priya could have jumped is below 3.85 m and every length Nadia could have jumped is at least 3.85 m, so Nadia jumped further whatever the exact lengths were. Answer: No — their possible jump lengths do not overlap. The distractors: the reason that a recorded jump is exactly the length jumped reaches the same verdict by treating a rounded record as exact, which is the idea this question tests; both jumps being 3.85 m would put 3.85 m inside Priya's range, when a jump of that length is recorded as 3.9 m; Priya jumping up to 3.9 m goes a whole 0.1 m above her record instead of half of it.
- (c) 18 — The error intervals are 35 ≤ distance < 45 and 2.5 ≤ time < 3.5. Average speed is distance ÷ time, and to make a quotient as large as possible you divide the largest possible numerator by the SMALLEST possible denominator: 45 ÷ 2.5 = 18 km/h. Using the upper bound of time as well as the upper bound of distance, 45 ÷ 3.5, gives roughly 12.9 km/h — dividing by a bigger number produces a smaller result, so this actually finds a value smaller than the true maximum. Using the lower bound of distance with the lower bound of time, 35 ÷ 2.5 = 14, mixes up which bound belongs to a maximum calculation. Using the lower bound of distance with the upper bound of time, 35 ÷ 3.5 = 10, is in fact the correct method for the MINIMUM speed, not the maximum.
- (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.
- (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.
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