Printable · GCSE Foundation · ages 14-16
Rounding, significant figures and error intervals worksheet — GCSE Foundation
Fifteen questions on "rounding, significant figures and error intervals" — DfE statement N15. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Rounding, significant figures and error intervals worksheet — GCSE Foundation
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- 1.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 2.The length of a pencil is 8.4 cm, correct to 1 decimal place. Using L for the length of the pencil in centimetres, write down the error interval for L.
- 3.Round 6.283 to 1 significant figure.
- 4.A measuring jug shows a volume of 340 ml, correct to the nearest 20 ml. Work out the smallest possible volume in the jug.
- 5.A recipe needs 0.485 kg of flour per cake. A bakery estimates its flour order by rounding this amount to 1 significant figure, then multiplying by the 60 cakes it plans to bake. Work out the bakery's estimate for the total flour needed, in kg.
- 6.Which of these numbers rounds to 0.048 when rounded to 2 significant figures?
- 7.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 8.A length, L cm, has the error interval 24.5 ≤ L < 25.5. Write down the degree of accuracy to which the length was measured.
- 9.A tin of paint has a mass of 5.672 kg. Round this mass to 1 decimal place.
- 10.A supermarket sells apples at £1.85 per kg. Anna buys 3.6 kg of apples. Estimate the cost by rounding each number to 1 significant figure before multiplying. Work out Anna's estimate.
- 11.To estimate the cost of buying 38.7 m of rope at £21.40 per metre, both numbers are first rounded to 1 significant figure. Work out the estimate.
- 12.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.
- 13.Round 0.0759 to 2 decimal places.
- 14.Sam rounds 0.070268 to 3 significant figures and writes 0.0703. Which of these statements is correct?
- 15.Round 24,681 to the nearest 1,000.
Answer key
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (a) 8.35 ≤ L < 8.45 — Method: a length rounded to 1 decimal place lies within half of 0.1 cm, that is 0.05 cm, of the value written down. The lower limit is included because it rounds up to that value, and the upper limit is excluded because it rounds up to the next value instead. Working: 8.4 − 0.05 = 8.35 and 8.4 + 0.05 = 8.45, so a length of 8.35 cm still rounds to 8.4 cm while a length of 8.45 cm rounds to 8.5 cm. Answer: 8.35 ≤ L < 8.45. The distractors: 8.3 ≤ L < 8.5 goes a whole 0.1 cm either side instead of half of it; 8.35 < L ≤ 8.45 has the two limits the wrong way round, excluding the length that does round to 8.4 cm and including the one that does not; 8.4 ≤ L < 8.5 is the interval for a length truncated to 1 decimal place, not one rounded to it.
- (c) 6 — Method: the first significant figure is the first non-zero digit; round using the digit after it to decide whether to round up or down. Working: the first significant figure of 6.283 is the 6; the next digit is 2, which rounds down, so 6.283 rounds to 6. 6.3 comes from rounding to 2 significant figures instead of 1. 10 comes from rounding up to the nearest 10 instead of finding 1 significant figure of the number itself. 0.6 comes from misplacing the decimal point after rounding. Answer: 6.
- (a) 330 ml — Correct to the nearest 20 ml means the true volume could be up to 10 ml (half of 20) either side of 340 ml. The smallest possible volume is 340 − 10 = 330 ml. A candidate who subtracted the full 20 ml instead of half of it worked out 340 − 20 = 320 ml. A candidate who added instead of subtracted, finding the largest possible volume instead of the smallest, worked out 340 + 10 = 350 ml. A candidate who halved the interval again by mistake, using 5 ml instead of 10 ml, worked out 340 − 5 = 335 ml.
- (d) 30 kg — Round 0.485 kg to 1 significant figure: 0.5 kg. Multiply by the 60 cakes: 0.5 × 60 = 30 kg. A candidate who rounded to 2 significant figures instead of 1 used 0.49 kg, giving 0.49 × 60 = 29.4 kg. A candidate who used the unrounded amount instead of the estimate worked out 0.485 × 60 = 29.1 kg. A candidate who rounded 0.485 down to 0.4 kg instead of up to 0.5 kg worked out 0.4 × 60 = 24 kg.
- (c) 0.0479 — Method: round each option to 2 significant figures and check which one gives 0.048. Working: for 0.0479, the first two significant figures are 4 and 7; the next digit is 9, so 7 rounds up to 8, giving 0.048. For 0.0485, the first two significant figures are 4 and 8; the next digit is 5, so 8 rounds up to 9, giving 0.049, not 0.048. 0.052 already has exactly 2 significant figures, 5 and 2, so it stays as 0.052 and does not round to 0.048 at all. 0.04 has only 1 significant figure, so it is already less precise than the 2 significant figures asked for. Answer: 0.0479.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (a) to the nearest centimetre — Method: the error interval of a rounded measurement runs from half a unit below the stated value to half a unit above it, so the width of the interval is one whole unit of the accuracy used. Working: the interval runs from 24.5 to 25.5, a width of 25.5 − 24.5 = 1, so the unit of accuracy is 1 cm; the stated value is the midpoint, 25 cm, and 25 correct to the nearest centimetre is exactly what gives 24.5 ≤ L < 25.5. Answer: to the nearest centimetre. To the nearest 0.5 cm comes from reading the half-unit, 0.5, as the accuracy itself instead of doubling it back to the full unit. To 1 decimal place comes from seeing the bounds written with one decimal place and taking that as the accuracy, but the bounds of a value given to 1 decimal place would be only 0.05 either side. To the nearest 10 cm comes from confusing the size of the value, about 25, with the unit it was rounded to; rounding to the nearest 10 cm would give an interval 5 cm either side of the stated value.
- (c) 5.7 kg — Method: to round to 1 decimal place, keep one digit after the decimal point and let the digit in the second decimal place decide whether that digit stays as it is or goes up. Working: 5.672 has 6 in the first decimal place and 7 in the second decimal place; 7 is 5 or more, so the 6 goes up to 7 and the digits beyond the first decimal place are dropped. Answer: 5.7 kg. The distractors: 5.6 kg comes from chopping the digits after the first decimal place off instead of rounding them, which is truncation rather than rounding; 6.0 kg comes from rounding to the nearest whole kilogram instead of to 1 decimal place; 5.0 kg comes from chopping everything after the decimal point off, so the mass is both truncated and given to the wrong degree of accuracy.
- (b) £8 — Rounding to 1 significant figure: £1.85 rounds to £2, and 3.6 kg rounds to 4 kg. The estimate is £2 × 4 = £8. A candidate who used the unrounded values instead of estimating worked out 1.85 × 3.6 = £6.66. A candidate who rounded only the mass and used the exact price worked out 1.85 × 4 = £7.40. A candidate who rounded the price to the nearest 10p instead of 1 significant figure worked out 1.9 × 4 = £7.60.
- (d) £800 — Rounding 38.7 to 1 significant figure gives 40, and rounding 21.40 to 1 significant figure gives 20. Multiplying the rounded values gives an estimate of 40 × 20 = £800. Rounding 21.40 to the nearest whole number instead of to 1 significant figure gives 21, and 40 × 21 = £840, one place value too fine for the price. Adding the rounded values instead of multiplying them gives 40 + 20 = £60. Rounding both numbers to 2 significant figures instead of 1, giving 39 and 21, produces 39 × 21 = £819.
- (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.
- (a) 0.08 — Method: decimal places are counted from the decimal point, including any zeros straight after it, so rounding to 2 decimal places is decided by the digit in the third decimal place. Working: 0.0759 has 7 in the second decimal place and 5 in the third, and 5 counts as rounding up, so the 7 goes up to 8. Answer: 0.08. The distractors: 0.07 comes from chopping the digits after the second decimal place off instead of rounding them; 0.076 is 0.0759 correct to 2 significant figures rather than 2 decimal places, because the zeros in front of the 7 are not significant figures; 0.1 is 0.0759 rounded to 1 decimal place, a coarser degree of accuracy than the question asks for.
- (c) Sam is correct — The leading zeros in 0.070268 are not significant, so the first three significant figures are 7, 0 and 2. The next digit along is 6, and since 6 is 5 or more, the third significant figure rounds up from 2 to 3, giving 0.0703. This means Sam's answer is correct. Writing 0.070 keeps only 2 significant figures, one short of what was asked. Writing 0.0702 ignores the digit 6 that follows and leaves the third figure unrounded. Writing 0.0704 rounds the third figure up twice, as if a later digit had also pushed it up.
- (d) 25,000 — Method: to round to the nearest 1,000, look at the hundreds digit; 5 or more sends the thousands digit up, less than 5 leaves it where it is, and every digit below the thousands becomes zero. Working: 24,681 has 4 in the thousands place and 6 in the hundreds place. As 6 is 5 or more, the 4 thousands go up to 5 thousands and the hundreds, tens and units are replaced by zeros. Answer: 25,000. The distractors: 24,000 comes from cutting the last three digits off instead of rounding, which leaves the thousands digit untouched; 24,700 is 24,681 rounded to the nearest 100, a finer degree of accuracy than the question asks for; 20,000 is 24,681 rounded to the nearest 10,000, a coarser degree of accuracy.
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