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.
Answer key: Rounding, significant figures and error intervals worksheet — GCSE Foundation
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- (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.
- (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.
- (a) 8.15 ≤ y < 8.25 — Rounding to 1 decimal place means the error interval spans half of 0.1, so 0.05, either side of 8.2: 8.2 − 0.05 = 8.15 and 8.2 + 0.05 = 8.25. The lower bound uses ≤ because 8.15 itself rounds to 8.2, but the upper bound uses < because 8.25 would round up to 8.3. So the error interval is 8.15 ≤ y < 8.25. A candidate who used the wrong rounding band gave 8.1 ≤ y < 8.2. A candidate who used a strict inequality at both ends wrote 8.15 < y < 8.25, wrongly excluding 8.15 itself. A candidate who added the full 0.1 instead of half of it wrote 8.2 ≤ y < 8.3.
- (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.
- (b) 245 ≤ m < 255 — Method: a mass given to the nearest 10 g lies within half of 10 g, that is 5 g, of the figure written down. The lower limit is included because it rounds up to that figure, and the upper limit is excluded because it rounds up to the next multiple of 10. Working: 250 − 5 = 245 and 250 + 5 = 255, and a mass of 245 g rounds to 250 g while a mass of 255 g rounds to 260 g. Answer: 245 ≤ m < 255. The distractors: 240 ≤ m < 260 goes a whole 10 g either side instead of half of it; 245 < m ≤ 255 has the two limits the wrong way round; 240 ≤ m < 250 treats the stated 250 g as the largest mass possible, as though rounding were always upwards.
- (c) 3.95 — The digit after the second decimal place is 7, which is 5 or more, so round the second decimal place up: 3.947 rounds to 3.95. A candidate who truncated instead of rounding, simply cutting off after 2 decimal places, wrote 3.94. A candidate who rounded to 1 decimal place instead of 2 wrote 3.9. A candidate who rounded up but mishandled the carry wrote 4.0.
- (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) 590 — To round to the nearest 10, decide which multiple of 10 the number is nearer to. 592.5 lies between 590 and 600. It is 592.5 − 590 = 2.5 above 590, but 600 − 592.5 = 7.5 below 600, so it is much nearer to 590. Equivalently, the units digit is 2, and 2 is less than 5, so round down: 592.5 rounds to 590. A candidate who wrote 600 rounded up because of the 5 in the tenths place, but that digit decides rounding to the nearest whole number, not to the nearest 10 — the units digit is what matters here. A candidate who wrote 595 rounded to the nearest 5 instead of the nearest 10. A candidate who wrote 500 cut the number down to its hundreds digit instead of rounding to the nearest 10.
- (d) 40 miles — Method: round each number to 1 significant figure first, then divide to estimate the daily distance. Working: 830 rounds to 800, and 19 rounds to 20, and 800 ÷ 20 = 40, so the estimate is 40 miles per day. 41.5 miles comes from rounding only the number of days and working out 830 ÷ 20 = 41.5, without rounding the distance too. 830 miles is the total distance for the whole trek, given as the answer without dividing by the number of days at all. 4 miles comes from working out 80 ÷ 20 = 4, misplacing a digit in the rounded distance. Answer: 40 miles.
- (b) 4,700 — To round to the nearest 100, look at the digit in the tens column, which decides whether the hundreds column rounds up or stays the same. In 4,685 that digit is 8, and since 8 is 5 or more, the 6 in the hundreds column rounds up to 7, giving 4,700. Simply changing the last two digits to zero without checking the tens digit gives 4,600, which rounds down when it should round up. Rounding to the nearest 10 instead of the nearest 100 gives 4,690. Rounding to the nearest 1,000 instead gives 5,000, one place value too coarse.
- (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.
- (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) 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.
- (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) 24.69 — To round to 2 decimal places, look only at the third decimal digit to decide whether the second decimal digit rounds up. In 24.6851 the third decimal digit is 5, and since 5 is 5 or more, the second decimal digit rounds up from 8 to 9, giving 24.69. Rounding to 1 decimal place instead of 2 gives 24.7, one place value too coarse. Keeping the third decimal digit rather than dropping it gives 24.685, which is 3 decimal places. Looking at the fourth decimal digit, 1, instead of the third one, and wrongly deciding that no rounding is needed, leaves the length unrounded at 24.68.
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